\input{"preamble.tex"}

\addbibresource{CharacteristicClasses.bib}

\let\Begin\begin
\let\End\end
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\def\ScaleWidthIfNeeded{%
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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Characteristic Classes
  }
    \\ {\normalsize Lectures by Akram Alishahi. University of Georgia,
Fall 2021} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in characteristic classes taught by Akram
Alishahi at the University of Georgia in Fall 2021. As such, any errors
or inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2021-12-02 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{thursday-august-19}{%
\section{Thursday, August 19}\label{thursday-august-19}}

\hypertarget{intro-and-overview}{%
\subsection{Intro and Overview}\label{intro-and-overview}}

\begin{remark}

Course website: \url{https://akramalishahi.github.io/CharClass.html}

Description from Akram's syllabus:

\begin{quote}
This course is about characteristic classes, which are cohomology
classes naturally associated to vector bundles or, more generally,
principal bundles. They are a key tool in modern \{algebraic,
differential\}×\{topology, geometry\}. The course starts with an
introduction to vector bundles and principal bundles. It then discusses
their main characteristic classes---the Euler class, Stiefel-Whitney
classes, Chern classes, and Pontryagin classes. The last part of the
class discusses some applications of characteristic classes to bordisms.
In the process, we will see some nice applications (e.g., to immersions)
and review some important parts of algebraic topology (e.g., obstruction
theory).
\end{quote}

\textbf{References}:

\begin{itemize}
\tightlist
\item
  \autocite{Hu} Husemoller, Fiber bundles.
\item
  \autocite{MS} Milnor and Stasheff, Characteristic classes.
\item
  \autocite{S} Steenrod, The topology of fibre bundles.
\item
  \autocite{Ha} Hatcher, Vector bundles and K-theory .
\item
  \autocite{BottTu} Bott and Tu, Differential forms in algebraic
  topology.
\end{itemize}

Prerequisites:

\begin{itemize}
\tightlist
\item
  Smooth manifolds: smooth maps and derivatives, differential forms.
\item
  Algebraic topology: homology and cohomology.
\end{itemize}

\end{remark}

\begin{remark}

An overview of what we'll cover:

\begin{itemize}
\item
  \textbf{General definitions and constructions related to \emph{vector
  bundles} and \emph{fiber bundles}.}

  Why bundles? For a bundle \(E \xrightarrow{\pi} B\), characteristic
  classes will be cohomology classes in \(H^*(B)\). Natural examples
  include

  \begin{itemize}
  \tightlist
  \item
    The tangent bundle \(TX\to X\), and vector fields will be sections.
  \item
    Exterior products \(\bigwedge\nolimits^n TX\), where differential
    forms live
  \item
    Normal bundles \(\nu\), giving directions an embedded submanifold
    can be deformed.
  \end{itemize}

  Also note that manifolds locally look like vectors spaces
  (\({\mathbb{R}}^n\)!) and so embedded manifolds locally look like
  vector bundles. In particular, if \(f: M^n \hookrightarrow N^k\) is an
  embedding, locally \(\nu\) is locally a \(k-n\) dimensional vector
  bundle over \({\mathbb{R}}^n\) (and globally a bundle of the form
  \(\nu: E \to f(M_n)\))

  \begin{figure}
  \centering
  \resizebox{\columnwidth}{!}{%
  \begin{tikzpicture}
  \fontsize{34pt}{1em} 
  \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-08-19_13-04.pdf_tex} };
  \end{tikzpicture}
  }
  \end{figure}
\item
  \textbf{Characteristic Classes: Euler, Stiefel-Whitney, Pontryagin,
  etc.}

  These package geometric information into algebraic invariants that are
  often computable. Some examples:

  \begin{itemize}
  \tightlist
  \item
    Stiefel-Whitney classes can detect if \(M^n = {{\partial}}M^{n+1}\)
    is a boundary (for smooth closed manifolds).
  \item
    Euler classes can prove the Hairy Ball theorem, i.e.~\(S^2\) admits
    no nonvanishing continuous vector fields, which can be generalized
    to \(S^{2n}\) and to splitting the tangent bundle.
  \item
    Pontryagin classes: Milnor used these to produce exotic \(S^7\)s!
    These are manifolds \(M^7\) which are homeomorphic but not
    diffeomorphic to \(S^7\).
  \item
    Chern classes.
  \end{itemize}
\end{itemize}

\end{remark}

\hypertarget{fiber-bundles}{%
\subsection{Fiber Bundles}\label{fiber-bundles}}

\begin{definition}[Fiber bundle]

A \textbf{fiber bundle} over \(B\) with fiber \(F\) is a continuous map
\(\pi: E\to B\) where each \(b\in B\) admits an open neighborhood
\(U \subseteq B\) and a homeomorphism \(\phi: \pi^{-1}(U)\to U\times F\)
such that the following diagram commutes in \({\mathsf{Top}}\):

\begin{center}
\begin{tikzcd}
    {\pi^{-1}(U)} && {U\times F} \\
    \\
    & U
    \arrow["\phi", from=1-1, to=1-3]
    \arrow["\pi"', from=1-1, to=3-2]
    \arrow["{p_2}", from=1-3, to=3-2]
\end{tikzcd}
\end{center}

Here the square is \([0, 1]^{\times 2}\).

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHBpXFxpbnYoVSkiXSxbMiwwLCJVXFxjcm9zcyBGIl0sWzEsMiwiVSJdLFswLDEsIlxccGhpIl0sWzAsMiwiXFxwaSIsMl0sWzEsMiwicF8yIl1d}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Note that this necessarily implies that all fibers are homeomorphic,
noting that
\(F_b \coloneqq\pi^{-1}(b) \xrightarrow{\phi} \left\{{b}\right\} \times F\).
We have inclusions: vector bundles \(\implies\) fiber bundles
\(\implies\) fibrations. For a fibration that's not a fiber bundle, one
can collapse a fiber in a trivial bundle, e.g.~

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{44pt}{1em}4
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-08-19_13-37.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{example}[?]

An \textbf{atlas bundle} for \(\pi:E\to B\) is a collection of charts
\(\left\{{(U_\alpha, \phi_\alpha)}\right\}_{\alpha\in I}\) such that
\(\left\{{U_\alpha}\right\}\rightrightarrows B\).

\end{example}

\begin{example}[?]

\envlist

\begin{itemize}
\item
  \(E \coloneqq B\times F \xrightarrow{p_2} F\) the trivial/product
  bundle.
\item
  \(\widehat{X} \to X\) any covering space. Note that the fibers are
  discrete.
\item
  The Möbius band:
\end{itemize}

\textless-- Xournal file:
/home/zack/SparkleShare/github.com/Notes/Class\_Notes/2021/Fall/CharacteristicClasses/sections/figures/2021-08-19\_13-41.xoj
--\textgreater{}

\includegraphics{figures/2021-08-19_13-44-14.png}

This is a fiber bundle with fibers \([0, 1]\). For a fiber bundle,
include the boundary, but to make this a vector bundle do not include
it!

\end{example}

\begin{remark}

Consider the following setup:

\begin{itemize}
\tightlist
\item
  \(B\in {\mathsf{Top}}\)
\item
  \(\pi:E\to B\) is a map of underlying sets
\item
  There is a bundle atlas \(\left\{{\phi_\alpha}\right\}\), each
  \(\phi_\alpha\) being a bijection.
\end{itemize}

Then there exists \emph{at most} one topology on \(E\) such that
\(\pi: E\to B\) is a fiber bundle with the given atlas.

\end{remark}

\begin{exercise}[?]

Find necessary conditions for at least one topology to exist!

\end{exercise}

\hypertarget{vector-bundles}{%
\subsection{Vector Bundles}\label{vector-bundles}}

\begin{definition}[Vector bundle]

An \(n{\hbox{-}}\)dimensional real (resp. complex) \textbf{vector
bundle} over \(B\) is a fiber bundle \(\pi:E\to B\) along with a real
vector space structure on each fiber \(F_b\) such that for each
\(b\in B\) there exists a neighborhood \(U \ni b\) and a chart
\((U, \phi: \pi^{-1}(U) \to U\times {\mathbb{R}}^n)\) (resp.
\({\mathbb{C}}^n\)) where
\({ \left.{{\phi}} \right|_{{F_b}} }: F_b \xrightarrow{\sim} {\mathbb{R}}^n\)
(resp. \({\mathbb{C}}^n\)) is an isomorphism of vector spaces.

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  The trivial (product) bundle
  \(B\times{\mathbb{R}}^n \xrightarrow{p_1} B\).
\item
  The tangent bundle \(T X\).
\item
  Identifying the Möbius band as \([0, 1] \times(0, 1)/\sim\) as
  \(I \times{\mathbb{R}}/ (0, t)\sim (1, -t)\) yields a 1-dimensional
  bundle \(M\to S^1\).
\end{itemize}

\end{example}

\begin{remark}

We have some natural operations:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Direct sums.
\end{enumerate}

For
\(E_1, E_2 \in { { {\mathsf{Bun}}\qty{\operatorname{GL}_r} }}_{/ {B}}\),
so \(E_1 \xrightarrow{\pi_1} B\) and \(E_2 \xrightarrow{\pi_2} B\), we
can form \(E_1 \oplus E_2 \xrightarrow{\pi} B\). As a set, take
\begin{align*}
E_1 \oplus E_2 \coloneqq\displaystyle\bigcup_{b\in B} F_{1, b} \oplus F_{2, b}
  \end{align*}
as a union of direct sums of vector spaces. For the bundle map, take
\(\pi(F_{1, b} \oplus F_{2, b}) \coloneqq\left\{{b}\right\}\). For
charts, for any \(b\in B\) pick individual charts about \(b\), say
\((U_1, \phi_1)\) for \(E_1\) and \((U_1, \phi_2)\) for \(E_2\), form
charts
\begin{align*}
\left\{{ (U_1 \cap U_2, \phi: \pi^{-1}(U_1 \cap U_2) \to {\mathbb{R}}^{n_1 + n_2})}\right\}
  \end{align*}
where \(n_1 \coloneqq\dim_{\mathbb{R}}F_{1, b}\) and
\(n_2 \coloneqq\dim_{\mathbb{R}}F_{2, b}\) and define
\((b, (v_1, v_2)) \xrightarrow{\phi} (\phi_1(v_1), \phi_2(v_2))\).

\end{remark}

\hypertarget{fiber-bundles-with-structure-and-principal-ghbox--bundles-tuesday-august-24}{%
\section{\texorpdfstring{Fiber Bundles with Structure and Principal
\(G{\hbox{-}}\) Bundles (Tuesday, August
24)}{Fiber Bundles with Structure and Principal G\{\textbackslash hbox\{-\}\} Bundles (Tuesday, August 24)}}\label{fiber-bundles-with-structure-and-principal-ghbox--bundles-tuesday-august-24}}

\begin{remark}

Setup:

\begin{itemize}
\tightlist
\item
  \(B \in {\mathsf{Top}}\) is a space.
\item
  \(\pi:E\to B\) is a map of sets with fibers/preimages
  \(F \coloneqq F_b \coloneqq\pi^{-1}(\left\{{b}\right\})\).
\item
  A \emph{bundle atlas} for \(\pi\) is \(\phi\) where
  \(\phi_U: \pi^{-1}(U) \to U \times F\) is a bijection of sets
\end{itemize}

Then there is at most one topology on \(E\) making \(\pi:E\to B\) into a
fiber bundle with the specified atlas.

\end{remark}

\begin{definition}[Dual of a vector bundle]

Given a vector bundle \(\pi:E\to B\), form the \textbf{dual bundle}
\(\pi {}^{ \vee }:E {}^{ \vee }\to B\) by setting

\begin{itemize}
\tightlist
\item
  \(E {}^{ \vee }\coloneqq{\textstyle\coprod}_{b\in B} F_b {}^{ \vee }\)
\item
  Set \(\pi {}^{ \vee }(F_b {}^{ \vee }) = \left\{{b}\right\} \in B\).
\item
  Given \(\phi: \pi^{-1}(U)\to U\times {\mathbb{R}}^n\), set
  \begin{align*}
  \phi {}^{ \vee }: (\pi {}^{ \vee })^{-1}(U) = {\textstyle\coprod}_{b\in U} F_b {}^{ \vee }\longrightarrow U\times({\mathbb{R}}^n) {}^{ \vee }\cong U\times{\mathbb{R}}^n
  .\end{align*}
\end{itemize}

Here \(A \subseteq \pi^{-1}(U)\) is open iff \(\phi_U(A)\) is open in
\(B\).

\end{definition}

\begin{remark}

Consider what happens on overlapping charts -- looking at maps fiberwise
yields:

\begin{center}
\begin{tikzcd}
    {\pi^{-1}(U)} && {\pi^{-1}(U\cap V)} && {\pi^{-1}(V)} \\
    && {} \\
    {U\times F} && {(U\cap V) \times F} && {V\times F}
    \arrow[hook', from=1-3, to=1-1]
    \arrow["{\varphi_U}", from=1-1, to=3-1]
    \arrow[hook', from=3-3, to=3-1]
    \arrow[hook, from=3-3, to=3-5]
    \arrow[hook, from=1-3, to=1-5]
    \arrow["{\varphi_V}"', from=1-5, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Starting at \((U \cap V)\times F\) and running the diagram clockwise
yields a map
\begin{align*}
\phi_V \circ \phi_U ^{-1}: (U \cap V ) \times F &\to (U \cap V ) \times F \\
(b, f) &\mapsto (b, \phi_{VU}(f) )
\end{align*}
where \(\phi_{VU}\) is the following continuous map, defining a
\textbf{transition function}:
\begin{align*}
\phi_{VU}: U \cap V &\to {\operatorname{Homeo}}(F)
,\end{align*}
where
\({\operatorname{Homeo}}(F) \coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(F, F)\)
with the compact-open topology.

\end{remark}

\begin{definition}[The compact-open topology]

Let
\(\mathop{\mathrm{Maps}}(X, Y)\coloneqq\mathop{\mathrm{Hom}}_{\mathsf{Top}}(X, Y)\)
be the set of continuous maps \(X\to Y\), then a map
\(Z\to \mathop{\mathrm{Maps}}(X, Y)\) is continuous iff the following
map is continuous:
\begin{align*}
Z\times X &\to Y \\
(z, x) &\mapsto f(z)(x)
.\end{align*}

If \(X\) is Hausdorff and locally compact then
\(\mathop{\mathrm{Maps}}(X, Y)\) will have this property for all \(Y\).
A subbasis for this topology will be given by taking \(C \subseteq X\)
compact, \(U \subseteq Y\) open and taking the basic opens to be
\begin{align*}
S(C, U) \coloneqq\left\{{f\in \mathop{\mathrm{Maps}}(X, Y){~\mathrel{\Big\vert}~}f(X) \subseteq U}\right\}
.\end{align*}
If \(Y\) has a metric, then this will coincide with the \emph{compact
convergence topology}, which has a basis
\begin{align*}
\left\{{ 
S(f, C, E) {~\mathrel{\Big\vert}~}C \subseteq X \text{ compact}, \forall {\varepsilon}>0,\,\forall f\in \mathop{\mathrm{Maps}}(X, Y)
}\right\}, \\
S(f, C, E) \coloneqq\left\{{ 
g\in \mathop{\mathrm{Maps}}(X, Y) {~\mathrel{\Big\vert}~}d( f(x), g(x) ) < {\varepsilon}\,\,\forall x\in C 
}\right\} 
.\end{align*}

\end{definition}

\begin{definition}[Structure Groups]

Let \(G \subseteq {\operatorname{Homeo}}(F)\), then a \textbf{fiber
bundle with structure group} \(G\) is a fiber bundle
\(F\to E \xrightarrow{\pi} B\) together with a bundle atlas such that
\(G \subseteq {\operatorname{Homeo}}(F)\).

\end{definition}

\begin{example}[?]

An \({\mathbb{R}}^n{\hbox{-}}\)bundle is just a bundle where
\(F = {\mathbb{R}}^n\) for all fibers, where we ignore the vector space
structure and only take transition functions to be homeomorphisms. An
\({\mathbb{R}}^n{\hbox{-}}\)bundle with a
\(G\coloneqq\operatorname{GL}_n({\mathbb{R}})\) is exactly a vector
bundle, where we can use the structure group to put a vector space
structure on the fibers. We have charts
\(\phi_U: \pi^{-1}(U)\to U \times{\mathbb{R}}^n\), so for all
\(b\in U\), writing \(F_b \coloneqq\pi^{-1}(\left\{{b}\right\})\) and
get \(\phi_U(F_b) = {~\mathrel{\Big\vert}~}{b} \times{\mathbb{R}}^n\).
We can then define addition and multiplication for \(w_1, w_2 \in F_b\)
as
\begin{align*}
cw_1+ w_2 \coloneqq\phi_U^{-1}\qty{ c\phi_U(w_1) + \phi_U(w_2) }
.\end{align*}
This is well-defined because for any other chart containing \(V\ni b\),
we have \(\phi_{VU}\in \operatorname{GL}_n({\mathbb{R}})\). This follows
by just setting \(A \coloneqq\phi_V \circ \phi_U ^{-1}\) and writing
\begin{align*}
\phi_V(w_1 + w_2) 
&= A \phi_U(w_1 + w_2) \\
&\coloneqq A\qty{\phi_U(w_1) + \phi_U(w_2) } \\
&= A\phi_U(w_1) + A\phi_U(w_2) \\
&= \phi_V(w_1) + \phi_V(w_2) \\
&\coloneqq\phi_V(w_1 + w_2)
.\end{align*}

\end{example}

\begin{example}[Bundles with structure]

An \({\mathbb{R}}^n{\hbox{-}}\)bundle with a
\(\operatorname{GL}_{n}^+({\mathbb{R}})\) structure is an orientable
vector bundle, where
\begin{align*}
\operatorname{GL}_n^+({\mathbb{R}}) = \left\{{ A \in \operatorname{GL}_n({\mathbb{R}}) {~\mathrel{\Big\vert}~}\operatorname{det}(A) > 0 }\right\} 
.\end{align*}

A \(G\coloneqq O_n({\mathbb{R}})\) structure yields vector bundles with
Riemannian metrics on fibers, where
\(O_n({\mathbb{R}}) \coloneqq\left\{{ A\in \operatorname{GL}_n({\mathbb{R}}) {~\mathrel{\Big\vert}~}AA^t = \operatorname{id}}\right\}\).
Here we use the fact that there is an equivalence between metrics
(symmetric bilinear pairings) and choices of an orthonormal basis,
e.g.~using that if \(\left\{{e_1, \cdots, e_n}\right\}\), one can
specify an inner product completely by writing
\begin{align*}
v\coloneqq\sum v_i e_i,\quad w \coloneqq\sum w_i e_i &&\implies {\left\langle {v},~{w} \right\rangle} = \sum v_i w_i
.\end{align*}

\end{example}

\begin{definition}[Principal $G\dash$bundles]

A \textbf{principal \(G{\hbox{-}}\)bundle} is a fiber bundle
\(\pi:P\to B\) with a right \(G{\hbox{-}}\)action
\(\psi: P\times G\to P\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\psi\qty{F_b} = F_b\), so the action preserves each fiber, and
\item
  \(\psi\) is free and transitive.
\end{enumerate}

\end{definition}

\hypertarget{principal-ghbox-bundles-thursday-august-26}{%
\section{\texorpdfstring{Principal \(G{\hbox{-}}\)bundles (Thursday,
August
26)}{Principal G\{\textbackslash hbox\{-\}\}bundles (Thursday, August 26)}}\label{principal-ghbox-bundles-thursday-august-26}}

\begin{remark}

Today: relating \(\mathop{\mathrm{Prin}}{\mathsf{Bun}}_{/ {G}}\) to
fiber bundles with a \(G{\hbox{-}}\)structure. Recall that a principal
\(G{\hbox{-}}\)bundle is a fiber bundle \(\pi:P\to B\) with a fiberwise
\(G{\hbox{-}}\)action \(P\times G\to P\) which induces a free and
transitive action on each fiber. Note that we assume
\(G\in{\mathsf{Top}}{\mathsf{Grp}}\). Any bundle in
\(\mathop{\mathrm{Prin}}{\mathsf{Bun}}_{/ {G}}\) is a fiber bundle with
fibers \(F\) homeomorphic to \(G\) and admits a
\(G{\hbox{-}}\)structure:
\begin{align*}
G &\hookrightarrow{\operatorname{Homeo}}(G) \\
g &\mapsto (h\mapsto gh)
.\end{align*}
Using that \(F\cong G\), taking charts \((U, \varphi), (V, \psi)\) for
\(\pi:P\to B\), we can identify

\begin{center}
\begin{tikzcd}
    {\pi^{-1}(U \cap V)} && {(U \cap V) \times G} && {(U \cap V) \times G} && {\pi^{-1}(U \cap V)} \\
    && {(b, 1)} && {(b, g)} \\
    && {(b, h)} && {(b, gh)}
    \arrow["{\phi_V \circ \phi_U^{-1}}", from=1-3, to=1-5]
    \arrow["{\phi_U, \cong}", from=1-1, to=1-3]
    \arrow["{\phi_V, \cong}"', from=1-7, to=1-5]
    \arrow[maps to, from=2-3, to=2-5]
    \arrow[maps to, from=3-3, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

So every transition function is given by left-multiplication by some
element in \(G\), as opposed to arbitrary homeomorphisms of \(G\) as a
topological group.

\end{remark}

\begin{example}[of principal bundles]

\envlist

\begin{itemize}
\item
  Trivial actions: \(B \times G \xrightarrow{p_1} B\).
\item
  Regular covering spaces \(\pi:\tilde X\to X\), then
  \(G = \mathop{\mathrm{Deck}}(\tilde X/X)\) with the discrete topology.
\item
  Given an \(n{\hbox{-}}\)dimensional vector bundle \(\pi: E\to B\),
  take
  \begin{align*}
  \mathop{\mathrm{Frame}}(F_b) \coloneqq\left\{{(e_1, \cdots, e_n) \in F_b}\right\} \subseteq F_b^{\times n}
  ,\end{align*}
  the collection of all ordered bases of \(F_b\). Then set
  \begin{align*}
  \mathop{\mathrm{Frame}}_n \coloneqq{\textstyle\coprod}_{b\in B} \mathop{\mathrm{Frame}}(F_b) \to B
  \end{align*}
  to get a principal \(G{\hbox{-}}\)bundle for
  \(G = \operatorname{GL}_n(F_b)\) under the following action: picking a
  framing \((e_1, \cdots, e_n)\) in \(F_b\), then for
  \(A\in \operatorname{GL}_n(F_b)\) regarded as a linear map, define
  \begin{align*}
  (\mathbf{e}_1, \cdots, \mathbf{e}_n) \cdot A \coloneqq\qty{\sum_i a_{i,1} \mathbf{e}_i, \sum_i a_{i, 2} \mathbf{e}_i, \cdots, \sum_i a_{i, n} \mathbf{e}_i }
  .\end{align*}
\item
  Given an \emph{oriented} \(n{\hbox{-}}\)dimensional vector bundle
  \(\pi:E\to B\), one gets a
  \(G\coloneqq\operatorname{GL}_n^+({\mathbb{F}}_b)\) by taking
  positively oriented frames.
\item
  Given a vector bundle with a Riemannian metric, we get a principal
  \({\mathcal{O}}_n({\mathbb{R}}){\hbox{-}}\)bundle by taking
  orthonormal frames.
\end{itemize}

\end{example}

\begin{definition}[?]

Given two principal \(G{\hbox{-}}\)bundles \(\pi: P\to B\) and
\(\pi': Q\to B\), an \textbf{isomorphism of principal bundles} is a
\(G{\hbox{-}}\)equivariant map \(P \xrightarrow{f} Q\) commuting over
\(B\):

\begin{center}
\begin{tikzcd}
    P && Q \\
    \\
    & B
    \arrow["\pi", from=1-1, to=3-2]
    \arrow["{\pi'}"', from=1-3, to=3-2]
    \arrow["f"', from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJQIl0sWzIsMCwiUSJdLFsxLDIsIkIiXSxbMCwyLCJcXHBpIl0sWzEsMiwiXFxwaSciLDJdLFswLDEsImYiLDJdXQ==}{Link
to Diagram}
\end{quote}

Here \emph{equivariant} means commuting with the \(G{\hbox{-}}\)action,
in the following precise sense: let \((U, \varphi)\) and \((V, \psi)\)
be charts for \(\pi, \pi'\), then consider the composition
\begin{align*}
F: \qty{ (U \cap V) \times F \xrightarrow{\phi^{-1}} \pi^{-1}(U \cap V) \xrightarrow{f} (\pi')^{-1}(U \cap V) \xrightarrow{\psi} (U \cap V) \times F}
.\end{align*}
Note that this fixes every point \(b\in U \cap V\), so we can regard
\(F: U \cap V \to {\operatorname{Homeo}}(F)\), using that \(f\) commutes
with the projection maps:
\begin{align*}
(b, ?) \mapsto \pi^{-1}(b) \mapsto (f\circ \pi^{-1})(b) = (\pi')^{-1}b \mapsto b
.\end{align*}
We say \(f\) is a \(G{\hbox{-}}\)isomorphism iff \(F\) sends everything
to \(G\).

\end{definition}

\hypertarget{sending-fiber-bundles-to-principal-ghbox-bundles}{%
\subsection{\texorpdfstring{Sending Fiber Bundles to Principal
\(G{\hbox{-}}\)bundles}{Sending Fiber Bundles to Principal G\{\textbackslash hbox\{-\}\}bundles}}\label{sending-fiber-bundles-to-principal-ghbox-bundles}}

\begin{remark}

Given a principal \(G{\hbox{-}}\)bundle \(\pi:P\to B\) and a
\(F\in {\mathsf{Top}}\) with a left \(G{\hbox{-}}\)action. Then define
\begin{align*}
P { \underset{\scriptscriptstyle {G} }{\times} } F / (pg, f)\sim (p, gf)
\end{align*}
as a fiber bundle over \(B\) using \(\pi\) as the projection. Note that
this looks like a tensor product, and this works in general for any
space \(P\) with a right \(G{\hbox{-}}\)action and \(F\) with a left
\(G{\hbox{-}}\)action. This will be a fiber bundle with fiber \(F\) and
structure group \(G \leq {\operatorname{Homeo}}(F)\).

Locally there is a homeomorphism:
\begin{align*}
(U\times G) \overset{\scriptscriptstyle {G} }{\times} F &\xrightarrow{\sim} U\times F \\
(p, g, f) &\mapsto (p, gf)
.\end{align*}
This is well defined since \((p, gh, f)\) and \((p, g, hf)\) map to
\((p, ghf)\). The inverse is \((p, f) \mapsto (p, 1, gf)\).

\end{remark}

\begin{exercise}[?]

Check that this is a fiber bundle with \(G{\hbox{-}}\)structure.

\end{exercise}

\hypertarget{tuesday-august-31}{%
\section{Tuesday, August 31}\label{tuesday-august-31}}

\begin{remark}

We want to show the equivalence between (isomorphism classes) of fiber
bundles with \(G\) structures with fiber \(F\) and principal
\(G{\hbox{-}}\)bundles. Recall that
\(\mathop{\mathrm{Prin}}{\mathsf{Bun}}_{/ {G}}\) are fiber bundles
\(P \xrightarrow{\pi} B\) with a right fiberwise \(G{\hbox{-}}\)action
which is free and transitive on each fiber.

To send fiber bundles to principal bundles, we used a \emph{mixing}
construction. Since \(G\curvearrowright F\), we get an identification
\(G \subseteq {\operatorname{Homeo}}(F, F)\). We constructed
\begin{align*}
P { \underset{\scriptscriptstyle {G} }{\times} } F \coloneqq(P\times F)/(pg, f)\sim (p, gf)
.\end{align*}
A lemma was that
\(P{ \underset{\scriptscriptstyle {G} }{\times} } F \to B\) is a fiber
bundle with fiber \(F\) and projection \(\pi(p, f) \coloneqq\pi(p)\).

Today we'll talk about the reverse direction. Note the composition of
sending \(E\) to \(\mathop{\mathrm{Prin}}{\mathsf{Bun}}_{/ {G}}\) and
then mixing recovers \(E\) when \(E\) is a vector bundle, but not
generally.
\begin{align*}
\left\{{\substack{
  \text{Fiber bundles with }G{\hbox{-}}\text{structures} \\
  \text{ and fiber }F
}}\right\}
\adjunction{\text{Clutching}}{\text{Mixing}}{}{}
\left\{{\substack{
  \text{Principal $G{\hbox{-}}$bundles}
}}\right\}
\end{align*}

\end{remark}

\begin{example}[?]

For \(E \xrightarrow{\pi} B\) a real vector bundle, we sent it to
\(\mathop{\mathrm{Frame}}(E)\), which is a principal
\(\operatorname{GL}_n({\mathbb{R}}){\hbox{-}}\)action Using a left
action \(\operatorname{GL}_n \curvearrowright{\mathbb{R}}^n\), we can
form
\(\mathop{\mathrm{Frame}}(E) { \underset{\scriptscriptstyle {\operatorname{GL}_n} }{\times} } {\mathbb{R}}^n\),
a fiber bundle with a \(G \coloneqq\operatorname{GL}_n\) structure,
i.e.~exactly a vector bundle.

\begin{exercise}[?]

Show that there is a homeomorphism
\begin{align*}
\mathop{\mathrm{Frame}}(E) { \underset{\scriptscriptstyle {\operatorname{GL}_n} }{\times} } {\mathbb{R}}^n \xrightarrow{\sim}E
.\end{align*}

\end{exercise}

For the reverse map, take a map \(f\) defined by
\((\mathbf{e}_1, \cdots, \mathbf{e}_n) \in \pi^{-1}(b) \subset \mathop{\mathrm{Frame}}(E)\)
and
\({\left[ {{ {b}_1, {b}_2, \cdots, {b}_{n}}} \right]}^t \in {\mathbb{R}}^n\)
to \(\sum_{i=1}^n b_i \mathbf{e}_i\). For this to be well-defined, one
needs to show the following:
\begin{align*}
f(( \mathbf{e}_1, \cdots, \mathbf{e}_n) A, \mathbf{b}) = f( (\mathbf{e}_1, \cdots, \mathbf{e}_n), A\mathbf{b}) && \forall A\in \operatorname{GL}_n({\mathbb{R}})
.\end{align*}
The left hand side is
\begin{align*}
b_1 (a_{1, 1} \mathbf{e}_1 + \cdots + a_{n, 1} \mathbf{e}_n ) + \cdots + b_n(a_{1, n} \mathbf{e}_1 + \cdots + a_{n, n} \mathbf{e}_n)
= \sum_{i=1}^n b_i \qty{ \sum_{j=1}^n a_{j, i} \mathbf{e}_j }
.\end{align*}

The right-hand side is
\begin{align*}
(a_{1, 1}b_1 + \cdots + a_{1, n}b_n)\mathbf{e}_1 + \cdots + (a_{n, 1} b_1 + \cdots + a_{n,n} b_n)\mathbf{e}_n
= \sum_{i=1}^n \qty{\sum_{j=1}^n a_{i, j} b_i } \mathbf{e}_i
,\end{align*}
and one can check that these sums match term by term.

\end{example}

\begin{remark}

Note that if we choose a basis for the fibers, we can set
\(A' \coloneqq{\left[ {\mathbf{e}_1, \cdots, \mathbf{e}_n} \right]}^t\)
to be the matrix with columns \(\mathbf{e}_i\), the map \(f\) is given
by \(f(A', \mathbf{b}) \coloneqq A'\mathbf{b}\), and we're showing that
\((A'A)\mathbf{b} = A'(A\mathbf{b})\). However, this involves choosing
an isomorphism between the abstract fibers and \({\mathbb{R}}^n\).

\end{remark}

\begin{remark}

What are local charts for a principal bundle? For
\(P{ \underset{\scriptscriptstyle {G} }{\times} } F\), pick charts
\((U, \phi)\) for \(P \xrightarrow{\pi} B\):
\begin{align*}
\phi: \pi^{-1}(U) &\to U \times G \\
x & \mapsto (\pi(x), \gamma(x))
.\end{align*}
Then a local chart for the principal bundle is of the form
\begin{align*}
\pi^{-1}(U) { \underset{\scriptscriptstyle {G} }{\times} } F &\xrightarrow{\tilde \phi} U \times F \\
(x, f) &\mapsto ( \pi(x), \gamma(x) f)
.\end{align*}
We also have
\begin{align*}
(U \times G){ \underset{\scriptscriptstyle {G} }{\times} } F &\to U \times F \\
( (x, g), f) &\mapsto (x, gf)
.\end{align*}
One can invert \(\tilde \phi\) using
\((a, f) \mapsto (\phi^{-1}(a, 1), f)\). This yields transition
functions: writing
\begin{align*}
\phi_V: \pi^{-1}(V) &\to V \times G \\ 
x &\mapsto (\pi(x), \psi(x) )
,\end{align*}
then
\begin{align*}
\phi_{VU} = \phi_V \circ \phi_U^{-1}: 
(a, f) & \\
&\xrightarrow{\phi_U^{-1}} ( \phi^{-1}_U(a, 1), \,\, f) \\
&\xrightarrow{\phi_V} (\pi \phi_U^{-1}(a, 1), \,\, \psi( \phi_U^{-1}(a, 1))f ) \\
&= (a,\,\, \psi(\phi_U^{-1}(a, 1)) f)
.\end{align*}
This says that \((a, 1) \mapsto \psi( \phi^{-1}_U(a, 1))\).

\end{remark}

\begin{remark}

In general, for a bundle \(E \xrightarrow{\pi} B\), taking local
trivializations \(\phi_U, \phi_V\), we get
\(\phi_{VU}: (U \cap V ) \times F {\circlearrowleft}\), or currying an
argument, \(\phi_{VU}: U \cap V \to {\operatorname{Homeo}}(F, F)\). If
the bundle satisfies the cocycle condition
\(\phi_{UW} = \phi_{VW} \circ \phi_{UV}\). Given a covering
\(\left\{{U_i}\right\}_{i\in I}\rightrightarrows B\), we get
\(\phi_{ij}: U_i \cap U_j \to G\) and a topological space \(F\) with
\(G \subseteq {\operatorname{Homeo}}(F, F)\) satisfying the cocycle
condition \(\phi_{ik} = \phi_{jk} \circ \phi_{ij}\), then we can build a
fiber bundle with fiber \(F\) and structure group \(G\) by setting
\(E = {\textstyle\coprod}_{i\in I} (U_i \times F)/\sim\) We then set for
\(b\in U_i \cap U_j\) the equivalence
\begin{align*}
(U_i \times F) \ni (b, f)\sim (b, \phi_{ij}(b) f) \in (U_j \times F)
.\end{align*}
This is an equivalence relation precisely when the cocycle condition
holds. This is referred to as \textbf{clutching data}.

\end{remark}

\begin{example}[The Mobius band is clutch]

Let \({\mathbb{Z}}/2\curvearrowright{\mathbb{R}}\) by \(t\mapsto -t\)
with \(U, V\) defined as follows:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{41pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-08-31_13-53.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Labeling the intersections as 1, 2, we set
\begin{align*}
\phi_{VU}: (U \cap V) = (U \cap V)_1 {\textstyle\coprod}(U \cap V)_2 &\to {\mathbb{Z}}/2 && \subseteq {\operatorname{Homeo}}({\mathbb{R}}) \\
x {\textstyle\coprod}y &\mapsto x {\textstyle\coprod}-y
.\end{align*}
This yields the open Mobius band.

\end{example}

\begin{question}

Actually, several questions. Assume \(F\) is a fixed fiber common to all
of the following constructions, since bundles with non-homeomorphic
fibers can't be isomorphic.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Given clutching data \(\left\{{\phi_{ij}}\right\}\), when is the
  resulting fiber bundle trivial?
\item
  Given two sets of clutching data \(\left\{{\phi_{ij}}\right\}\) and
  \(\left\{{\psi_{ij}}\right\}\) with the same open cover
  \(\left\{{U_i}\right\} \rightrightarrows X\), when are the
  corresponding bundles \(G{\hbox{-}}\)isomorphic?
\item
  Given two sets of clutching data \(\left\{{\phi_{ij}}\right\}\) and
  \(\left\{{\psi_{ij}}\right\}\) with the \emph{different} open cover
  \(\left\{{U_i}\right\} \rightrightarrows X\) and
  \(\left\{{V_i}\right\} \rightrightarrows X\), when are the
  corresponding bundles \(G{\hbox{-}}\)isomorphic?
\end{enumerate}

\end{question}

\begin{lemma}[?]

The fiber bundle obtained from \(\phi_{ij}\) is trivial iff there exists
a map \(\gamma_i: U_i \to G\) such that
\(\phi_{ij} = \gamma_i \gamma_j^{-1}\).

\end{lemma}

\begin{proof}[?]

The trivial bundle is \(B\times F\to B\), so if we have \(E\to B\), we
can take a map
\begin{align*}
U_i \times F &\to U_i \times F \\
(b, f) &\mapsto (b, \gamma_i(b) f)
.\end{align*}
Use that \(B\times F\) is a trivial bundle, so it is its own
trivialization.

\begin{quote}
To be continued next time.
\end{quote}

\end{proof}

\hypertarget{thursday-september-02}{%
\section{Thursday, September 02}\label{thursday-september-02}}

\begin{remark}

Recall that we have a correspondence
\begin{align*}
\left\{{\substack{
  \text{Vector bundles }E
}}\right\}
&\adjunction{\text{clutching}}{\text{mixing}}{}{}
\left\{{\substack{
  \text{Principal $\operatorname{GL}_n{\hbox{-}}$bundles $\mathop{\mathrm{Frame}}(E)$}
}}\right\}
\end{align*}
We saw that
\(E \cong \mathop{\mathrm{Frame}}(E) { \underset{\scriptscriptstyle {\operatorname{GL}_n({\mathbb{R}})} }{\times} } {\mathbb{R}}^n\).
If we take \(\mathop{\mathrm{Frame}}(E)\), mix, and apply the clutching
construction, is the result bundle-isomorphic to the frame bundle?

\end{remark}

\begin{remark}

Recall the clutching construction: we take a cover
\(\left\{{U_i}\right\}_{i\in I}\) and \(\phi_{ij}: U_i \cap U_j \to G\)
satisfying the cocycle condition \(\phi_{ij}\phi_{jk} = \phi_{ik}\),
then \(G \subseteq {\operatorname{Homeo}}(F, F)\) and we construct a
fiber bundle \(\displaystyle\bigcup_{i\in I} U_i \times F / \sim\) where
for \(b\in (U_i \cap U_j )\) and
\begin{align*}
(b, f) \in (U_i \cap U_j )\times F \subseteq U_i\times F
,\end{align*}
we send this to
\begin{align*}
(b, \phi_{ji}(b) f ) \in (U_i \cap U_j)\times F \subseteq U_j \times F
.\end{align*}
This will be a fiber bundle with fiber \(F\) and structure group \(G\).
Moreover, if \(F=G\), this will be a principal \(G{\hbox{-}}\)bundle
using right-multiplication.

\end{remark}

\begin{question}

How can we tell when two fiber bundles constructed via clutching are
isomorphic?

\end{question}

\begin{lemma}[when clutched bundles are trivial]

The bundle formed by the clutching data \(\left\{{\phi_{ij}}\right\}\)
is trivial (so isomorphic to the trivial bundle) iff there exist
\(\gamma_i: U_i\to G\) such that
\(\phi_{ji} = \gamma_j \circ \gamma_i^{-1}\).

