# Thursday, August 19 ## Intro and Overview ::: {.remark} Course website: Description from Akram's syllabus: > 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). **References**: - [@Hu] Husemoller, Fiber bundles. - [@MS] Milnor and Stasheff, Characteristic classes. - [@S] Steenrod, The topology of fibre bundles. - [@Ha] Hatcher, Vector bundles and K-theory . - [@BottTu] Bott and Tu, Differential forms in algebraic topology. Prerequisites: - Smooth manifolds: smooth maps and derivatives, differential forms. - Algebraic topology: homology and cohomology. ::: ::: {.remark} An overview of what we'll cover: - **General definitions and constructions related to *vector bundles* and *fiber bundles*.** Why bundles? For a bundle $$E \xrightarrow{\pi} B$$, characteristic classes will be cohomology classes in $$H^*(B)$$. Natural examples include - The tangent bundle $$TX\to X$$, and vector fields will be sections. - Exterior products $$\bigwedge\nolimits^n TX$$, where differential forms live - Normal bundles $$\nu$$, giving directions an embedded submanifold can be deformed. 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 $$\nu: E \to f(M_n)$$) {=tex} \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}  - **Characteristic Classes: Euler, Stiefel-Whitney, Pontryagin, etc.** These package geometric information into algebraic invariants that are often computable. Some examples: - Stiefel-Whitney classes can detect if $$M^n = {{\partial}}M^{n+1}$$ is a boundary (for smooth closed manifolds). - 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. - Pontryagin classes: Milnor used these to produce exotic $$S^7$$s! These are manifolds $$M^7$$ which are homeomorphic but not diffeomorphic to $$S^7$$. - Chern classes. ::: ## Fiber Bundles ::: {.definition title="Fiber bundle"} A **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 $${\mathsf{Top}}$$: {=tex} \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}  Here the square is $$[0, 1]^{\times 2}$$. > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHBpXFxpbnYoVSkiXSxbMiwwLCJVXFxjcm9zcyBGIl0sWzEsMiwiVSJdLFswLDEsIlxccGhpIl0sWzAsMiwiXFxwaSIsMl0sWzEsMiwicF8yIl1d) ::: ::: {.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.  {=tex} \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}  ::: ::: {.example title="?"} An **atlas bundle** for $$\pi:E\to B$$ is a collection of charts $$\left\{{(U_\alpha, \phi_\alpha)}\right\}_{\alpha\in I}$$ such that $$\left\{{U_\alpha}\right\}\rightrightarrows B$$. ::: ::: {.example title="?"} {=tex} \envlist  - $$E \coloneqq B\times F \xrightarrow{p_2} F$$ the trivial/product bundle. - $$\widehat{X} \to X$$ any covering space. Note that the fibers are discrete. - The Möbius band: \<-- Xournal file: /home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures/2021-08-19_13-41.xoj --\> ![](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! ::: ::: {.remark} Consider the following setup: - $$B\in {\mathsf{Top}}$$ - $$\pi:E\to B$$ is a map of underlying sets - There is a bundle atlas $$\left\{{\phi_\alpha}\right\}$$, each $$\phi_\alpha$$ being a bijection. Then there exists *at most* one topology on $$E$$ such that $$\pi: E\to B$$ is a fiber bundle with the given atlas. ::: ::: {.exercise title="?"} Find necessary conditions for at least one topology to exist! ::: ## Vector Bundles ::: {.definition title="Vector bundle"} An $$n{\hbox{-}}$$dimensional real (resp. complex) **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. ::: ::: {.example title="?"} {=tex} \envlist  - The trivial (product) bundle $$B\times{\mathbb{R}}^n \xrightarrow{p_1} B$$. - The tangent bundle $$T X$$. - 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 $$M\to S^1$$. ::: ::: {.remark} We have some natural operations: 1. Direct sums. 