\end{lemma}

\begin{remark}

For principal bundles, these \(\gamma_i\) will give sections assembling
to a global section obtained from clutching data:

\begin{center}
\begin{tikzcd}
    P && {U_i \times G} \\
    \\
    B && {U_i}
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow["{s(b) = (b, \gamma_i(b))}"', curve={height=24pt}, from=3-3, to=1-3]
    \arrow["{\psi_B}"', from=3-1, to=3-3]
    \arrow["\psi"', from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJVX2kgXFx0aW1lcyBHIl0sWzIsMiwiVV9pIl0sWzAsMiwiQiJdLFswLDAsIlAiXSxbMywyXSxbMCwxXSxbMSwwLCJzKGIpID0gKGIsIFxcZ2FtbWFfaShiKSkiLDIseyJjdXJ2ZSI6NH1dLFsyLDEsIlxccHNpX0IiLDJdLFszLDAsIlxccHNpIiwyXV0=}{Link
to Diagram}
\end{quote}

The map on \(U_i \to \displaystyle\bigcup_i U_i \times F\) will be
\((b, \gamma_i(b))\), and we can use that
\begin{align*}
(b, \gamma_i(b)) \sim (b, \phi_{ji}(b)\gamma_i(b) ) \sim (b, \gamma_j(b))
,\end{align*}
so these agree on overlaps.

\end{remark}

\begin{lemma}[?]

If a principal bundle \(P\to B\) has a global section, then \(P\) is
trivial, so \(P\cong B\times G\) as bundles. The idea:

\begin{center}
\begin{tikzcd}
    {(b, s(b)g)} &&&& {(b, g)} \\
    P &&&& {B\times G} \\
    \\
    && B \\
    && b
    \arrow["{p_1}"', from=2-5, to=4-3]
    \arrow["\pi", from=2-1, to=4-3]
    \arrow[dashed, from=2-1, to=2-5]
    \arrow["s", curve={height=-24pt}, from=4-3, to=2-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwxLCJQIl0sWzIsMywiQiJdLFs0LDEsIkJcXHRpbWVzIEciXSxbMiw0LCJiIl0sWzAsMCwiKGIsIHMoYilnKSJdLFs0LDAsIihiLCBnKSJdLFsyLDEsInBfMSIsMl0sWzAsMSwiXFxwaSJdLFswLDIsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDAsInMiLDAseyJjdXJ2ZSI6LTR9XV0=}{Link
to Diagram}
\end{quote}

\end{lemma}

\begin{proof}[of lemma about when clutched bundles are trivial]

\(\implies\):

If \(E\) is trivial, we have an isomorphism

\begin{center}
\begin{tikzcd}
E 
  \ar[rd, "f"]
  \ar[rr, "\pi"] 
& 
& 
P\times G
  \ar[ld, "p_1"] 
\\
& 
B 
& 
\end{tikzcd}
\end{center}

We have a \(G{\hbox{-}}\)isomorphism \(E_1 \xrightarrow{f} E_2\), and so
a composition

\begin{center}
\begin{tikzcd}
    {(U\cap V) \times F} && {\pi^{-1}(U \cap V)} && {\pi^{-1}(U \cap V)} && {(U\cap V) \times F}
    \arrow["{\phi_U}", from=1-3, to=1-1]
    \arrow["f"', from=1-3, to=1-5]
    \arrow["{\phi_V}"', from=1-5, to=1-7]
    \arrow["F", curve={height=-30pt}, from=1-1, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCIoVVxcaW50ZXJzZWN0IFYpIFxcdGltZXMgRiJdLFsyLDAsIlxccGlcXGludihVIFxcaW50ZXJzZWN0IFYpIl0sWzQsMCwiXFxwaVxcaW52KFUgXFxpbnRlcnNlY3QgVikiXSxbNiwwLCIoVVxcaW50ZXJzZWN0IFYpIFxcdGltZXMgRiJdLFsxLDAsIlxccGhpX1UiXSxbMSwyLCJmIiwyXSxbMiwzLCJcXHBoaV9WIiwyXSxbMCwzLCJGIiwwLHsiY3VydmUiOi01fV1d}{Link
to Diagram}
\end{quote}

Here we've used that \(f\) commutes with the projection maps. We want to
show \(\operatorname{im}(F) \subseteq G\). We have a composite

\begin{center}
\begin{tikzcd}
    {U\times F} && {\pi^{-1}(U_i)} && {U\times F}
    \arrow["{\phi_i}", from=1-3, to=1-1]
    \arrow["f"', from=1-3, to=1-5]
    \arrow["{\gamma_i = \phi_i^{-1}\circ f: U\to G}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJVXFx0aW1lcyBGIl0sWzQsMCwiVVxcdGltZXMgRiJdLFsyLDAsIlxccGlcXGludihVX2kpIl0sWzIsMCwiXFxwaGlfaSJdLFsyLDEsImYiLDJdLFswLDEsIlxcZ2FtbWFfaSA9IFxccGhpX2lcXGludiBcXGNpcmMgZjogVVxcdG8gRyIsMCx7ImN1cnZlIjotNX1dXQ==}{Link
to Diagram}
\end{quote}

We can fill this in to a commutative diagram:

\begin{center}
\begin{tikzcd}
    {(U_i\cap U_j)\times F} && {\phi^{-1}(U_i \cap U_j)} && {(U_i\cap U_j)\times F} \\
    \\
    {\phi^{-1}(U_i \cap U_j)} && {\phi^{-1}(U_i \cap U_j)} && {(U_i\cap U_j)\times F}
    \arrow["f", from=1-3, to=1-5]
    \arrow["{\phi_i}"', from=1-3, to=1-1]
    \arrow["{\phi_j}"', from=3-3, to=3-1]
    \arrow["f", from=3-3, to=3-5]
    \arrow[Rightarrow, no head, from=1-5, to=3-5]
    \arrow[Rightarrow, no head, from=1-3, to=3-3]
    \arrow["{\gamma_i}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, from=1-1, to=1-5]
    \arrow["{\gamma_j}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=30pt}, from=3-1, to=3-5]
    \arrow["{\therefore \phi_{ji}}", color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

The converse direction proceeds similarly!

\end{proof}

\begin{lemma}[?]

A \(G{\hbox{-}}\)isomorphism between the bundles \(E_1, E_2\) obtained
from clutching data \(\left\{{\phi_{ij}}\right\}\) and
\(\left\{{\psi_{ij}}\right\}\) respectively with the same cover
\(\left\{{U_i}\right\}_{i\in I}\) give maps \(\gamma_i: U_i\to G\) such
that
\begin{align*}
\gamma_j \phi_{ji} \gamma_i^{-1}= \psi_{ji}
.\end{align*}

\end{lemma}

\begin{proof}[?]

We can form the composite

\begin{center}
\begin{tikzcd}
    {U_i\times F} && {\pi_1^{-1}(U_i)} && {\pi_2^{-1}(U_i)} && {U_i \times F}
    \arrow["{\phi_i}", from=1-3, to=1-1]
    \arrow["f"', from=1-3, to=1-5]
    \arrow["{\psi_j}"', from=1-5, to=1-7]
    \arrow["{\gamma_i: U_i\to G}", curve={height=30pt}, from=1-1, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJVX2lcXHRpbWVzIEYiXSxbMiwwLCJcXHBpXzFcXGludihVX2kpIl0sWzQsMCwiXFxwaV8yXFxpbnYoVV9pKSJdLFs2LDAsIlVfaSBcXHRpbWVzIEYiXSxbMSwwLCJcXHBoaV9pIl0sWzEsMiwiZiIsMl0sWzIsMywiXFxwc2lfaiIsMl0sWzAsMywiXFxnYW1tYV9pOiBVX2lcXHRvIEciLDAseyJjdXJ2ZSI6NX1dXQ==}{Link
to Diagram}
\end{quote}

And then assemble a commuting diagram:

\begin{center}
\begin{tikzcd}
    {(U_i \cap U_j)\times F} && {\pi_1^{-1}(U_i \cap U_j)} && {\pi_2^{-1}(U_i \cap U_j)} && {(U_i \cap U_j)\times F} \\
    \\
    {(U_i \cap U_j)\times F} && {\pi_1^{-1}(U_i \cap U_j)} && {\pi_2^{-1}(U_i \cap U_j)} && {\pi_2^{-1}(U_i \cap U_j)}
    \arrow["{\phi_i}", from=1-3, to=1-1]
    \arrow["{\psi_j}", from=1-5, to=1-7]
    \arrow["f", from=1-3, to=1-5]
    \arrow["{\gamma_i}", curve={height=-30pt}, no head, from=1-1, to=1-7]
    \arrow["{\psi_{ji}}", from=1-7, to=3-7]
    \arrow["{\psi_j}"', from=3-5, to=3-7]
    \arrow[Rightarrow, no head, from=1-5, to=3-5]
    \arrow["f"', from=3-3, to=3-5]
    \arrow[Rightarrow, no head, from=1-3, to=3-3]
    \arrow["{\phi_i}"', from=3-3, to=3-1]
    \arrow["{\phi_{ji}}"', from=1-1, to=3-1]
    \arrow["{\gamma_j}"', curve={height=30pt}, from=3-1, to=3-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{proof}

\hypertarget{nonabelian-ux10dech-cohomology}{%
\subsection{Nonabelian Čech
Cohomology}\label{nonabelian-ux10dech-cohomology}}

\begin{definition}[Čech complex]

Let
\({\mathcal{U}}\coloneqq\left\{{U_i}\right\}_{i\in I}\rightrightarrows B\)
an open cover, and define
\begin{align*}
{\check{C}}^0({\mathcal{U}}; G) \coloneqq\left\{{ \left\{{\gamma_i: U_i\to G}\right\}_{i\in I} }\right\}
,\end{align*}
which is a group under pointwise multiplication. Define
\begin{align*}
{\check{C}}^2({\mathcal{U}}; G) &\coloneqq\left\{{ \left\{{\phi_{ij}: U_i \cap U_j \to G}\right\}_{i, j\in I} }\right\}\\
{\check{C}}^3({\mathcal{U}}; G) &\coloneqq\left\{{ \left\{{\phi_{ijk}: U_i \cap U_j \cap U_k \to G}\right\}_{i,j,k \in I} }\right\}
,\end{align*}
and boundary maps
\begin{align*}
\delta^{0}: {\check{C}}^0({\mathcal{U}}; G) &\to {\check{C}}^1({\mathcal{U}}; G)\\
\left\{{\gamma_i: U_i\to G}\right\} &\mapsto \left\{{\phi_{ji} \coloneqq\gamma_j\gamma_i^{-1}: U_i \cap U_j \to G }\right\}
,\end{align*}

\begin{align*}
\delta^{1}: {\check{C}}^1({\mathcal{U}}; G) &\to {\check{C}}^2({\mathcal{U}}; G)\\
\left\{{\phi_{ij}: U_i \cap U_j \to G}\right\} &\mapsto \left\{{\eta{ijk} \coloneqq\phi_{ij} \phi_{jk} \phi_{ik}^{-1}: U_i \cap U_j \cap U_k \to G }\right\}
.\end{align*}

\end{definition}

\begin{remark}

One can check that \(\delta^1 \circ \delta^0 = 0\) is trivial. And
1-cocycle will yield a fiber bundle.

\end{remark}

\begin{lemma}[1]

A bundle is trivial iff it is a 1-coboundary, where we take
\(Z^1({\mathcal{U}}; G) \coloneqq\ker \delta^1\),
\(B^1({\mathcal{U}}; G) \coloneqq\operatorname{im}\delta^0\).

\end{lemma}

\begin{warnings}

We'd like to define homology as \(Z/B\), but since these aren't abelian
groups, the coboundaries \(B\) may not be normal in \(Z\) and the
quotient may not yield a group.

\end{warnings}

\begin{definition}[First Čech cohomology]

There is an action of
\({\check{C}}^0({\mathcal{U}}; G)\curvearrowright{\check{C}}^1({\mathcal{U}}; G)\)
given by taking \(\gamma \coloneqq\left\{{\gamma_i}\right\}_{i\in I}\)
and setting \((\gamma \phi)_{ij} = \gamma_i \phi_{ij} \gamma_j^{-1}\),
which descends to an action on \(Z^1\). We can take the quotient by this
action to define
\begin{align*}
{\check{H}}^1({\mathcal{U}}; G) \coloneqq Z^1({\mathcal{U}}; G) / \sim
.\end{align*}

\end{definition}

\begin{lemma}[2]

Two bundles are isomorphic iff they yield the same element in
\({\check{H}}^1({\mathcal{U}}; G)\).

\end{lemma}

\begin{remark}

This works when bundles have the same open cover, and if not, we can
take a common refinement.

\end{remark}

\hypertarget{tuesday-september-07}{%
\section{Tuesday, September 07}\label{tuesday-september-07}}

\begin{remark}

Recall that given a \(B\in {\mathsf{Top}}\) and
\({\mathcal{U}}\rightrightarrows B\), we defined
\({\check{H}}_1({\mathcal{U}}; G)\) which classified isomorphism classes
of fiber bundles \(E \xrightarrow{\pi} B\) with fiber \(F\),
\(G \subseteq {\operatorname{Homeo}}(F)\), and structure group \(G\),
given by clutching data using \({\mathcal{U}}\). The cochains were given
by the following:

\begin{align*}
{\check{C}}^0({\mathcal{U}}; G) &= \left\{{\left\{{ \gamma_i: U_i \to G}\right\}_{i\in I}}\right\}\\
{\check{C}}^1({\mathcal{U}}; G) &= \left\{{\left\{{ \phi_{ij}: U_i \cap U_j \to G }\right\}_{i, j \in I}}\right\}\\
{\check{C}}^2({\mathcal{U}}; G) &= \left\{{\left\{{ \eta_{ijk}: U_i \cap U_j \cap U_k \to G }\right\}_{i,j,k\in I}}\right\}
\end{align*}
with boundary maps \(\delta_i: {\check{C}}^{i-1} \to {\check{C}}^i\):
\begin{align*}
(\delta_1 \gamma)_{ij} &= \gamma_i \gamma_j^{-1}\\
(\delta_2 \phi)_{ijk} &= \phi_{ij} \phi_{jk} \phi_{ik}^{-1}
.\end{align*}
Note that

\begin{itemize}
\tightlist
\item
  \(\delta_2 \circ \delta_1 = 0\)
\item
  \(\ker \delta_2 = Z^1({\mathcal{U}}; G)\) yields clutching data,
  i.e.~a fiber bundle with fiber \(F\),
\item
  \(\operatorname{im}\delta_1\) yields trivial bundles,
\item
  \({\check{H}}^1({\mathcal{U}}; G) \coloneqq Z^1({\mathcal{U}}; G) / \operatorname{im}({\check{C}}^0({\mathcal{U}}; G) \to Z^1({\mathcal{U}}; G))\).
\end{itemize}

We'll see that
\((\gamma \varphi)_{ij} = \gamma_i \varphi_{ij} \gamma_j^{-1}\), and by
a lemma this will prove the above claim about classifying isomorphism
classes.

\end{remark}

\begin{definition}[Refinement of covers]

We say a cover \({\mathcal{V}}\coloneqq\left\{{V_j}\right\}_{j\in J}\)
is a \textbf{refinement} of
\({\mathcal{U}}\coloneqq\left\{{U_i}\right\}_{i\in I}\) iff there exists
a function \(f:J\to I\) between the index sets where
\(V_j \subseteq U_{f(j)}\) for all \(j\).

\begin{quote}
DZG: I'll write \({\mathcal{V}}\leq {\mathcal{U}}\) if \({\mathcal{V}}\)
is a refinement of \({\mathcal{U}}\).
\end{quote}

\end{definition}

\begin{remark}

Since any two covers have a common refinement, we'll assume
\({\mathcal{V}}\leq {\mathcal{U}}\) is always a refinement. We can then
restrict clutching data from \({\mathcal{U}}\) to \({\mathcal{V}}\):
given \(\left\{{\phi_{ij}}\right\}_{i,j\in I}\), we can set
\(\psi_{ij} \coloneqq{ \left.{{ \phi_{f(i), f(j)}}} \right|_{{V_i \cap V_j}} }\),
noting that if \(V_j \subseteq U_{f(j)}\) and \(V_i \subseteq U_{f(i)}\)
then \(V_i \cap V_j \subseteq U_{f(i)} \cap U_{f(j)}\). These yield maps
\(\psi_{ij}: V_i \cap V_j \to G\) satisfying the cocycle condition, so
\(\psi_{ij} \in Z^1({\mathcal{V}}; G)\). This means that we have map
\(Z^1({\mathcal{U}}; G)\to Z^1({\mathcal{V}}; G)\) which respects the
actions of
\({\check{C}}^0({\mathcal{U}}; G), {\check{C}}^0({\mathcal{V}}; G)\)
respectively. Since the category of covers with morphisms given by
refinements come from a preorder, we can take a colimit to define
\begin{align*}
{\check{H}}^1(B; G) \coloneqq\colim_{{\mathcal{U}}\rightrightarrows B} {\check{H}}^1({\mathcal{U}}; G)
.\end{align*}

\end{remark}

\begin{lemma}[?]

There is a bijection
\begin{align*}
\left\{{\substack{
  \text{Fiber bundles with fiber }F \\
  \text{and structure group }G
}}\right\}_{/ {\sim}} 
&\rightleftharpoons
{\check{H}}^1(B; G)
\end{align*}
In particular, these classes are independent of \(F\).

\end{lemma}

\begin{corollary}[?]

There is an equivalence of categories
\begin{align*}
\left\{{\substack{
  \text{Fiber bundles with fiber $F$} \\
  \text{and structure group }G \\
  \text{over }B
}}\right\}_{/ {\sim}} 
&\rightleftharpoons
\mathop{\mathrm{Prin}}{\mathsf{Bun}}(G) _{/ {B}} 
,\end{align*}
where the right-hand side are principal \(G{\hbox{-}}\)bundles.

\end{corollary}

\begin{definition}[$G\dash$structures]

Given a map \(G \to {\operatorname{Homeo}}(F)\), a
\textbf{\(G{\hbox{-}}\)structure} on an \(F{\hbox{-}}\)bundle
\(E \xrightarrow{\pi} B\) is the following data: given clutching data
\(\phi_{ij}\), lifts of the following form that again satisfy the
cocycle condition:

\begin{center}
\begin{tikzcd}
    && G \\
    \\
    {U_i\cap U_j} && {{\operatorname{Homeo}}(F)}
    \arrow["{\phi_{ij}}"', from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow["{\tilde \phi_{ij}}", dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwyLCJVX2lcXGludGVyc2VjdCBVX2oiXSxbMiwwLCJHIl0sWzIsMiwiXFxIb21lbyhGKSJdLFswLDIsIlxccGhpX3tpan0iLDJdLFsxLDJdLFswLDEsIlxcdGlsZGUgXFxwaGlfe2lqfSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Note that we need to impose the cocycle condition, since lifts may not
be unique and some choices may not glue correctly!

\end{remark}

\begin{example}[$\Spin_n\dash$structures]

Using the known \({\operatorname{Spin}}\) double covers, we can form the
composition
\begin{align*}
{\operatorname{Spin}}_n({\mathbb{R}}) \xrightarrow{\times 2}  {\operatorname{SO}}_n({\mathbb{R}}) \hookrightarrow{\operatorname{Homeo}}({\mathbb{R}}^n)
.\end{align*}
Then a \({\operatorname{Spin}}_n{\hbox{-}}\)structure on any
\({\mathbb{R}}^n{\hbox{-}}\)bundle is a lift of transition functions
from \({\operatorname{Homeo}}({\mathbb{R}}^n)\) to
\({\operatorname{Spin}}_n\) satisfying the cocycle condition.

\end{example}

\begin{definition}[Fiber products]

We can fill in a commutative square in the following way:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{X{ \underset{\scriptscriptstyle {B} }{\times} }Z} && Z \\
    \\
    X && B
    \arrow["\pi", from=1-3, to=3-3]
    \arrow["f"', from=3-1, to=3-3]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=3-1]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, color={rgb,255:red,92;green,92;blue,214}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJYIl0sWzIsMiwiQiJdLFsyLDAsIloiXSxbMCwwLCJYXFxmaWJlcnByb2R7Qn1aIixbMjQwLDYwLDYwLDFdXSxbMiwxLCJcXHBpIl0sWzAsMSwiZiIsMl0sWzMsMCwiIiwyLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsMiwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsMSwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d}{Link
to Diagram}
\end{quote}

Here we can construct the fiber product as
\begin{align*}
X{ \underset{\scriptscriptstyle {B} }{\times} } Z = \left\{{(x, e) {~\mathrel{\Big\vert}~}\pi(e) = f(x)}\right\}
.\end{align*}

It satisfies the following universal property:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,214;green,92;blue,92}{W} \\
    \\
    && {X{ \underset{\scriptscriptstyle {B} }{\times} }Z} && Z \\
    \\
    && X && B
    \arrow["\pi", from=3-5, to=5-5]
    \arrow["f"', from=5-3, to=5-5]
    \arrow[from=3-3, to=5-3]
    \arrow[from=3-3, to=3-5]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-3, to=5-5]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-5]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=5-3]
    \arrow["{\exists!}", color={rgb,255:red,214;green,92;blue,92}, dashed, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiw0LCJYIl0sWzQsNCwiQiJdLFs0LDIsIloiXSxbMiwyLCJYXFxmaWJlcnByb2R7Qn1aIl0sWzAsMCwiVyIsWzAsNjAsNjAsMV1dLFsyLDEsIlxccGkiXSxbMCwxLCJmIiwyXSxbMywwXSxbMywyXSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwyLCIiLDAseyJjb2xvdXIiOlswLDYwLDYwXX1dLFs0LDAsIiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdfV0sWzQsMywiXFxleGlzdHMhIiwwLHsiY29sb3VyIjpbMCw2MCw2MF0sInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX0sWzAsNjAsNjAsMV1dXQ==}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{lemma}[?]

If \(\pi: P\to X\) is a principal \(G{\hbox{-}}\)bundle and \(f:Y\to X\)
is a continuous map, then the following highlighted portion of the
pullback is again a principal \(G{\hbox{-}}\)bundle:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{f^* p \coloneqq Y{ \underset{\scriptscriptstyle {X} }{\times} } P} && P \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{Y} && X
    \arrow["{{\operatorname{pr}}_1}", color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=3-1]
    \arrow["\pi", from=1-3, to=3-3]
    \arrow["f"', from=3-1, to=3-3]
    \arrow["{\exists \tilde f}", dashed, from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJmXiogcCBcXGRhIFlcXGZpYmVycHJvZHtYfSBQIixbMjQwLDYwLDYwLDFdXSxbMCwyLCJZIixbMjQwLDYwLDYwLDFdXSxbMiwyLCJYIl0sWzIsMCwiUCJdLFswLDEsIlxccHJfMSIsMCx7ImNvbG91ciI6WzI0MCw2MCw2MF19LFsyNDAsNjAsNjAsMV1dLFszLDIsIlxccGkiXSxbMSwyLCJmIiwyXSxbMCwzLCJcXGV4aXN0cyBcXHRpbGRlIGYiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwyLCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

We in fact obtain
\({\operatorname{pr}}_1^{-1}(y) = \pi^{-1}(f(y))\cong G\), and there
will be a right \(G{\hbox{-}}\)action on each fiber. Behold this gnarly
diagram:

\begin{center}
\begin{tikzcd}
    {f^{-1}(U)\times G} \\
    {{\operatorname{pr}}_1^{-1}(f^{-1}(U))} & {f^* P} && P & {\pi^{-1}(U)} && {U\times G} \\
    \\
    & Y && X & U \\
    {f^{-1}(U)}
    \arrow["\phi", from=2-5, to=2-7]
    \arrow[dotted, hook', from=2-5, to=2-4]
    \arrow[dotted, hook', from=4-5, to=4-4]
    \arrow["\pi"', from=2-4, to=4-4]
    \arrow["f"', curve={height=30pt}, dotted, maps to, from=5-1, to=4-5]
    \arrow["f"', from=4-2, to=4-4]
    \arrow["{{\operatorname{pr}}_1}"', from=2-7, to=4-5]
    \arrow["\pi"', dotted, from=2-5, to=4-5]
    \arrow[from=2-2, to=2-4]
    \arrow["{{\operatorname{pr}}_1}"', from=2-2, to=4-2]
    \arrow[curve={height=-30pt}, dotted, from=1-1, to=2-7]
    \arrow[dotted, hook, from=1-1, to=2-2]
    \arrow[dotted, hook, from=5-1, to=4-2]
    \arrow["{{\operatorname{pr}}_1}"', dotted, from=2-1, to=5-1]
    \arrow[dotted, hook, from=2-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

If \(P\to X\) is trivial, this says the pullback will be trivial and
\(U\times G\mapsto f^{-1}(U)\times G\) will be a homeomorphism.

\end{lemma}

\begin{remark}

So the functor \(X\mapsto \mathop{\mathrm{Prin}}_G(X)\) is contravariant
functor.

\end{remark}

\begin{theorem}[Bundle homotopy lemma]

Suppose \(B\) is paracompact and Hausdorff, then there is a principal
\(G{\hbox{-}}\)bundle \(P \xrightarrow{\pi} I \times B\). Consider the
fiber bundle
\(P_0 \coloneqq{ \left.{{P}} \right|_{{\left\{{0}\right\}\times B}} } \to B\),
then there is a diagram:

\begin{center}
\begin{tikzcd}
    {P_0} && {{ \left.{{P}} \right|_{{0\times B}} }} \\
    \\
    B && 0\times B
    \arrow["\operatorname{id}", from=1-1, to=1-3]
    \arrow["\pi"', from=1-1, to=3-1]
    \arrow["\operatorname{id}"', from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJQXzAiXSxbMiwwLCJcXHJve1B9ezBcXHRpbWVzIEJ9Il0sWzAsMiwiQiJdLFsyLDIsIjBcXHRpbWVzQiJdLFswLDEsIlxcaWQiXSxbMCwyLCJcXHBpIiwyXSxbMiwzLCJcXGlkIiwyXSxbMSwzXV0=}{Link
to Diagram}
\end{quote}

This extends to an isomorphism \(I\times P_0 \to I\times B\) and
\(P\to I\times B\):

\begin{center}
\begin{tikzcd}
    & {P_0\times I} && P \\
    & {} \\
    & {B\times I} && I\times B \\
    {P_0} && {{ \left.{{P}} \right|_{{0\times B}} }} \\
    \\
    B && 0\times B
    \arrow["\operatorname{id}", from=4-1, to=4-3]
    \arrow["\pi"', from=4-1, to=6-1]
    \arrow["\operatorname{id}"', from=6-1, to=6-3]
    \arrow[from=4-3, to=6-3]
    \arrow["\cong"', from=1-2, to=1-4]
    \arrow[from=3-2, to=3-4]
    \arrow[from=1-2, to=3-2]
    \arrow["\pi"', from=1-4, to=3-4]
    \arrow[dashed, from=4-1, to=1-2]
    \arrow[dashed, from=6-1, to=3-2]
    \arrow[dashed, from=4-3, to=1-4]
    \arrow[dashed, from=6-3, to=3-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{theorem}

\begin{corollary}[?]

\(P_1 = { \left.{{P}} \right|_{{1\times B}} } \cong P_0\).

\end{corollary}

\begin{corollary}[?]

If \(f_0 \sim f_1: Y\to X\) are homotopic and \(P\to X\), then
\(f_0^*P \cong f_1^* P\).

\end{corollary}

\begin{proof}[?]

Use the homotopy lifting property to get a map \(h\):

\begin{center}
\begin{tikzcd}
    {h^*P} && P \\
    \\
    {I\times Y} && Y
    \arrow["h", from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\pi"', from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJoXipQIl0sWzAsMiwiSVxcdGltZXMgWSJdLFsyLDAsIlAiXSxbMiwyLCJZIl0sWzEsMywiaCJdLFswLDFdLFsyLDMsIlxccGkiLDJdLFswLDJdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

Then
\({ \left.{{h^* P}} \right|_{{0\times Y}} } \simeq{ \left.{{h^*P}} \right|_{{1\times Y}} } \cong f_1^*P\).

\end{proof}

\hypertarget{thursday-september-09}{%
\section{Thursday, September 09}\label{thursday-september-09}}

\hypertarget{corollaries-of-the-homotopy-bundle-lemma}{%
\subsection{Corollaries of the homotopy bundle
lemma}\label{corollaries-of-the-homotopy-bundle-lemma}}

\begin{remark}

Last time: the bundle homotopy lemma. If
\(P\to I\times X \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)\), then
there is a bundle isomorphism

\begin{center}
\begin{tikzcd}
    {I\times P_0} &&& P \\
    \\
    {I\times X} &&& {I\times X}
    \arrow[""{name=0, anchor=center, inner sep=0}, from=1-1, to=3-1]
    \arrow[""{name=1, anchor=center, inner sep=0}, from=1-4, to=3-4]
    \arrow["f\cong", shorten <=20pt, shorten >=20pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJJXFx0aW1lcyBQXzAiXSxbMCwyLCJJXFx0aW1lcyBYIl0sWzMsMCwiUCJdLFszLDIsIklcXHRpbWVzIFgiXSxbMCwxXSxbMiwzXSxbNCw1LCJmXFxjb25nIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==}{Link
to Diagram}
\end{quote}

where \({ \left.{{f}} \right|_{{0\times P_0}} }\) is the identity.

\end{remark}

\begin{corollary}[?]

If \(P\to I\times X \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)\) then
\(P_0 \cong P_1\) where
\(P_i \coloneqq{ \left.{{P}} \right|_{{i\times X}} }\).

\end{corollary}

\begin{corollary}[?]

If \(f_0, f_1: Y\to X\) with \(P \xrightarrow{\pi} X\), then
\(f_0^* P \cong f_1^* P\) are isomorphic bundles.

\end{corollary}

\begin{corollary}[?]

If \(X\) is contractible, then any
\(P\in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)_{/ {X}}\) is trivial.

\end{corollary}

\begin{proof}[?]

Consider the two maps

\begin{center}
\begin{tikzcd}
    x && {x_0} \\
    X && X \\
    x && x
    \arrow["{f_1}"', shift right=2, curve={height=6pt}, from=2-1, to=2-3]
    \arrow["{f_0}", shift left=3, curve={height=-6pt}, from=2-1, to=2-3]
    \arrow[maps to, from=3-1, to=3-3]
    \arrow[maps to, from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwxLCJYIl0sWzIsMSwiWCJdLFswLDAsIngiXSxbMiwwLCJ4XzAiXSxbMCwyLCJ4Il0sWzIsMiwieCJdLFswLDEsImZfMSIsMix7Im9mZnNldCI6MiwiY3VydmUiOjF9XSxbMCwxLCJmXzAiLDAseyJvZmZzZXQiOi0zLCJjdXJ2ZSI6LTF9XSxbNCw1LCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV0sWzIsMywiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

Then \(f_0 \simeq f_1\), and conclude by noting that
\begin{align*}
f_0^*P = X{ \underset{\scriptscriptstyle {x_0} }{\times} } P = X\times \pi^{-1}(x_0) = X\times G
\end{align*}
and \(f_1^* P = P\).

\end{proof}

\hypertarget{existenceuniqueness-of-metrics}{%
\subsection{Existence/Uniqueness of
Metrics}\label{existenceuniqueness-of-metrics}}

\begin{definition}[Riemannian metrics]

A \textbf{Riemannian metric} on a vector bundle
\(E \xrightarrow{\pi} X\) is a continuous map
\(E{ \underset{\scriptscriptstyle {X} }{\times} } E\to {\mathbb{R}}\)
which restricts to an inner product on each fiber.

\end{definition}

\begin{proposition}[?]

A Riemannian metric on \(E\) corresponds to a restriction of the
structure group from \(\operatorname{GL}_n({\mathbb{R}})\) to
\({\operatorname{O}}_n({\mathbb{R}})\).

\end{proposition}

\begin{proposition}[?]

Every vector bundle over a paracompact \(X\) has a unique Riemannian
metric.

\end{proposition}

\begin{proof}[?]

\textbf{Existence:} Cover \(X\) by charts and choose a locally
finite\footnote{Here \emph{locally finite} means every point is covered
  by finitely many opens in the cover.} refinement
\({\mathcal{U}}= \left\{{U_i}\right\}_{i\in I}\) and pick a partition of
unity \(\left\{{\chi_i}\right\}_{i\in I}\) subordinate to
\({\mathcal{U}}\).

Define an inner product \(g_i\) on \(\pi^{-1}(U_i)\) where
\(\phi_i: \pi^{-1}(U_i)\to U_i \times {\mathbb{R}}^n\) by pulling back
the inner product on \({\mathbb{R}}^n\), i.e.~taking
\(e_1\xrightarrow{\phi_i} (p_1, \mathbf{v}_1)\) and
\(e_2 \xrightarrow{\phi}(p_2, \mathbf{v}_2)\) and setting
\begin{align*}
g_i(e_1, e_2) \coloneqq{\left\langle {\mathbf{v}_1},~{\mathbf{v}_2} \right\rangle}_{{\mathbb{R}}^n}
.\end{align*}
Then define
\begin{align*}
g_p({-}, {-}) \coloneqq\sum_i \chi_i(p) g_i({-}, {-})
.\end{align*}

\textbf{Uniqueness:} Consider two inner products
\(g_0({-}, {-}), g_1({-}, {-})\) on the bundle
\(E \xrightarrow{\pi} X\), then define
\begin{align*}
g_t({-}, {-}) = tg_0({-}, {-})+ (1-t)g_1({-}, {-})
.\end{align*}
Then \(I\times E \xrightarrow{\operatorname{id}, \pi} I\times X\) is a
bundle, and \(g_t\) is a Riemannian metric on \(I\times E\). Consider
its corresponding principal bundle
\begin{align*}
P\to I\times X \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n({\mathbb{R}}))
\end{align*}
These correspond to restricting \(I\times E\) to \(0, 1\), yielding
\(P_0, P_1\) with Riemannian metrics \(g_0, g_1\). But \(P_0 \cong P_1\)
are isomorphic principal bundles, and using the correspondence between
bundles with metric and bundles with structure group
\({\operatorname{O}}_n\), this shows the two bundles with metric are
isomorphic.

\end{proof}

\begin{definition}[Universal $G\dash$bundles]

A \textbf{universal \(G{\hbox{-}}\)bundle} is a principal
\(G{\hbox{-}}\)bundle \(\pi: EG\to {\mathbf{B}}G\) such that
\(\pi_i EG = 0\) for all \(i\) (so \(EG\) is \emph{weakly
contractible}).

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\item
  \(\qty{{\mathbb{R}}\to S^1}\in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\mathbb{Z}})_{/ {S^1}}\)
  since all of the regular covers are principal bundles. Since
  \({\mathbb{R}}\) is contractible, this is the universal
  \({\mathbb{Z}}{\hbox{-}}\)bundle, so
  \(S^1 \simeq{\mathbf{B}}{\mathbb{Z}}\).
\item
  \(\qty{S^\infty \to {\mathbb{RP}}^\infty} \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(C_2)\)
  is a universal \(C_2{\hbox{-}}\)bundle, so
  \({\mathbb{RP}}^\infty \simeq{\mathbf{B}}C_2\)
\item
  \(\qty{S^\infty \to {\mathbb{CP}}^\infty}\) is a universal
  \(S^1 = U_1\) bundle, so
  \({\mathbb{CP}}^\infty \simeq{\mathbf{B}}U_1 \simeq{\mathbf{B}}S^1 \simeq{\mathbf{B}}{\mathbb{C}}^{\times}\):

  \begin{center}
  \begin{tikzcd}
    {F_z = \left\{{\lambda z {~\mathrel{\Big\vert}~}\lambda \neq 0}\right\}} &&& {{\left[ {z_1, \cdots, z_n} \right]}} \\
    {\mathbb{C}}^{\times}&& {{\mathbb{C}}^n} \\
    \\
    && {{\mathbb{CP}}^{n-1}} & {z \coloneqq{\left[ {z_1: \cdots : z_n} \right]}}
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=4-3]
    \arrow[maps to, from=1-4, to=4-4]
    \arrow[maps to, from=1-1, to=1-4]
  \end{tikzcd}
  \end{center}

  \begin{quote}
  \href{https://q.uiver.app/?q=WzAsNixbMiwxLCJcXENDXm4iXSxbMiwzLCJcXENQXntuLTF9Il0sWzAsMSwiXFxDQ1xcdW5pdHMiXSxbMywwLCJcXHR2e3pfMSwgXFxjZG90cywgel9ufSJdLFszLDMsInogXFxkYSBcXHR2e3pfMTogXFxjZG90cyA6IHpfbn0iXSxbMCwwLCJGX3ogPSBcXHRze1xcbGFtYmRhIHogXFxzdCBcXGxhbWJkYSBcXG5lcSAwfSJdLFsyLDBdLFswLDFdLFszLDQsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbNSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d}{Link
  to Diagram}
  \end{quote}
\end{itemize}

\end{example}

\begin{theorem}[?]

If \(X\in {\mathsf{CW}}\subseteq {\mathsf{Top}}\) and
\(EG \xrightarrow{\pi} {\mathbf{B}}G \in {\mathsf{Bun}}_G\) is
universal, then there is a bijection
\begin{align*}
\mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)_{/ {X}}  &\rightleftharpoons[X, {\mathbf{B}}G] \\
f*EG &\mapsfrom f
.\end{align*}

\end{theorem}

\begin{lemma}[?]

If \(E \xrightarrow{\pi} X\) is a fiber bundle with fiber \(F\) and
\(X\) is weakly contractible then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\pi\) admits a section, and
\item
  Any two sections are homotopic (through other sections).
\end{enumerate}

\end{lemma}

\begin{proof}[of lemma, part 1]

\textbf{Step 1:} build a section inductively.

\begin{itemize}
\tightlist
\item
  Define a section over the 0-skeleton arbitrarily.
\item
  Inductively, suppose the section is defined on the \(n-1\) skeleton,
  so it's defined over every \(n{\hbox{-}}\)cell boundary
  \({{\partial}}e^n\).
\item
  Write \({ \left.{{E}} \right|_{{e_n}} } = e^n\times F\), which is
  contractible since \(e_n\) is contractible.
\item
  Then \(s:{{\partial}}e^n = S^{n-1} \to F\) with \(\pi_n(F) = 0\), so
  the section extends:
\end{itemize}

\begin{center}
\begin{tikzcd}
    && {{ \left.{{E}} \right|_{{e^n}} }} && {e^n\times F} \\
    && {} \\
    {{{\partial}}e^n} && {e^n}
    \arrow[from=1-3, to=3-3]
    \arrow["\cong", from=1-3, to=1-5]
    \arrow["{{\operatorname{pr}}_1}", from=1-5, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow["s", from=3-1, to=1-3]
    \arrow["{\exists \tilde s}", curve={height=-12pt}, dashed, from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwwLCJcXHJve0V9e2Vebn0iXSxbNCwwLCJlXm5cXHRpbWVzIEYiXSxbMiwxXSxbMiwyLCJlXm4iXSxbMCwyLCJcXGJkIGVebiJdLFswLDNdLFswLDEsIlxcY29uZyJdLFsxLDMsIlxccHJfMSJdLFs0LDNdLFs0LDAsInMiXSxbMywwLCJcXGV4aXN0cyBcXHRpbGRlIHMiLDAseyJjdXJ2ZSI6LTIsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\textbf{Step 2:} Build the homotopy between sections inductively
cell-by-cell as in part (1).

\end{proof}

\begin{proof}[of theorem]

We want to show that the assignment \(f\mapsto f^* EG\) is bijective.

\textbf{Surjectivity}: Note that \(EG\) has a left \(G{\hbox{-}}\)action
defined by \(g\cdot e \coloneqq eg^{-1}\). Recall that we can use the
mixing construction:

\begin{center}
\begin{tikzcd}
    F && P && EG && {P{ \underset{\scriptscriptstyle {G} }{\times} } EG} \\
    &&& {} & {} & {} \\
    && X &&&& X
    \arrow["\pi", from=1-3, to=3-3]
    \arrow[from=1-7, to=3-7]
    \arrow[from=1-7, to=3-7]
    \arrow[from=1-5, to=1-7]
    \arrow["{\text{mixing}}", squiggly, from=2-4, to=2-6]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMiwwLCJQIl0sWzIsMiwiWCJdLFs2LDAsIlBcXGZpYmVycHJvZHtHfSBFRyJdLFs2LDIsIlgiXSxbNCwwLCJFRyJdLFszLDFdLFs0LDFdLFs1LDFdLFswLDAsIkYiXSxbMCwxLCJcXHBpIl0sWzIsM10sWzIsM10sWzQsMl0sWzUsNywiXFx0ZXh0e21peGluZ30iLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJzcXVpZ2dseSJ9fX1dLFs4LDBdXQ==}{Link
to Diagram}
\end{quote}

Sections of the mixed bundle biject with \(G{\hbox{-}}\)equivariant maps
\(P\to EG\). Writing
\(s(x) = [P, e] \sim [Pg, g\cdot e] \coloneqq[Pg, g^{-1}e]\), so given
\(p\in \pi^{-1}(x)\) we can send \(p\mapsto e\in EG\) such that
\([p, e]\in s(x)\). This is essentially currying an argument.
Conversely, given a \(G{\hbox{-}}\)equivariant map
\begin{align*}
P\to EG\\
p\mapsto e
,\end{align*}
we can define \(s(x) \coloneqq[p, e]\) where \(x = \pi(p)\). This is
well-defined: if \(x = \pi(pg)\), then \(s(x) = [pg, eg] = [p, e]\). Now
note that \(EG\) is weakly contractible, so
\(EG\to P{ \underset{\scriptscriptstyle {G} }{\times} } EG \to X\) has a
section \(s: X\to P{ \underset{\scriptscriptstyle {G} }{\times} }EG\)
and this we get a \(G{\hbox{-}}\)equivariant map \(P\to EG\) which
induces a map \(P/G \xrightarrow{h} EG/G\), where \(P/G = X\) and
\(EG/G = {\mathbf{B}}G\).

\begin{center}
\begin{tikzcd}
    P &&&& {h^* EG} && EG \\
    &&&&& {} \\
    &&&& X && {{\mathbf{B}}G}
    \arrow["h", from=3-5, to=3-7]
    \arrow[from=1-7, to=3-7]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-5, to=3-5]
    \arrow["f", curve={height=-30pt}, from=1-1, to=1-7]
    \arrow["{\exists p \xrightarrow{\sim} (\pi(p), f(p))}", dashed, from=1-1, to=1-5]
    \arrow["\pi"', from=1-1, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbNiwwLCJFRyJdLFs2LDIsIlxcQiBHIl0sWzQsMiwiWCJdLFswLDAsIlAiXSxbNCwwLCJoXiogRUciXSxbNSwxXSxbMiwxLCJoIl0sWzAsMV0sWzQsMF0sWzQsMl0sWzMsMCwiZiIsMCx7ImN1cnZlIjotNX1dLFszLDQsIlxcZXhpc3RzIHAgXFxtYXBzdmlhe1xcc2ltfSAoXFxwaShwKSwgZihwKSkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMywyLCJcXHBpIiwyXV0=}{Link
to Diagram}
\end{quote}

\begin{exercise}[?]

Show that this map is an isomorphism.

\end{exercise}

\end{proof}

\hypertarget{universal-bundles-thursday-september-16}{%
\section{Universal Bundles (Thursday, September
16)}\label{universal-bundles-thursday-september-16}}

\begin{definition}[Universal $G\dash$bundles]

A \textbf{universal \(G{\hbox{-}}\)bundle} is a principal
\(G{\hbox{-}}\)bundle \(EG \xrightarrow{\pi} {\mathbf{B}}G\) such that
\(EG\) is weakly contractible, i.e.~\(\pi_*(EG) = 0\).

\end{definition}

\begin{remark}

We looked at a theorem stating the correspondence
\begin{align*}
\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}  \rightleftharpoons[X, {\mathbf{B}}G]
.\end{align*}

\end{remark}

\begin{proof}[of surjectivity in theorem, continued]

We showed surjectivity of the following map:
\begin{align*}
[X, {\mathbf{B}}G] &\twoheadrightarrow\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}  \\
(f \in [X, {\mathbf{B}}G]) &\mapsto f^* EG
.\end{align*}

Given a principal \(G{\hbox{-}}\)bundle \(P \xrightarrow{\pi} B\), the
mixing construction used an action \(G\curvearrowright F\) to construct
the fiber bundle
\(P\overset{\scriptscriptstyle {G} }{\times} F \xrightarrow{\pi} B\).
Then the data of an equivariant map \(f: P\to F\), so
\(f(pg) = f(p)\cdot g \coloneqq g^{-1}f(p)\) is equivalent to a section
\(s: B\to P\overset{\scriptscriptstyle {G} }{\times} F\). Note that
fixing the first coordinate \(p\) in \([p, y]\) also fixes the second
coordinate. For \(b\in B\), we can set
\(s(b) = [p, y] \sim [pg, g^{-1}y]\) (noting that these are equivalent
in the mixed space), and we can define \(f(p) = y\) to get an
equivariant map since \(f(pg) = g^{-1}y = g^{-1}f(p)\)

So send
\(P \xrightarrow{\pi} X \in \mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}\)
to \(P\overset{\scriptscriptstyle {G} }{\times} EG \to X\). We proved
that this has a section
\(s: X \to P\overset{\scriptscriptstyle {G} }{\times} EG\) and any two
sections are homotopic, so from this we extract a
\(G{\hbox{-}}\)equivariant map \(f:P\to EG\). Modding out the \(G\)
action yields \(h: P/G\to EG/G\). But \(P/G\cong X\) and
\(EG/G\cong {\mathbf{B}}G\), so \(h: X\to {\mathbf{B}}G\), and moreover
\(h^* EG \cong P\).

\end{proof}

\begin{proof}[of injectivity in theorem]

Suppose \(h_0, h_1\in [X, {\mathbf{B}}G]\), then
\(h_0^* EG \xrightarrow{\phi, \cong} h_1^* EG\). We can construct the
following section \(s_0\):

\begin{center}
\begin{tikzcd}
    {(x, y)} && {y \coloneqq f_0([x, y])} \\
    {h_0^* EG} && EG &&& {h_0^* EG \overset{\scriptscriptstyle {G} }{\times} EG} & {[(x, y), y]} \\
    &&& \leadsto \\
    X && {{\mathbf{B}}G} &&& X & x
    \arrow["{f_0}", dashed, from=2-1, to=2-3]
    \arrow[from=2-3, to=4-3]
    \arrow["{h_0}"', from=4-1, to=4-3]
    \arrow[from=2-1, to=4-1]
    \arrow[from=2-6, to=4-6]
    \arrow[maps to, from=1-1, to=1-3]
    \arrow["{s_0}"', curve={height=30pt}, dashed, from=4-6, to=2-6]
    \arrow[maps to, from=4-7, to=2-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTEsWzIsMSwiRUciXSxbMiwzLCJcXEIgRyJdLFswLDMsIlgiXSxbMCwxLCJoXzBeKiBFRyJdLFszLDIsIlxcbGVhZHN0byJdLFs1LDEsImhfMF4qIEVHIFxcbWl4e0d9IEVHIl0sWzUsMywiWCJdLFswLDAsIih4LCB5KSJdLFsyLDAsInkgXFxkYSBmXzAoW3gsIHldKSJdLFs2LDMsIngiXSxbNiwxLCJbKHgsIHkpLCB5XSJdLFszLDAsImZfMCIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDFdLFsyLDEsImhfMCIsMl0sWzMsMl0sWzUsNl0sWzcsOCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFs2LDUsInNfMCIsMix7ImN1cnZlIjo1LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbOSwxMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

We can build another section \(s_1\) in the following way: use the
isomorphism \(\phi: h_0^* EG \to h_1^* EG\) to construct the composite

\begin{center}
\begin{tikzcd}
    {(x, y)} && {(x, \phi(y))} && {\phi(y)} \\
    \\
    {h_0^* EG} && {h_1^* EG} && EG \\
    \\
    && X && {{\mathbf{B}}G}
    \arrow["{f_1}", from=3-3, to=3-5]
    \arrow["{f_1\circ \phi}", curve={height=-30pt}, from=3-1, to=3-5]
    \arrow["\phi", from=3-1, to=3-3]
    \arrow[from=3-5, to=5-5]
    \arrow[from=3-3, to=5-3]
    \arrow[from=5-3, to=5-5]
    \arrow[from=3-1, to=5-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwyLCJoXzBeKiBFRyJdLFsyLDIsImhfMV4qIEVHIl0sWzQsMiwiRUciXSxbMCwwLCIoeCwgeSkiXSxbMiwwLCIoeCwgXFxwaGkoeSkpIl0sWzQsMCwiXFxwaGkoeSkiXSxbMiw0LCJYIl0sWzQsNCwiXFxCIEciXSxbMSwyLCJmXzEiXSxbMCwyLCJmXzFcXGNpcmMgXFxwaGkiLDAseyJjdXJ2ZSI6LTV9XSxbMCwxLCJcXHBoaSJdLFsyLDddLFsxLDZdLFs2LDddLFswLDZdXQ==}{Link
to Diagram}
\end{quote}

So we have
\begin{align*}
s_0(x) &\coloneqq[x,y,y] \\
s_1(x) &\coloneqq[x,y,\phi(y)] 
.\end{align*}

By the lemma, \(s_0\simeq s_1\) through sections, so there is a homotopy
\begin{align*}
s: I\times X &\to I \times h_0^* EG\overset{\scriptscriptstyle {G} }{\times} EG \\
(t, x) &\mapsto (t, x, y, z)
.\end{align*}
But this is a section of a principle \(EG{\hbox{-}}\)bundle over a CW
complex, which yields a \(G{\hbox{-}}\)equivariant map
\begin{align*}
f: I \times h_0^* EG &\to EG \\
(t, x, y) &\mapsto z
.\end{align*}

Then

\begin{itemize}
\tightlist
\item
  At \(t=0\), we have \((0, x)\mapsto (0, x,y,y)\), so
  \(f(0, x, y) = y\),
\item
  At \(t=1\), we have \((1, x)\mapsto (1, x,y,\phi(y))\), so
  \(f(1,x,y) = \phi(y)\).
\end{itemize}

Since \(f\) is \(G{\hbox{-}}\)equivariant, we can quotient both sides by
\(G\) to get a map
\begin{align*}
h: I\times X &\to {\mathbf{B}}G \\
(0, x) &\mapsto h_0(x) \\
(1, x) &\mapsto h_1(x) 
.\end{align*}

\end{proof}

\begin{exercise}[?]

Try this proof yourself!

\end{exercise}

\begin{remark}

The same proof shows the following:

\end{remark}

\begin{lemma}[?]

If \(F\to E \xrightarrow{\pi} B\) is a fiber bundle and
\(B\in {\mathsf{CW}}\), if \(\pi_{0\leq i\leq n}(F) = 0\) then we can
inductively build sections over skeleta \(B_{(k)}\) fir \(k\leq n\) to
construct a section over \(B_{(n+1)}\). Moreover, any two sections over
the \(n{\hbox{-}}\)skeleton are homotopic.

\end{lemma}

\begin{proposition}[?]

If
\(P \xrightarrow{\pi} B \in \mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}\)
and \(\pi_{0\leq i \leq n} P = 0\) (so \(B\) is a ``weak universal
bundle'') then
\([X, B] \twoheadrightarrow\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}\)
for any \(X\in {\mathsf{CW}}\) with \(\dim(X) \leq n+1\), and it is
bijective if \(\dim X \leq n\). Here the map is again
\(h \mapsto h^* P\).

\end{proposition}

\hypertarget{existence-of-universal-bundles}{%
\subsection{Existence of Universal
Bundles}\label{existence-of-universal-bundles}}

\begin{theorem}[Milnor, 1966]

For any group \(G \in {\mathsf{Top}}{\mathsf{Grp}}\), there is a
universal bundle \(EG\to {\mathbf{B}}G\).

\end{theorem}

\begin{remark}

Uniqueness up to homotopy:

\begin{center}
\begin{tikzcd}
    {h_*E'} & E && {E'} & {h'_*E} \\
    \\
    & B && {B'}
    \arrow["f", from=1-2, to=1-4]
    \arrow["h"', from=3-2, to=3-4]
    \arrow[from=1-2, to=3-2]
    \arrow[from=1-4, to=3-4]
    \arrow["{f'}"', curve={height=30pt}, from=1-4, to=1-2]
    \arrow["{h'}", curve={height=-30pt}, from=3-4, to=3-2]
    \arrow["\sim"', from=1-1, to=1-2]
    \arrow["\sim"', from=1-5, to=1-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwwLCJFIl0sWzEsMiwiQiJdLFszLDAsIkUnIl0sWzMsMiwiQiciXSxbMCwwLCJoXypFJyJdLFs0LDAsImgnXypFIl0sWzAsMiwiZiJdLFsxLDMsImgiLDJdLFswLDFdLFsyLDNdLFsyLDAsImYnIiwyLHsiY3VydmUiOjV9XSxbMywxLCJoJyIsMCx7ImN1cnZlIjotNX1dLFs0LDAsIlxcc2ltIiwyXSxbNSwyLCJcXHNpbSIsMl1d}{Link
to Diagram}
\end{quote}

Then since \((h'h^{-1})^* E \cong E\),
\(h' h^{-1}\simeq\operatorname{id}\) and
\(h (h')^{-1}\simeq\operatorname{id}\), so we get a homotopy equivalence
\(B \simeq B'\).

\end{remark}

\begin{exercise}[?]

Show \(E \simeq E'\).

\end{exercise}

\begin{remark}

We'll prove this theorem using Segal's construction. For discrete groups
\(G\), the construction is covered in Hatcher 1B. Hatcher constructs
\(K(G, 1)\), a space with \(\pi_1 = G\) and contractible universal
cover. Then the universal cover \(\widehat{X}\to X\) is a principle
\(G{\hbox{-}}\)bundle, and since \(\widehat{X}\) is contractible, it is
universal:

\begin{center}
\begin{tikzcd}
    && {\tilde X\simeq{\operatorname{pt}}} \\
    &&& {\in \mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}}  \text{ universal}} \\
    {K(G, 1)} && X
    \arrow["\pi", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJLKEcsIDEpIl0sWzIsMiwiWCJdLFsyLDAsIlxcdGlsZGUgWFxcaG9tb3RvcGljIFxccHQiXSxbMywxLCJcXGluIFxcUHJpblxcQnVuZ1xcc2xpY2UgWCBcXHRleHR7IHVuaXZlcnNhbH0iXSxbMiwxLCJcXHBpIl0sWzAsMV1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{definition}[Nerve of a category]

Given a category \(\mathsf{C}\), the \textbf{nerve}
\({ \mathcal{N}({\mathsf{C}}) }\) is the following
\(\Delta{\hbox{-}}\)complex:

\begin{itemize}
\tightlist
\item
  \(0{\hbox{-}}\)simplices are objects of \(\mathsf{C}\)
\item
  \(n{\hbox{-}}\)simplices for \(n\geq 1\) are sequences of composable
  morphisms
  \begin{align*}
  x_0 \xrightarrow{f_0} x_1 \xrightarrow{f_1} \cdots \xrightarrow{f_{n-1}} x_n
  .\end{align*}
\item
  Gluing data for 1-simplices: for \(x_0 \xrightarrow{f} x_1\), set
  \({{\partial}}_1(f) = x_1, {{\partial}}_0(f) = x_0\).
\item
  Gluing data for \(n{\hbox{-}}\)simplices: the \(i\)th boundary maps
  are given by dropping vertex \(i\):
  \begin{align*}
  {{\partial}}_i(f_1, \cdots, f_n)
  =
  \begin{cases}
  (f_2, f_3, \cdots, f_n) & i=0 \\
  (f_1, f_2, \cdots, f_{i+1} \circ f_{i}, \cdots, f_n) & i=0 \\
  (f_1, f_2, \cdots, f_{n-1}) & i=n.
  \end{cases}
  \end{align*}
\end{itemize}

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-09-16_13-50.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\hypertarget{thursday-september-16}{%
\section{Thursday, September 16}\label{thursday-september-16}}

\begin{remark}

Let \(G\in {\mathsf{Grp}}\), and consider the following two categories.
\({\mathbf{B}}G\) will be the category:

\begin{itemize}
\tightlist
\item
  \({\operatorname{Ob}}({\mathbf{B}}G) = \left\{{{\operatorname{pt}}}\right\}\).
\item
  \(\mathop{\mathrm{Mor}}_{{\mathbf{B}}G}({\operatorname{pt}}, {\operatorname{pt}}) = \left\{{g\in G}\right\}\),
  i.e.~there is one morphism for every group element, with composition
  \(g_1 \circ g_2 \coloneqq g_1g_2\) given by group multiplication.
\end{itemize}

\(E G\) will be the category:

\begin{itemize}
\tightlist
\item
  \({\operatorname{Ob}}(EG) = \left\{{g\in G}\right\}\), one object for
  each element of \(G\),
\item
  \(\mathop{\mathrm{Mor}}(g, h)=\left\{{g^{-1}h}\right\}\), a single
  (conveniently labeled!) morphism for each ordered pair \((g, h)\).
\end{itemize}

Note that \(G\curvearrowright EG\):

\begin{center}
\begin{tikzcd}
    {g_0} && {g_1} \\
    \\
    {gg_0} && {gg_1}
    \arrow[""{name=0, anchor=center, inner sep=0}, "{g_0^{-1}g_1}", from=1-1, to=1-3]
    \arrow[""{name=1, anchor=center, inner sep=0}, "{g\cdot g_0^{-1}g_1}", from=3-1, to=3-3]
    \arrow[shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJnXzAiXSxbMiwwLCJnXzEiXSxbMCwyLCJnZ18wIl0sWzIsMiwiZ2dfMSJdLFswLDEsImdfMFxcaW52IGdfMSJdLFsyLDMsImdcXGNkb3QgZ18wXFxpbnYgZ18xIl0sWzQsNSwiIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==}{Link
to Diagram}
\end{quote}

This induces an action on \({ \mathcal{N}({EG}) } \in {\mathsf{Top}}\),
where the 0-simplices correspond to elements of \(G\). and
\(n{\hbox{-}}\)simplices are chains
\begin{align*}
g_0 \xrightarrow{g_0^{-1}g_1} g_1 \xrightarrow{g_1^{-1}g_2} g_2 \to \cdots \to g_n
.\end{align*}
Acting on this by \(G\) yields
\begin{align*}
gg_0 \xrightarrow{g_0^{-1}g_1} gg_1 \xrightarrow{g_1^{-1}g_2} gg_2 \to \cdots \to gg_n
,\end{align*}
noting we leave the morphism labeling unchanged, and that uniqueness of
morphisms makes the simplicial boundary map behave nicely.

\end{remark}

\begin{exercise}[?]

Show that
\begin{align*}
{ \mathcal{N}({EG}) }/G = { \mathcal{N}({{\mathbf{B}}G}) }
.\end{align*}

\end{exercise}

\begin{remark}

Note that
\begin{align*}
{ \mathcal{N}({{\mathbf{B}}G}) } &= \Delta^0 {\textstyle\coprod}\Delta^1 \times G {\textstyle\coprod}\Delta^2 \times G{ {}^{ \scriptscriptstyle\times^{2} }  } {\textstyle\coprod}\Delta^3 \times G{ {}^{ \scriptscriptstyle\times^{3} }  }\cdots \\
{ \mathcal{N}({E G}) } &= \Delta^0 \times G {\textstyle\coprod}\Delta^1 \times G{ {}^{ \scriptscriptstyle\times^{2} }  } {\textstyle\coprod}\Delta^2 \times G{ {}^{ \scriptscriptstyle\times^{3} }  }\cdots 
,\end{align*}
where the gluing data for \({ \mathcal{N}({{\mathbf{B}}G}) }\) is given
by
\begin{align*}
{{\partial}}_n: 
\Delta^n \times G{ {}^{ \scriptscriptstyle\times^{n} }  } &\to
\Delta^{n-1} \times G{ {}^{ \scriptscriptstyle\times^{n-1} }  } \\
(\mathbf{x}, \mathbf{g}) &\mapsto (\mathbf{x}\setminus\left\{{x_n}\right\}, \mathbf{g} \setminus\left\{{g_n}\right\})
\end{align*}
and for \({ \mathcal{N}({EG}) }\) is
\begin{align*}
{{\partial}}_n: \Delta^n \times G{ {}^{ \scriptscriptstyle\times^{n+1} }  } &\to
\Delta \times G{ {}^{ \scriptscriptstyle\times^{n} }  } \\
(\mathbf{x}, \mathbf{g}) &\mapsto (\mathbf{x}\setminus\left\{{x_n}\right\}, \mathbf{g} \setminus\left\{{g_n}\right\})
.\end{align*}

The action \(G\curvearrowright EG\) is the following:
\begin{align*}
g\cdot (\mathbf{x}, \mathbf{g}) \mapsto (\mathbf{x}, {\left[ {gg_0, gg_1,\cdots, gg_n} \right]} )
.\end{align*}

\end{remark}

\begin{example}[?]

Take \(G = C_4, G' = C_2{ {}^{ \scriptscriptstyle\times^{2} } }\), and
\([ (x_0, x_1, x_2), (a, a^2)] \in \Delta^2 \times G{ {}^{ \scriptscriptstyle\times^{2} } }\).
Then its faces are
\begin{align*}
[(0, x_1, x_2), (a, a^2)] &\sim [(x_1, x_2), (a^2)] \\
[(x_0, 0, x_2), (a, a^2)] &\sim [(x_0, x_2), (a)] \\
[(x_0, x_1, 0), (a, a^2)] &\sim [(x_0, x_1), (a)] \\
.\end{align*}
These describe a 2-simplex mapping into \({\mathbf{B}}C_4\) by
\(a \to a^2 \to a^3\), yielding the relation \(a\cdot a^2 = a^3\). One
can check that in \({\mathbf{B}}G\), these groups yield distinct higher
simplices:

\includegraphics{figures/2021-09-16_16-59-53.png}

\end{example}

\begin{lemma}[?]

If \(\mathsf{C}\) has an initial or terminal object, then
\({ \mathcal{N}({\mathsf{C}}) }\) is contractible.

\end{lemma}

\begin{remark}

This clearly holds for \(E G\), since every object is initial and
terminal.

\end{remark}

\begin{proof}[?]

Suppose \(y\in \mathsf{C}\) is terminal and any other object
\(x\in \mathsf{C}\), denote \(f_x: x\to y\) the unique morphism.

Then for any sequence ending in \(y\), deformation retract to \(y\):
\(x_0 \xrightarrow{f_0} \to x_1 \xrightarrow{f_1} \cdots \xrightarrow{f_x} y \leadsto y\).
If a sequence doesn't end in \(y\), add it on:
\(x_0 \xrightarrow{f_0} x_1 \cdots \to x_n {\color{green} \xrightarrow{f_{x_n}} y} \leadsto y\).

\end{proof}

\begin{corollary}[?]

\({ \mathcal{N}({EG}) }\) is contractible, and the quotient
\({ \mathcal{N}({EG}) }\to { \mathcal{N}({{\mathbf{B}}G}) }\) is a
universal \(G{\hbox{-}}\)bundle.

\end{corollary}

\begin{exercise}[?]

Construct \(E G\) and \({\mathbf{B}}G\) for
\(G = C_4, C_2{ {}^{ \scriptscriptstyle\times^{2} } }\) and explicitly
compare their 3-skeleta.

\end{exercise}

\hypertarget{tuesday-september-21}{%
\section{Tuesday, September 21}\label{tuesday-september-21}}

\begin{remark}

Today: a short discussion on generalizations of \({\mathbf{B}}G\) to
topological groups.

\end{remark}

\begin{definition}[Topological categories]

A \textbf{topological category} is a category where the objects are
topological spaces and morphisms form topological spaces in a coherent
way, i.e.~the following maps should be continuous:

\begin{itemize}
\tightlist
\item
  \(\mathrm{source}, \mathrm{target}: \mathop{\mathrm{Mor}}_{\mathsf{C}} \to {\operatorname{Ob}}(\mathsf{C})\),
\item
  \(\operatorname{id}: {\operatorname{Ob}}(\mathsf{C})\to \mathop{\mathrm{Mor}}_{\mathsf{C}}\),
\item
  Composition:
  \(\mathsf{C}(x, y) \times \mathsf{C}(y,z) \to \mathsf{C}(x, z)\).
\end{itemize}

I.e. it is a category enriched over topological spaces (plus
conditions).

\end{definition}

\begin{example}[?]

If \(G\in {\mathsf{Top}}{\mathsf{Grp}}\), then \({\mathcal{B}}G\) is a
topological category since the morphism space
\(\mathop{\mathrm{Mor}}({\operatorname{pt}}, {\operatorname{pt}}) = G\)
has a topology. Similarly \({\mathcal{E}}G\) is a topological category.

\end{example}

\begin{remark}

We can take nerves of topological categories; this just requires
tracking the additional enrichment (i.e.~the various topologies). The
same proof will yield a principal \(G{\hbox{-}}\)bundle
\({ \mathcal{N}({{\mathcal{E}}G}) } \xrightarrow{\pi} { \mathcal{N}({{\mathcal{B}}G}) }\),
noting that \(G\) again acts on \({ \mathcal{N}({{\mathcal{E}}G}) }\).

\end{remark}

\begin{definition}[Absolute Neighborhood Retract]

A space is called an \textbf{absolute neighborhood retract} (ANR) if for
any \(X\hookrightarrow Y\) (as a closed subspace) into a metric space,
\(X\) is a retract of a neighborhood in \(Y\).

\end{definition}

\begin{example}[?]

Every CW complex is an ANR. This is also true if every point of \(X\)
has a contractible neighborhood.

\end{example}

\begin{lemma}[?]

If \(G\) is ANR, then
\(EG = { \mathcal{N}({{\mathcal{E}}G}) }\to { \mathcal{N}({{\mathcal{B}}G}) } = {\mathbf{B}}G \in \mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }\).

\end{lemma}

\begin{proof}[?]

Note that \({\mathbf{B}}G\) is a \(\Delta{\hbox{-}}\)complex, so we'll
try to build bundle charts by inducting over the skeleta. Each graded
piece of the complex is of the form
\(\Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } }\), so pick an
interior point \(((x_0, \cdots, x_i), (g_1,\cdots, g_i))\) so
\(x_i\neq 0\) for every \(i\). Define a map
\begin{align*}
\Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} }  } \times G &\to E G \\
( (x_0,\cdots, x_i), (g_1, \cdots, g_i), g) &\mapsto (\operatorname{id}(\cdots), (g, gg_1, gg_1g_2, \cdots, gg_1\cdots g_i))
,\end{align*}
which corresponds to the sequence of composable morphisms
\begin{align*}
(g \xrightarrow{g_1} gg_1 \xrightarrow{g_2} g g_1 g_2 \to \cdots \to gg_1\cdots g_i) 
.\end{align*}

\begin{exercise}[?]

Show that this is not compatible with the gluing!

\end{exercise}

Write
\(p: \Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } } \to {{\mathbf{B}}G}\)
for the quotient attaching map, so we can write the
\(m{\hbox{-}}\)skeleton as
\({{\mathbf{B}}G}^{(m)} = \displaystyle\bigcup_{i\leq m} p(\Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } })\).
Now suppose \((U_m, \phi_m)\) is a chart for
\({ \left.{{{\mathsf{E} G}}} \right|_{{{{\mathbf{B}}G}^{(m)}}} } \to{{\mathbf{B}}G}^{(m)}\),
we want to extend this to a chart or \({{\mathbf{B}}G}^{(m+1)}\). We
have a retraction \(r: U_{m+1}\to U_m\) where
\(U_{m+1} \subseteq {{\mathbf{B}}G}^{(m+1)}\) is an open inclusion. We
construct a trivialization of \(\pi^{-1}(U_{m+1}) \to U_{m+1}\):

\begin{center}
\begin{tikzcd}
    {\pi^{-1}(p^{-1}(U_{m+1}))} && {\pi^{-1}(p^{-1}(U_m))} && {\pi^{-1}(U_m)} && {U_m\times G} \\
    \\
    {p^{-1}(U_{m+1})} && {p^{-1}(U_m)} && {U_m} \\
    {p^{-1}(U_{m+1})} &&&& {U_{m+1}}
    \arrow["{\phi_m}", from=1-5, to=1-7]
    \arrow["\pi"', from=1-5, to=3-5]
    \arrow["p", from=3-3, to=3-5]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-3, to=3-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
    \arrow[hook, from=4-5, to=3-5]
    \arrow[hook, from=3-1, to=3-3]
    \arrow["r"', curve={height=18pt}, from=3-3, to=3-1]
    \arrow[from=1-1, to=3-1]
    \arrow["{\exists\tilde r}", dashed, from=1-1, to=1-3]
    \arrow["p", from=4-1, to=4-5]
    \arrow[Rightarrow, no head, from=3-1, to=4-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
    \arrow["{\phi_{m+1}}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, from=1-1, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

This extends the chart to \({{\mathbf{B}}G}^{(m+1)}\), noting that \(p\)
is a quotient map and thus preserves open sets.

\end{proof}

\begin{remark}

We can't necessarily extend over the entire \(m+1\) skeleton! But here
extending it over a retractable neighborhood was enough, so we needed
\(G\) to be an ANR in order for \({{\mathbf{B}}G}\) to be an ANR. Why:
consider
\begin{align*}
p^{-1}(U_m) \subseteq \displaystyle\bigcup_{i\leq m} \Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} }  } \subseteq \displaystyle\bigcup_{i\leq m-1} \Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} }  }
.\end{align*}
If \(G\) is an ANR, use that \(\Delta^i\) is an ANR and so their product
will be, then pick a neighborhood and apply \(p\) to get the required
open.

\end{remark}

\hypertarget{building-mathbfboperatornameo_n-and-mathsfe-operatornameo_n}{%
\subsection{\texorpdfstring{Building
\({{\mathbf{B}}{\operatorname{O}}}_n\) and
\({\mathsf{E} {\operatorname{O}}}_n\)}{Building \{\{\textbackslash mathbf\{B\}\}\{\textbackslash operatorname\{O\}\}\}\_n and \{\textbackslash mathsf\{E\} \{\textbackslash operatorname\{O\}\}\}\_n}}\label{building-mathbfboperatornameo_n-and-mathsfe-operatornameo_n}}

\begin{remark}

We'll assume all spaces paracompact from this point forward! We have a
correspondence
\begin{align*}
\left\{{\substack{
  \text{$n{\hbox{-}}$dimensional CW complexes }
}}\right\}
\rightleftharpoons
\left\{{\substack{
  \text{$n{\hbox{-}}$dimensional vector bundles }
  \\
  \text{with an ${\operatorname{O}}_n{\hbox{-}}$structure}
}}\right\}
\rightleftharpoons
\mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)_{/ {X}} 
\rightleftharpoons
[X, {\mathbf{B}}{\operatorname{O}}_n]
\end{align*}
Our next goal is to construct \({{\mathbf{B}}{\operatorname{O}}}_n\) and
\({\mathsf{E} {\operatorname{O}}}_n\) as spaces. Let
\(V_n({\mathbb{R}}^k) \coloneqq\left\{{(\mathbf{v}_1, \cdots, \mathbf{v}_n) \text{orthonormal} }\right\}\).
Note that \({\operatorname{O}}_n\curvearrowright V_n({\mathbb{R}}^k)\)
by
\begin{align*}
\qty{\mathbf{v}_1, \cdots, \mathbf{v}_n} 
\cdot A = 
\qty{\sum_i a_{i, 1} \mathbf{v}_i,
\sum_i a_{i, 2} \mathbf{v}_i,
\cdots,
\sum_i a_{i, n} \mathbf{v}_i}
.\end{align*}

There is a projection

\begin{center}
\begin{tikzcd}
    {F_{\mathbf{v}_1} = V_{n-1}({\mathbb{R}}^{k-1})} && {V_n({\mathbb{R}}^k)} & {(\mathbf{v}_1, \cdots, \mathbf{v}_1)} \\
    \\
    && {V_1({\mathbb{R}}^k)} & {\mathbf{v}_1}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[maps to, from=1-4, to=3-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwwLCJWX24oXFxSUl5rKSJdLFsyLDIsIlZfMShcXFJSXmspIl0sWzMsMCwiKFxcdmVjdG9yIHZfMSwgXFxjZG90cywgXFx2ZWN0b3Igdl8xKSJdLFszLDIsIlxcdmVjdG9yIHZfMSJdLFswLDAsIkZfe1xcdmVjdG9yIHZfMX0gPSBWX3tuLTF9KFxcUlJee2stMX0pIl0sWzQsMF0sWzAsMV0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

We'll use the fact that \(V_1({\mathbb{R}}^k)\) is
\((k-2){\hbox{-}}\)connected, since it's homotopy equivalent to
\(S^{k-1}\).

\end{remark}

\begin{lemma}[?]

\(V_n({\mathbb{R}}^{k})\) is \((k-n-1){\hbox{-}}\)connected.

\end{lemma}

\begin{proof}[?]

Induct on \(n\) using the homotopy LES for the fiber bundle:

\begin{center}
\begin{tikzcd}
    && \cdots && {\pi_{i+1} V_{n-1} {\mathbb{R}}^{k-1} \cong \pi_{i+1}(S^{k-1})} \\
    \\
    {\pi_i V_{n-1} {\mathbb{R}}^{k-1} = 0} && \textcolor{rgb,255:red,92;green,92;blue,214}{\pi_i V_{n} {\mathbb{R}}^{k}} && {\pi_i V_{1} {\mathbb{R}}^{k} \cong \pi_iS^{k-2} = 0} \\
    { \substack{(k-n-1){\hbox{-}}\text{connected} \\ i\leq k-n-1} } && {\therefore\text{zero}} && {i\leq k-n-1 \implies i\leq k-3 }
    \arrow[from=1-5, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{proof}

\begin{remark}

Using the inclusions
\(V_n({\mathbb{R}}^k) \hookrightarrow V_n({\mathbb{R}}^{k+1})\), we can
define
\(V_n({\mathbb{R}}^\infty) = \colim_k V_n({\mathbb{R}}^k) = \displaystyle\bigcup_{k\geq 0}V_n ({\mathbb{R}}^k)\).
We equip it with the \textbf{weak topology},
i.e.~\(U \subseteq V_n({\mathbb{R}}^\infty)\) is open iff
\(U \cap V_n({\mathbb{R}}^k)\) is open for all \(k\).

\end{remark}

\begin{lemma}[?]

\begin{align*}
\pi_* V_n({\mathbb{R}}^\infty) = 0
.\end{align*}

\end{lemma}

\begin{proof}[?]

By compactness, any sphere \(S^m\) maps to \(V_n({\mathbb{R}}^k)\) for
some large \(k\), and using
\(V_n({\mathbb{R}}^k) \hookrightarrow V_n({\mathbb{R}}^\ell)\) with
\(\ell-n-1 > m\) where \(\pi_n V_n({\mathbb{R}}^\ell) = 0\) to make the
map nullhomotopic.

\end{proof}

\begin{definition}[?]

\begin{align*}
V_n({\mathbb{R}}^\infty) / {\operatorname{O}}_n = {\operatorname{Gr}}_n({\mathbb{R}}^\infty)
.\end{align*}

\end{definition}

\begin{remark}

It will turn out that
\({\mathsf{E} {\operatorname{O}}}_n = V_n({\mathbb{R}}^\infty)\),
sometimes referred to as the \emph{Stiefel manifold} of
\(n{\hbox{-}}\)frames in \({\mathbb{R}}^\infty\).

\end{remark}

\hypertarget{thursday-september-23}{%
\section{Thursday, September 23}\label{thursday-september-23}}

\begin{remark}

Last time: we were trying to construct
\({\mathsf{E} {\operatorname{O}}}_n\) and
\({{\mathbf{B}}{\operatorname{O}}}_n\), and we defined
\(V_n({\mathbb{R}}^\infty) = \directlim_{k} V_n({\mathbb{R}}^k)\), where
\(V_n({\mathbb{R}}^k)\) was the space of \(n\) orthonormal vectors in
\({\mathbb{R}}^k\). There is a map
\(V_n({\mathbb{R}}^\infty)\to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\),
which will be our candidate for
\({\mathsf{E} {\operatorname{O}}}_n \to {{\mathbf{B}}{\operatorname{O}}}_n\).

\end{remark}

\begin{lemma}[?]

\(V_n({\mathbb{R}}^\infty) \xrightarrow{\pi} {\operatorname{Gr}}_n({\mathbb{R}}^\infty) \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)\).

\end{lemma}

\begin{proof}[?]

We'll show this directly in charts. Let
\(W\in{\operatorname{Gr}}_n({\mathbb{R}}^\infty)\) be an
\(n{\hbox{-}}\)dimensional plane, the consider an open neighborhood
\begin{align*}
U_W \coloneqq\left\{{W' \in {\operatorname{Gr}}_n({\mathbb{R}}^\infty) {~\mathrel{\Big\vert}~}W^\perp \cap W' = 0}\right\}
.\end{align*}
For any such \(W'\), we have a map \(W'\to W\) given by orthogonal
projection, which is an isomorphism since \(W^\perp \cap W' = 0\).

\begin{claim}

\begin{align*}
\pi^{-1}(U_W) \cong U_W \times {\operatorname{O}}_n
.\end{align*}

\end{claim}

Fix some \(\alpha\in \pi^{-1}(U_W)\) (an orthonormal basis for \(W\)),
apply \(f^{-1}\) to get \(f^{-1}( \alpha)\), then apply Gram-Schmidt to
get \(\tilde a\), an orthonormal basis for \(W'\). Define \(F_{W'}\) to
be this composition; this yields a bijection
\(\pi^{-1}(W) { { \, \xrightarrow{\sim}\, }}\pi^{-1}(W')\) for all
\(W'\in U_W\), namely
\begin{align*}
U_w \times {\operatorname{O}}_n &\to \pi^{-1}(W) \\
(W', A) &\mapsto F_{W'}( \alpha) \cdot A
.\end{align*}
The claim is that this trivializes the bundle, since this constructs a
local section using \({\operatorname{O}}_n\) translations:

\begin{center}
\begin{tikzcd}
    {s(W')\cdot A} && {(W', A)} \\
    {\pi^{-1}(U_W)} && {U_w \times {\operatorname{O}}_n} \\
    \\
    {U_W} && {W'}
    \arrow["\pi", from=2-1, to=4-1]
    \arrow["s"', curve={height=18pt}, from=4-1, to=2-1]
    \arrow["\cong", from=2-3, to=2-1]
    \arrow[dashed, maps to, from=1-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwxLCJcXHBpXFxpbnYoVV9XKSJdLFsyLDEsIlVfdyBcXHRpbWVzIFxcT3J0aF9uIl0sWzIsMCwiKFcnLCBBKSJdLFswLDAsInMoVycpXFxjZG90IEEiXSxbMCwzLCJVX1ciXSxbMiwzLCJXJyJdLFswLDQsIlxccGkiXSxbNCwwLCJzIiwyLHsiY3VydmUiOjN9XSxbMSwwLCJcXGNvbmciXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

Summary: pick an orthonormal basis for \(W\), say \(\alpha\), then
\(s(W) = \alpha\) and we define \(s(W')\) by sending \(\alpha\) to a
basis for \(W'\) by \(P^{-1}\) and the applying Gram-Schmidt to get an
orthonormal basis for \(W'\).