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 $E_1 \oplus E_2 \coloneqq\displaystyle\bigcup_{b\in B} F_{1, b} \oplus F_{2, b}$ 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 $\left\{{ (U_1 \cap U_2, \phi: \pi^{-1}(U_1 \cap U_2) \to {\mathbb{R}}^{n_1 + n_2})}\right\}$ where $$n_1 \coloneqq\dim_{\mathbb{R}}F_{1, b}$$ and $$n_2 \coloneqq\dim_{\mathbb{R}}F_{2, b}$$ and define $$(b, (v_1, v_2)) \xrightarrow{\phi} (\phi_1(v_1), \phi_2(v_2))$$. ::: # Fiber Bundles with Structure and Principal $$G{\hbox{-}}$$ Bundles (Tuesday, August 24) {#fiber-bundles-with-structure-and-principal-ghbox--bundles-tuesday-august-24} ::: {.remark} Setup: - $$B \in {\mathsf{Top}}$$ is a space. - $$\pi:E\to B$$ is a map of sets with fibers/preimages $$F \coloneqq F_b \coloneqq\pi^{-1}(\left\{{b}\right\})$$. - A *bundle atlas* for $$\pi$$ is $$\phi$$ where $$\phi_U: \pi^{-1}(U) \to U \times F$$ is a bijection of sets Then there is at most one topology on $$E$$ making $$\pi:E\to B$$ into a fiber bundle with the specified atlas. ::: ::: {.definition title="Dual of a vector bundle"} Given a vector bundle $$\pi:E\to B$$, form the **dual bundle** $$\pi {}^{ \vee }:E {}^{ \vee }\to B$$ by setting - $$E {}^{ \vee }\coloneqq{\textstyle\coprod}_{b\in B} F_b {}^{ \vee }$$ - Set $$\pi {}^{ \vee }(F_b {}^{ \vee }) = \left\{{b}\right\} \in B$$. - Given $$\phi: \pi^{-1}(U)\to U\times {\mathbb{R}}^n$$, set $\phi {}^{ \vee }: (\pi {}^{ \vee })^{-1}(U) = {\textstyle\coprod}_{b\in U} F_b {}^{ \vee }\longrightarrow U\times({\mathbb{R}}^n) {}^{ \vee }\cong U\times{\mathbb{R}}^n .$ Here $$A \subseteq \pi^{-1}(U)$$ is open iff $$\phi_U(A)$$ is open in $$B$$. ::: ::: {.remark} Consider what happens on overlapping charts -- looking at maps fiberwise yields: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCJcXHBpXFxpbnYoVSkiXSxbMiwwLCJcXHBpXFxpbnYoVVxcaW50ZXJzZWN0IFYpIl0sWzQsMCwiXFxwaVxcaW52KFYpIl0sWzAsMiwiVVxcdGltZXMgRiJdLFs0LDIsIlZcXHRpbWVzIEYiXSxbMiwyLCIoVVxcaW50ZXJzZWN0IFYpIFxcdGltZXMgRiJdLFsyLDFdLFsxLDAsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoiYm90dG9tIn19fV0sWzAsMywiXFx2YXJwaGlfVSJdLFs1LDMsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoiYm90dG9tIn19fV0sWzUsNCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDQsIlxcdmFycGhpX1YiLDJdXQ==) Starting at $$(U \cap V)\times F$$ and running the diagram clockwise yields a map $\phi_V \circ \phi_U ^{-1}: (U \cap V ) \times F &\to (U \cap V ) \times F \\ (b, f) &\mapsto (b, \phi_{VU}(f) )$ where $$\phi_{VU}$$ is the following continuous map, defining a **transition function**: $\phi_{VU}: U \cap V &\to {\operatorname{Homeo}}(F) ,$ where $${\operatorname{Homeo}}(F) \coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(F, F)$$ with the compact-open topology. ::: ::: {.definition title="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: $Z\times X &\to Y \\ (z, x) &\mapsto f(z)(x) .$ 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 $S(C, U) \coloneqq\left\{{f\in \mathop{\mathrm{Maps}}(X, Y){~\mathrel{\Big\vert}~}f(X) \subseteq U}\right\} .$ If $$Y$$ has a metric, then this will coincide with the *compact convergence topology*, which has a basis $\left\{{ S(f, C, E) {~\mathrel{\Big\vert}~}C \subseteq X \text{ compact}, \forall {\varepsilon}>0,\,\forall f\in \mathop{\mathrm{Maps}}(X, Y) }\right\}, \\ S(f, C, E) \coloneqq\left\{{ g\in \mathop{\mathrm{Maps}}(X, Y) {~\mathrel{\Big\vert}~}d( f(x), g(x) ) < {\varepsilon}\,\,\forall x\in C }\right\} .