\end{proof}

\begin{remark}

\envlist

\begin{itemize}
\tightlist
\item
  Replace \({\operatorname{O}}_n\) with \(U_n\) and \({\mathbb{R}}\)
  with \({\mathbb{C}}\) to get
  \({\operatorname{Gr}}_n({\mathbb{C}}^\infty) = {\mathbf{B}}U_n\).
\item
  \(V_n({\mathbb{R}}^\infty)/ {\operatorname{SO}}_n = {\mathbf{B}}{\operatorname{SO}}_n\)
  yields the Grassmannian of \emph{oriented} planes.
\item
  For \(H\leq G\), we have \({\mathbf{E}}H = {\mathbf{E}}G\) and
  \({\mathbf{B}}H = {\mathbf{E}}G/H\).
\end{itemize}

\end{remark}

\begin{quote}
Question: can you get \({\mathbf{B}}{\operatorname{Spin}}_n\) this way?
\end{quote}

\begin{remark}

An alternative description of \({\mathsf{E} {\operatorname{O}}}_n\) and
\({{\mathbf{B}}{\operatorname{O}}}_n\), due to Milnor-Stasheff: write
\({{\mathbf{B}}{\operatorname{O}}}_n = {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\)
and define the \textbf{canonical bundle} \(\gamma\). Recall that every
principal \({\operatorname{O}}_n\) bundle is a pullback of the following
form:

\begin{center}
\begin{tikzcd}
    {P = f^* {\mathsf{E} {\operatorname{O}}}_n} && {{\mathsf{E} {\operatorname{O}}}_n = V_n({\mathbb{R}}^\infty)} \\
    \\
    X && {{{\mathbf{B}}{\operatorname{O}}}_n}
    \arrow["f", from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[dashed, from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJQID0gZl4qIFxcRU9fbiJdLFswLDIsIlgiXSxbMiwwLCJcXEVPX24gPSBWX24oXFxSUl5cXGluZnR5KSJdLFsyLDIsIlxcQk9fbiJdLFsxLDMsImYiXSxbMCwxXSxbMCwyLCIiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiwzXV0=}{Link
to Diagram}
\end{quote}

Moreover,
\(\mathop{\mathrm{Prin}}({\operatorname{O}}_n)_{/ {X}} = [X, {{\mathbf{B}}{\operatorname{O}}}_n] = [X, {\operatorname{Gr}}_n({\mathbb{R}}^\infty)]\).
Then
\(\gamma_n \xrightarrow{\pi} {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\)
is the \({\mathbb{R}}^n{\hbox{-}}\)bundle where \(\pi^{-1}(v) = v = V\),
regarded as a plane in \({\mathbb{R}}^\infty\). Another description
comes from the mixing construction:
\(\gamma_n = V_n({\mathbb{R}}^\infty) \overset{\scriptscriptstyle {{\operatorname{O}}_n} }{\times} {\mathbb{R}}^n \to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\).

\begin{center}
\begin{tikzcd}
    {[\mathbf{v}_1, \cdots, \mathbf{v}_n] \times [t_1,\cdots, t_n]} && {\sum t_i \mathbf{v}_i} \\
    {V_n({\mathbb{R}}^\infty) \times {\mathbb{R}}^n} && {\gamma_n} \\
    \\
    {{\operatorname{Gr}}_n({\mathbb{R}}^\infty)}
    \arrow["{\pi'}", from=2-1, to=4-1]
    \arrow["\pi"', from=2-3, to=4-1]
    \arrow["{\exists \cong}"', from=2-1, to=2-3]
    \arrow[maps to, from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwxLCJWX24oXFxSUl5cXGluZnR5KSBcXHRpbWVzIFxcUlJebiJdLFsyLDEsIlxcZ2FtbWFfbiJdLFswLDAsIltcXHZlY3RvciB2XzEsIFxcY2RvdHMsIFxcdmVjdG9yIHZfbl0gXFx0aW1lcyBbdF8xLFxcY2RvdHMsIHRfbl0iXSxbMiwwLCJcXHN1bSB0X2kgXFx2ZWN0b3Igdl9pIl0sWzAsMywiXFxHcl9uKFxcUlJeXFxpbmZ0eSkiXSxbMCw0LCJcXHBpJyJdLFsxLDQsIlxccGkiLDJdLFswLDEsIlxcZXhpc3RzIFxcY29uZyIsMl0sWzIsMywiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{definition}[Subbundles]

\(E' \leq E\) is a \textbf{subbundle} iff there is an embedding
\(E' \hookrightarrow E\) over \(X\):

\begin{center}
\begin{tikzcd}
    {E'} && E \\
    \\
    & X
    \arrow["{\pi'}"', from=1-1, to=3-2]
    \arrow["\pi", from=1-3, to=3-2]
    \arrow["f", hook, from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJFJyJdLFsyLDAsIkUiXSxbMSwyLCJYIl0sWzAsMiwiXFxwaSciLDJdLFsxLDIsIlxccGkiXSxbMCwxLCJmIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=}{Link
to Diagram}
\end{quote}

We also require that restrictions to fibers
\(f_x: { \left.{{E'}} \right|_{{x}} } \to { \left.{{E}} \right|_{{x}} } \in \operatorname{Mat}(m\times n; {\mathbb{R}})\)
is a linear map to an \(n{\hbox{-}}\)dimensional subspace
\({ \left.{{E}} \right|_{{x}} }\), where
\(\dim_{\mathbb{R}}{ \left.{{E'}} \right|_{{x}} } = n\) and
\(\dim_{\mathbb{R}}{ \left.{{E}} \right|_{{x}} } = m\).

\end{definition}

\begin{lemma}[?]

Every vector bundle \(E \xrightarrow{\pi} X\) with
\(X\in {\mathsf{CW}}\) compact is a subbundle of a trivial bundle.

\end{lemma}

\begin{proof}[?]

Take \(\left\{{(U_i, \phi_i)}\right\}_{i=1}^m\rightrightarrows X\) a
finite cover by charts, and choose a subordinate partition of unity
\(\left\{{\chi_i}\right\}_{i=1}^m\) such that
\(\mathop{\mathrm{supp}}\chi_i \subseteq U_i\). Then define
\begin{align*}
\psi: E &\to {\mathbb{R}}^{nm} =
{\mathbb{R}}^n \times {\mathbb{R}}^n \times \cdots \times {\mathbb{R}}^n
\\
v &\mapsto {\left[ {\chi_1(v) \phi_1(v),  \chi_2(v) \phi_2(v), \cdots, \chi_m(v) \phi_m(v)} \right]} 
.\end{align*}
This exhibits \((E \to X) \leq (X\times {\mathbb{R}}^{nm} \to X)\) as a
subbundle.

\end{proof}

\begin{lemma}[?]

Every
\((E\to X)\in { { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} } }_{/ {X}}\)
for \(X\in {\mathsf{CW}}\) compact is a pullback of the canonical bundle
along some map \(f:X\to {{\mathbf{B}}{\operatorname{O}}}_n\).

\end{lemma}

\begin{example}[?]

For \(E \xrightarrow{\pi} X\) and
\(f: X\to {{\mathbf{B}}{\operatorname{O}}}_n\),
\(\psi: E\to {\mathbb{R}}^{nm} \subseteq {\mathbb{R}}^\infty\) and we
can take
\(f(x) \coloneqq\psi(\pi^{-1}(x)) \in {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\)
to get \(f^* \gamma_n \cong E\).

\end{example}

\begin{lemma}[?]

If \(f^* \gamma_n \cong E\) and \(g^* \gamma_n \cong E\), then
\(f \simeq g\).

\end{lemma}

\begin{proof}[?]

Corresponding to \(f^* \gamma_n \cong E\) we get a map
\(\tilde f: E\to {\mathbb{R}}^{\infty}\) which restricts to an embedding
on all fibers, and similarly \(g^*\gamma_n \cong E\) yields
\(\tilde g: E\to {\mathbb{R}}^\infty\). So take
\begin{align*}
L_t: {\mathbb{R}}^\infty &\to {\mathbb{R}}^\infty \\
\mathbf{x} &\mapsto (t-1){\left[ {x_1,x_2,\cdots} \right]} + t{\left[ {x_1, 0, x_2, 0, x_3, 0, \cdots} \right]}
,\end{align*}
which is a homotopy between identity and the self-embedding that maps
into only odd coordinates. Composing \(L_t \circ \tilde f\) yields a
homotopy between \(\tilde f\) and a map \(F'\) whose image has only odd
coordinates. Similarly, we can construct a \(G_t\) for \(\tilde g\) to
get a homotopy between \(\tilde g\) and a map \(G'\) whose images has
only even coordinates. Now take a linear homotopy \(F'\to G'\), this
yields a homotopy through embeddings (where we've first made them
``transverse'').

\end{proof}

\hypertarget{vector-bundle-classification-theorem-tuesday-september-28}{%
\section{Vector Bundle Classification Theorem (Tuesday, September
28)}\label{vector-bundle-classification-theorem-tuesday-september-28}}

\begin{quote}
See homework posted on the website! Turn in 2 total problem sets, one by
mid-October.
\end{quote}

\begin{theorem}[?]

\begin{align*}
[X, {{\mathbf{B}}{\operatorname{O}}}_n] \cong [X, {\operatorname{Gr}}_n({\mathbb{R}}^\infty)] &{ { \, \xrightarrow{\sim}\, }}{ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X) \\
f & \mapsto f^* \gamma_n
.\end{align*}

\end{theorem}

\begin{remark}

We proved surjectivity last time for \(X \in {\mathsf{CW}}\) compact,
using compactness to embed any bundle into a trivial bundle. We proved
injectivity in the form of
\(f^* \gamma_n \cong g^* \gamma_n \implies f\simeq g\), again for
\(X\in {\mathsf{CW}}\) compact. So we need to handle the case of \(X\)
not compact.

\end{remark}

\begin{lemma}[?]

Let \(\pi: E\to X\) be a vector bundle over \(X\). Suppose for
\(B\in {\mathsf{CW}}\) compact with \(B \subseteq X\), we have
\(f: B\to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\) such that
\(f^* \gamma_n \cong { \left.{{E}} \right|_{{B}} }\). Suppose also that
there exists a \(g: X\to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\)
with \(g^* \gamma_n \cong E\). Then there exists an
\(h: X\to{\operatorname{Gr}}_n({\mathbb{R}}^\infty)\) such that
\({ \left.{{h}} \right|_{{B}} } = f\) and \(h^* \gamma_n \cong E\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-09-28_12-57.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{lemma}

\begin{remark}

Idea: write \(X = \colim_n X^{(n)}\) as a limit of compact finite
skeleta, define maps
\(f_n: X^{(n)} \to {{\mathbf{B}}{\operatorname{O}}}_n\) and
\(f_{n+1}: X^{(n+1)}\to {{\mathbf{B}}{\operatorname{O}}}_n\), then
modify \(\tilde f_{n+1} \simeq f_{n+1}\) to extend \(f_n\) in such a way
that \(f_n^* \gamma_i = { \left.{{E}} \right|_{{B_n}} }\).

\end{remark}

\begin{proof}[?]

For \(g^* \gamma_n \cong E\) with
\(({ \left.{{g}} \right|_{{B}} })^* \gamma_n \cong { \left.{{E}} \right|_{{B}} }\)
and \(f^* \gamma_n \cong { \left.{{E}} \right|_{{B}} }\), then
\(f\simeq{ \left.{{g}} \right|_{{B}} }\) by injectivity for compact
\(B\). We can then extend the homotopy
\(H: I\times X\to {{\mathbf{B}}{\operatorname{O}}}_n\) where \(H_0 = g\)
and \(h\coloneqq H_1\) with \({ \left.{{h}} \right|_{{B}} } = f\).

\end{proof}

\hypertarget{characteristic-classes}{%
\subsection{Characteristic Classes}\label{characteristic-classes}}

\begin{definition}[Characteristic classes and representability]

Let \(F, G\) be two contravariant functors with source \(\mathsf{C}\). A
\textbf{characteristic class of \(F\) valued in \(\mathsf{G}\)} is a
natural transformation \(c: F\to G\). \(F\) is \textbf{representable} if
there exists an object \({\mathbf{B}}F\) such that
\(F(X) = [X, {\mathbf{B}}F]\) for every
\(X\in {\operatorname{Ob}}(\mathsf{C})\).

\begin{quote}
Note: we aren't requiring the target categories to coincide!
\end{quote}

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(F(X) \coloneqq\mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)_{/ {X}} = { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} } {}_{/ {X}} = [X {{\mathbf{B}}{\operatorname{O}}}_n]\)
  is a contravariant functor
  \({\mathsf{ho}}{\mathsf{CW}}^{\operatorname{op}}\to {\mathsf{Set}}\),
  where contravariance is due to pullbacks.
\item
  \(G(X) \coloneqq H^j_{\mathrm{sing}}(X)\), which is representable: for
  any \(X\in {\mathsf{CW}}\) and any
  \(G\in {\mathsf{Ab}}{\mathsf{Grp}}\), we have
  \(H^j(X; M) = [X, K(G, j)]\). This comes from taking the sphere that
  generates \(\pi_j K(G, j) = \left\langle{\alpha}\right\rangle\) and
  pulling any \(f: X\to K(G, j)\) back to \(f^* \alpha\).
\end{itemize}

\end{example}

\begin{lemma}[?]

If \(F = [{-}, {\mathbf{B}}F]\) is representable, then characteristic
classes of \(F\) valued in a functor \(G\) biject with \(G(B)\)

\end{lemma}

\begin{remark}

We can write \(F(B) = [B, B] \ni \operatorname{id}_B\), and it turns out
that the characteristic class is determined by where
\(\operatorname{id}_B\) is sent:

\begin{center}
\begin{tikzcd}
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{\operatorname{id}_B\in F} && \textcolor{rgb,255:red,92;green,92;blue,214}{c(\operatorname{id}_B)} \\
    X &&& {F(B) = [B, B]} && {G(B)} \\
    \\
    B &&& {F(X) = [X, B]} && {G(X)} \\
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{\psi} && \textcolor{rgb,255:red,92;green,92;blue,214}{c(\psi) = G(\psi)(c(\operatorname{id}_B))}
    \arrow["c", from=2-4, to=2-6]
    \arrow["{\psi \in [X, B]}", from=2-1, to=4-1]
    \arrow["c", from=4-4, to=4-6]
    \arrow["{F(\psi)}", from=2-4, to=4-4]
    \arrow["{G(\psi)}", from=2-6, to=4-6]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-4, to=1-6]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, from=1-4, to=5-4]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=5-4, to=5-6]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, dashed, from=1-6, to=5-6]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

For us, taking \(B\coloneqq{\operatorname{Gr}}_n({\mathbb{R}}^\infty)\)
and
\(G(B) = H^j({\operatorname{Gr}}_n({\mathbb{R}}^\infty)) \ni \alpha = c(\operatorname{id}_{B})\),
so we can pullback to define \(c(f) = f^*\alpha \in H^j(X)\).

\end{remark}

\begin{example}[?]

Take
\(F(X) = \mathop{\mathrm{Prin}}{\mathsf{Bun}}(U_n)_{/ {X}} = [X {\operatorname{BU}}_n]\)
and \(G(X) = H^j(X)\), then \(\alpha\in H^j({\operatorname{BU}}_n)\)
maps to to \(c_\alpha(E) = f^*(\alpha)\) for any
\(f\in [X, {\operatorname{BU}}_n]\).

\end{example}

\begin{example}[?]

Take \(F(X) = H^n(X; M)\) and \(G(X) = H^m(X; N)\) with
\(M, N \in {\mathsf{Ab}}{\mathsf{Grp}}\). Then \(F(X) = [X, K(M, n)]\),
and taking \(G(K(M, n)) = H^m(K(M, n), N) \ni \alpha\) yields a map
\begin{align*}
H^n(X; M) &\to H^m(X; N) \\
(f: X\to K(M, n)) &\mapsto f^* \alpha
,\end{align*}
i.e.~a cohomology operation. If for example
\begin{align*}
\phi \in \mathop{\mathrm{Hom}}_{\mathsf{Grp}}(M, N) 
= \mathop{\mathrm{Hom}}_{\mathsf{Grp}}( H_n(K(M, n); {\mathbb{Z}}), N)
= H^n( K(M, n); N)
,\end{align*}
using that \(H_n(K(M, n); {\mathbb{Z}})\cong M\). This yields a change
of coefficient morphism
\begin{align*}
H^n(X; M) \to H^n(X; N)
,\end{align*}
which turns out to be the same map as above. So any element in
\(H^m(K(M, n), N)\) yields a map \(H^n(X, M)\to H^m(X, N)\) by sending
\(f:X\to K(M, n)\) to \(f^*\alpha\). Taking \(n=m\) yields
\(H^n(X; M)\to H^n(X; N)\).

\end{example}

\hypertarget{thursday-september-30}{%
\section{Thursday, September 30}\label{thursday-september-30}}

\hypertarget{line-bundles-chern-and-stiefel-whitney-classes}{%
\subsection{Line Bundles: Chern and Stiefel-Whitney
Classes}\label{line-bundles-chern-and-stiefel-whitney-classes}}

\begin{remark}

Last time: defining characteristic classes. Recall that given
\(F, G \in {\mathsf{Fun}}({\mathsf{Top}}, {-})\), a characteristic class
with values in \(G\) is a natural transformation
\(c: F { { \, \xrightarrow{\sim}\, }}G\). If \(F\) is representable,
then characteristic classes \(c: F\to G\) is of the form
\(c(\operatorname{id}_B) \in G(B)\) for
\(\operatorname{id}_B \in [B, B] \cong F(B)\), since \(c\) is determined
by where it sends \(\operatorname{id}_B\). Recall that Eilenberg-MacLane
spaces \(K(G, n)\) represent \(H^n({-}; G)\) for
\(G\in {\mathsf{Ab}}{\mathsf{Grp}}\), and are characterized by
\(\pi_i(K(G, n)) = G\) only in degree \(i=n\).

\end{remark}

\begin{example}[?]

\envlist

\begin{itemize}
\item
  \({ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X) { { \, \xrightarrow{\sim}\, }}[X, {{\mathbf{B}}{\operatorname{O}}}_n]\)
  and we realized
  \({{\mathbf{B}}{\operatorname{O}}}_n \simeq{\operatorname{Gr}}_n({\mathbb{R}}^\infty)\).
  For
  \(\alpha\in H^j({{\mathbf{B}}{\operatorname{O}}}_n; {\mathbb{Z}})\),
  we can take a homotopy class
  \(f:X\to {{\mathbf{B}}{\operatorname{O}}}_n\) and pullback to get
  \(f^*(\alpha) \in H^j(X; {\mathbb{Z}})\).
\item
  \({ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}}, X) { { \, \xrightarrow{\sim}\, }}[X, {\operatorname{BU}}_n]\),
  where
  \({\operatorname{BU}}_n \simeq{\operatorname{Gr}}_n({\mathbb{C}}^\infty)\)
  and we can again pullback cohomology classes.
\end{itemize}

\end{example}

\begin{example}[?]

For line bundles, we can identify
\({\operatorname{BU}}_1 \simeq{\operatorname{Gr}}_1({\mathbb{C}}^\infty) \simeq{\mathbb{CP}}^\infty\),
so
\({ {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{C}}, X) { { \, \xrightarrow{\sim}\, }}[X, {\mathbb{CP}}^\infty]\).
The claim is that line bundles are uniquely characterized by their first
Chern classes. Using that
\(H^2({\mathbb{CP}}^\infty; {\mathbb{Z}}) \cong {\mathbb{Z}}= \left\langle{\alpha}\right\rangle\),
where we've chosen a positive generator, we obtain the \textbf{first
Chern class} \(c_1 \coloneqq f^*(\alpha) \in H^2(X; {\mathbb{Z}})\).
Note that \({\mathbb{CP}}^\infty \simeq K({\mathbb{Z}}, 2)\), and
\([X, K(G, n)] { { \, \xrightarrow{\sim}\, }}H^n(X; M)\) where
\(f\mapsto f^*(\alpha)\) for
\(H^n(K(G, n); {\mathbb{Z}}) = \left\langle{\alpha}\right\rangle\), so
there is an isomorphism
\begin{align*}
[X, {\mathbb{CP}}^\infty] &{ { \, \xrightarrow{\sim}\, }}H^2(X; {\mathbb{Z}}) \\
f &\mapsto f^*(\alpha),\quad \left\langle{\alpha}\right\rangle = H^2({\mathbb{CP}}^\infty)
.\end{align*}
So the set of bundles is an affine space over \(H^2(X)\).

\end{example}

\begin{corollary}[?]

There is a bijection
\begin{align*}
c_1: { {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{R}}, X) { { \, \xrightarrow{\sim}\, }}H^2(X)
,\end{align*}

\end{corollary}

\begin{example}[?]

For \({ {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{R}}, X)\)
we can identify
\({{\mathbf{B}}{\operatorname{O}}}_1 \simeq{\operatorname{Gr}}_1({\mathbb{R}}^\infty) = {\mathbb{RP}}^\infty\),
so for
\(\left\langle{\alpha}\right\rangle = H^2({\mathbb{RP}}^\infty; {\mathbb{Z}}/2) \cong {\mathbb{Z}}/2\),
we obtain the \textbf{first Stiefel-Whitney class} \(w_1\) and a
bijection
\begin{align*}
w_1: [X, {\mathbb{RP}}^\infty] &{ { \, \xrightarrow{\sim}\, }}H^1(X; {\mathbb{Z}}/2) \\
f &\mapsto f^*(\alpha)
.\end{align*}

\end{example}

\begin{remark}

We can define \(c_1\) for vector bundles of any dimension by taking a
top exterior power to get a line bundle:

\begin{center}
\begin{tikzcd}
    {{ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}}, X)} && {{ {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{C}}, X)} && {H^2(X; {\mathbb{Z}})} \\
    E && {\bigwedge\nolimits^n E} \\
    \\
    X && X
    \arrow["{\bigwedge\nolimits^n({-})}", from=1-1, to=1-3]
    \arrow[from=2-1, to=4-1]
    \arrow[from=2-3, to=4-3]
    \arrow["{c_1}", from=1-3, to=1-5]
    \arrow[maps to, from=2-3, to=1-5]
    \arrow["{\bigwedge\nolimits^n({-})}"', from=2-1, to=2-3]
    \arrow["{\coloneqq c_1 E}"{description}, dashed, from=2-1, to=1-5]
    \arrow["\operatorname{id}"', from=4-1, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMCwwLCJcXFZlY3RCdW5kbGVfbihcXENDLCBYKSJdLFsyLDAsIlxcVmVjdEJ1bmRsZV8xKFxcQ0MsIFgpIl0sWzAsMSwiRSJdLFswLDMsIlgiXSxbMiwxLCJcXEV4dGFsZ15uIEUiXSxbMiwzLCJYIl0sWzQsMCwiSF4yKFg7IFxcWlopIl0sWzAsMSwiXFxFeHRhbGdebihcXHdhaXQpIl0sWzIsM10sWzQsNV0sWzEsNiwiY18xIl0sWzQsNiwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFsyLDQsIlxcRXh0YWxnXm4oXFx3YWl0KSIsMl0sWzIsNiwiXFxkYSBjXzEgRSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDUsIlxcaWQiLDJdXQ==}{Link
to Diagram}
\end{quote}

So
\begin{align*}
c_1(E) \coloneqq c_1(\bigwedge\nolimits^n E)
.\end{align*}

\end{remark}

\begin{remark}

There is a natural isomorphism \(c(f^*(E)) \cong f^*(c(E))\), since we
can take iterated pullbacks:

\begin{center}
\begin{tikzcd}
    {f^* E} && {E = g^*\gamma_n} && {\gamma_n} \\
    \\
    Y && X && {{\operatorname{Gr}}_n({\mathbb{R}}^{\infty})}
    \arrow["g"', from=3-3, to=3-5]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[dashed, from=1-1, to=1-3]
    \arrow["f"', from=3-1, to=3-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
    \arrow[from=1-5, to=3-5]
\end{tikzcd}
\end{center}

So we can identify \(c(f^*(E)) = ( g\circ f)^* \alpha\) and
\(f^*(c(E)) = f^*(g^* \alpha) = (g\circ f)^* \alpha\).

\end{remark}

\hypertarget{euler-classes-and-the-thom-isomorphism}{%
\subsection{Euler Classes and the Thom
Isomorphism}\label{euler-classes-and-the-thom-isomorphism}}

\begin{remark}

Note that any vector bundle with a Riemannian metric admits a unit disc
bundle.

\end{remark}

\begin{definition}[Oriented disc bundles]

A unit disc bundle \(D \xrightarrow{\pi} B\) is \textbf{oriented} if
there is a locally coherent choice of a generator of
\(H^n(D_b \coloneqq\pi^{-1}(b), S^b \coloneqq{{\partial}}D_b; {\mathbb{Z}})\).

\end{definition}

\begin{example}[?]

The unit disc bundle for an oriented
\(E \in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X)\)
with a Riemannian metric will be oriented as a disc bundle.

\end{example}

\begin{remark}

Given a bundle \(E\to X\) and taking its disc bundle
\({\mathbb{D}}E\to X\), taking boundaries on fibers yields a sphere
bundle \({\mathbb{S}}E\to X\), so
\({\mathbb{S}}E_b \coloneqq{{\partial}}{\mathbb{D}}E_b\) on fibers. Note
the \({\mathbb{D}}E \simeq X\) by a deformation retraction.

\end{remark}

\begin{theorem}[Thom Isomorphism Theorem]

Let
\({\mathbb{D}}E\to X\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X)\)
be an oriented disc bundle and \({\mathbb{S}}E \to X\) its corresponding
sphere bundle. Then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(H^{i< n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) = 0\)
\item
  There exists a generator, the \textbf{Thom class}
  \(u_E \in H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})\) mapping to
  a distinguished generator of
  \(H^n({\mathbb{D}}E_x, {\mathbb{S}}E_x)\), inducing isomorphisms
  \begin{align*}
  ({-}) \smile u_E : H^j( {\mathbb{D}}E) &{ { \, \xrightarrow{\sim}\, }}H^{j+n}({\mathbb{D}}E, {\mathbb{S}}E) && \forall j\geq 0 \\
  \eta &\mapsto \eta \smile u_E
  ,\end{align*}
\end{enumerate}

\end{theorem}

\begin{proof}[of theorem]

\envlist

\begin{itemize}
\item
  \textbf{Step 1}: Look locally to see why we might expect this result!
  If \({\mathbb{D}}E\to X\) is trivial, then the claims hold.
\item
  \textbf{Step 2}: If the claims hold for
  \({ \left.{{{\mathbb{D}}E}} \right|_{{U}} }, { \left.{{{\mathbb{D}}E}} \right|_{{V}} }, { \left.{{{\mathbb{D}}E}} \right|_{{U \cap V}} }\),
  then it holds for
  \({ \left.{{{\mathbb{D}}E}} \right|_{{U \cup V}} }\). As a corollary,
  the claims hold for compact \(X\).
\item
  \textbf{Step 3}: Prove claims for \(H^*({-}; k)\) for
  \(k\in \mathsf{Field}\)
\item
  \textbf{Step 4}: Prove claims for \(H^*({-}; {\mathbb{Z}})\).
\end{itemize}

\end{proof}

\begin{proof}[step 1]

Trivial bundles are products, and we have formulas for cohomology of
products. Write \({\mathbb{D}}E = X \times {\mathbb{D}}^n\), and note
that \(H^*({\mathbb{D}}^n, S^{n-1}; {\mathbb{Z}}) = {\mathbb{Z}}[n]\),
which is always torsionfree. Thus
\begin{align*}
H^k({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) 
&= H^k({\mathbb{D}}^n\times X, S^{n-1} \times X; {\mathbb{Z}}) \\
&\cong \bigoplus_{0\leq i\leq k} H^i(X; {\mathbb{Z}}) \otimes_{\mathbb{Z}}H^{k-i}({\mathbb{D}}^n, S^{n-1}; {\mathbb{Z}}) \\
&\cong \bigoplus_{0\leq i\leq k} H^i(X; {\mathbb{Z}}) \otimes_{\mathbb{Z}}\qty{{\mathbb{Z}}[i+n ]}  \\
&= H^{k-n}(X; {\mathbb{Z}}) & \quad \text{ when } k>n, \text{ else } 0
.\end{align*}

So pick \(u\in H^n({\mathbb{D}}^n, S^{n-1})\) be the generator specified
by the orientation, and take
\(u_E \in H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})\) to be the
corresponding generator pulled back along the Kunneth isomorphism (which
recall was induced by a cup product).

\end{proof}

\begin{proof}[step 2]

Use Mayer-Vietoris:

\includegraphics{figures/2021-10-03_21-35-21.png}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwyLCJIXmsoXFxERCBFXFxtaWRfe1UgXFx1bmlvbiBWfSwgXFxTUyBFXFxtaWRfe1UgXFx1bmlvbiBWfSApIl0sWzIsMiwiSF5rKFxcREQgRVxcbWlkX3tVfSwgXFxTUyBFXFxtaWRfe1V9ICkgXFxvcGx1cyBIXmsoXFxERCBFXFxtaWRfe1Z9LCBcXFNTIEVcXG1pZF97Vn0gKSJdLFs0LDIsIkheayhcXEREIEVcXG1pZF97VSBcXGludGVyc2VjdCBWfSwgXFxTUyBFXFxtaWRfe1UgXFxpbnRlcnNlY3QgVn0gKSJdLFs0LDAsIkhee2stMX0oXFxERCBFXFxtaWRfe1UgXFxpbnRlcnNlY3QgVn0sIFxcU1MgRVxcbWlkX3tVIFxcaW50ZXJzZWN0IFZ9ICkiXSxbMCw0LCJcXGNkb3RzIl0sWzAsMV0sWzEsMl0sWzMsMF0sWzIsNF1d}{Link
to Diagram}
\end{quote}

\begin{itemize}
\tightlist
\item
  If \(k< n\), the union terms vanish in degree \(k\), since they're
  surrounded by zeros.
\item
  If \(k=n\), the kernel of \(\oplus \to \cap\) is isomorphic to
  \({\mathbb{Z}}\), so pick a generator
  \(u_{U \cup V} = { \left.{{u_E}} \right|_{{U \cup V}} }\) that lifts
  \({ \left.{{u_E}} \right|_{{U}} }\) and
  \({ \left.{{u_E}} \right|_{{V}} }\).
\end{itemize}

Next time: we'll show that \(u_{U \cup V} \smile({-})\) yields the
isomorphism in part 2 of the theorem statement.

\end{proof}

\hypertarget{tuesday-october-05}{%
\section{Tuesday, October 05}\label{tuesday-october-05}}

\begin{remark}

Recall that we were proving the Thom isomorphism theorem. Some
motivation:

\end{remark}

\begin{corollary}[The Gysin Sequence]

Consider an oriented sphere bundle:

\begin{center}
\begin{tikzcd}
S^{n-1} 
  \ar[r] 
& 
{\mathbb{S}}E
  \ar[d] 
\\
& 
X 
\end{tikzcd}
\end{center}

Then there is a LES induced by the Euler class \(e\in H^n(X)\)

\begin{center}
\begin{tikzcd}
    \cdots \\
    \\
    {H^{j-n}(X)} && {H^j(X)} && {H^{j}({\mathbb{S}}E)} \\
    &&&& {} \\
    && \cdots && {H^{j-1}({\mathbb{S}}E)}
    \arrow[from=5-5, to=3-1]
    \arrow["{e\smile({-})}", from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=1-1]
    \arrow[from=5-3, to=5-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbNCwzXSxbMCwyLCJIXntqLW59KFgpIl0sWzQsMiwiSF57an0oXFxTUyBFKSJdLFsyLDIsIkheaihYKSJdLFs0LDQsIkhee2otMX0oXFxTUyBFKSJdLFswLDAsIlxcY2RvdHMiXSxbMiw0LCJcXGNkb3RzIl0sWzQsMV0sWzEsMywiZVxcY3VwcHJvZChcXHdhaXQpIl0sWzMsMl0sWzIsNV0sWzYsNF1d}{Link
to Diagram}
\end{quote}

\end{corollary}

\begin{proof}[of corollary]

Corresponding to \({\mathbb{S}}E \xrightarrow{\pi} X\), the mapping cone
of \(\pi\) is \({\mathbb{D}}E\). So consider the LES in relative
homology:

\begin{center}
\begin{tikzcd}
    \cdots \\
    \textcolor{rgb,255:red,214;green,92;blue,92}{u_E} && \textcolor{rgb,255:red,214;green,92;blue,92}{e} \\
    {H^j({\mathbb{D}}E, {\mathbb{S}}E)} && {H^j({\mathbb{D}}E)} && {H^j({\mathbb{S}}E)} \\
    &&&& {} \\
    &&&& {H^{j-1}({\mathbb{S}}E)} \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-n}({\mathbb{D}}E)} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^j(X)} \\
    {} \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-n}(X)}
    \arrow[curve={height=6pt}, from=5-5, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow["{u_E \smile({-}), \cong}", color={rgb,255:red,92;green,92;blue,214}, from=7-1, to=3-1]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=9-1, to=7-1]
    \arrow["\cong", color={rgb,255:red,92;green,92;blue,214}, from=7-3, to=3-3]
    \arrow["{e\smile({-})}"', color={rgb,255:red,92;green,92;blue,214}, dashed, from=9-1, to=7-3]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, maps to, from=2-1, to=2-3]
    \arrow[curve={height=6pt}, from=3-5, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

After identifying terms, we see that
\(u_E \in H^n({\mathbb{D}}E,{\mathbb{S}}E)\) maps to the Euler class
\(e\in H^n(X)\), which carries interesting geometric information.

\end{proof}

\begin{proof}[of theorem, for fields]

Suppose the claim holds for compact \(X\), and write
\(X = \displaystyle\bigcup_i C_i\) for \(C_i\) compact CW skeleta. Then
\(H_i(X) = \colim_j H_i(C_j)\) since simplices are also compact. Note
that
\begin{align*}
H^i(X; k) 
&\cong H_i(X; k) {}^{ \vee }\\
&\coloneqq\mathop{\mathrm{Hom}}( H_i(X; k), k) \\
&= \mathop{\mathrm{Hom}}( \colim_j H_i(C_j;k); k) \\
&= \cocolim_j \mathop{\mathrm{Hom}}( H_i(C_j; k), k) \\
&= \cocolim_j H^i(C_j; k)
.\end{align*}

Similarly,
\begin{align*}
H^i({\mathbb{D}}E, {\mathbb{S}}E; k) { { \, \xrightarrow{\sim}\, }}\cocolim_j H^i\qty{ { \left.{{{\mathbb{D}}E}} \right|_{{ C_j}} }, { \left.{{{\mathbb{S}}E}} \right|_{{C_j}} }; k } =
\begin{cases}
0 & i<n 
\\
? & i\geq n.
\end{cases}
.\end{align*}
Assembling the maps and isomorphisms we have:

\begin{center}
\begin{tikzcd}
    {H^i({\mathbb{D}}E; k)} &&& {H^m({\mathbb{D}}E, {\mathbb{S}}E; k)} \\
    \\
    {H^i({ \left.{{{\mathbb{D}}E}} \right|_{{C_j}} }; k)} &&& {H^i({ \left.{{{\mathbb{D}}E}} \right|_{{C_j}} }, { \left.{{{\mathbb{S}}E}} \right|_{{ C_j}} } ; k)}
    \arrow["{{ \left.{{u_E}} \right|_{{{\mathbb{D}}E}} }\smile({-}), \cong}", from=3-1, to=3-4]
    \arrow["{u_E \smile({-})}", from=1-1, to=1-4]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-4, to=3-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfTsgaykiXSxbMywyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfSwgXFxyb3tcXFNTIEV9eyBDX2p9IDsgaykiXSxbMCwwLCJIXmkoXFxERCBFOyBrKSJdLFszLDAsIkhebShcXEREIEUsIFxcU1MgRTsgaykiXSxbMCwxLCJcXHJve3VfRX17XFxERCBFfVxcY3VwcHJvZChcXHdhaXQpLCBcXGNvbmciXSxbMiwzLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwwXSxbMywxXV0=}{Link
to Diagram}
\end{quote}

Now use that the vertical maps become isomorphisms after a colimit, so
the top map must become an isomorphism as well.

\end{proof}

\begin{proof}[of theorem, for arbitrary rings]

Consider
\(H_i({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) \to \cocolim_j H_i( { \left.{{{\mathbb{D}}E}} \right|_{{C_j}} }, { \left.{{{\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}})\)
Using universal coefficients, we have
\begin{align*}
H^i({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) 
\cong \mathop{\mathrm{Hom}}( H_i({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}), {\mathbb{Z}}) 
\oplus \operatorname{Ext} ( H_{i-1}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}), {\mathbb{Z}}) = 0 && i < n
,\end{align*}
since each summand will be zero. For \(i=n\), the Ext term vanishes, and
we have
\begin{align*}
H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) 
&\cong \mathop{\mathrm{Hom}}( H_n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}), {\mathbb{Z}}) \\
&\cong \mathop{\mathrm{Hom}}( \colim_j H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}), {\mathbb{Z}}) \\
&\cong \cocolim_j \mathop{\mathrm{Hom}}( H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}), {\mathbb{Z}}) \\
&\cong \cocolim_j H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}) \\
&\cong \left\langle{u_E}\right\rangle \cong {\mathbb{Z}}
,\end{align*}
using the distinguished generator
\(u_E \in H^n({\mathbb{D}}E, {\mathbb{S}}E)\). So we can define a chain
map
\begin{align*}
u_E \frown({-}): C_{j+n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) \to C_j({\mathbb{D}}E)
,\end{align*}
which shifts degree by \(-n\) by capping against \(u_E\). This induces
the cup product
\(u_E \smile({-}):H^j({\mathbb{D}}E; {\mathbb{Z}}) \to H^{j+n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})\).

\begin{lemma}[Milnor-Stasheff 10.6]

Given chain complexes of \({\mathbb{Z}}{\hbox{-}}\)modules \(C_*\) and
\(D_*\) and a chain map \(f:C_* \to D_*\), if
\(f^*: H^*(D_*; k) \to H_*(C_*; k)\) are isomorphisms for every
\(k\in \mathsf{Field}\), then \(f_*, f^*\) are isomorphisms over any
\(R\in \mathsf{Ring}\).

\end{lemma}

We showed that \(u_E \smile({-})\) was an isomorphism for all \(k\), so
now we get an isomorphism over every \(R\) and this completes the proof.

\end{proof}

\begin{remark}

Without the oriented assumption, this theorem still holds with \(C_2\)
coefficients.

Note also that we have naturality for characteristic classes: given a
pullback

\begin{center}
\begin{tikzcd}
    {f^* E} && E \\
    \\
    Y && X
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJFIl0sWzIsMiwiWCJdLFswLDIsIlkiXSxbMCwwLCJmXiogRSJdLFszLDBdLFswLDFdLFsyLDFdLFszLDJdLFszLDEsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

Then
\begin{align*}
c(f^* E) = f^*c(E)
.\end{align*}
Note that we can always pull back the canonical:

\begin{center}
\begin{tikzcd}
    {f^* \gamma_n} && {\gamma_n} \\
    \\
    X && {{\operatorname{Gr}}_n({\mathbb{R}}^\infty)}
    \arrow[from=1-3, to=3-3]
    \arrow["f"', from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJmXiogXFxnYW1tYV9uIl0sWzIsMCwiXFxnYW1tYV9uIl0sWzIsMiwiXFxHcl9uKFxcUlJeXFxpbmZ0eSkiXSxbMCwyLCJYIl0sWzEsMl0sWzMsMiwiZiIsMl0sWzAsMV0sWzAsM10sWzAsMiwiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d}{Link
to Diagram}
\end{quote}

If the Euler class \(e\) is natural, then
\(e(f^* \gamma_n) = f^* e(\gamma_n)\) where
\(e(\gamma_n) \in H^n({\operatorname{Gr}}_n({\mathbb{R}}^\infty))\), so
\(e\) is the characteristic class defined by \(e(\gamma_n)\).

\end{remark}

\begin{lemma}[?]

The Euler class \(e\) is natural with respect to pullback and thus a
characteristic class.

\end{lemma}

\begin{proof}[?]

We'll check naturality for the Thom class \(u_E\). Consider pulling back
a disc bundle:

\begin{center}
\begin{tikzcd}
    {D'\coloneqq f^* {\mathbb{D}}E} && {D \coloneqq{\mathbb{D}}E} \\
    \\
    Y && X
    \arrow["f", from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJEJ1xcZGEgZl4qIFxcREQgRSJdLFsyLDAsIkQgXFxkYSBcXEREIEUiXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFszLDIsImYiXSxbMSwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

Recall that \(u_E \in H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})\)
such that
\({ \left.{{u_E}} \right|_{{x}} }: H^n( { \left.{{{\mathbb{D}}E}} \right|_{{x}} }, { \left.{{{\mathbb{S}}E}} \right|_{{x}} }; {\mathbb{Z}})\)
is the generator. We get an element in the fibers of the pullback in the
following way:

\begin{center}
\begin{tikzcd}
    {u_E} &&& {u'} \\
    {H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})} &&& {H^n(f^* {\mathbb{D}}E, f^* {\mathbb{S}}E; {\mathbb{Z}})} \\
    \\
    {H^n({\mathbb{D}}E{ \left.{{}} \right|_{{f(x)}} }, {\mathbb{S}}E{ \left.{{}} \right|_{{f(x)}} }; {\mathbb{Z}})} &&& {H^n(f^* {\mathbb{D}}E{ \left.{{}} \right|_{{f(x)}} }, f^* {\mathbb{S}}E{ \left.{{}} \right|_{{f(x)}} }; {\mathbb{Z}})} \\
    {{ \left.{{u_E}} \right|_{{x}} }} &&& {{ \left.{{u'}} \right|_{{x}} }}
    \arrow["\cong", from=4-1, to=4-4]
    \arrow[from=2-4, to=4-4]
    \arrow[from=2-1, to=4-1]
    \arrow[from=2-1, to=2-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwzLCJIXm4oXFxERCBFXFxyb3t9e2YoeCl9LCBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzMsMywiSF5uKGZeKiBcXEREIEVcXHJve317Zih4KX0sIGZeKiBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzAsMSwiSF5uKFxcREQgRSwgXFxTUyBFOyBcXFpaKSJdLFszLDEsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFswLDAsInVfRSJdLFszLDAsInUnIl0sWzAsNCwiXFxyb3t1X0V9e3h9Il0sWzMsNCwiXFxyb3t1J317eH0iXSxbMCwxLCJcXGNvbmciXSxbMywxXSxbMiwwXSxbMiwzXV0=}{Link
to Diagram}
\end{quote}

We then get naturality of the Euler class from the following:

\begin{center}
\begin{tikzcd}
    {u_E} &&& e \\
    {H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})} &&& {H^n({\mathbb{D}}E; {\mathbb{Z}}) \cong H^n(X)} \\
    \\
    {H^n(f^* {\mathbb{D}}E, f^* {\mathbb{S}}E; {\mathbb{Z}})} &&& {H^n(Y)} \\
    {u'} &&& {e'}
    \arrow[from=2-4, to=4-4]
    \arrow[from=4-1, to=4-4]
    \arrow[from=2-1, to=2-4]
    \arrow[from=2-1, to=4-1]
    \arrow[from=1-1, to=1-4]
    \arrow[from=5-1, to=5-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwxLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMSwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYKSJdLFswLDMsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFszLDMsIkhebihZKSJdLFswLDAsInVfRSJdLFszLDAsImUiXSxbMCw0LCJ1JyJdLFszLDQsImUnIl0sWzEsM10sWzIsM10sWzAsMV0sWzAsMl0sWzQsNV0sWzYsN11d}{Link
to Diagram}
\end{quote}

\end{proof}

\begin{example}[?]

\(e({\mathbb{CP}}^1)\) is a generator of \(H^2({\mathbb{CP}}^1)\) Apply
the Gysin sequence, taking the canonical line bundle \({\mathbb{S}}E\)
over \({\mathbb{CP}}^1\):

\begin{center}
\begin{tikzcd}
    {H^{j-1}({\mathbb{S}}E)} && {H^{j-2}({\mathbb{CP}}^1)} && {H^{j}({\mathbb{CP}}^1)} && {H^j({\mathbb{CP}}^1)} && {H^j({\mathbb{S}}E)} \\
    &&&&&& {}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbNiwxXSxbMiwwLCJIXntqLTJ9KFxcQ1BeMSkiXSxbNCwwLCJIXntqfShcXENQXjEpIl0sWzYsMCwiSF5qKFxcQ1BeMSkiXSxbMCwwLCJIXntqLTF9KFxcU1MgRSkiXSxbOCwwLCJIXmooXFxTUyBFKSJdLFs0LDFdLFsxLDJdLFsyLDNdLFszLDVdXQ==}{Link
to Diagram}
\end{quote}

The claim is that the total space here is the Hopf fibration:

\begin{center}
\begin{tikzcd}
S_1\to {\mathbb{S}}E \cong S^3 
  \ar[r, "\subseteq"] 
  \ar[dr]
& 
E
  \ar[d] 
\\
& 
{\mathbb{CP}}^1\cong S^2 
\end{tikzcd}
\end{center}

What is the Hopf fibration? Write
\({\mathbb{CP}}^1 = \left\{{{\left[ {z_0: z_1} \right]} {~\mathrel{\Big\vert}~}z_0^2 + z_1^2 = 1}\right\}/\sim\),
then realize
\(S^3 = \left\{{z_0^2 + z_1^2 = 1}\right\} \subseteq {\mathbb{C}}^2\).
Then take a map \(S^3\to {\mathbb{CP}}^1\), whose fibers are
\(\left\{{\lambda \in {\mathbb{C}}{~\mathrel{\Big\vert}~}{\left\lvert {\lambda } \right\rvert}= 1}\right\} = {\mathbb{C}}^{\times}\cong S^1\).
Then identifying elements and maps in the Gysin sequence yields the
following:

\begin{center}
\begin{tikzcd}
    & {H^j({\mathbb{S}}E)} && \cdots \\
    {j=3:} & \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j}({\mathbb{S}}^3)} \\
    \\
    & {H^{j-1}({\mathbb{S}}E)} && {H^{j-2}({\mathbb{CP}}^1)} && {H^{j}({\mathbb{CP}}^1)} \\
    {j=2:} & \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-1}(S^3) = 0} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0({\mathbb{CP}}^1)} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^2({\mathbb{CP}}^1)} && {}
    \arrow[from=4-2, to=4-4]
    \arrow[from=4-4, to=4-6]
    \arrow["{e\smile({-})}", color={rgb,255:red,92;green,92;blue,214}, from=5-4, to=5-6]
    \arrow[from=4-6, to=1-2]
    \arrow[from=1-2, to=1-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

So \(e\) is a generator of \(H^2({\mathbb{CP}}^1)\)

\end{example}

\begin{exercise}[?]

Check that \({\mathbb{S}}E\to {\mathbb{CP}}^1\) with \({\mathbb{S}}E\)
the canonical is the Hopf fibration.

\end{exercise}

\begin{exercise}[?]

Try to compute \(e({\mathbf{T}}S^2)\)! You may need to add on a bundle
to trivialize it.

\end{exercise}

\hypertarget{thursday-october-07}{%
\section{Thursday, October 07}\label{thursday-october-07}}

\hypertarget{the-euler-class}{%
\subsection{The Euler Class}\label{the-euler-class}}

\begin{remark}

Let \(E\to X\) be an orientable bundle, then

\begin{center}
\begin{tikzcd}
    {H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})} &&& {H^n({\mathbb{D}}E; {\mathbb{Z}}) \cong H^n(X; {\mathbb{Z}})} \\
    {u_{-E} = -u_E} &&& {e(E)}
    \arrow[maps to, from=2-1, to=2-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMCwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYOyBcXFpaKSJdLFswLDEsInVfey1FfSA9IC11X0UiXSxbMywxLCJlKEUpIl0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{lemma}[?]

\(e(-E) = -e(C)\).

\end{lemma}

\begin{proof}[?]

Using naturality:

\begin{center}
\begin{tikzcd}
    {E = f^*({\mathbb{R}}^n)} && {{\mathbb{R}}^n} \\
    \\
    X && {\operatorname{pt}}
    \arrow["\pi", from=1-3, to=3-3]
    \arrow["f"', from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJFID0gZl4qKFxcUlJebikiXSxbMiwwLCJcXFJSXm4iXSxbMiwyLCJcXHB0Il0sWzAsMiwiWCJdLFsxLDIsIlxccGkiXSxbMywyLCJmIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

Then \(e({\mathbb{R}}^n\to {\operatorname{pt}}) = 0\), and
\(0 = f^*(0) = e(f^*({\mathbb{R}}^n\to {\operatorname{pt}})) = e(E)\).

\end{proof}

\begin{lemma}[?]

\(e(E) = 0\) if \(E\) is the trivial bundle.

\end{lemma}

\begin{proof}[?]

Using the exact sequence for the pair
\(({\mathbb{D}}E, {\mathbb{S}}E)\):

\begin{center}
\begin{tikzcd}
    {u_E} && e \\
    {H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})} && {H^n({\mathbb{D}}E; {\mathbb{Z}})} && {H^n({\mathbb{S}}E; {\mathbb{Z}})} \\
    \\
    && {\cong H^n(X\times {\mathbb{D}}^n; {\mathbb{Z}}) \cong H^n(X) \otimes H^n({\mathbb{D}}; {\mathbb{Z}})} && {\cong H^n(X\times S^{n-1}; {\mathbb{Z}})} \\
    && {H^n(X)} && {H^n(X)\oplus H^1(X)} \\
    && \alpha && {(\alpha, 0)}
    \arrow[hook, from=5-3, to=5-5]
    \arrow[from=6-3, to=6-5]
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=2-5]
    \arrow[maps to, from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

So the second map is the inclusion \(\alpha \mapsto (\alpha, 0)\), which
is thus injective. The first map is 0, so \(e(E) = 0\).

\end{proof}

\begin{example}[?]

Compute \(e( {\mathbf{T}}S^2)\).

\end{example}

\begin{lemma}[?]

If \(E\) is an odd dimensional vector bundle, then \(2e(E) = 0\)

\end{lemma}

\begin{remark}

Check that \(\alpha \smile\beta = (-1)^{ij} \beta \smile\alpha\).

\end{remark}

\begin{proof}[First way]

The antipodal map will give an isomorphism \(E\cong -E\), so
\(e(E) = e(-E) = -e(E)\) and thus \(2e(E) = 0\).

\end{proof}

\begin{proof}[Second way]

Use the map:

\begin{center}
\begin{tikzcd}
    {H^n({\mathbb{D}}E, {\mathbb{S}}E;{\mathbb{Z}})} &&& {H^{2n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})} \\
    {u_E} &&& {u_E\smile u_E = - u_E \smile u_E}
    \arrow["{u_E \smile({-})}", from=1-1, to=1-4]
    \arrow[maps to, from=2-1, to=2-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7XFxaWikiXSxbMywwLCJIXnsybn0oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzAsMSwidV9FIl0sWzMsMSwidV9FXFxjdXBwcm9kIHVfRSA9IC0gdV9FIFxcY3VwcHJvZCB1X0UiXSxbMCwxLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d}{Link
to Diagram}
\end{quote}

Then \(2u_E \smile u_E = 0\), making \(2u_E = 0\) and thus
\(2e(E) = 0\).

\end{proof}

\begin{lemma}[?]

Given \(X\to X\) and \(F\to Y\), then we can form the bundle
\(E\times F\to X\times Y\). Writing \(\pi_1:X\times Y\to X\) and
\(\pi_2:X\times Y\to Y\), we have
\begin{align*}
e(E\times F) = \pi_1^*(e(E)) \smile\pi_2^*(e(F)) = e(E) \otimes e(F)
.\end{align*}

\end{lemma}

\begin{exercise}[?]

Prove this using the Kunneth formula.

\end{exercise}

\begin{corollary}[?]

Let \(E,E'\to X\) be two vector bundles. Then
\(e(E \oplus E') = e(E) \smile e(E')\).

\end{corollary}

\begin{proof}[?]

Consider the diagonal
\(\Delta: X\to X{ {}^{ \scriptscriptstyle\times^{2} } }\), then take the
pullback:

\begin{center}
\begin{tikzcd}
    {\Delta^*(E\times E')\cong E \oplus E'} && {E\times E'} \\
    \\
    {X{ {}^{ \scriptscriptstyle\times^{2} }  }} && X
    \arrow["\Delta", from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXERlbHRhXiooRVxcdGltZXMgRScpXFxjb25nIEUgXFxvcGx1cyBFJyJdLFsyLDAsIkVcXHRpbWVzIEUnIl0sWzIsMiwiWCJdLFswLDIsIlhcXGNhcnRwb3dlcnsyfSJdLFszLDIsIlxcRGVsdGEiXSxbMCwzXSxbMSwyXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

Then
\begin{align*}
e(E \oplus E')
&= \Delta^*( e(E\times E'))\\
&= \Delta^*(\pi_1^*(e(E)) \smile\pi_2^*(e(E')) ) \\
&= \Delta^* \pi_1^* (e(E))\smile\Delta^* \pi_2^* e(E') \\
&= e(E) \smile e(E')
.\end{align*}

\end{proof}

\begin{corollary}[?]

If \(E' = L \oplus E\) for \(L\) a trivial line bundle, \(e(E') = 0\).

\end{corollary}

\begin{proof}[?]

\begin{align*}
e(E') = e(L) \smile e(E) = 0
.\end{align*}

\end{proof}

\hypertarget{obstruction-theory}{%
\subsection{Obstruction Theory}\label{obstruction-theory}}

\begin{remark}

The Euler class is the \textbf{obstruction} to finding a nonvanishing
section over the \(n{\hbox{-}}\)skeleton of \(X\). Note that if we have
a line bundle, we automatically have a section. Conversely, a
nonvanishing section will produce a line bundle summand.

\end{remark}

\begin{exercise}[?]

Describe the correspondence between line subbundles and nonvanishing
global section.

\end{exercise}

\begin{remark}

Some review: let \(X\in {\mathsf{CW}}\) and write
\(H^*(X; {\mathbb{Z}})\) for cellular homology. Then
\(C^k_{{ \mathrm{cell}} }(X) = \mathop{\mathrm{Hom}}(H_k(X^{(k)}, X^{(k-1)} ), {\mathbb{Z}}) = \mathop{\mathrm{Hom}}(C_k^{{ \mathrm{cell}} }, {\mathbb{Z}})\).
To produce the differential (green), use that the LES for the pairs (red
and blue) are intertwined:

\begin{center}
\begin{tikzcd}
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{\vdots} \\
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^k(X^{(k+1)}, X^{(k)}; {\mathbb{Z}})} \\
    \\
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^k(X^{(k+1)}; {\mathbb{Z}})} \\
    \\
    \textcolor{rgb,255:red,214;green,92;blue,92}{\cdots} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k-1)}; {\mathbb{Z}})} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k)}; {\mathbb{Z}})} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k)}, X^{(k-1)}; {\mathbb{Z}})} & \textcolor{rgb,255:red,214;green,92;blue,92}{\cdots} \\
    \\
    &&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^{k+1}(X^{(k+1)}, X^{(k)}; {\mathbb{Z}})}
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-4, to=4-4]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=6-4]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, from=6-6, to=6-4]
    \arrow["\delta", color={rgb,255:red,92;green,214;blue,92}, from=6-6, to=8-4]
    \arrow["\delta", color={rgb,255:red,92;green,92;blue,214}, from=6-4, to=8-4]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, from=6-4, to=6-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Thus we have a formula
\begin{align*}
{\left\langle { \delta \alpha},~{ [\Delta^{k+1}] } \right\rangle} = {\left\langle { \alpha},~{ {{\partial}}[\Delta^{k+1}] } \right\rangle}
.\end{align*}

\end{remark}

\begin{remark}

Recall that if the fibers of a bundle were \(n{\hbox{-}}\)connected, we
could construct sections on \(X^{(n-1)}\). Given any section, we can use
a Riemannian metric to project onto norm 1 elements to get a section for
the sphere bundle, say \(s: X^{(n-1)} \to {\mathbb{S}}E\). Note that
\(\pi_i(S^{n-1}) = 0\) for \(i\leq n-2\), so we can construct such a
section. Let \(i: \Delta^n\to X\) be the attaching map for some
\(n{\hbox{-}}\)cell, then form the following pullback to get a trivial
bundle, where we can pull back the section:

\begin{center}
\begin{tikzcd}
    {\Delta^n \times S^{n-1}} && {i^* {\mathbb{S}}E} && {{\mathbb{S}}E} \\
    \\
    && {\Delta^n} && X && {X^{(n-1)}}
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-5, to=3-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-3, to=1-5]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
    \arrow["\cong"{description}, no head, from=1-1, to=1-3]
    \arrow["{\pi_1}"', from=1-1, to=3-3]
    \arrow[hook', from=3-7, to=3-5]
    \arrow["s"', color={rgb,255:red,214;green,92;blue,92}, from=3-7, to=1-5]
    \arrow["{\tilde s}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-30pt}, from=3-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMiwwLCJpXiogXFxTUyBFIl0sWzIsMiwiXFxEZWx0YV5uIl0sWzQsMiwiWCJdLFs0LDAsIlxcU1MgRSJdLFswLDAsIlxcRGVsdGFebiBcXHRpbWVzIFNee24tMX0iXSxbNiwyLCJYXnsobi0xKX0iXSxbMSwyXSxbMywyXSxbMCwxXSxbMCwzXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwwLCJcXGNvbmciLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzQsMSwiXFxwaV8xIiwyXSxbNSwyLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6ImJvdHRvbSJ9fX1dLFs1LDMsInMiLDIseyJjb2xvdXIiOlswLDYwLDYwXX0sWzAsNjAsNjAsMV1dLFsxLDQsIlxcdGlsZGUgcyIsMCx7ImN1cnZlIjotNSwiY29sb3VyIjpbMCw2MCw2MF19LFswLDYwLDYwLDFdXV0=}{Link
to Diagram}
\end{quote}

So \(s\) yields a map
\({{\partial}}\Delta^n \cong S^{n-1} \to S^{n-1}\), so it has a degree.
Then a cocycle can be defined as
\([\Delta^n] \xrightarrow{e_s} \deg( {{\partial}}\Delta^n \to S^{n-1})\in {\mathbb{Z}}\).

\end{remark}

\begin{lemma}[?]

\(e_S\) is independent of the choice of trivialization for the pullback.

\end{lemma}

\begin{proof}[?]

Since \(\Delta^n\) is contractible, any two trivializations are
homotopic. Thus the sections
\(s_T, s_{T'}: {{\partial}}\Delta^n \to S^{n-1}\) corresponding to two
trivializations are homotopic.

\end{proof}

\begin{lemma}[?]

If \(s'\) is a different section over \(X^{(n-1)}\), then
\(e_{s'} - e_s = \delta(\alpha)\) for some
\(\alpha\in C^{n-1}_{{ \mathrm{cell}} }(X)\).

\end{lemma}

\begin{proof}[?]

See phone notes.

\end{proof}

\hypertarget{tuesday-october-12}{%
\section{Tuesday, October 12}\label{tuesday-october-12}}

\hypertarget{more-euler-class}{%
\subsection{More Euler Class}\label{more-euler-class}}

\begin{quote}
Review how to construct the Euler and Thom classes.
\end{quote}

\begin{remark}

Recall that the Euler class is the obstruction to finding a nowhere
vanishing section on the \(n{\hbox{-}}\)skeleton. Given
\(E\to X \in { {\mathsf{Bun}}\qty{\operatorname{GL}_r} }(X)^{\dim = n}\),
we can form the sphere bundle \(S^{n-1}\to {\mathbb{S}}E \to X\). Define
a section \(s: X\to {\mathbb{S}}E\) over \(X{ {}^{ (0) } }\), then
inductively if \(s\) is defined over \(X{ {}^{ (i-1) } }\) for \(i<n\),
let \(j: \Delta^i \to X{ {}^{ (i) } }\) be the attaching map for an
\(i{\hbox{-}}\)cell. Note that \(\Delta^i\) is contractible, and we can
form a pullback

\begin{center}
\begin{tikzcd}
\Delta^i \times S^{n-1} 
  \ar[r] 
  \ar[d]
& 
{\mathbb{S}}E
  \ar[d] 
\\
\Delta^i
  \ar[r, "j"]
& 
X{ {}^{ (i) } } 
\end{tikzcd}
\end{center}

If \(s\) is defined over
\(X{ {}^{ (i-1) } } \subseteq X{ {}^{ (i) } }\), we have
\({ \left.{{s}} \right|_{{{{\partial}}\Delta^i}} }: {{\partial}}\Delta^i \to S^{n-1}\).
But \({{\partial}}\Delta^{i} \cong S^{i-1}\), and if \(i\leq n-1\),
\(\pi_i S^{n-1} = 0\) so \(s\) extends over \(\Delta^i\). If \(i=n\),
note that \(\pi_{n-1} S^{n-1} \cong {\mathbb{Z}}\), so we define
\(e_s(\Delta^n)\) to be the corresponding homotopy class of \(s\),
i.e.~\(e_s(\Delta^n)\) is the degree of this map between spheres.

We showed that

\begin{itemize}
\tightlist
\item
  \(e_s\) doesn't depend on the trivialization chosen, and
\item
  If \(s'\) is another nowhere vanishing section on
  \(X{ {}^{ (n-1) } }\), then
  \(e_s - e_{s'} \in \operatorname{im}\delta\) for
  \(\delta: C^{n-1}(X) \to C^n(X)\).
\item
  If \(\delta e_s = 0\), then \([e_s] \in H^n(X; {\mathbb{Z}})\). The
  content here is that the cochain descends to a cocycle.
\end{itemize}

For \(i:\Delta^{-1} \to X{ {}^{ (n-1) } }\) and
\({\mathbb{S}}E\to X{ {}^{ (n-1) } }\), form the pullback
\(i^* {\mathbb{S}}E\). Generally given two sections
\(s, s': \Delta^{n-1} \to i^* {\mathbb{S}}E \cong \Delta^{n-1} \times S^{n-1}\),
we can glue them to define a map
\(-\Delta^{n-1} {\textstyle\coprod}_{{{\partial}}} \Delta^{n-1} \xrightarrow{-s {\textstyle\coprod}s'} S^{n-1}\).
The glued space is an \(S^{n-1}\), so the degree of this map will be
\(\alpha^{\Delta^{n-1}}\)

\end{remark}

\begin{lemma}[?]

For any \(\alpha\in C^{n-1}(X)\), there exists a section \(s'\) on
\(X{ {}^{ (n-1) } }\) such that \(e_{s'} - e_s = \delta \alpha\).

\end{lemma}

\begin{exercise}[?]

Prove this!

\end{exercise}

\begin{proposition}[?]

\([e_s] \in H^n(X)\) agrees with the Euler class.

\end{proposition}

\begin{proof}[?]

We can equivalently define \(e_s\) as follows: extend \(s\) to any
section \(\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu\)
of the disc bundle \({\mathbb{D}}E \to X\). Then for any
\(n{\hbox{-}}\)cell \(\Delta^n\), we can form the composition:

\begin{center}
\begin{tikzcd}
    {(\Delta^n, {{\partial}}\Delta^n)} && {(\Delta^n \times {\mathbb{D}}^n, ({{\partial}}\Delta^n)\times S^{n-1})} && {({\mathbb{D}}^n, S^{n-1})} \\
    \\
    {{{\partial}}\Delta^n} &&&& {S^{n-1}}
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu}", from=1-1, to=1-3]
    \arrow["{p}", from=1-3, to=1-5]
    \arrow["\psi", curve={height=-30pt}, dashed, from=1-1, to=1-5]
    \arrow[hook, from=3-1, to=1-1]
    \arrow["{\psi'}"', from=3-1, to=3-5]
    \arrow[hook, from=3-5, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

We can then define \(e_s(\Delta^n) \coloneqq\deg \psi\), and we claim
that this also equals \(\deg \psi'\). Look at the LES of a pair:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,214;green,92;blue,92}{H_n({\mathbb{D}}^n)=0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_n({\mathbb{D}}^n) = 0} \\
    \\
    {H_n({\mathbb{D}}^n; S^{n-1})\cong {\mathbb{Z}}} &&& {H_n({\mathbb{D}}^n; S^{n-1})\cong {\mathbb{Z}}} \\
    \\
    {H_{n-1}(S^{n-1})\cong {\mathbb{Z}}} &&& {H_{n-1}(S^{n-1})\cong {\mathbb{Z}}} \\
    \\
    \textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}({\mathbb{D}}^n) = 0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}({\mathbb{D}}^n) = 0}
    \arrow["{\psi^*}", from=3-1, to=3-4]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-4, to=3-4]
    \arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-1, to=5-1]
    \arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-4, to=5-4]
    \arrow["{(\psi')^*}"{description}, from=5-1, to=5-4]
    \arrow[from=5-1, to=7-1]
    \arrow[from=5-4, to=7-4]
    \arrow[from=7-1, to=7-4]
    \arrow[from=1-1, to=1-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

We may assume that \(X\) is \(n{\hbox{-}}\)dimensional since for
\((X, X{ {}^{ (n) } })\) we have
\begin{align*}
H^n(X, X{ {}^{ (n) } }) = 0 \to H^n(X) \hookrightarrow H^n(X^n)
,\end{align*}
so anything equal in \(H^n(X{ {}^{ (n) } })\) must be equal in
\(H^n(X)\). Fix a positive generator
\(\left\langle{x}\right\rangle = H^n({\mathbb{D}}^n, S^{n-1})\) and
\(\left\langle{y}\right\rangle = H_n(\Delta^n, {{\partial}}\Delta^n)\)
to be the fundamental class (positive generator). Then
\begin{align*}
e_s(\Delta^n) = {\left\langle {\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* p^* x},~{y} \right\rangle}
.\end{align*}
Consider the map
\((\Delta^n \times {\mathbb{D}}^n, {{\partial}}\Delta^n \times S^{n-1}) \xrightarrow{p} ({\mathbb{D}}^n, S^{n-1})\),
we have \(p(\Delta^n\times S^{n-1}) = S^{n-1}\). Then we claim that
\(p^*(x) = u\) will be the Thom class. Using the attaching map
\(i: \Delta^n \to X{ {}^{ (n) } }\), we obtain
\begin{align*}
H^n({\mathbb{D}}E, {\mathbb{S}}E) \xrightarrow{i^*} H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1})
.\end{align*}

Use that
\(\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu: X\to {\mathbb{D}}E\)
induces an isomorphism
\(\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*: H^n({\mathbb{D}}E)\to H^n(X)\),
inducing the same isomorphism as the zero section. So
\((s')^* = \mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*\),
i.e.~any two sections of the disc bundle will be homotopic and thus
induce equal maps in homology. Now doing exactly what we did for the
Euler class, we get a diagram:

\begin{center}
\begin{tikzcd}
    && {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} \\
    & {H^n({\mathbb{D}}E)} & {} & {H^n({\mathbb{D}}E, {\mathbb{S}}E) } && {H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1})} \\
    & {H^n(X)} \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{e} &&& {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} && {H^n(\Delta^n, {{\partial}}\Delta^n)}
    \arrow["{i^*}", from=2-4, to=2-6]
    \arrow["{i_*}", from=4-4, to=4-6]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=2-6, to=4-6]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-4, to=4-4]
    \arrow["LES"', from=2-4, to=2-2]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-2, to=3-2]
    \arrow["LES"', from=4-4, to=3-2]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, maps to, from=1-4, to=4-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

More directly, we can map the LESs to get a commutative square:

\begin{center}
\begin{tikzcd}
    {H^n({\mathbb{D}}E, {\mathbb{S}}E)} && {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} \\
    \\
    {H^n({\mathbb{D}}E)} && {H^n(X{ {}^{ (n) } })}
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=1-1, to=1-3]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEUpIl0sWzAsMiwiSF5uKFxcREQgRSkiXSxbMiwwLCJIXm4oWFxcc2tlbCBuLCBYXFxza2Vse24tMX0pIl0sWzIsMiwiSF5uKFhcXHNrZWwgbikiXSxbMCwyLCJcXGJhcntzfV4qIl0sWzEsMywiXFxiYXJ7c31eKiJdLFswLDFdLFsyLDNdXQ==}{Link
to Diagram}
\end{quote}

In other words, regard \(e\in H^n_{ \mathrm{cell}} (X)\), so \(e\)
corresponds to
\([c]\in \mathop{\mathrm{Hom}}_{{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}}(C^n(X), {\mathbb{Z}})\).
We then claim that for \(c\in H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })\),
we have \(e_s(\Delta^n) = i^* c(\Delta^n, {{\partial}}\Delta^n)\) using
\(i: (\Delta^n, {{\partial}}\Delta^n)\to (X{ {}^{ (n) } }, X{ {}^{ (n-1) } })\).
Final diagram:

\begin{center}
\begin{tikzcd}
    && {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} && \textcolor{rgb,255:red,92;green,92;blue,214}{p^*x = i^* u} \\
    & {H^n({\mathbb{D}}E)} & {} & {H^n({\mathbb{D}}E, {\mathbb{S}}E) } && {H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1})} \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{e} & {H^n(X)} \\
    &&& {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} && {H^n(\Delta^n, {{\partial}}\Delta^n)} & \textcolor{rgb,255:red,92;green,92;blue,214}{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* p^* x = \mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* i^* u}
    \arrow["{i^*}", from=2-4, to=2-6]
    \arrow["{i_*}", from=4-4, to=4-6]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=2-6, to=4-6]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-4, to=4-4]
    \arrow["LES"', from=2-4, to=2-2]
    \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-2, to=3-2]
    \arrow["LES"', two heads, from=4-4, to=3-2]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=30pt}, dashed, maps to, from=1-4, to=3-1]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, maps to, from=1-4, to=1-6]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, maps to, from=1-6, to=4-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{proof}

\hypertarget{computations-on-smooth-manifolds}{%
\subsection{Computations on Smooth
Manifolds}\label{computations-on-smooth-manifolds}}

\begin{remark}

Recall that a smooth structure on a manifold \(M\) is a collection
\((U, \phi_U)\) where \(U \subseteq M\) is open,
\(\phi_U: U\to {\mathbb{R}}^n\) is a homeomorphism onto an open subset
of \({\mathbb{R}}^n\), and for all \(U, V\) we have
\begin{align*}
\psi_{VU} \coloneqq\phi_V^{-1}\circ \phi_U: \phi_U(U \cap V) \to \phi_V(U \cap V) \in C^\infty
,\end{align*}
so there are all diffeomorphisms of open subsets of \({\mathbb{R}}^n\).
A smooth atlas is maximal if not contained in any other smooth atlas,
and two atlases are compatible if their union is again a smooth atlas.
We say two smooth structures are equivalent if they are compatible.

\end{remark}

\begin{exercise}[?]

Show that any smooth manifold has a unique maximal smooth atlas.

\end{exercise}

\begin{remark}

Recall that for \(f: {\mathbb{R}}^n\to {\mathbb{R}}^m\),
\(df\in \operatorname{Mat}(m\times n; {\mathbb{R}})\) is given by
\((df)_{ij} = {\frac{\partial f_i}{\partial x_j}\,}\). For any
\(p\in U \cap V\), we have
\(d_{\phi_U(p)} \psi_{VU} \in \operatorname{GL}_n({\mathbb{R}})\), so we
get a map \(d\psi_{VU}: U \cap V\to \operatorname{GL}_n({\mathbb{R}})\).
By the chain rule they satisfying the cocycle definition, so these glue
to a vector bundle \({\mathbf{T}}M\to M\).

\end{remark}

\begin{exercise}[?]

Show that every other definition of \({\mathbf{T}}M\) coincides with
this one.

\end{exercise}

\hypertarget{thursday-october-14}{%
\section{Thursday, October 14}\label{thursday-october-14}}

\begin{remark}

Some background from smooth manifolds: a map \(f:M\to N\) of manifolds
is \textbf{smooth} if for any smooth charts \((U, \phi_U)\) and
\((V, \psi_V)\) on \(M, N\) respectively, the transition map
\(\psi_V \circ f \circ \phi_U ^{-1}{\mathbb{R}}^m\to{\mathbb{R}}^n\) is
smooth.

\end{remark}

\begin{exercise}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Any smooth map \(f:M\to N\) induces a bundle map
  \(df: {\mathbf{T}}M\to f^* {\mathbf{T}}N\).
\item
  There is a canonical isomorphism
  \({\mathbf{T}}{\mathbb{R}}^n \to {\mathbb{R}}^n\times {\mathbb{R}}^n\).
\item
  Let \((U, \phi_U)\) be a chart on \(M\), then show that there are
  trivializing charts for \({\mathbf{T}}M\):
  \begin{align*}
  d\phi_U: { \left.{{{\mathbf{T}}M}} \right|_{{U}} } \to { \left.{{ \phi_U^* }} \right|_{{ {\mathbf{T}}{\mathbb{R}}^n }} }{U} \cong U \times {\mathbb{R}}^n
  .\end{align*}
  In particular, the following diagram commutes:
\end{enumerate}

\begin{center}
\begin{tikzcd}
    {{ \left.{{{\mathbf{T}}M}} \right|_{{U}} }} && {U\times {\mathbb{R}}^n} \\
    \\
    U
    \arrow["\pi", from=1-1, to=3-1]
    \arrow["{d\phi_U}", from=1-1, to=1-3]
    \arrow[from=1-3, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve1xcVCBNfXtVfSJdLFswLDIsIlUiXSxbMiwwLCJVXFx0aW1lcyBcXFJSXm4iXSxbMCwxLCJcXHBpIl0sWzAsMiwiZFxccGhpX1UiXSxbMiwxXV0=}{Link
to Diagram}
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that there is a canonical isomorphism
\begin{align*}
T_p M { { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Der}}(M, {\mathbb{R}}) \leq C^\infty(M, {\mathbb{R}})
,\end{align*}
where a derivation at \(p\) is a smooth functional
\(v: C^\infty(M, {\mathbb{R}})\to {\mathbb{R}}\) such that
\(v(fg) = v(f) g(p) + f(p) v(g)\).

\end{exercise}

\begin{remark}

\(N \subseteq M\) is a \textbf{smooth submanifold} if for any \(p\in N\)
there exists a smooth chat \((U, \varphi)\) on \(M\) such that
\(p\in U\) and
\(\phi(U \cap N) \subseteq {\mathbb{R}}^k \times\left\{{0}\right\} \subseteq {\mathbb{R}}^n\).

\end{remark}

\begin{exercise}[?]

Show that this is equivalent to \(N = f(\tilde N)\) where \(\tilde N\)
is some smooth manifold and \(f:\tilde N\to M\) is a smooth embedding
and \(d_p f: {\mathbf{T}}_p \tilde N\hookrightarrow F_{f(p)} M\) is
injective for all \(p\in \tilde N\).

\end{exercise}

\begin{definition}[?]

Given a Riemannian metric on \({\mathbf{T}}M\) and a smooth submanifold
\(N \subseteq M\), let \(\nu N\) denote
\(({\mathbf{T}}N)^\perp \subseteq { \left.{{ {\mathbf{T}}M}} \right|_{{N}} }\)
for the orthogonal complement of \({\mathbf{T}}N\). This is a vector
bundle \(\nu N \to N\).

\end{definition}

\begin{exercise}[?]

Show that up to a canonical isomorphism, this is independent of the
choice of Riemannian metric.

\end{exercise}

\begin{definition}[?]

Given a curve \(\gamma: I\to M\), then
\(\gamma'(t) \in {\mathbf{T}}_{\gamma(t)} M\) is the following
derivation:
\begin{align*}
\gamma'(t) . f \coloneqq\lim_{h\to 0} { f(\gamma(t+h)) - f(\gamma(t)) \over h}
.\end{align*}
A \textbf{vector field} on \(M\) is a section of \({\mathbf{T}}M\), and
a vector field along \(\gamma\) is a section of
\(\gamma^* {\mathbf{T}}M\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{43pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-15.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{example}[?]

\(\gamma'(t)\) is a vector field along \(\gamma\).

\end{example}

\begin{remark}

A \textbf{Lie group} is a group \(G\) with the structure of a smooth
manifold where multiplication and inversion are smooth
self-diffeomorphisms.

\end{remark}

\begin{definition}[?]

Given a Lie group \(G\) and smooth principal \(G{\hbox{-}}\)bundle
\(P \xrightarrow{\pi} M^n\), a \textbf{connection} on \(P\) is a choice
of subspaces \(\xi_p \subseteq T_p P\) for all \(p\in P\) such that
\(d\pi: \xi_p \to {\mathbf{T}}_{\pi(p)} M\) is an isomorphisms for all
\(p\), where \(\xi_p\) is the horizontal subspace:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-25.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{remark}

Given a connection, curves \(\gamma\) in \(M\) can be lifted to curves
\(\tilde \gamma\) in \(P\) in such a way that tangents of
\(\tilde \gamma\) are projected to tangents of \(\gamma\):

\end{remark}

\begin{definition}[?]

Given a connection on \(P\) and a smooth path \(\gamma: I\to M\), a
horizontal lift \(\tilde \gamma\) of \(\gamma\) is a path
\(\tilde \gamma: I\to P\) such that
\(\tilde \gamma'(t) \in \xi_{\tilde \gamma(t)}\) and
\(\pi \circ \tilde \gamma = \gamma\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-32.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{lemma}[?]

Given a smooth path \(\gamma\) and \(\tilde \gamma(0)\), there is a
unique horizontal lift \(\tilde \gamma\) starting at
\(\tilde \gamma(0)\).

\end{lemma}

\begin{remark}

Consider \(\mathop{\mathrm{Frame}}({\mathbf{T}}M)\), then recall that
\({{\mathbf{T}}M} = \mathop{\mathrm{Frame}}({\mathbf{T}}M) \overset{\scriptscriptstyle {\operatorname{GL}_n({\mathbb{R}})} }{\times} {\mathbb{R}}^n\)
by the mixing construction. Given a connection on
\(\mathop{\mathrm{Frame}}({\mathbf{T}}M)\), we can parallel transport
vectors in \({\mathbf{T}}M\) along curves. This comes from taking
\([F_0, v_0]\) and evolving \(F_0\) along
\(F_t \coloneqq\tilde \gamma(t)\), choosing pairs \([F_t, v_0]\) for all
\(t\), and ending at \([F_t, v_0]\).

\end{remark}

\begin{exercise}[?]

Show that parallel transport yields a well-defined map
\({\mathbf{T}}_{\gamma(0)}M \to {\mathbf{T}}_{\gamma(1))}M\).

\end{exercise}

\begin{theorem}[?]

Given a Riemannian metric on \({\mathbf{T}}M\), there is a canonical
connection \(\nabla\), the \textbf{Levi-Cevita} connection, which is
torsionfree (i.e.~\(\nabla_X Y - \nabla_Y X = [X, Y]\)) for \(X, Y\)
vector fields and \(\nabla_X Y\) denotes parallel transporting \(Y\)
along \(X\).

\end{theorem}

\begin{definition}[?]

A curve \(\gamma\) is a \textbf{geodesic} if \(\gamma'(t)\) is parallel.

\end{definition}

\begin{theorem}[?]

For any \(v\in {{\mathbf{T}}M}\), there is a unique geodesic
\(\gamma_v\) with \(\gamma_v'(0) = v\).

\end{theorem}

\begin{definition}[Exponential map]

There is a map
\begin{align*}
\exp: T_p M &\to M \\
v &\mapsto \gamma_v(1)
.\end{align*}
This is well-defined for \(v\) of small norm, and for all \(v\) if \(M\)
is closed.

\end{definition}

\begin{theorem}[?]

If \(N \leq M\) is a submanifold, then \(\exp\) defines a diffeomorphism
from a neighborhood of the zero section in \(\nu N\) to a neighborhood
of \(N\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_14-04.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{theorem}

\hypertarget{tuesday-october-19}{%
\section{Tuesday, October 19}\label{tuesday-october-19}}

\begin{remark}

Today: computations involving the Euler class.

\end{remark}

\begin{theorem}[?]

Let \(M\in {\mathsf{sm}}{\mathsf{Mfd}}^{\dim =n}\) be closed and
oriented, then noting that
\(e({\mathbf{T}}M) \in H^n(M; {\mathbb{Z}})\), \footnote{\({\mathbb{Z}}\)
  can be replaced with \({\mathbb{Z}}/2\) if \(M\) is not oriented.}

\begin{align*}
{\left\langle { e({\mathbf{T}}M)},~{ [M] } \right\rangle} = \chi_{\mathsf{Top}}(M)
,\end{align*}
the topological Euler characteristic of \(M\).

\end{theorem}

\begin{theorem}[?]

Let \(M\in {\mathsf{sm}}{\mathsf{Mfd}}^{\dim = n}\) closed and
\(E \xrightarrow{\pi} M \in { {\mathsf{Bun}}\qty{\operatorname{GL}_{k}} }(X)\)
oriented. Suppose a generic section \(s\) of \(E\),
i.e.~\(\operatorname{im}s \coloneqq\left\{{s(x) \in E {~\mathrel{\Big\vert}~}x\in M}\right\} \subseteq E\),
intersects \(M \subseteq E\) transversally. Then

\begin{itemize}
\tightlist
\item
  \(z\coloneqq s^{-1}(0)\) is a submanifold of \(M\) is a submanifold of
  \(M\)
\item
  \({ \left.{{ds}} \right|_{{Z}} }: \nu Z { { \, \xrightarrow{\sim}\, }}{ \left.{{E}} \right|_{{Z}} }\)
\end{itemize}

\end{theorem}

\begin{exercise}[?]

Prove this.

\end{exercise}

\begin{corollary}[?]

\(e(E) = \mathrm{PD}([Z])\), where \([Z] \in H_{n-k}(M)\).

\end{corollary}

\begin{claim}

The second theorem implies the first.

\end{claim}

\begin{proof}[?]

Note that a section of \({\mathbf{T}}M\) is a vector field. Use Morse
functions \(f: M\to {\mathbb{R}}\):

\begin{itemize}
\tightlist
\item
  \(\operatorname{crit}(f) = \left\{{p\in M {~\mathrel{\Big\vert}~}df_p = 0}\right\}\).
\item
  \(p\in \operatorname{crit}(f)\) is \textbf{nondegenerate} if the
  Hessian
  \(H_p(f) \coloneqq\qty{ {\partial^2 \over \partial x_i \partial x_j } } (p)\)
  is nonsingular, i.e.~\(\operatorname{det}H_p\neq 0\).
\item
  \(f\) is \textbf{Morse} if all critical points are nondegenerate.
\item
  If \(p\in \operatorname{crit}(f)\) is nondegenerate, then
  \(\operatorname{Ind}_p(f)\) is the number of negative eigenvalues of
  \(H_p\).
\item
  \(f\) Morse on \(M\) induces a CW complex structure with exactly one
  \(k{\hbox{-}}\)cell corresponding to each index \(k\) critical point
  \(p\in \operatorname{Crit}(f)_k\). Thus we can compute
  \(\chi(M) = \sum_k (-1)^k \dim C_k^{ \mathrm{cell}} = \sum_k (-1)^k \# \operatorname{Crit}(f)_k\).
\end{itemize}

\end{proof}

\begin{proof}

\(f\) Morse induces a gradient vector field on \(M\): picking a
Riemannian metric \(g\) on \(M\), define
\(df({-}) \coloneqq g( \operatorname{grad}f, {-})\) to get a section
\(\operatorname{grad}f: M\to {\mathbf{T}}M\). Note that if
\(p\in \operatorname{crit}(f)\) then \(d_p f = 0\) since
\(\operatorname{grad}_p f = 0\). So the vector field \(df\) vanishes at
\(\operatorname{crit}(f)\) and \((\operatorname{grad}f)^{-1}(0)\)

\begin{exercise}[?]

Show that the sign of a zero of the gradient vector field is
\((-1)^{\operatorname{Ind}_p(f)}\).

\end{exercise}

So taking \(Z = \operatorname{crit}(f)\), we can write
\begin{align*}
e({\mathbf{T}}M) = \mathrm{PD} { \left[ { \sum_{p\in \operatorname{crit}(f) } (-1)^{\operatorname{Ind}_p(f)} [x] } \right] } 
,\end{align*}
where \([x]\) is the dual of a generator of \(H_0(M)\) Then
\begin{align*}
{\left\langle { e({\mathbf{T}}M)},~{ [M] } \right\rangle} = \sum_{p\in \operatorname{crit}(f) } (-1)^{\operatorname{Ind}_p(f)} = \# \operatorname{crit}_{\text{even}}(f) - \# \operatorname{crit}_{\text{odd}}(f) = \chi_{\mathsf{Top}}(M)
.\end{align*}

\end{proof}

\begin{remark}

Given \(N^n \hookrightarrow M^m\) an oriented closed submanifold with
\(M\) closed, we can consider Thom class of the disc/sphere bundles. We
identify the disc bundle \({\mathbb{D}}\nu N\) with a tubular
neighborhood of \(N\) in \(M\) and apply excision to get the following:

\begin{center}
\begin{tikzcd}
    {u_{\nu N} \in H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu n)} && {H^{m-n}(M, M\setminus N)} && {H^{m-n}(M)} \\
    \\
    && {H^{m-n}({\mathbb{D}}\nu N, {\mathbb{D}}\nu N \setminus N)}
    \arrow["\cong", from=1-1, to=1-3]
    \arrow["\cong", from=1-1, to=3-3]
    \arrow["{\text{excision} \cong}", from=1-3, to=3-3]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJ1X3tcXG51IE59IFxcaW4gSF57bS1ufShcXEREIFxcbnUgTiwgXFxTUyBcXG51IG4pIl0sWzIsMCwiSF57bS1ufShNLCBNXFxzbSBOKSJdLFsyLDIsIkhee20tbn0oXFxERCBcXG51IE4sIFxcREQgXFxudSBOIFxcc20gTikiXSxbNCwwLCJIXnttLW59KE0pIl0sWzAsMSwiXFxjb25nIl0sWzAsMiwiXFxjb25nIl0sWzEsMiwiXFx0ZXh0e2V4Y2lzaW9ufSBcXGNvbmciXSxbMSwzXV0=}{Link
to Diagram}
\end{quote}

We can also consider the composition:

\begin{center}
\begin{tikzcd}
    {[N]\in H_n(N)} && {H_n(M)} && {H^{m-n}(M)}
    \arrow["{i_*}", from=1-1, to=1-3]
    \arrow["\mathrm{PD}", from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJbTl1cXGluIEhfbihOKSJdLFsyLDAsIkhfbihNKSJdLFs0LDAsIkhee20tbn0oTSkiXSxbMCwxLCJpXyoiXSxbMSwyLCJcXFBEIl1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{claim}

These two classes are equal in \(H^{m-n}(M)\).

\end{claim}

\begin{proof}[?]

Consider the following:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu N}} \\
    {\left\langle{u_{\nu N}}\right\rangle = H^{m-n}({\mathbb{D}}\nu n, {\mathbb{S}}\nu n)} && {H_n({\mathbb{D}}\nu N)} \\
    \\
    {H^{m-n}(M, M\setminus N)} && {H_n(N) = \left\langle{[N]}\right\rangle} & \textcolor{rgb,255:red,92;green,92;blue,214}{\pm [N]} \\
    \\
    {H^{m-n}(M)} && {H_m(M)}
    \arrow["{\mathrm{PD}, \cong}", tail reversed, from=2-1, to=2-3]
    \arrow["{\mathrm{PD}({-}) = [M]\frown({-}), \cong}", tail reversed, from=6-1, to=6-3]
    \arrow["\cong", from=2-3, to=4-3]
    \arrow["\cong"', from=2-1, to=4-1]
    \arrow[from=4-1, to=6-1]
    \arrow[from=4-3, to=6-3]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, dashed, from=1-1, to=4-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

The claim is that \(u_{\nu N}\) is mapped to the \emph{positive}
generator, so the both paths from
\(H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N)\to H^{m-n}(M)\) agree.

\end{proof}

\begin{remark}

An aside on cup/cap products. The cap product is a map:
\begin{align*}
\frown: H_m(X) \otimes_{\mathbb{Z}}H^i(X) &\to H_{m-i}(X) \\
.\end{align*}
On chains, it's given by
\begin{align*}
\frown C_m(X) \otimes_{\mathbb{Z}}C^i(X) &\to C_{m-i}(X) \\
(\sigma, \phi) &\mapsto \phi\qty{ { \left.{{\sigma}} \right|_{{{\left[ {v_0,\cdots, v_i} \right]}}} } } { \left.{{\sigma}} \right|_{{v_{i+1}, \cdots, v_m}} }
.\end{align*}
Then \(\mathrm{PD}({-}) \coloneqq[X]\frown({-})\).

The cup product is a map
\begin{align*}
\smile: H^i(X) \otimes_{\mathbb{Z}}H^j(X) &\to H^{i+j}(X) \\
C^i(X) \otimes_{\mathbb{Z}}C^j(X) &\to C^{i+j}(X) \\
(\phi, \psi) &\mapsto \qty{\sigma \mapsto \phi\qty{{ \left.{{\sigma}} \right|_{{[v_0,\cdots, v_i]}} }} \psi\qty{{ \left.{{\sigma}} \right|_{{[v_{i+1}, \cdots, v_{i+j}}} }} }
.\end{align*}
Then fixing any element yields a map
\(\alpha \smile({-}): C^j(X) \to C^{i+j}(X)\), which is induced by a map
\(\phi\frown({-}): C_{i+j}(X) \to C_i(X)\).

\end{remark}

\hypertarget{thursday-october-21}{%
\section{Thursday, October 21}\label{thursday-october-21}}

\begin{remark}

Recall that if \(N^n \leq M^n\) is a submanifold, we have the following
diagram:

\begin{center}
\begin{tikzcd}
    {u_{\nu N}} & {H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N)} && {H^{m-n}(M, M\setminus N)} && {H^{m-n}(M)} \\
    \\
    & {H_n({\mathbb{D}}\nu N)} && {H_n(N)} && {H_n(M)} \\
    &&& {[N]}
    \arrow["\mathrm{PD}", from=3-6, to=1-6]
    \arrow["\cong", from=3-2, to=3-4]
    \arrow["{i_*}", from=3-4, to=3-6]
    \arrow["{j^*}"', from=1-4, to=1-6]
    \arrow["{\cong, \text{ excision}}"', from=1-2, to=1-4]
    \arrow["\mathrm{PD}"', from=3-2, to=1-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMSwwLCJIXnttLW59KFxcREQgXFxudSBOLCBcXFNTIFxcbnUgTikiXSxbMywwLCJIXnttLW59KE0sIE1cXHNtIE4pIl0sWzUsMCwiSF57bS1ufShNKSJdLFs1LDIsIkhfbihNKSJdLFsxLDIsIkhfbihcXEREIFxcbnUgTikiXSxbMywyLCJIX24oTikiXSxbMywzLCJbTl0iXSxbMCwwLCJ1X3tcXG51IE59Il0sWzMsMiwiXFxQRCJdLFs0LDUsIlxcY29uZyJdLFs1LDMsImlfKiJdLFsxLDIsImpeKiIsMl0sWzAsMSwiXFxjb25nLCBcXHRleHR7IGV4Y2lzaW9ufSIsMl0sWzQsMCwiXFxQRCIsMl1d}{Link
to Diagram}
\end{quote}

Then the Thom class
\(u_{\nu N}\in H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N) \to H^{m-n}({\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x)\)
is mapped to the generator specified by the orientations on fibers.

\end{remark}

\begin{theorem}[?]

For \(A^i \pitchfork B^j \leq X\) are smooth oriented submanifolds
intersecting transversally, then
\begin{align*}
\mathrm{PD}([A]) \smile\mathrm{PD}([B]) &= \mathrm{PD}([A \frown B]) \\
H^{n-i}(X) \times H^{n-j}(X) &\to H^{2n-i-j}(X)
.\end{align*}

\end{theorem}

\begin{remark}

Then
\begin{align*}
[( {\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) \frown N] = [\left\{{x}\right\}]
,\end{align*}
which is the positive generator. So
\begin{align*}
\mathrm{PD}[ ({\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) ] \smile\mathrm{PD}[N] = \mathrm{PD}[x]
.\end{align*}
Now we can cap this to obtain
\begin{align*}
[{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\qty{ \mathrm{PD}[ ({\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) ] \smile\mathrm{PD}[N]} 
&=
\qty{ [{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\mathrm{PD}[{\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x] } \frown\mathrm{PD}[N] \\
&= 
[{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\mathrm{PD}[x] \\
&= 1
,\end{align*}
where we've used that
\({\left\langle {\mathrm{PD}[x]},~{ [{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] } \right\rangle}\).
So \(\mathrm{PD}[N]\) is \(j^*\) of the Thom class of \(\nu N\).

\end{remark}

\begin{theorem}[?]

Let \(M \in {\mathsf{sm}}{\mathsf{Mfd}}^n\) be closed and oriented and
\(E \xrightarrow{\pi} M\) a \(k{\hbox{-}}\)dimensional oriented vector
bundle. Consider a generic section \(s\) of \(E\), so
\(\operatorname{im}(s) \pitchfork M\) in \(E\). Then
\begin{align*}
e(E) = \mathrm{PD}[Z] && Z \coloneqq\operatorname{im}(s) \cap M = s^{-1}(0)
.\end{align*}

\end{theorem}

\begin{remark}

Recall that \(s^{-1}(0) \leq M\) is a smooth submanifold, and
\({ \left.{{ds}} \right|_{{Z}} }: \nu Z { { \, \xrightarrow{\sim}\, }}{ \left.{{E}} \right|_{{Z}} }\),
and since this orients \(\nu Z\) this orients \(Z\) as well.

\end{remark}

\begin{proof}[?]

Let \(N\) be a tubular neighborhood of \(Z\) in \(M\), such that
\(N \cong {\mathbb{D}}\nu Z\). By the lemma, we have two maps

\begin{center}
\begin{tikzcd}
    {H^k(N, N\setminus Z)} && {H^k(M, M\setminus Z)} && {H^k(M)} \\
    {u_{\nu Z}} &&&& {\mathrm{PD}[Z]}
    \arrow["\cong", from=1-1, to=1-3]
    \arrow["{j^*}", from=1-3, to=1-5]
    \arrow["{\text{Lemma}}"', curve={height=30pt}, from=1-1, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJIXmsoTiwgTlxcc20gWikiXSxbMiwwLCJIXmsoTSwgTVxcc20gWikiXSxbNCwwLCJIXmsoTSkiXSxbMCwxLCJ1X3tcXG51IFp9Il0sWzQsMSwiXFxQRFtaXSJdLFswLDEsIlxcY29uZyJdLFsxLDIsImpeKiJdLFswLDIsIlxcdGV4dHtMZW1tYX0iLDIseyJjdXJ2ZSI6NX1dXQ==}{Link
to Diagram}
\end{quote}

on the other hand, since \(Z \xhookrightarrow{i} M\), we get

\begin{center}
\begin{tikzcd}
    {{ \left.{{E}} \right|_{{Z}} }} && E \\
    \\
    & N
    \arrow["\iota", from=1-1, to=1-3]
    \arrow["{{ \left.{{ds}} \right|_{{Z}} }}", from=3-2, to=1-1]
    \arrow["s"', from=3-2, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve0V9e1p9Il0sWzIsMCwiRSJdLFsxLDIsIk4iXSxbMCwxLCJcXGlvdGEiXSxbMiwwLCJcXHJve2RzfXtafSJdLFsyLDEsInMiLDJdXQ==}{Link
to Diagram}
\end{quote}

\begin{exercise}[?]

Show \({ \left.{{ds}} \right|_{{Z}} } \simeq s\) are homotopic sections.

\end{exercise}

Then \(({ \left.{{ds}} \right|_{{Z}} })^* \circ i^* (u_E) = u_{\nu Z}\),
so \(u_{\nu Z} = s^* u_E\), and we have

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{u_E} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu Z}} \\
    & {H^k(E, E\setminus M)} && {H^k(M, M\setminus Z)} \\
    \\
    & {H^k(E)} && {H^k(M)} \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{?} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\mathrm{PD}[Z]} \\
    &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{e(E)}
    \arrow["{\text{LES}}", from=2-2, to=4-2]
    \arrow["{s^*}", from=2-2, to=2-4]
    \arrow["{s^*, \cong}", from=4-2, to=4-4]
    \arrow[from=2-4, to=4-4]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=1-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-5, to=5-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=5-1, to=6-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=5-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Since the diagram commutes, we get \(e(E) = \mathrm{PD}[Z]\).

\end{proof}

\begin{remark}

The Euler class for a bundle \(E\to X\in {\mathsf{CW}}\) is the
obstruction to finding a nowhere vanishing section on
\(X{ {}^{ (n) } }\) for \(n\coloneqq\dim E\), and \(e(E) = 0\) iff there
is a nowhere vanishing section on \(X{ {}^{ (n) } }\). For a smooth
manifold \(M\) and \(E \xrightarrow{\pi} M\) with
\(\dim M = n, \dim E = k\) and \(s:M\to E\) a section, then
\(\mathrm{PD}[s^{-1}(0)] = e[E]\) since \(\dim s^{-1}(0) = n-k\). If
\(E = {\mathbf{T}}M\), then \(e({\mathbf{T}}M) = 0\) implies
\(\chi(M) = 0\) and there exists a nowhere vanishing vector field.

\end{remark}

\hypertarget{tuesday-october-26}{%
\section{Tuesday, October 26}\label{tuesday-october-26}}

\begin{remark}

For \(F\to E \xrightarrow{\pi} B\) a fiber bundle with a
\(G{\hbox{-}}\)structure, we have maps:

\begin{center}
\begin{tikzcd}
    && G \\
    \\
    {\phi_{ij}(U_i \cap U_j)} && {{\operatorname{Homeo}}(F, F)}
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMiwwLCJHIl0sWzIsMiwiXFxIb21lbyhGLCBGKSJdLFswLDIsIlxccGhpX3tpan0oVV9pIFxcaW50ZXJzZWN0IFVfaikiXSxbMCwxXSxbMiwxXSxbMiwwXV0=}{Link
to Diagram}
\end{quote}

A vector bundle was an \({\mathbb{R}}^n\) bundle with a
\(\operatorname{GL}_n{\hbox{-}}\)structure, having a Riemannian metric
meant having an \({\operatorname{O}}_n\) structure, and orientability
meant having a \(\operatorname{GL}^+_n\) structure. A \textbf{principal}
\(G{\hbox{-}}\)bundle was \(P\to B\) where \(P\) has a
\(G{\hbox{-}}\)action acting freely and transitively on each fiber.
Taking the frame bundle sent vector bundles to principal
\(\operatorname{GL}_n{\hbox{-}}\)bundles.

We had a construction sending fiber bundles to principal
\(G{\hbox{-}}\)bundles, namely the \textbf{clutching} construction:
given \({\mathcal{U}}\rightrightarrows X\) and
\(\phi_{ij}: U_i \cap U_j \to G\) satisfying the cocycle condition
\(\phi_{ij} \phi_{jk} = \phi_{ik}\), we get a fiber bundle with fiber
\(F\) and transition functions for any
\(F\in{G{\hbox{-}}\mathsf{Spaces}}\).

\end{remark}

\begin{remark}

On universal bundles and classifying spaces: for \(X\in {\mathsf{CW}}\),
\begin{align*}
\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }(X) { { \, \xrightarrow{\sim}\, }}[X, {\mathbf{B}}G] \\
f^* EG &\mapsfrom f
.\end{align*}

We noted

\begin{itemize}
\tightlist
\item
  \(G\) discrete implies \({\mathbf{B}}G \simeq K(G, 1)\)
\item
  \(K(C_2, 1) = {\mathbb{RP}}^{\infty}\) with \(EC_2 = S^{\infty}\)
\item
  \({\mathbf{B}}U_1 = {\mathbb{CP}}^{\infty}\) with
  \(EU_1 = S^{\infty}\)
\item
  \({{\mathbf{B}}{\operatorname{O}}}_n = {\operatorname{Gr}}_n({\mathbb{R}}^{\infty})\),
  with
  \({\mathsf{E} {\operatorname{O}}}_n = V_n({\mathbb{R}}^{\infty})\) the
  Stiefel manifold of \(n{\hbox{-}}\)dimensional frames.
\item
  \({\mathbf{B}}{\operatorname{SO}}_n\) are oriented
  \(n{\hbox{-}}\)planes in \({\mathbb{R}}^\infty\).
\item
  For \(H\leq G\), \(EH = EG\) and \({\mathbf{B}}H = EG/H\).
\end{itemize}

We had a canonical bundle
\(\gamma_n \to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\) whose fiber
above \(W\leq {\mathbb{R}}^{\infty}\) was exactly \(W\), so
\(\gamma_n = \left\{{(W, w\in W) \subseteq {\operatorname{Gr}}_n({\mathbb{R}}^\infty) \times {\mathbb{R}}^{\infty}}\right\}\).
Every vector bundle \(E\to X\) is of the form \(f^* \gamma_n\to X\) for
some \(f\in [X, {\operatorname{Gr}}_n({\mathbb{R}}^\infty)]\), and
similarly \({\operatorname{O}}_n\) bundles are pullbacks of
\(V_n({\mathbb{R}}^\infty)\to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)\).

\end{remark}

\begin{example}[?]

A useful application: characteristic classes. For any
\(c\in H^d({{\mathbf{B}}{\operatorname{O}}}_n)\) can be pulled back:
\begin{align*}
H^d(X) &\to { \mathsf{Vect} }_n(X) \\
f^*c &\mapsfrom f\in [X, {{\mathbf{B}}{\operatorname{O}}}_n]
.\end{align*}

Noting that \({\operatorname{BU}}_1 ={\mathbb{CP}}^\infty\) and
\(H^2({\mathbb{CP}}^\infty) = {\mathbb{Z}}\left\langle{c_1}\right\rangle\),
so any line bundle \(L\to X\) \(f:X\to {\operatorname{BU}}_1\) yields
\(c_1(L) \coloneqq f^* c_1\in H^2(X)\), the \textbf{first Chern class}.
Noting
\(H^2(X; {\mathbb{Z}}) \cong [X, K({\mathbb{Z}}, 2)] \cong [X, {\mathbb{CP}}^\infty]\).

\end{example}

\begin{remark}

Note: why \(c_1=0\) in symplectic settings, related to Maslov index and
ensures that the dimension of the relevant moduli space is zero.

\end{remark}

\hypertarget{thursday-october-28}{%
\section{Thursday, October 28}\label{thursday-october-28}}

\begin{remark}

The Euler class is \emph{natural} in the following sense: for \(E\to X\)
with \(\dim E = n\) and \(X\in {\mathsf{CW}}\), we can write

\begin{center}
\begin{tikzcd}
    {E\cong f^* \gamma_n} && {\gamma_n} \\
    \\
    X && {{{\mathbf{B}}{\operatorname{O}}}_n}
    \arrow[from=1-1, to=3-1]
    \arrow["f", from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJFXFxjb25nIGZeKiBcXGdhbW1hX24iXSxbMCwyLCJYIl0sWzIsMiwiXFxCT19uIl0sWzIsMCwiXFxnYW1tYV9uIl0sWzAsMV0sWzEsMiwiZiJdLFszLDJdLFswLDNdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

Then naturality is the following equality:
\begin{align*}
e(E) = e(f^* \gamma_n) = f^*(e(\gamma_n))
.\end{align*}

Recall the Thom isomorphism theorem: for an oriented
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_r} }^n\), we have a disk
bundle \({\mathbb{D}}E\) and sphere bundle \({\mathbb{S}}E\), and
\begin{align*}
H^j({\mathbb{D}}E; {\mathbb{S}}E) \cong
\begin{cases}
0 &  j< n
\\
{\mathbb{Z}}& j=n
\\
H^{j-n}({\mathbb{D}}E) & j> n
\end{cases}
,\end{align*}
noting that we can take
\(H^n({\mathbb{D}}E, {\mathbb{S}}E) \to H^n ({\mathbb{D}}E_x, {\mathbb{S}}E_x) { { \, \xrightarrow{\sim}\, }}H^n(D^n, S^{n-1})\),
where the target has a canonical positive generator. The preimage of
this generator is \(u_E\), the Thom class. The isomorphisms in the range
\(j>n\) are given by \({-}\smile u_E\).

We had a claim:
\begin{align*}
H^n({\mathbb{D}}E, {\mathbb{S}}E) &\to H^n({\mathbb{D}}E) \cong H^n(X) \\
u_E &\mapsto e(E)
.\end{align*}

\end{remark}

\begin{remark}

On the \textbf{Gysin sequence}: use the bundle
\(S^{n-1} \hookrightarrow{\mathbb{S}}E \to X\) and the LES
\begin{align*}
\cdots \to H^{j-1}({\mathbb{S}}E) \xrightarrow{\delta} H^{j-n}(X) \xrightarrow{({-}) \smile e(E)} H^j(X) \to H^j({\mathbb{S}}E) \to \cdots
.\end{align*}
The connecting map \(\delta\) comes from the Thom isomorphisms
\(H^j({\mathbb{D}}E, {\mathbb{S}}E){ { \, \xrightarrow{\sim}\, }}H^{j-n}({\mathbb{D}}E)\)
and splicing the LES of the pair \(({\mathbb{D}}E, {\mathbb{S}}E)\).

\end{remark}

\begin{proposition}[?]

If \(E\) is odd dimensional, then \(2 e(E) = 0\). Note that
\(e(E_1 \oplus E_2) - e(E_1) \smile e(E_2)\), and if \(E\) has a
nonvanishing section then \(e(E) = 0\).

\end{proposition}

\begin{remark}

So \(e(E)\) is the obstruction to finding a nonvanishing section over
the \(n{\hbox{-}}\)skeleton, where \(\dim E = n\). Given a nonvanishing
section over \(X{ {}^{ (k-1) } }\), consider extending it over
\(X{ {}^{ (k) } }\). We can use the cellular attaching maps to write

\begin{center}
\begin{tikzcd}
    {\Delta^k \times S^{n-1}} && {i^* {\mathbb{S}}E} && {{\mathbb{S}}E} \\
    \\
    && {\Delta^k} && X
    \arrow["i", from=3-3, to=3-5]
    \arrow[from=1-5, to=3-5]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-3]
    \arrow["{s: {{\partial}}\Delta^k \to S^{n-1} \in \pi_{k}^{S^{n-1}}}", curve={height=-18pt}, dashed, from=3-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwyLCJcXERlbHRhXmsiXSxbNCwyLCJYIl0sWzQsMCwiXFxTUyBFIl0sWzIsMCwiaV4qIFxcU1MgRSJdLFswLDAsIlxcRGVsdGFeayBcXHRpbWVzIFNee24tMX0iXSxbMCwxLCJpIl0sWzIsMV0sWzMsMl0sWzMsMF0sWzQsM10sWzQsMF0sWzAsNCwiczogXFxiZCBcXERlbHRhXmsgXFx0byBTXntuLTF9IFxcaW4gXFxwaV97a31ee1Nee24tMX19IiwwLHsiY3VydmUiOi0zLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}

\begin{itemize}
\tightlist
\item
  If \(s \in \pi_k(S^n)\) and \(k<n\), it is nullhomotopic and can thus
  be extended from \({{\partial}}\Delta^k\) to \(\Delta^k\).
\item
  If \(k=n\), we get
  \(s_*: H_{n-1}({{\partial}}\Delta^n) \to H_{n-1}(S^{n-1})\) where
  \(a\mapsto na\) with \(n\coloneqq\deg(s_*)\). This yields an cellular
  \(k{\hbox{-}}\)cochain \(C^k \to {\mathbb{Z}}\) where
  \(\Delta^n \mapsto \deg s_*\), and this represents the Euler class.
\end{itemize}

\end{remark}

\begin{remark}

For smooth manifolds, we have natural bundles \({\mathbf{T}}X\) and
\(\nu X\) to work with. Then
\begin{align*}
{\left\langle {e({\mathbf{T}}M)},~{ [M] } \right\rangle} = \chi(M)
,\end{align*}
using that \(e({\mathbf{T}}M) \in H^n(M) \cong {\mathbb{Z}}\). In
general, so \(E\to M^n\) with a generic section \(s\), writing
\(Z\coloneqq s^{-1}(0) \leq E\) a submanifold of dimension \(n-k\) when
\(\dim Z = k\), we have \([Z] \in H_{n-k}(M)\) and
\(\mathrm{PD}[Z]= e(E)\).

\end{remark}

\begin{example}[?]

Consider \(N^n \hookrightarrow M^{2n}\), then
\(\nu N\to N \in { {\mathsf{Bun}}\qty{\operatorname{GL}_r} }^n(N)\). To
compute \(e(\nu N)\), pick a generic section:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-28_13-35.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then \(e(\nu N) = N\cdot N \in H_n(M)\) is the self-intersection number.

\end{example}

\begin{remark}

For \(N^n \subseteq M^m\), we have

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu N}} & {H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N)} && {H^{m-n}(M, M\setminus N)} \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{e(u_{\nu N})} & {H^{m-n}(N) \cong H^{m-n}({\mathbb{D}}\nu N)} && {H^{m-n}(M)} & \textcolor{rgb,255:red,92;green,92;blue,214}{\mathrm{PD}[N]}
    \arrow[from=1-4, to=3-4]
    \arrow["\operatorname{res}", from=3-4, to=3-2]
    \arrow["{\text{excision}}"', from=1-4, to=1-2]
    \arrow[from=1-2, to=3-2]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-6pt}, dashed, from=1-1, to=3-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, from=3-5, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

So we can compute \(e(E)\) for \({\mathbf{T}}M\) and \(\nu N\).

\end{remark}

\begin{remark}

Next topics: Chern and Stiefel-Whitney classes.

\end{remark}

\begin{theorem}[?]

For \(E\to X\) a real vector bundle, there are characteristic classes
\(w_i(E) \in H^i(X; C_2)\) and
\(w \coloneqq 1 + w_1(E) + w_2(E) + \cdots \in H^*(E)\) , the
\textbf{Stiefel-Whitney classes}, satisfying the following properties:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Naturality: \(w_i(f^*(E)) = f^*(w_i(E))\)
\item
  \(w(E_1 \bigoplus E_2) = w(E_1 )\smile w(E_2)\)
\item
  \(w_i(E) = 0\) if \(i>\dim_{\mathbb{R}}E\).
\item
  If \(E\to {\mathbb{RP}}^\infty\) is the canonical line bundle, then
  \(w_1(E) = \alpha\) for
  \(\left\langle{\alpha}\right\rangle = H^1({\mathbb{RP}}^\infty; C_2)\).
\end{enumerate}

Moreover, these properties characterize \(w_i(E)\) uniquely. For complex
vector bundles, there are \(c_i(E) \in H^{2i}(X; {\mathbb{Z}})\) and
\(c_(E)\), the \textbf{Chern classes}, which satisfy the same properties
with \({\mathbb{C}}\) instead of \({\mathbb{R}}\) and
\(\left\langle{\alpha}\right\rangle = H^2({\mathbb{CP}}^\infty; {\mathbb{Z}})\).

\end{theorem}

\begin{remark}

Next time: existence and uniqueness.

\end{remark}

\hypertarget{stiefel-whitney-and-chern-classes-tuesday-november-02}{%
\section{Stiefel-Whitney and Chern Classes (Tuesday, November
02)}\label{stiefel-whitney-and-chern-classes-tuesday-november-02}}

\begin{remark}

The Stiefel-Whitney classes \(w_i \in H^i(X; C_2)\) will be defined for
\({ {\mathsf{Bun}}\qty{\operatorname{GL}_r} }(X)_{/ {{\mathbb{R}}}}\)
while Chern classes \(c_i \in H^{2i}(X; {\mathbb{Z}})\) will be defined
for
\({ {\mathsf{Bun}}\qty{\operatorname{GL}_r} }(X)_{/ {{\mathbb{C}}}}\).
We setting \(w(E) \coloneqq\sum_i w_i(E) \in H^*(X; C_2)\), we mentioned
several properties:

\begin{itemize}
\tightlist
\item
  \(w_i(f^* E) = f^* w_i(E)\)
\item
  \(w(E_1 \oplus E_2) = w(E_1) \smile w(E_2)\)
\item
  \(w_{> \dim E} E = 0\)
\item
  \(w_1(\gamma)\) is the generator of
  \(H^1({\mathbb{RP}}^\infty; C_2) = C_2\) where
  \(\gamma \to {\mathbb{RP}}^\infty\) is the canonical line bundle. For
  complex bundles, \(c_1(\gamma)\) is the positive generator of
  \(H^2({\mathbb{CP}}^\infty; {\mathbb{Z}}) = {\mathbb{Z}}\).
\end{itemize}

These properties characterize the \(w_i\) and \(c_i\) uniquely.

\end{remark}

\begin{remark}

On why we need \(C_2\) coefficients: we're pulling back
\(\gamma \to {{\mathbf{B}}{\operatorname{O}}}_1\) and
\({{\mathbf{B}}{\operatorname{O}}}_1 \simeq{\mathbb{RP}}^\infty\) where
\(H^1({\mathbb{RP}}^\infty; {\mathbb{Z}}) = 0\)! For line bundles, we'll
automatically have \(w_{\geq 2} = 0\) for any such class, so we only
have \(w_1\) the work with, and this we need to pull back something
nonzero to get anything interesting.

\end{remark}

\begin{corollary}[?]

Some immediate consequences:

\begin{itemize}
\item
  \(w_i(E \oplus {\mathbb{R}}) = w_i(E)\), this if \(E\) is trivial then
  \(w_i(E) = 0\) and \(w(E) = 1\). Moreover
  \(w(E \oplus {\mathbb{R}}) = w(E) w({\mathbb{R}}) = w(E)\) since
  \(w({\mathbb{R}}) = 1\).
\item
  If \(E\) has \(k\) linearly independent nonvanishing sections, then
  there is a splitting
  \(E = E' \oplus R{ {}^{ \scriptscriptstyle\oplus^{k} } }\) where
  \(\dim E' = n-k\). Thus \(w_i(E) = 0\) for \(i> n-k\).
\item
  If \(E_1 \oplus E_2\) is trivial, then \(w(E_1) \smile w(E_2) = 1\).
\end{itemize}

\end{corollary}

\begin{example}[?]

If \(M \subseteq {\mathbb{R}}^N\), then
\(w({\mathbf{T}}M) \smile w(\nu M) = 1\). If
\(S^n \subseteq {\mathbb{R}}^{n+1}\), then \(w({\mathbf{T}}S^n) = 1\)
since \(w(\nu S^n) = 1\). So for example \(w({\mathbf{T}}S^2) = 1\), but
\({\mathbf{T}}S^2\) has no nonvanishing sections.

\end{example}

\begin{remark}

Consider inverting a formal power series \(\sum_{i\geq 0} a_i\):
\begin{align*}
(1 + a_1 + a_2 + a_3 + \cdots)( 1 - a_1 + (a_1^2 + a_2) + (a_1^3 + 2a_1 a_2 - a_3)  + \cdots) = 1
.\end{align*}

So if \(w(E_1)w(E_2) = 1\), we can solve for \(w(E_1)\) in terms of
\(w(E_2)\).

\end{remark}

\begin{proposition}[?]

Note that
\(H^*({\mathbb{RP}}^n; C_2) = {\mathbb{F}}_2[a]/\left\langle{a^{n+1}}\right\rangle\)
where \(H^1({\mathbb{RP}}^n; C_2) = \left\langle{a}\right\rangle\).
Claim: \(w({\mathbf{T}}{\mathbb{RP}}^n) = (1+a)^{n+1}\)

\end{proposition}

\begin{proof}[?]

Let \(\gamma\to {\mathbb{RP}}^n\) be the canonical line bundle, which is
a pullback of \(\gamma^\infty \to {\mathbb{RP}}^\infty\) and
\(\left\langle{ w_1(\gamma^\infty) }\right\rangle = H^1({\mathbb{RP}}^\infty; C_2)\).
By the naturality property, \(w_1(\gamma) = \alpha\) in
\(H^1({\mathbb{RP}}^n; C_2)\).

\begin{lemma}[?]

\begin{align*}
{\mathbf{T}}{\mathbb{RP}}^n = \mathop{\mathrm{Hom}}(\gamma, \gamma^\perp)
.\end{align*}

\end{lemma}

Recall that
\(\gamma \subseteq {\mathbb{RP}}^n \times {\mathbb{R}}^{n+1}\) as a
subbundle over \({\mathbb{RP}}^n\), and
\(\gamma = \left\{{ ([\mathbf{x}], \lambda \mathbf{x}) {~\mathrel{\Big\vert}~}\lambda \in {\mathbb{R}}}\right\} = \pi^{-1}[\mathbf{x}]\).
We can write
\({\mathbf{T}}{\mathbb{RP}}^n = {\mathbf{T}}S^n/ (\mathbf{x}, \mathbf{v})\sim (- \mathbf{x}, - \mathbf{v})\),
so
\begin{align*}
{\mathbf{T}}{\mathbb{RP}}^n 
= \left\{{ [(x, v), (- \mathbf{x}, - \mathbf{v})] {~\mathrel{\Big\vert}~}v\in {\mathbf{T}}_x S^n }\right\}
= \left\{{ [(x, v), (- \mathbf{x}, - \mathbf{v})] {~\mathrel{\Big\vert}~}v\in {\mathbf{T}}_x S^n \iff {\left\langle {x},~{v} \right\rangle} = 0 }\right\}
.\end{align*}

So define a map
\begin{align*}
{\mathbf{T}}{\mathbb{RP}}^n &\to \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp) \\
[(\mathbf{x}, \mathbf{v}), (-\mathbf{x}, \mathbf{v})] &\mapsto \qty{\lambda \mathbf{x} \xrightarrow{\ell} \lambda \mathbf{v} }
,\end{align*}
and one can check that this is well-defined.

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-11-02_13-45.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

We can thus write
\begin{align*}
{\mathbf{T}}{\mathbb{RP}}^n \oplus {\mathbb{R}}
&= \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp) \oplus \mathop{\mathrm{Hom}}( \gamma, \gamma) \\
&= \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp \oplus \gamma) \\
&= \mathop{\mathrm{Hom}}(\gamma, {\mathbb{R}}^{n+1}) \\
&= \mathop{\mathrm{Hom}}(\gamma, {\mathbb{R}}){ {}^{ \scriptscriptstyle\oplus^{(n+1)} }  } 
.\end{align*}

Note that \(\mathop{\mathrm{Hom}}( \gamma, {\mathbb{R}}) \cong \gamma\),
so
\(w({\mathbf{T}}{\mathbb{RP}}^n \oplus {\mathbb{R}}) = w(\gamma)^{n+1} = (1+a)^{n+1}\)
and
\(w({\mathbf{T}}{\mathbb{RP}}^n \oplus {\mathbb{R}}) = w({\mathbf{T}}{\mathbb{RP}}^n)\).

\end{proof}

\begin{corollary}[?]

\(w({\mathbf{T}}{\mathbb{RP}}^n) = 1\) iff \(n+1=2^r\) for some \(r\).

\end{corollary}

\begin{proof}[?]

\(\impliedby\): By induction, \((1+a)^{2^r} = 1 + a^{2^r}\), using that
\((1+a)^2 = 1 + a^2 + 2a\) and \(2a=0\) when we have \(C_2\)
coefficients. Now write
\begin{align*}
(1+a)^{2^r} = ( (1+a)^{2^{r-1}} )^2 \overset{IH}{=} (1 + a^{2^{r-1}})^2 = 1 + a^{2^r}
.\end{align*}
Now use that \(\dim {\mathbf{T}}{\mathbb{RP}}^n = n\), so
\((1+a)^{n+1} = 1\) since \(a^{2^r} = a^{n+1} = 0\).

\(\implies\): Suppose \(n+1 = m2^r\) with \(m\) odd, then
\begin{align*}
(1 + a)^{ m2^{r}}
= ((1+a)^{2^r})^m \\
= (1+a^{2^r})^m \\
= 1 + ma^{2^r} + \cdots
,\end{align*}
where the first nontrivial term doesn't vanish since \(a^{2^r}\neq 0\).

\end{proof}

\begin{lemma}[?]

If \({\mathbb{RP}}^{2^r}\) admits an immersion into \({\mathbb{R}}^N\),
then \(N\geq 2^{r+1} - 1\).

\end{lemma}

\begin{proof}[?]

Why?
\begin{align*}
w({\mathbf{T}}{\mathbb{RP}}^{2^r}) = (1+a)^{2^r+1} = (1+a)^{2^r}(1+a) \\
= (1+a^{2^r})(1+a) \\
= 1 +a + a^{2^r} + a^{2^r + 1} \\
= 1 +a + a^{2^r} 
.\end{align*}
Try to invert this: let \(n\coloneqq 2^r\), then
\begin{align*}
(1+a+a^n)(1 + a + a^2 + \cdots + a^{n-1}) = 1
.\end{align*}

Then
\begin{align*}
w(\nu {\mathbb{RP}}^n) = \sum_{0\leq i \leq n-1} a^i
,\end{align*}
so \(\dim \nu {\mathbb{RP}}^n \geq n-1\).

\end{proof}

\hypertarget{thursday-november-04}{%
\section{Thursday, November 04}\label{thursday-november-04}}

\begin{remark}

Recall that \(w({\mathbf{T}}{\mathbb{RP}}^n) = (1+a)^{n+1}\) where
\(\left\langle{a}\right\rangle = H^1({\mathbb{RP}}^n; C_2)\), and as an
application, if \({\mathbb{RP}}^{2^r} \hookrightarrow{\mathbb{R}}^N\),
then \(N \geq 2^{r+1} - 1\). Similarly,
\(c({\mathbf{T}}{\mathbb{CP}}^n) = (1+a)^{n+1}\) with
\(\left\langle{a}\right\rangle = H^2({\mathbb{CP}}^\infty; {\mathbb{Z}})\)
the positive generator.

\end{remark}

\begin{theorem}[Leray-Hirsch]

Fix a commutative ring \(R\) and a fiber bundle
\(F\hookrightarrow E \to B\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(H^j(F; R) \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is finitely
  generated and free,
\item
  For any \(j\), there exist \(c_{j_k} \in H^j(E; R)\) such that the
  restrictions of \(\left\{{c_{j_k}}\right\}_{k\geq 0}\) is a basis for
  \(H^j(E_x; R) = H^j(F; R)\).
\end{enumerate}

Then there is an isomorphism
\begin{align*}
\Phi: H^*(B; R)\otimes_R H^*(F; R){ { \, \xrightarrow{\sim}\, }}H^*(E; R) \\
b_i \otimes i^*(c_{j_k}) \mapsto \pi^*(b_i) \smile c_{j_k}
.\end{align*}

\end{theorem}

\begin{example}[?]

Let \({\mathbb{P}}(E)\) be the projectivization of \(E\), so each vector
space fiber \(V\) is replaced with \({\mathbb{P}}(V)\). Let
\(\gamma \subseteq \pi^*(E)\) be the canonical over \({\mathbb{P}}(E)\),
so the fibers are
\(\gamma_{(\gamma, [\gamma])} = \left\{{w\in E_x {~\mathrel{\Big\vert}~}w\in [v]}\right\}\).
For
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}})_{/ {X}}\),
pick a metric on \(E\), and define \(E' \coloneqq\gamma^\perp\). Now
take the pullback:

\begin{center}
\begin{tikzcd}
    && \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{C}}^n} \\
    & { \pi^*E = \gamma \oplus E'} && E \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{CP}}^n} \\
    & E && X
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-1, to=4-2]
    \arrow[from=4-2, to=4-4]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-3, to=2-4]
    \arrow[from=2-4, to=4-4]
    \arrow[from=2-2, to=4-2]
    \arrow[from=2-2, to=2-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwxLCIgXFxwaV4qRSA9IFxcZ2FtbWEgXFxvcGx1cyBFJyJdLFszLDEsIkUiXSxbMywzLCJYIl0sWzEsMywiRSJdLFsyLDAsIlxcQ0NebiIsWzI0MCw2MCw2MCwxXV0sWzAsMiwiXFxDUF5uIixbMjQwLDYwLDYwLDFdXSxbNSwzLCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdfV0sWzMsMl0sWzQsMSwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX1dLFsxLDJdLFswLDNdLFswLDFdXQ==}{Link
to Diagram}
\end{quote}

So we have a bundle
\({\mathbb{CP}}^{n-1} \hookrightarrow{\mathbb{P}}(E) \to X\), we'll now
try to apply the theorem. Note that
\(H^*({\mathbb{CP}}^{n-1}; {\mathbb{Z}}) = {\mathbb{Z}}[u]/\left\langle{u^n}\right\rangle\),
so \(H^i\) is generated by \(u^i\), fulfilling condition 1.

Let \(\iota_x: E_x \hookrightarrow E\) be the inclusion of a fiber. If
we restrict \({\mathbb{P}}(E)\) to a fiber \({\mathbb{P}}(E)_x\), this
yields the canonical over \({\mathbb{CP}}^{n-1}\):

\begin{center}
\begin{tikzcd}
    {\gamma_x} && \gamma \\
    \\
    {{\mathbb{P}}(E)_x = {\mathbb{CP}}^{n-1}} && {{\mathbb{P}}(E)}
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow["{\iota_x^*}"{description}, hook, from=3-1, to=3-3]
    \arrow[hook, from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzIsMiwiXFxQUChFKSJdLFswLDIsIlxcUFAoRSlfeCA9IFxcQ1Bee24tMX0iXSxbMCwwLCJcXGdhbW1hX3giXSxbMywyXSxbMCwxXSxbMiwxLCJpX3giLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}

Then \(\iota_x^* c_1(\gamma) = c_1(\gamma_x) = e(\gamma_x) = u\), so we
have condition 2.

\begin{exercise}[?]

For
\(L \in { {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{C}})\),
\begin{align*}
c_1(L {}^{ \vee }) = c_1(L)
.\end{align*}

\end{exercise}

So we have classes
\(1, c_1(\gamma), c_1(\gamma)^2,\cdots, c_1( \gamma)^{n-1}\), and we can
take duals to obtain
\(1, c_1(\gamma {}^{ \vee }), c_1(\gamma {}^{ \vee })^2,\cdots, c_1( \gamma {}^{ \vee })^{n-1}\).
There exist \(c_1(E), c_2(E), \cdots, c_n(E) \in H^*(X)\) such that we
can write the former as a linear combination of the latter:
\begin{align*}
c_1( \gamma {}^{ \vee })^n + c_1(E) c_1( \gamma {}^{ \vee })^{n-1} + \cdots + c_n(E) = 0
,\end{align*}
and the \(c_i(E)\) are called the \textbf{Chern classes}.

\end{example}

\begin{remark}

Write \(c(E) = 1 + c_1(E) + \cdots + c_n(E)\), so
\begin{align*}
\pi^* c(E) 
&= c(\gamma) c(E') = (1 +c_1(\gamma))(1 + c_1(E') + \cdots + c_{n-1}(E'))\\
&= 1 + (c_1(\gamma) + c_1(E'))
+ (c_1( \gamma)c_1(E') + c_2(E'))
+ \cdots 
+ (c_1(\gamma) c_{i-1}(E') + c_1(E') )
.\end{align*}

Plugging this into the LHS above yields
\begin{align*}
c_1( \gamma {}^{ \vee })^n + 
(c_1(\gamma) + c_1(E')) c_1(\gamma {}^{ \vee })^{n-1} +
(c_1(\gamma)c_1(E') + c_2(E'))c_1(\gamma {}^{ \vee })^{n-2}
+ \cdots +
c_1(\gamma)c_{n-1}(E') 
&=
(c_1 (\gamma) + c_1(\gamma {}^{ \vee }))( c_1( \gamma {}^{ \vee })^{n-1} + c_1(E') c_1( \gamma {}^{ \vee })^{n-2} + c_2(E') c_1( \gamma {}^{ \vee })^{n-2} + \cdots + c_{n-1}(E') )
.\end{align*}
Since \(\gamma {}^{ \vee }\) and \(\gamma\) are dual, the first term is
zero, so this entire expression is zero.

\end{remark}

\begin{exercise}[?]

Show that \(c_1(E {}^{ \vee }) = -c_1(E)\).

\begin{quote}
Hint: consider an explicit description in terms of transition functions.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that given \(b_1 = 1, b_2 = x+a_1, b_3 = xa_1 + a_2, \cdots\), then
\(b_1 (-x)^n + b_2 (-x)^{n-1} + \cdots = 0\).

\end{exercise}

\begin{remark}

Consider the following pullbacks:

\begin{center}
\begin{tikzcd}
    {f^* E} && {\gamma \oplus E'} && E \\
    \\
    {{\mathbb{S}}E} && {{\mathbb{P}}(E)} && X
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=3-5]
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow["f"{description}, curve={height=30pt}, tail, from=3-1, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMiwwLCJcXGdhbW1hIFxcb3BsdXMgRSciXSxbNCwwLCJFIl0sWzQsMiwiWCJdLFsyLDIsIlxcUFAoRSkiXSxbMCwyLCJcXFNTIEUiXSxbMCwwLCJmXiogRSJdLFs1LDBdLFswLDFdLFsxLDJdLFszLDJdLFswLDNdLFs0LDNdLFs1LDRdLFs0LDIsImYiLDEseyJjdXJ2ZSI6NSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibW9ubyJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

What is \(f^* E\)? There is a pullback

\begin{center}
\begin{tikzcd}
    {\varphi^* \gamma} && \gamma \\
    \\
    {S^{2n-1}} && {{\mathbb{CP}}^{n-1}}
    \arrow["\varphi", from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJTXnsybi0xfSJdLFsyLDIsIlxcQ1Bee24tMX0iXSxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiXFx2YXJwaGleKiBcXGdhbW1hIl0sWzAsMSwiXFx2YXJwaGkiXSxbMiwxXSxbMywwXSxbMywyXV0=}{Link
to Diagram}
\end{quote}

Here \(\phi^* \gamma\) will be trivial: we have
\(\gamma \subseteq {\mathbb{CP}}^{n-1} \times {\mathbb{C}}^n\), so
\(\phi^* \gamma \subseteq S^{2n-1} \times {\mathbb{C}}^n\) and
\(\dim_{\mathbb{R}}\phi^* \gamma = 2\). Then the fibers are
\(\phi^*\gamma_x = \left\{{(x, v) {~\mathrel{\Big\vert}~}v = \lambda x,\, \lambda \in {\mathbb{C}}}\right\}\).
It turns out that \(\phi^* E = {\mathbb{C}}\oplus q^* E'\) where
\({\mathbb{S}}E \xrightarrow{q} {\mathbb{P}}(E)\). Writing
\(\pi: {\mathbb{P}}(E) \to X\), we have
\begin{align*}
(\pi \circ q)^* c_i(E) = c_i(E) && i < n-1
,\end{align*}
and \(c_n(E) = e(E)\). To find this we'll need \(\pi \circ q\) to be
injective.

Consider

\begin{center}
\begin{tikzcd}
S^{2n-1} 
  \ar[r] 
& 
{\mathbb{S}}E
  \ar[d, "\pi \circ q"] 
\\
& 
X 
\end{tikzcd}
\end{center}

By the Gysin sequence, for \(j<2n-1\) we'll have
\(H^j(X) \cong H^j({\mathbb{S}}E)\) since the red terms vanish:

\begin{center}
\begin{tikzcd}
    {H^j({\mathbb{S}}E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n+1}(X)} && {H^{j+1}(X)} \\
    \\
    {H^{j-1}({\mathbb{S}}E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n}(X)} && {H^j(X)}
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=1-1]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwyLCJIXntqLTF9KFxcU1MgRSkiXSxbMCwwLCJIXmooXFxTUyBFKSJdLFsyLDIsIkhee2otMm59KFgpIixbMCw2MCw2MCwxXV0sWzQsMiwiSF5qKFgpIl0sWzIsMCwiSF57ai0ybisxfShYKSIsWzAsNjAsNjAsMV1dLFs0LDAsIkhee2orMX0oWCkiXSxbMCwyXSxbMiwzXSxbMywxXSxbMSw0XSxbNCw1XV0=}{Link
to Diagram}
\end{quote}

This uses that for \(j<2n\) that \(H^{j-2n}(X) = 0\). For \(i\leq n-1\),
\(2i<2n-i<2n-1\), and so the Chern classes for \(X\) and
\({\mathbb{S}}E\) are isomorphic via
\((\pi \circ q)^*: H^j(X) \to H^j({\mathbb{S}}E)\) in the LES. By
induction, we can define \(c_i(E)\) for \(i\leq n-1\).

\end{remark}

\begin{remark}

Idea: pull back to split off a line bundle, pull back further to split
off another line bundle, and continue. This is why we only need \(c_1\)
to determine a line bundle.

\end{remark}

\hypertarget{tuesday-november-09}{%
\section{Tuesday, November 09}\label{tuesday-november-09}}

\begin{remark}

Today: proving/checking the axioms for Chern classes. Recall that given
\(E \xrightarrow{\pi} X\), we can form a pullback:

\begin{center}
\begin{tikzcd}
    & {\pi^* E \cong \gamma \oplus E'} && E \\
    {{\mathbb{CP}}^{n-1}} \\
    & {{\mathbb{P}}(E)} && X
    \arrow["\pi", from=1-4, to=3-4]
    \arrow["\pi"', from=3-2, to=3-4]
    \arrow[from=1-2, to=3-2]
    \arrow[from=1-2, to=1-4]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=3-4]
    \arrow[hook, from=2-1, to=3-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMSwwLCJcXHBpXiogRSBcXGNvbmcgXFxnYW1tYSBcXG9wbHVzIEUnIl0sWzMsMCwiRSJdLFszLDIsIlgiXSxbMSwyLCJcXFBQKEUpIl0sWzAsMSwiXFxDUF57bi0xfSJdLFsxLDIsIlxccGkiXSxbMywyLCJcXHBpIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwzLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

Here \(\gamma {}^{ \vee }\) is the dual of \(\gamma\) and
\(c_1( \gamma {}^{ \vee }) = -_1( \gamma)\). By Leray-Hirsch, we have
\begin{align*}
1 + c_1( \gamma {}^{ \vee }) + c_1(\gamma {}^{ \vee })^2 + \cdots + c_1( \gamma {}^{ \vee })^{n-1} \divides (c_1(\gamma {}^{ \vee }))^n
\end{align*}
and for \(c_i(E)\) the \(i\)th Chern class of \(E\),
\begin{align*}
c_1(\gamma {}^{ \vee })^n + c_1(E) c_1(\gamma {}^{ \vee })^{n-1} + \cdots + c_n(E) = 0
.\end{align*}

\end{remark}

\begin{proposition}[?]

These satisfy some axioms:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Naturality: \(c_1(f^* E) = f^* c_1(E)\)
\item
  Homomorphism: \(c_(E_1 \oplus E_2) = c(E_1) \smile c(E_2)\).
\item
  \(c_{>\dim E}(E) = 0\).
\item
  For \(\gamma\to {\mathbb{CP}}^\infty\) the canonical,
  \(c_1(\gamma)\in H^2({\mathbb{CP}}^\infty)\) is the positive
  generator.
\end{enumerate}

\end{proposition}

\begin{proof}[?]

Here (3) is clear by definition, since we don't even define \(c_{n+1}\)
or higher if \(\dim E = n\). Number (4) isn't bad either, since
\(c_1(\gamma) = e(\gamma)\), using that
\(c({\mathcal{L}}) = c_1({\mathcal{L}})\) for line bundles. For (1),
consider a pullback:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(f)^* \gamma \oplus {\mathbb{P}}(f)^* E' } &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\gamma \oplus E'} \\
    & {f^* E} &&&& E \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(f^* E)} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(E)} \\
    & {X'} &&&& X
    \arrow["f", from=5-2, to=5-6]
    \arrow[from=2-6, to=5-6]
    \arrow[from=2-2, to=2-6]
    \arrow[from=2-2, to=5-2]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=5-6]
    \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-5, to=5-6]
    \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-1, to=5-2]
    \arrow["{\exists {\mathbb{P}}(f)}", color={rgb,255:red,92;green,92;blue,214}, from=4-1, to=4-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=1-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=4-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=4-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, color={rgb,255:red,92;green,92;blue,214}, draw=none, from=1-1, to=4-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

We then use that on fibers,
\(\gamma_{{\mathbb{P}}(f)^* E} \cong {\mathbb{P}}(f)^* \gamma_{{\mathbb{P}}(E)}\),
so the canonical pulls back to something isomorphic to the canonical.
Then
\(c_1(\gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E}) = {\mathbb{P}}(f)^* c_1(\gamma {}^{ \vee }_{{\mathbb{P}}(E)})\),
so
\begin{align*}
(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n}
+ c_1(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n-1}
+ c_2(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n-2}
+ \cdots + 
c_n(E)
,\end{align*}
and applying \({\mathbb{P}}(f)^*\) to this entire expression yields
zero. Thus
\begin{align*}
(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n}
{\mathbb{P}}(f)^* c_1(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-1}
{\mathbb{P}}(f)^* c_2(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-2}
+ \cdots + 
+ {\mathbb{P}}(f)^* c_n(E) 
= 0
,\end{align*}
and we can note that by definition this equals
\begin{align*}
(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n}
c_1 f^* E \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-1}
c_2 f^* E \cdot c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-2}
+ \cdots + 
c_n f^* E
= 0
.\end{align*}

\end{proof}

\begin{remark}

Recall the alternative description of Chern classes:

\begin{center}
\begin{tikzcd}
    {\pi^* E \cong {\mathbb{R}}\oplus E'} && E \\
    \\
    {{\mathbb{S}}E} && X
    \arrow[from=1-3, to=3-3]
    \arrow["\pi"', from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJcXFNTIEUiXSxbMiwyLCJYIl0sWzIsMCwiRSJdLFswLDAsIlxccGleKiBFIFxcY29uZyBcXFJSIFxcb3BsdXMgRSciXSxbMiwxXSxbMCwxLCJcXHBpIiwyXSxbMywwXSxbMywyXV0=}{Link
to Diagram}
\end{quote}

By the Gysin sequence, we obtained \(H^j(X) \cong H^j({\mathbb{S}}E)\)
for \(j\leq n-2\), and there are \(c_1(E'), \cdots, c_{n-2}(E')\) living
in
\(H^2({\mathbb{S}}E), H^4({\mathbb{S}}E), \cdots, H^{2n-2}({\mathbb{S}}E)\).
So we defined \(c_i(E)\) such that \(\pi^* c_i(E) = c_i(E')\) for
\(i\leq n-1\), so \(c_n(E) = e(E)\).

\end{remark}

\hypertarget{splitting-principle}{%
\subsection{Splitting Principle}\label{splitting-principle}}

\begin{proposition}[?]

Given an
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}\),
there exists \(Y\in {\mathsf{Top}}\) with \(f:Y\to X\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(f^* E = L_1 \oplus \cdots \oplus L_n\) with \(L_i\) line bundles.
\item
  \(f^*: H^*(X) \hookrightarrow H^*(Y)\) is injective, so classes in
  \(H^*(Y)\) can be uniquely pulled back.
\end{enumerate}

\end{proposition}

\begin{proof}[?]

By induction, it suffices to find \(Y\) where
\(f^* E \cong E' \oplus E''\) splits nontrivially and \(f^*\) is
injective. Taking the projectivization does exactly this, so take
\(Y \coloneqq{\mathbb{P}}(E)\) and \(f: {\mathbb{P}}(E)\to X\) to be the
projection. Then \(f^* E = \gamma \oplus E'\), so checking the 2nd
condition in Leray-Hirsch, we get that \(H^*(Y)\) is generated over
\(H^*(X)\) by
\(1, c_1(\gamma),c_1(\gamma)^2, \cdots, c_1(\gamma)^{n-1}\).
Equivalently, the following map is injective:
\begin{align*}
f^* : H^*(X) &\to H^*(Y) \\
\alpha & \mapsto 1\cdot \alpha
.\end{align*}
Here being generated means that
\begin{align*}
\beta\in H^*(Y) \implies \beta = \sum_{k=0}^{n-1} c_1(\gamma)^k \alpha_k, \quad \alpha_k \in H^*(X)
.\end{align*}

\end{proof}

\begin{corollary}[?]

If a polynomial identity holds for Chern classes in
\(E \coloneqq\bigoplus_i L_i\) for \(L_i\) line bundles, then it holds
for any bundle.

\end{corollary}

\begin{lemma}[?]

For line bundles \(L_1,\cdots, L_n\),
\begin{align*}
c(\bigoplus L_i) = \prod c(L_i) = \prod(1 + c_1(L_i)) = 1 + \qty{\sum c_1(L_i) } + \cdots + \qty{\prod c_1(L_i)}
,\end{align*}
so for \(E \coloneqq\bigoplus L_i\), \(c_i(E)\) is the \(i\)th symmetric
polynomial in the \(c_1(L)\). In other words, writing
\(\sigma_i(x_1, \cdots, x_n)\) as the \(i\)th symmetric polynomial,
\(c_i(E) = \sigma_i(c_1(L_1), c_1(L_2), \cdots, c_1(L_n))\).

\end{lemma}

\begin{corollary}[?]

The direct sum formula \(c(E \oplus E') = c(E) c(E')\). Take a pullback

\begin{center}
\begin{tikzcd}
    {\bigoplus_{i} L_i \oplus \bigoplus_j L_j} && {E \oplus E'} \\
    \\
    Y && X
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJFIFxcb3BsdXMgRSciXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFswLDAsIlxcYmlnb3BsdXNfe2l9IExfaSBcXG9wbHVzIFxcYmlnb3BsdXNfaiBMX2oiXSxbMywwXSxbMCwxXSxbMiwxXSxbMywyXSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

Then
\begin{align*}
c( \bigoplus_i L_i \oplus \bigoplus_j L_j) 
= \qty{\prod_i c(L_i)} \cdot \qty{\prod_j c(L_j)} 
= c\qty{ \bigoplus_i L_i} \cdot c\qty{\bigoplus L_j}
= f^* c(E) \cdot f^* c(E')
,\end{align*}
so
\(f^*c(E \oplus E') = f^* c(E) \cdot f^* c(E') = f^*(c(E) \cdot c(E'))\),
and by injectivity, \(c(E \oplus E') = c(E) c(E')\).

\end{corollary}

\begin{proof}[of lemma]

Write \(E = L_1 \oplus \cdots \oplus L_n\) where \(E\to X\). Then pull
back along \(\pi:{\mathbb{P}}E\to X\) to get
\(\pi^* E = \pi^* L_1 \oplus \cdots \oplus \pi^* L_n\). Note that
\(\gamma \subseteq \pi^* E\).

\begin{claim}

\begin{align*}
\prod_{i=1}^n \qty{ c_1(\gamma {}^{ \vee }) + c_1( \pi^* L_i) } = 0
.\end{align*}

\end{claim}

Using the claim, we get
\begin{align*}
0 = 
c_1( \gamma {}^{ \vee })^n + 
\qty{ \sum c_1 \pi^* L_i }(c_1(\gamma {}^{ \vee }))^{n-1} + \cdots + \prod c_1(\pi^* L_i) \cdot c_1(\gamma {}^{ \vee })^1
= c_1(\gamma {}^{ \vee })^{n} + \pi^* c_1(E) \cdot (c_1(\gamma {}^{ \vee }))^{n-1} + \cdots + \pi^* c_n(E) \cdot c_1(\gamma {}^{ \vee })^1
,\end{align*}
which proves e.g.~that \(c_1(E) = \sigma_1(c_1L_1, \cdots, c_1 L_n)\).
The idea is that e.g.~for the first term,
\(c_1( \gamma {}^{ \vee }) = -c_1(\gamma)\) and
\(c_1\pi^* L_1 = c_1( \gamma)\), yielding zero. If we can cover \(E\) by
opens where this happens for at least one term, then the entire product
must be zero on \(E\).

Take a SES and apply \(\mathop{\mathrm{Hom}}(\gamma, {-})\):

\begin{center}
\begin{tikzcd}
    0 && \gamma && {\bigoplus_i \pi^* L_i} && {\gamma^\perp} && 0 \\
    \\
    && {\mathop{\mathrm{Hom}}(\gamma, \gamma)} && {\mathop{\mathrm{Hom}}\qty{\gamma, \bigoplus_i \pi^* L_i} \cong \bigoplus_i \mathop{\mathrm{Hom}}(\gamma, \pi^* L_i)}
    \arrow[from=1-1, to=1-3]
    \arrow[""{name=0, anchor=center, inner sep=0}, "\iota", from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
    \arrow[""{name=1, anchor=center, inner sep=0}, "{\iota_*}", from=3-3, to=3-5]
    \arrow["{\mathop{\mathrm{Hom}}(\gamma,{-})}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzIsMCwiXFxnYW1tYSJdLFs0LDAsIlxcYmlnb3BsdXNfaSBcXHBpXiogTF9pIl0sWzYsMCwiXFxnYW1tYV5cXHBlcnAiXSxbOCwwLCIwIl0sWzIsMiwiXFxIb20oXFxnYW1tYSwgXFxnYW1tYSkiXSxbNCwyLCJcXEhvbVxccXR5e1xcZ2FtbWEsIFxcYmlnb3BsdXNfaSBcXHBpXiogTF9pfSBcXGNvbmcgXFxiaWdvcGx1c19pIFxcSG9tKFxcZ2FtbWEsIFxccGleKiBMX2kpIl0sWzAsMV0sWzEsMiwiXFxpb3RhIl0sWzIsM10sWzMsNF0sWzUsNiwiXFxpb3RhXyoiXSxbOCwxMSwiXFxIb20oXFxnYW1tYSxcXHdhaXQpIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==}{Link
to Diagram}
\end{quote}

Now note that we can get a splitting:

\begin{center}
\begin{tikzcd}
    {\mathop{\mathrm{Hom}}(\gamma, \gamma)} && {\bigoplus_i \mathop{\mathrm{Hom}}(\gamma, \pi^* L_i)} \\
    \\
    & X
    \arrow[from=1-3, to=3-2]
    \arrow[from=1-1, to=3-2]
    \arrow[from=1-1, to=1-3]
    \arrow["s"{description}, curve={height=-24pt}, dashed, from=3-2, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXEhvbShcXGdhbW1hLCBcXGdhbW1hKSJdLFsyLDAsIlxcYmlnb3BsdXNfaSBcXEhvbShcXGdhbW1hLCBcXHBpXiogTF9pKSJdLFsxLDIsIlgiXSxbMSwyXSxbMCwyXSxbMCwxXSxbMiwwLCJzIiwxLHsiY3VydmUiOi00LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}

So take \(X = {\mathbb{P}}E\) and pick a nonvanishing section \(s\) such
that \(\iota_* \circ s = (s_1, s_2,\cdots s_n)\) is a sum of sections.
Then let
\(U_i = \left\{{x\in {\mathbb{P}}E{~\mathrel{\Big\vert}~}s_i(x) \neq 0}\right\} \subseteq {\mathbb{P}}E\)
by the complement of \(Z(s_i)\), then
\(\displaystyle\bigcup_i U_i = {\mathbb{P}}E\) since \(s\) is
nonvanishing. So
\({ \left.{{\mathop{\mathrm{Hom}}(\gamma, \pi^* L_i)}} \right|_{{U_i}} }\)
is trivial, this yields
\({ \left.{{\gamma}} \right|_{{U_i}} } \cong { \left.{{\pi^* L_i}} \right|_{{U_i}} }\),
making the restrictions of \(c({-})\) equal, so
\(c_1(\pi^* L_i) - c_1(\gamma) = 0\) on \(U_i\), and this is equal to
\(c_1(\pi^* L_i) + c_1( \gamma {}^{ \vee })\). Since
\(\left\{{U_i}\right\}\rightrightarrows{\mathbb{P}}E\), this concludes
the proof.

\end{proof}

\hypertarget{chern-and-stiefel-whitney-classes-thursday-november-11}{%
\section{Chern and Stiefel-Whitney classes (Thursday, November
11)}\label{chern-and-stiefel-whitney-classes-thursday-november-11}}

\begin{remark}

Recall that we were proving the splitting principle: given \(E\to X\) of
dimension \(n\), there is a space \(Y\) with \(f:Y\to X\) where

\begin{itemize}
\tightlist
\item
  \(f^* E = \bigoplus L_i\) splits as a direct sum of line bundles.
\item
  \(f^*: H^*(X) \to H^*(Y)\) is injective.
\end{itemize}

\end{remark}

\begin{lemma}[?]

Assuming the exercise from last time that
\(c_1(L {}^{ \vee }) = - c_1(L)\)
\begin{align*}
c_i(E {}^{ \vee }) = (-1)^i c_i(E)
.\end{align*}

\end{lemma}

\begin{proof}[?]

It suffices to show this for \(E = \bigoplus L_i\). Write
\(E {}^{ \vee }= \bigoplus L_i {}^{ \vee }\), so
\begin{align*}
c(E {}^{ \vee }) = \prod c(L_i {}^{ \vee }) 
&= \prod (1- c_1(L_i)) \\
&= 1 - \qty{\sum c_1(L_i)} + \sigma_2(c_1 L_1,\cdots, c_1 L_n) - \cdots \pm \prod c_1(L_i) \\
&= 1 - c_1(E) + c_2(E) \cdots \pm c_1(L_n)
.\end{align*}

\end{proof}

\begin{theorem}[?]

For
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}})_{/ {X}}\),
the top Chern class equals the Euler class:
\begin{align*}
c_n(E) = e(E)
.\end{align*}
Similarly, for
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}\),
\begin{align*}
w_n(E) \equiv e(E) \operatorname{mod}2
.\end{align*}

\end{theorem}

\begin{proof}[?]

Both classes satisfy the Whitney sum formula, so
\begin{align*}
E &= \bigoplus L_i 
\implies c(E) = \prod c(L_i) \\
\implies e(E) 
&= \prod e(L_i) \\
&= \prod (1 + c_1(L_i)) \\
&= 1 + \sigma_1(c_1 L_1, \cdots, c_1 L_n) + \cdots + \prod c_1(L_i) \\
&\coloneqq 1 + \sigma_2(c_1 L_1, \cdots, c_1 L_n)+ \cdots + c_n(E)
.\end{align*}
Then \(e(E) = \prod e(L_i) = \prod c_1(L_i) = c_n(E)\).

Now the claim follows from the splitting principle and naturality of
characteristic classes.

\end{proof}

\begin{exercise}[?]

Show that for any
\(E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}})_{/ {E}}\),
\begin{align*}
c_i(E) \equiv w_{2i} (E) \operatorname{mod}2
.\end{align*}

\end{exercise}

\begin{lemma}[?]

For
\(E\in{ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}\),\\
Then
\begin{align*}
c_1(E) = c_1(\bigwedge\nolimits^n E)
,\end{align*}
noting that \(\bigwedge\nolimits^n E\) is a line bundle. A similar claim
holds for \(c_1\).

\end{lemma}

\begin{fact}

\begin{align*}
c_1(L_1\otimes L_2) = c_1(L_1) + c_1(L_2)
.\end{align*}

\end{fact}

:::\{.remark\} ETS this holds for \(E = \bigoplus L_i\), in which case
\(\bigwedge\nolimits^n E = \bigotimes_{i=1}^n L_i\). By the above fact,
\(c_1(\bigwedge\nolimits^n E) = \sum c_1(L_i)\). Since
\(c_i(E) = \sigma_i(c_1 L_1, \cdots, c_1 L_n)\), we have
\begin{align*}
c_1(E) = \sigma_1(c_1 L_1,\cdots, c_1 L_n) = \sum c_1 L_i
.\end{align*}
::

\begin{remark}

Recall that the Euler class is the obstruction to finding an obstruction
to extending a section over the \(n{\hbox{-}}\)skeleton.

\end{remark}

\hypertarget{obstruction-theory-1}{%
\subsection{Obstruction Theory}\label{obstruction-theory-1}}

\begin{lemma}[?]

\(w_1 E\) is the obstruction to orienting \(E\) in the following sense:
for any \(S^1 \xrightarrow{f} E\), write \(f_*[S]\) for the image of the
fundamental class, then
\begin{align*}
{\left\langle {w_1(E)},~{f_*[S^1] } \right\rangle} = \chi_{f^* E \text{ is orientable}}  =
\begin{cases}
1 & f^*E \text{ is nonorientable} 
\\
0 & f^* E \text{ is orientable}.
\end{cases}
.\end{align*}

\end{lemma}

\begin{remark}

Why? For example, consider \(E\to S^1\). Trivialize over
\(S^1\setminus\left\{{{\operatorname{pt}}}\right\}\), then glue the ends
by some element \(A\in \operatorname{GL}_n({\mathbb{R}})\). If
\(\operatorname{det}(A) > 0\), this will be orientable, and
\(\operatorname{det}(A) < 0\) will be nonorientable.

\end{remark}

\begin{proof}[?]

If \(E\) is a line bundle, define \(\tilde w_1(E) \in H^1(X)\) by the
following
\begin{align*}
{\left\langle {\tilde w_1(E)},~{ f_* [S^1] } \right\rangle} = \chi_{f^* E \text{ nonorientable}}
.\end{align*}
This is natural under pullback, consider the following diagram:

\begin{center}
\begin{tikzcd}
    {f^* h^* E} && {h^*E} && E \\
    \\
    {S^1} && Y && X
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=3-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["f", from=3-1, to=3-3]
    \arrow["h", from=3-3, to=3-5]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMiwwLCJoXipFIl0sWzQsMCwiRSJdLFs0LDIsIlgiXSxbMCwyLCJTXjEiXSxbMiwyLCJZIl0sWzAsMCwiZl4qIGheKiBFIl0sWzAsMV0sWzEsMl0sWzAsNF0sWzUsMF0sWzMsNCwiZiJdLFs0LDIsImgiXSxbNSwzXSxbNSw0LCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbMCwyLCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

We have
\begin{align*}
&{\left\langle { \tilde w_1 h^* E},~{ f_* [S^1] } \right\rangle}
&= {\left\langle { \tilde w_1 h^* E},~{ (h\circ f)_* [S^1] } \right\rangle} \\
&= {\left\langle { h^*\tilde w_1 h^* E},~{ f_* [S^1] } \right\rangle} \\
&= \tilde w_1 (h^* E) \\
&= h^* \tilde w_1(E)
.\end{align*}

We need to show \(\tilde w_1(\gamma) = w_1(\gamma)\) where
\(\gamma\to {\mathbb{RP}}^\infty\) is the canonical. Write
\(H^1({\mathbb{RP}}^\infty; C_2) = C_2 = \left\langle{a}\right\rangle\),
so \(w_1( \gamma) = a\). Take the pullback:

\begin{center}
\begin{tikzcd}
    {i^*\gamma} && \gamma \\
    \\
    {{\mathbb{RP}}^1 \cong S^1} && {{\mathbb{RP}}^\infty}
    \arrow["i", from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiaV4qXFxnYW1tYSJdLFsyLDIsIlxcUlBeXFxpbmZ0eSJdLFswLDIsIlxcUlBeMSBcXGNvbmcgU14xIl0sWzMsMiwiaSJdLFswLDJdLFsxLDNdLFsxLDBdLFsxLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

Then \(i^* \gamma\) is not orientable since locally this looks like a
Mobius band, so we have
\begin{align*}
{\left\langle {\tilde w_1( \gamma)},~{ i_* [{\mathbb{RP}}^1] } \right\rangle} = 1
.\end{align*}

Now \(w_1(E) = w_1(\bigwedge\nolimits^n E)\) and we have to bundles:

\begin{center}
\begin{tikzcd}
    {f^*(\bigwedge\nolimits^n E)} && {f^* E} \\
    \\
    & {S^1}
    \arrow[from=1-1, to=3-2]
    \arrow[from=1-3, to=3-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJmXiooXFxFeHRhbGdebiBFKSJdLFsyLDAsImZeKiBFIl0sWzEsMiwiU14xIl0sWzAsMl0sWzEsMl1d}{Link
to Diagram}
\end{quote}

These are either simultaneously orientable or simultaneously
nonorientable, by considering
\(\operatorname{det}A_1, \operatorname{det}A_2\) for the gluing maps
between trivializations. So
\(w_1(E) = {\left\langle {\bigwedge\nolimits^n E},~{f_* [S^1] } \right\rangle} = \chi_{f^* \bigwedge\nolimits^n E \text { nonorientable}}\).

\end{proof}

\begin{theorem}[?]

\(w_k(E)\) is the mod 2 reduction of the obstruction to extending
\(n-k+1\) linearly independent sections of \(E\) over
\(X{ {}^{ (k) } }\) when \(\dim_{\mathbb{R}}E = n\).

\end{theorem}

\begin{remark}

For \(k=n\), \(w_n(E)\) corresponds to 1 linearly independent section of
\(X{ {}^{ (n) } }\) since \(e(E) \equiv w_n(E) \operatorname{mod}2\).
For \(k=1\), \(w_1(E)\) corresponds to \(n\) linearly independent
sections over \(X{ {}^{ (1) } }\).

\end{remark}

\hypertarget{pontryagin-classes-tuesday-november-23}{%
\section{Pontryagin Classes (Tuesday, November
23)}\label{pontryagin-classes-tuesday-november-23}}

\hypertarget{complexification}{%
\subsection{Complexification}\label{complexification}}

\begin{question}

How to we get integral cohomology classes when we don't have a complex
structure?

\end{question}

\begin{definition}[Complexification]

Given a real vector bundle \(E\), its \textbf{complexification} is
defined as
\begin{align*}
E\otimes_{\mathbb{R}}{\mathbb{C}}
\coloneqq\displaystyle\coprod_{x\in X} E_x\otimes_{\mathbb{R}}{\mathbb{C}}
.\end{align*}
Then
\(\dim_{\mathbb{R}}E\otimes_{\mathbb{R}}{\mathbb{C}}= 2\dim_{\mathbb{R}}E\).

\end{definition}

\begin{remark}

In terms of transition functions (i.e.~Čech cohomology data), this is
the inclusion
\(\operatorname{GL}_n({\mathbb{R}}) \hookrightarrow\operatorname{GL}_n({\mathbb{C}})\).
We can now consider the Chern classes \(c_i(E\otimes{\mathbb{C}})\).

\end{remark}

\begin{definition}[Conjugate bundle]

Given a complex vector bundle \(E\), there is a \textbf{conjugate
bundle} \(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu\)
with the complex structure
\(i_{\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu}\cdot V\coloneqq-i_{E}\cdot V\).
This corresponds to conjugating Čech cohomology data, i.e.~replacing
transition functions \(f_{ij}\) with
\(\mkern 1.5mu\overline{\mkern-1.5muf_{ij}\mkern-1.5mu}\mkern 1.5mu\).

\end{definition}

\begin{lemma}[?]

For any real vector bundle \(E\),
\begin{align*}
E\otimes{\mathbb{C}}\cong \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{E}}\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu
.\end{align*}

\end{lemma}

\begin{proof}[?]

The transition functions for the former are in
\(\operatorname{GL}_n({\mathbb{R}})\), which is fixed under conjugation.

\end{proof}

\begin{remark}

Note that \(E\otimes{\mathbb{C}}\cong E \oplus E\), and we can map
\begin{align*}
{\mathbb{R}}{ {}^{ \scriptscriptstyle\oplus^{2} }  }\otimes{\mathbb{C}}&{ { \, \xrightarrow{\sim}\, }}{\mathbb{C}}{ {}^{ \scriptscriptstyle\oplus^{2} }  } \\
(x\oplus y) \otimes\left\langle{1, i}\right\rangle &\mapsto (x, ix) \oplus (y, iy)
.\end{align*}

\end{remark}

\begin{exercise}[?]

Show that if \(E\) is a complex vector bundle,
\begin{align*}
E\otimes{\mathbb{C}}\cong E \oplus \mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu
,\end{align*}
and
\begin{align*}
c_i(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = (-1)^i c_i(E)
.\end{align*}

\end{exercise}

\hypertarget{pontryagin-classes}{%
\subsection{Pontryagin Classes}\label{pontryagin-classes}}

\begin{lemma}[?]

For \(i\) odd, \(c_i(E\otimes{\mathbb{C}})\) is 2-torsion.

\end{lemma}

\begin{proof}[?]

On one hand,
\begin{align*}
E\otimes{\mathbb{C}}\cong \mkern 1.5mu\overline{\mkern-1.5muE\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu \implies
c_i(E\otimes{\mathbb{C}}) = c_i(\mkern 1.5mu\overline{\mkern-1.5muE\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu) = (-1)^i c_i(E\otimes{\mathbb{C}})
.\end{align*}
So if \(i\) is odd, \(c_i = -c_i\) and thus \(2c_i = 0\).

\end{proof}

\begin{definition}[Pontryagin classes]

The \textbf{\(i\)th Pontryagin class} is
\begin{align*}
p_i(E) = (-1)^{i} c_{2i} (E\otimes{\mathbb{C}}) \in H^{4i}(X)
.\end{align*}

\end{definition}

\begin{remark}

Suppose \(R\) is a coefficient ring where 2 is invertible (or not a zero
divisor). Some properties of the \(p_i\):

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(p_i\) is natural, so \(p_i (f^* E) = f^* p_i(E)\).
\item
  \(p_{i> \dim(E)/2}\).
\item
  \(p(E \bigoplus E') = p(E) \smile p(E')\). Why? Note
  \(p(E) = 1 - c_2 + c_4 - c_6 + \cdots\), so multiplying two such
  things yields
  \begin{align*}
    p(E) p(E') = 1 - (c_2 + c_2') + (c_4 + c_2c_2' + c_4') - (c_6 + c_4 c_2' + c_2 c_4' + c_6') - \cdots = p(E \oplus E')
    .\end{align*}
\item
  If \(\dim_{\mathbb{R}}E = 2n\) then \(p_n(E) = e(E)^2\). Why?
  \begin{align*}
    p_n(E) = (-1)^n c_{2n}(E\otimes{\mathbb{C}}) = (-1)^n e(E\otimes{\mathbb{C}}) = (-1)^n (-1)^{2n(2n+1) \over 2} = (-1)^{n + n(2n+1)} e(E)^2 = (-1)^{n(2n+2)}e(E)^2 = e(E)^2
    .\end{align*}
\item
  If \(E\) is a complex line bundle,
  \begin{align*}
  c(E \oplus \mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = c(E\otimes{\mathbb{C}}) = 1 - p_1(E) + p_2(E) - \cdots = c(E) c(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = (1 + c_1(E) + c_2(E) + \cdots)(1 - c_1(E) + c_2(E) - \cdots)
  .\end{align*}
\end{enumerate}

\end{remark}

\begin{example}[?]

We can compute \(p({\mathbf{T}}{\mathbb{CP}}^5)\). Recall that
\(c({\mathbf{T}}{\mathbb{CP}}^n) = (1+a)^{n+1}\) where
\(\left\langle{a}\right\rangle = H^2({\mathbb{CP}}^n; {\mathbb{Z}})\) is
the positive generator. So
\(c({\mathbf{T}}{\mathbb{CP}}^5) = (1+a)^6 = 1 + {6\choose 1}a + {6\choose 2}a + \cdots\).
Using (e) above, we have
\(c(E) c(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu)= (1+a)^6(1-a)^6 = (1-a^2)^6 = 1-6a^2 + 15a^4\).
So \(p({\mathbf{T}}{\mathbb{CP}}^5) = 1 + 6a^2 + 15a^4\).

\end{example}

\begin{corollary}[?]

If \({\mathbb{CP}}^5\hookrightarrow{\mathbb{R}}^N\) immerses, then
\(p(\nu {\mathbb{CP}}^5) = {1\over 1 + 6a^2 + 15a^4} = 1 - 6a^2 + 21a^4\).
So \(p_2(\nu {\mathbb{CP}}^5) \neq 0\), so
\(\dim(\nu)/2 \geq 2 \implies \dim(\nu) \geq 4\) and this forces
\(n\geq 14\). Note that here we used that \(p_i(E) = 0\) for
\(i > \dim E/2\).

\end{corollary}

\begin{theorem}[?]

\begin{align*}
H^*({{\mathbf{B}}{\operatorname{SO}}}_{2n+1}; R) &{ { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_n] \\
H^*({{\mathbf{B}}{\operatorname{SO}}}_{2n}; R) &{ { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_{n-1}, e] \\
H^*({{\mathbf{B}}{\operatorname{O}}}_{2n+1}; R) &{ { \, \xrightarrow{\sim}\, }}H^*({{\mathbf{B}}{\operatorname{O}}}_{2n}; R) { { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_{n}] \\
,\end{align*}
where \(p_i\) are the Pontryagin classes for the respective canonical
bundles.

\end{theorem}

\hypertarget{bordism-tuesday-november-30}{%
\section{Bordism (Tuesday, November
30)}\label{bordism-tuesday-november-30}}

\begin{definition}[Bordism]

Let \(M_1, M_2\in {\mathsf{sm}}{\mathsf{Mfd}}^n\) be closed, then
\(M_1\) is \textbf{bordant} to \(M_2\) iff
\(\exists W\in {\mathsf{sm}}{\mathsf{Mfd}}^{n+1}\) with
\({{\partial}}W = M_1 {\textstyle\coprod}M_2\).

\end{definition}

\begin{remark}

This defines an equivalence relation on
\({\mathsf{sm}}{\mathsf{Mfd}}^n\), and yields a ring
\((\prod_n {\mathcal{N}}^n, {\textstyle\coprod}, \times)\). We'll add
extra structure to get refinements of this ring, for which we'll need a
bit about stable normal bundles and the Whitney embedding theorem.

\end{remark}

\begin{theorem}[Whitney Embedding]

Any \(M\in {\mathsf{sm}}{\mathsf{Mfd}}^n\) (possibly with boundary)
embeds in \({\mathbb{R}}^{2n+1}\), and any two such embeddings into
\({\mathbb{R}}^{2n+3}\) are isotopic.

\end{theorem}

\begin{remark}

Moreover if \(i, i': M\hookrightarrow{\mathbb{R}}^n\) with
\(n\geq 2n+3\), then the normal bundles \(\nu M, (\nu M)'\) are bundle
isomorphic.

\end{remark}

\begin{corollary}[?]

Every such \(M\) has a stable normal bundle \(\nu_M\) which is
well-defined up to adding trivial line bundles.

\end{corollary}

\begin{remark}

Note that adding trivial line bundles yields a tower
\({{\mathbf{B}}{\operatorname{O}}}_1 \to {{\mathbf{B}}{\operatorname{O}}}_{2}\to \cdots\),
where the classifying maps of \(E\) and \(E \oplus L\) are classified by
maps to \({{\mathbf{B}}{\operatorname{O}}}_k\) and
\({{\mathbf{B}}{\operatorname{O}}}_{k+1}\) respectively. We can define
lifts of extra structures by looking for commutative towers over \(X_k\)
for
e.g.~\(X = {{\mathbf{B}}{\operatorname{O}}}, {\operatorname{Spin}}\),
etc:

\begin{center}
\begin{tikzcd}
    & \cdots && {X_k} && {X_{k+1}} && \cdots \\
    \\
    & \cdots && {{{\mathbf{B}}{\operatorname{O}}}_k} && {{{\mathbf{B}}{\operatorname{O}}}_{k+1}} && \cdots \\
    M
    \arrow[from=1-2, to=1-4]
    \arrow[from=1-4, to=1-6]
    \arrow[from=1-6, to=1-8]
    \arrow[from=3-2, to=3-4]
    \arrow[from=3-4, to=3-6]
    \arrow[from=3-6, to=3-8]
    \arrow[from=1-4, to=3-4]
    \arrow[from=1-6, to=3-6]
    \arrow["{\nu_M}"{description}, curve={height=6pt}, from=4-1, to=3-4]
    \arrow[curve={height=24pt}, from=4-1, to=3-6]
    \arrow[curve={height=12pt}, dashed, from=4-1, to=1-4]
    \arrow[curve={height=12pt}, dashed, from=4-1, to=1-6]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMSwwLCJcXGNkb3RzIl0sWzMsMCwiWF9rIl0sWzUsMCwiWF97aysxfSJdLFs3LDAsIlxcY2RvdHMiXSxbMywyLCJcXEJPX2siXSxbNSwyLCJcXEJPX3trKzF9Il0sWzcsMiwiXFxjZG90cyJdLFsxLDIsIlxcY2RvdHMiXSxbMCwzLCJNIl0sWzAsMV0sWzEsMl0sWzIsM10sWzcsNF0sWzQsNV0sWzUsNl0sWzEsNF0sWzIsNV0sWzgsNCwiXFxudV9NIiwxLHsiY3VydmUiOjF9XSxbOCw1LCIiLDEseyJjdXJ2ZSI6NH1dLFs4LDEsIiIsMSx7ImN1cnZlIjoyLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbOCwyLCIiLDEseyJjdXJ2ZSI6Miwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{definition}[Structures]

An \(X{\hbox{-}}\)structure on \(M\) is a family of lifts for large
\(k\):

\begin{center}
\begin{tikzcd}
    && {X_k} \\
    \\
    M && {{{\mathbf{B}}{\operatorname{O}}}_k}
    \arrow[from=1-3, to=3-3]
    \arrow["{\nu_M^k}"', from=3-1, to=3-3]
    \arrow[dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMiwwLCJYX2siXSxbMCwyLCJNIl0sWzIsMiwiXFxCT19rIl0sWzAsMl0sWzEsMiwiXFxudV9NXmsiLDJdLFsxLDAsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

We also require these to be compatible with the morphism
\({{\mathbf{B}}{\operatorname{O}}}_k\to {{\mathbf{B}}{\operatorname{O}}}_{k+1}\)
and \(X_k\to X_{k+1}\).

\end{definition}

\begin{example}[?]

Some examples:

\begin{itemize}
\tightlist
\item
  \({{\mathbf{B}}{\operatorname{SO}}}_k\) for orientation,
\item
  \(V_k\) (the Stiefel manifold) for framings.
\end{itemize}

\end{example}

\begin{definition}[$X\dash$bordism]

Given \(M_i\) with \(X{\hbox{-}}\)structures, we say \(N\) is an
\textbf{\(X{\hbox{-}}\)bordism} if

\begin{itemize}
\tightlist
\item
  \({{\partial}}N = M_1 {\textstyle\coprod}M_2\)
\item
  There is an \(X{\hbox{-}}\)structure on \(N\) that restricts to the
  \(X{\hbox{-}}\)structures on \(M_1\) and \(M_2\).
\end{itemize}

\end{definition}

\begin{remark}

For orientations, we get a bordism ring \(\Omega_n\), and for framings
we get \(\Omega_n^{\mathop{\mathrm{Fr}}}\).

\end{remark}

\begin{theorem}[Identification of framed bordism group]

\begin{align*}
\Omega_n^{\mathop{\mathrm{Fr}}} \cong \pi_n^s
,\end{align*}
the stable homotopy groups of spheres \(\colim_k \pi_{n+k} S^n\) (which
stabilize for \(k>n+1\)).

\end{theorem}

\begin{definition}[Stiefel-Whitney numbers]

Given \(M\in {\mathsf{sm}}{\mathsf{Mfd}}^n\) and
\(I \coloneqq\left\{{r_1,\cdots, r_n}\right\}\) such that
\(\sum_{1\leq k \leq n} kr_k = n\), consider
\(W_n \coloneqq\prod_{1\leq k \leq n} w_k({\mathbf{T}}M)^{r_k} \in H^n(M; C_2)\).
We can evaluate this to get
\begin{align*}
w_I(M) \coloneqq{\left\langle {W_n},~{[M]} \right\rangle} \in C_2
.\end{align*}
Note that the condition on \(I\) guarantees that \(W_n\in H^n\).

\end{definition}

\begin{remark}

For \(n=3\), the only possibilities for \(I\) are

\begin{itemize}
\tightlist
\item
  \((3,0,0) \leadsto W_n = w_1^3({\mathbf{T}}M)\)
\item
  \((1, 1, 0 \leadsto W_n = w_1({\mathbf{T}}M)w_2({\mathbf{T}}M) )\)
\item
  \((0,0,1) \leadsto W_n = w_3({\mathbf{T}}M)\)
\end{itemize}

\end{remark}

\begin{definition}[Pontryagin number]

Given an \emph{oriented} \(M \in {\mathsf{sm}}{\mathsf{Mfd}}^n\), the
\textbf{Pontryagin number} of \(M\) is given by taking
\(I \coloneqq\left\{{r_1,\cdots, r_{n\over 4}}\right\}\) such that
\(4r_1 + 8r_2 + \cdots + nr_{n\over 4} = n\), setting
\(P_n \coloneqq\prod_k p_k({\mathbf{T}}M)^{r_k}\), and evaluating
\begin{align*}
p_I(M) \coloneqq{\left\langle {P_n},~{[M]} \right\rangle}
.\end{align*}

\end{definition}

\begin{corollary}[?]

If any Pontryagin number is nonzero, then \(M\) can not admit an
orientation reversing diffeomorphism.

\end{corollary}

\begin{proof}[?]

Use that
\begin{align*}
P_I(-M) = {\left\langle {\prod p_1 {{\mathbf{T}}M} },~{[-M]} \right\rangle} = -P_I(M)
,\end{align*}
but also
\begin{align*}
P_I(M)
&= {\left\langle {\prod p_i {\mathbf{T}}M},~{[M]} \right\rangle} \\
&= {\left\langle {f^* \qty{ \prod p_i {\mathbf{T}}M} },~{[M]} \right\rangle} \\
&= {\left\langle {\prod p_i {\mathbf{T}}M},~{[f_* M]} \right\rangle} \\
&= {\left\langle {\prod p_i {\mathbf{T}}M},~{[- M]} \right\rangle} \\
&= P_I(-M)
.\end{align*}

\end{proof}

\begin{example}[?]

Some examples:

\begin{itemize}
\tightlist
\item
  \(P_I({\mathbb{CP}}^{2n}) \neq 0\) by a computation.
\item
  \(P_I({\mathbb{CP}}^{2n+1}) = 0\) since conjugating the complex
  structure \(J\mapsto -J\) is an orientation reversing diffeomorphism.
\end{itemize}

\end{example}

\begin{theorem}[Pontryagin, Thom]

\(M=0\) in \({\mathcal{N}}_n\) iff \(w_I(M) = 0\) for all \(I\).

\end{theorem}

\begin{proof}[?]

\(\implies\): Easy, a simple algebraic topology calculation.

\(\impliedby\): Difficult and omitted!

\end{proof}

\begin{corollary}[?]

There is an injective group morphism
\begin{align*}
{\mathcal{N}}_n ({\mathcal{N}}_{I_1}, \cdots, {\mathcal{N}}_{I_k}) \to C_2{ {}^{ \scriptscriptstyle\times^{k} }  }
.\end{align*}

\end{corollary}

\begin{theorem}[Pontryagin-Thom]

If \(M=0\) in \(\Omega_n\), then \(P_I(M) = 0\) for all \(I\).
Conversely, if \(P_I(M)=0\) for all \(I\) then \(M\) is torsion in
\(\Omega_n\), so there is some \(k\) such that
\(M{ {}^{ \scriptscriptstyle\coprod^{k} } } = {{\partial}}W\) for some
\(W\).

\end{theorem}

\begin{remark}

Milnor and Wall: the only torsion in \(\Omega_n\) is order 2, and an
oriented manifold \(M\) is 0 in \(\Omega_n \iff w_I(M), P_I(M) = 0\) for
all \(I\). For a proof (for at least the first statement), see
Milnor-Stasheff.

\end{remark}

\begin{remark}

We can kill torsion to get an injective map
\begin{align*}
\Omega_n \otimes{\mathbb{Q}}\xrightarrow{(P_{I_1}, \cdots, P_{i_k} ) } {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{k} }  }
.\end{align*}

\end{remark}

\begin{theorem}[Pontryagin-Thom]

The ring \({\mathcal{N}}_n\) is a polynomial ring over
\({\mathbb{F}}_2\) with one generator in each dimension:
\begin{align*}
{\mathcal{N}}_n \cong {\mathbb{F}}_2[\left\{{v_i {~\mathrel{\Big\vert}~}i\neq 2^i-1}\right\}]
.\end{align*}

\end{theorem}

\begin{theorem}[Pontryagin-Thom]

The ring \(\Omega_* \otimes{\mathbb{Q}}\) is polynomial ring over
\({\mathbb{Q}}\) generated by \({\mathbb{CP}}^n\) for \(n=1,2,\cdots\):
\begin{align*}
\Omega_* \cong \cong {\mathbb{Q}}[v_1,v_2,\cdots] && {\left\lvert {v_i} \right\rvert} = i
.\end{align*}

\end{theorem}

\hypertarget{thursday-december-02}{%
\section{Thursday, December 02}\label{thursday-december-02}}

\begin{remark}

Today: the Hirzebruch signature theorem and exotic \(S^7\). We saw that
\(\Omega_* \otimes{\mathbb{Q}}\) is the polynomial algebra on
\({\mathbb{Q}}[v_1, v_2\cdots]\) where \(v_2\) corresponds to
\({\mathbb{CP}}^{2n}\). This was because
\(\ker\qty{\Omega_n \xrightarrow{\left\{{p_{I_n}}\right\}} {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{n} } } }\)
can only be torsion.

\end{remark}

\begin{definition}[Signature]

For \(B\) a symmetric bilinear form,
\(\operatorname{sig}(B) = \dim V^+ - \dim V^-\), the dimensions of
positive/negative definite subspaces respectively, or equivalently the
difference in the number of positive and negative eigenvalues. For
\(M\in {\mathsf{sm}}{\mathsf{Mfd}}^{4n}\), there is a middle-dimensional
pairing
\begin{align*}
H^{2n}(M; {\mathbb{Z}}){ {}^{ \scriptstyle\otimes_{{\mathbb{Z}}}^{2} }  } &\to {\mathbb{Z}}\\
a\otimes b &\mapsto {\left\langle {a \smile b},~{[M]} \right\rangle}
,\end{align*}
and we write \(\sigma(M)\) for the signature of this form.

\end{definition}

\begin{theorem}[?]

The signature induces a ring morphism
\begin{align*}
\sigma: (\Omega_*, {\textstyle\coprod}, \times) \to ({\mathbb{Z}}, +, \cdot)
,\end{align*}
so

\begin{itemize}
\tightlist
\item
  \(\sigma(M {\textstyle\coprod}N) = \sigma(M) + \sigma(N)\),
\item
  \(\sigma(M\times N) = \sigma(M) \sigma(N)\),
\item
  \(\sigma(M) = 0\) if \(M = {{\partial}}M'\) for some \(M'\).
\end{itemize}

\end{theorem}

\begin{lemma}[?]

If there is a half-dimensional subspace
\(L \subseteq H^{2n}(M; {\mathbb{Q}})\) such that
\({\left\langle {a \smile},~{[M]} \right\rangle} = 0\) for all
\(a,b\in L\) then \(\sigma(M) = 0\).

\end{lemma}

\begin{proof}[?]

Use Poincare duality to get an isomorphism:

\begin{center}
\begin{tikzcd}
    {H^{2n}(N; {\mathbb{Q}})} && {H^{2n}(M; {\mathbb{Q}})} && {H^{2n+1}(N, M; {\mathbb{Q}})} \\
    \\
    && {H_{2n}(M; {\mathbb{Q}})} && {H_{2n}(N; {\mathbb{Q}})}
    \arrow["{i_*}", from=3-3, to=3-5]
    \arrow["{\cong, \mathrm{PD}}"', from=1-5, to=3-5]
    \arrow["\cong", from=1-3, to=3-3]
    \arrow["{f}", from=1-3, to=1-5]
    \arrow["{i^*}", from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbNCwwLCJIXnsybisxfShOLCBNOyBcXFFRKSJdLFs0LDIsIkhfezJufShOOyBcXFFRKSJdLFsyLDAsIkheezJufShNOyBcXFFRKSJdLFsyLDIsIkhfezJufShNOyBcXFFRKSJdLFswLDAsIkheezJufShOOyBcXFFRKSJdLFszLDEsImlfKiJdLFswLDEsIlxcY29uZywgXFxQRCIsMl0sWzIsMywiXFxjb25nIl0sWzIsMF0sWzQsMiwiaV4qIl1d}{Link
to Diagram}
\end{quote}

Note that \(\operatorname{rank}i^* = \operatorname{rank}i_*\) since
these spaces are dual, and this has the same rank as \(f\) since the
sides are isomorphisms. Then
\(\dim H^{2n}(M) = \dim \operatorname{im}i^* + \dim \operatorname{coker}i^*\)
and \(\dim \operatorname{coker}i^* = \dim \operatorname{im}i_*\), so
\(\dim H^{2n}(M) = 2\dim \operatorname{im}i^* \leq 2\dim \operatorname{im}i^*\).
Then writing \(i^* \alpha = a, i^* \beta = b\in \operatorname{im}i^*\),
\begin{align*}
{\left\langle { i^*(a\smile b)},~{[M]} \right\rangle} = {\left\langle { \alpha\smile\beta},~{i_*[M]} \right\rangle} =0
,\end{align*}
since \(i_*[M] = 0\). Thus \(\sigma(M) = 0\).

\end{proof}

\begin{corollary}[?]

One can compute the signature in terms of Pontryagin numbers, i.e.~there
is a formula for \(\sigma(M)\) in terms of
\(\left\{{p_{I_n}(M)}\right\}\). The explicit formula is the
\textbf{Hirzebruch signature theorem}.

\end{corollary}

\begin{example}[?]

Write
\(\Omega_4 \otimes{\mathbb{Q}}= \left\langle{{\mathbb{CP}}^2}\right\rangle\)
and consider \(M \in {\mathsf{sm}}{\mathsf{Mfd}}^4\). Note
\({\mathbb{CP}}^2\) has only 1 Pontryagin number and
\(c({\mathbb{CP}}^n) = (1+x)^{n+1}\), so we have a formula
\begin{align*}
1 - p_1({\mathbb{CP}}^2) 
&= c({\mathbb{CP}}^2) \cdot c(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu) \\
&= (1+x)^3(1-x)^3 \\
&= (1-x^2)^3 \\
&= 1 + -3x^2 + 3x^4 - x^6 \\
&= 1 - 3x^2
,\end{align*}
where we've used that \(H^{\geq 4} = 0\). So
\(p_1({\mathbb{CP}}^2)= 3x^2\), which implies
\begin{align*}
{\left\langle { p_1 {\mathbb{CP}}^2},~{[{\mathbb{CP}}^2]} \right\rangle} = {\left\langle {3x^2},~{[{\mathbb{CP}}^2]} \right\rangle} = 3
.\end{align*}
Using that \(\sigma({\mathbb{CP}}^2) = 1\), we can deduce
\(\sigma(M) = p_1(M)/3\) by considering eigenvalues of the intersection
pairing.

\end{example}

\begin{example}[?]

Consider \(M^8\) and
\(\Omega_8 \otimes{\mathbb{Q}}= \left\langle{{\mathbb{CP}}^4, {\mathbb{CP}}^2\times {\mathbb{CP}}^2}\right\rangle\).
Then
\(\Sigma({\mathbb{CP}}^4) = \sigma({\mathbb{CP}}^2{ {}^{ \scriptscriptstyle\times^{2} } }) = 1\),
what are the Pontryagin numbers? Write
\(v_2 = {\mathbb{CP}}^2, v_2^2 = {\mathbb{CP}}^2 \times {\mathbb{CP}}^2, v_4 = {\mathbb{CP}}^4\),then
\begin{align*}
1 - p_1(v_4) + p_2(v_4) 
&= c(v_4)c(\mkern 1.5mu\overline{\mkern-1.5muv\mkern-1.5mu}\mkern 1.5mu_4) \\
&= (1+x)^5(1-x)^5 \\
&= (1-x^2)^5 \\
&= 1-5x^2 + 10x^4
,\end{align*}
so \(p_1(v_4) = 5x^2\) and \(p_2(v_4) = 10x^4\). Then
\begin{align*}
{\left\langle {p_1^2 v_4},~{[v_4]} \right\rangle} &= {\left\langle {25x^4},~{[v_4]} \right\rangle} = 25 \\
{\left\langle {p_2 v_4},~{[v_4]} \right\rangle} &= {\left\langle {10x^4},~{[v_4]} \right\rangle} = 10 
.\end{align*}
Similarly,
\begin{align*}
1 + p_1(v_4^2) + p_2(v_4^2) = (1+ 3x^2)(1+3y^2) = 1 + 3x^2 + 3y^2 + 9x^2y^2
,\end{align*}
so
\begin{align*}
{\left\langle {p_1^2 v_2^2},~{[v_2^2]} \right\rangle} 
= {\left\langle {(3x^2 + 3y^2)^2 },~{[v_2^2]} \right\rangle}
= {\left\langle {9x^4 + 9y^4 + 18x^2 y^2 },~{[v_2^2]} \right\rangle}
= 18
,\end{align*}
since the cohomology ring is
\({\mathbb{F}}[x]/\left\langle{x^2}\right\rangle \otimes{\mathbb{F}}[y]/\left\langle{y^2}\right\rangle\),
and a similar calculation shows
\begin{align*}
{\left\langle {p_2(v_2^2)},~{[v_2^2]} \right\rangle} = 9
.\end{align*}

Summarizing, where we abuse notation and identity classes with numbers,

\begin{itemize}
\tightlist
\item
  \(p_1^2(v_4) = 25\),
\item
  \(p_2(v_4) = 10\),
\item
  \(p_1^2(v_2^2) = 18\),
\item
  \(p_2(v_2^2) = 9\).
\end{itemize}

Writing \(\sigma = ap_1^2 + b p_2\),
\begin{align*}
1 = \sigma(v_4) &= 25 a + 10 b \\
1 = \sigma(v_2^2) &= 18a + 9b
.\end{align*}

This yields \(b = {1\over 9}-2a\) and
\(1 = 25a + 10\qty{{1\over 9} - 2a}\) yields \(a=-1/45\) and \(b=7/45\).
Thus
\begin{align*}
\sigma = {1\over 45}\qty{2p_2 - p_1^2}
.\end{align*}

\end{example}

\begin{theorem}[Thom]

\begin{align*}
\Omega_7 = 0
.\end{align*}

\end{theorem}

\begin{definition}[?]

Given \(M^7\) with \(H^3(M) = H^4(M)\) and suppose
\(M = {{\partial}}B^*\). Let \(i: H^4(B; M) \to H^4(B)\) and define
\begin{align*}
\lambda(M) \coloneqq 2 {\left\langle { (i^{-1}p_1 {\mathbf{T}}B)^2 },~{[B]} \right\rangle} - \sigma(B)
.\end{align*}
Here \(\sigma(B)\) is the signature of
\begin{align*}
H^4(B, M ; {\mathbb{Q}}){ {}^{ \scriptstyle\otimes_{{\mathbb{Z}}}^{2} }  } &\to {\mathbb{Z}}\\
a\otimes b &\mapsto {\left\langle {a \smile b},~{[B]} \right\rangle} 
.\end{align*}

\end{definition}

\begin{proposition}[?]

\(\lambda(M)\) is independent of \(B\).

\end{proposition}

\begin{remark}

Let \(C \coloneqq B_1 { \displaystyle\coprod_{M} } B_2\) where
\({{\partial}}B_i = M\), then there is a relation expressing
\({\left\langle {p_1(C)^2},~{[C]} \right\rangle}\) in terms of pairings
of \(p(B_i)\) against \([B_i]\), and
\(\sigma(C) = \sigma(B_1) - \sigma(B_2)\). We have
\begin{align*}
45\sigma(C) + {\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\
\implies -40\sigma(C) + {\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\
\implies \sigma(C) + 2{\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\
\implies (\sigma(B_1) - \sigma(B_2) ) - 2 \qty{ {\left\langle {(i^{-1}p_1 B_1)^2 },~{[B_1]} \right\rangle} \cdots }
&= \lambda_{B_1}(M) - \lambda_{B_2}(M)
.\end{align*}

\end{remark}

\begin{remark}

If the \(\lambda\)s differ, the manifolds can not be diffeomorphic.
Constructing \(S^7\)s: take \(S^3\) bundles over \(S^4\). Pick any map
\(S^3\hookrightarrow{\operatorname{SO}}_4({\mathbb{R}}) \subseteq {\operatorname{Homeo}}(S^3)\)
to get clutching data, which are elements of
\(\pi_3({\operatorname{SO}}_4) = {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{2} } }\).
Label these bundles with parameters
\((m, n) \in {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{2} } }\) so
that \(\xi_{1, 0}\) corresponds to lifting
\([S^3, {\operatorname{SO}}_3]\) to \([S^3, {\operatorname{SO}}_4]\)
fixing the \(x{\hbox{-}}\)axis, and \(\xi_{1, 1}\) is the canonical for
\({\mathbb{H}}{\mathbb{P}}^1\). If \(M_k\) is the total space of
\(\xi_{k, 1}\), then \(\lambda(M_k) \equiv k^2-1 \operatorname{mod}7\),
so aren't diffeomorphic. Then one can show that the \(M_k\) are
homeomorphic by finding a Morse function with exactly 2 critical points.

\end{remark}

\hypertarget{problem-set-1}{%
\section{Problem Set 1}\label{problem-set-1}}

\hypertarget{section}{%
\subsection{1}\label{section}}

\begin{problem}[?]

With the definition of a vector bundle from class, show that the vector
space operations define continuous maps:

\begin{align*}
&+: E { \underset{\scriptscriptstyle {B} }{\times} } E \rightarrow E \\
&\times: \mathbb{R} \times E \rightarrow E
\end{align*}

\end{problem}

\begin{remark}

Definition of vector bundle: need charts \((U, \phi)\) with
\(\phi: \pi^{-1}(U) \to U\times{\mathbb{R}}^n\) which when restricted to
a fiber \(F_b\) yields an isomorphism
\(F_b { { \, \xrightarrow{\sim}\, }}{\mathbb{R}}^n\).

What are these maps??

\end{remark}

\hypertarget{section-1}{%
\subsection{2.}\label{section-1}}

\begin{problem}[?]

Suppose you are given the following data:

\begin{itemize}
\item
  Topological spaces \(B\) and \(F\)
\item
  A set \(E\) and a map of sets \(\pi: E \rightarrow B\)
\item
  An open cover \(\mathcal{U}=\left\{U_{i}\right\}\) of \(B\) and for
  each \(i\) a bijection
  \(\phi: \pi^{-1}\left(U_{i}\right) \rightarrow U_{i} \times F\) so
  that \(\pi \circ \phi_{i}=\pi\).
\end{itemize}

Give conditions on the maps \(\phi_{i}\) so that there is a topology on
\(E\) making \(\phi: E \rightarrow B\) into a fiber bundle with
\(\left\{\left(U_{i}, \phi_{i}\right)\right\}\) as an atlas.

\end{problem}

\hypertarget{section-2}{%
\subsection{3.}\label{section-2}}

\begin{problem}[?]

An \emph{oriented \(n\)-dimensional vector bundle} is a vector bundle
\(\pi: E \rightarrow B\) together with an orientation of each fiber
\(E_{b}\), so that these orientations are continuous in the following
sense.

For each \(b \in B\) there is a chart \((U, \phi)\) with \(b \in U\) and
\(\phi: \pi^{-1}(U) \rightarrow U \times \mathbb{R}^{n}\) so that for
all \(b^{\prime} \in U\),
\begin{align*}
\left.\phi\right|_{E_{b^{\prime}}}: E_{b^{\prime}} \rightarrow \mathbb{R}^{n}
\end{align*}
is orientation-preserving.

Show that given an oriented \(n\)-dimensional vector bundle there is an
induced principal \(G L_{+}\left(\mathbb{R}^{n}\right)\)-bundle (the
``bundle of oriented frames''), and conversely given a principal
\(G L_{+}\left(\mathbb{R}^{n}\right)\)-bundle there is an induced
oriented \(n\)-plane bundle.

\end{problem}

\hypertarget{section-3}{%
\subsection{4.}\label{section-3}}

\begin{problem}[?]

A Riemannian metric on a vector bundle \(\pi: E \rightarrow B\) is an
inner product \(\langle\cdot, \cdot\rangle_{b}\) on each fiber \(E_{b}\)
of \(E\), which is continuous in the sense that the induced map
\(E \oplus E=E \times_{B} E \rightarrow \mathbb{R}\) is continuous.

Show that given a Riemannian metric on a vector bundle, there is an
induced principal \(O(n)\)-bundle (the ``bundle of orthonormal
frames''), and conversely given a principal \(O(n)\)-bundle there is an
induced vector bundle with Riemannian metric.

\end{problem}

\hypertarget{section-4}{%
\subsection{5 .}\label{section-4}}

\begin{problem}[?]

What operation on principal \(O(n)\)-bundles corresponds to dualizing a
vector bundle? What about the direct sum of vector bundle?

\end{problem}

\hypertarget{section-5}{%
\subsection{6.}\label{section-5}}

\begin{problem}[?]

For nice spaces \(X\) (e.g.~CW complexes) and abelian groups \(G\),
there is a canonical isomorphism
\begin{align*}
\check{H}^{i}(X ; G) \cong H^{i}(X ; G)
\end{align*}
between Čech and singular cohomology of \(X\) with coefficients in
\(G\).

\begin{quote}
A nice, readable proof can be found in Frank Warner's Foundations of
Differential Manifolds and Lie Groups, Chapter 5. In the rest of this
problem, cohomology either means Čech cohomology or singular cohomology
after applying this isomorphism.
\end{quote}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Let \(\pi: E \rightarrow B\) be an \(n\)-dimensional vector bundle, or
  equivalently, a principal \(G L(n, \mathbb{R})\)-bundle, given by a
  Čech cocycle \(\phi \in H^{1}(B ; G L(n, \mathbb{R}))\). Show that the
  sign of the determinant
  \begin{align*}
  \operatorname{sgn}\operatorname{det}: \operatorname{GL}_n(\mathbb{R}) \rightarrow\{\pm 1\} \cong \mathbb{Z} / 2 \mathbb{Z}
  \end{align*}
  induces a map
  \begin{align*}
    \check{H}^{1}(B ; G L(n, \mathbb{R})) \rightarrow \check{H}^{1}(B ; \mathbb{Z} / 2 \mathbb{Z})
    ,\end{align*}
  and so \(\phi\) induces an element
  \(w_{1}(E) \in H^{1}(B ; \mathbb{Z} / 2 \mathbb{Z})\).
\item
  Compute \(w_{1}\) for the trivial line bundle (1-dimensional vector
  bundle) over the circle and for the Möbius band.
\item
  Prove that (for nice spaces) a line bundle \(\pi: E \rightarrow B\) is
  trivial if and only if \(w_{1}(E)=0 \in\)
  \(H^{1}(B ; \mathbb{Z} / 2 \mathbb{Z})\)
\end{enumerate}

\end{problem}

\hypertarget{section-6}{%
\subsection{7.}\label{section-6}}

\begin{problem}[?]

Show that the exact sequence of abelian topological groups

\begin{align*}
0 \rightarrow \mathbb{Z} \rightarrow \mathbb{R} \rightarrow S^{1}=G L(1, \mathbb{C}) \rightarrow 0
\end{align*}

induces an exact sequence in Čech cohomology

\begin{align*}
\check{H}^{1}(B, \mathbb{Z}) \rightarrow \breve{H}^{1}(B, \mathbb{R}) \rightarrow \check{H}^{1}\left(B ; S^{1}\right) \stackrel{\delta}{\rightarrow} \check{H}^{2}(B ; \mathbb{Z})
\end{align*}

Given a complex line bundle (principal \(G L(1, \mathbb{C})\)-bundle)
\(\pi: E \rightarrow B\) coming from the cocycle data
\(\phi \in H^{1}(B ; G L(1, \mathbb{C}))\), let
\(c_{1}(E)=\delta(\phi)\). Compute \(c_{1}(E)\) for some complex line
bundle over \(S^{2}\).

\end{problem}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
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\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
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\addsec{Theorems}
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\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
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\addsec{Exercises}
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\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
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\addsec{Figures}
\listoffigures
\cleardoublepage


\printbibliography[title=Bibliography]


\end{document}