$ ::: ::: {.definition title="Structure Groups"} Let $$G \subseteq {\operatorname{Homeo}}(F)$$, then a **fiber bundle with structure group** $$G$$ is a fiber bundle $$F\to E \xrightarrow{\pi} B$$ together with a bundle atlas such that $$G \subseteq {\operatorname{Homeo}}(F)$$. ::: ::: {.example title="?"} 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 $cw_1+ w_2 \coloneqq\phi_U^{-1}\qty{ c\phi_U(w_1) + \phi_U(w_2) } .$ 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 $\phi_V(w_1 + w_2) &= A \phi_U(w_1 + w_2) \\ &\coloneqq A\qty{\phi_U(w_1) + \phi_U(w_2) } \\ &= A\phi_U(w_1) + A\phi_U(w_2) \\ &= \phi_V(w_1) + \phi_V(w_2) \\ &\coloneqq\phi_V(w_1 + w_2) .$ ::: ::: {.example title="Bundles with structure"} An $${\mathbb{R}}^n{\hbox{-}}$$bundle with a $$\operatorname{GL}_{n}^+({\mathbb{R}})$$ structure is an orientable vector bundle, where $\operatorname{GL}_n^+({\mathbb{R}}) = \left\{{ A \in \operatorname{GL}_n({\mathbb{R}}) {~\mathrel{\Big\vert}~}\operatorname{det}(A) > 0 }\right\} .$ 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 $v\coloneqq\sum v_i e_i,\quad w \coloneqq\sum w_i e_i &&\implies {\left\langle {v},~{w} \right\rangle} = \sum v_i w_i .$ ::: ::: {.definition title="Principal $G\\dash$bundles"} A **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 1. $$\psi\qty{F_b} = F_b$$, so the action preserves each fiber, and 2. $$\psi$$ is free and transitive. ::: # Principal $$G{\hbox{-}}$$bundles (Thursday, August 26) {#principal-ghbox-bundles-thursday-august-26} ::: {.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: $G &\hookrightarrow{\operatorname{Homeo}}(G) \\ g &\mapsto (h\mapsto gh) .$ Using that $$F\cong G$$, taking charts $$(U, \varphi), (V, \psi)$$ for $$\pi:P\to B$$, we can identify {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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. ::: ::: {.example title="of principal bundles"} {=tex} \envlist  - Trivial actions: $$B \times G \xrightarrow{p_1} B$$. - Regular covering spaces $$\pi:\tilde X\to X$$, then $$G = \mathop{\mathrm{Deck}}(\tilde X/X)$$ with the discrete topology. - Given an $$n{\hbox{-}}$$dimensional vector bundle $$\pi: E\to B$$, take $\mathop{\mathrm{Frame}}(F_b) \coloneqq\left\{{(e_1, \cdots, e_n) \in F_b}\right\} \subseteq F_b^{\times n} ,$ the collection of all ordered bases of $$F_b$$. Then set $\mathop{\mathrm{Frame}}_n \coloneqq{\textstyle\coprod}_{b\in B} \mathop{\mathrm{Frame}}(F_b) \to B$ 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 $(\mathbf{e}_1, \cdots, \mathbf{e}_n) \cdot A \coloneqq\qty{\sum_i a_{i,1} \mathbf{e}_i, \sum_i a_{i, 2} \mathbf{e}_i, \cdots, \sum_i a_{i, n} \mathbf{e}_i } .$ - Given an *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. - Given a vector bundle with a Riemannian metric, we get a principal $${\mathcal{O}}_n({\mathbb{R}}){\hbox{-}}$$bundle by taking orthonormal frames. ::: ::: {.definition title="?"} Given two principal $$G{\hbox{-}}$$bundles $$\pi: P\to B$$ and $$\pi': Q\to B$$, an **isomorphism of principal bundles** is a $$G{\hbox{-}}$$equivariant map $$P \xrightarrow{f} Q$$ commuting over $$B$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJQIl0sWzIsMCwiUSJdLFsxLDIsIkIiXSxbMCwyLCJcXHBpIl0sWzEsMiwiXFxwaSciLDJdLFswLDEsImYiLDJdXQ==) Here *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 $F: \qty{ (U \cap V) \times F \xrightarrow{\phi^{-1}} \pi^{-1}(U \cap V) \xrightarrow{f} (\pi')^{-1}(U \cap V) \xrightarrow{\psi} (U \cap V) \times F} .$ 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: $(b, ?) \mapsto \pi^{-1}(b) \mapsto (f\circ \pi^{-1})(b) = (\pi')^{-1}b \mapsto b .$ We say $$f$$ is a $$G{\hbox{-}}$$isomorphism iff $$F$$ sends everything to $$G$$. ::: ## Sending Fiber Bundles to Principal $$G{\hbox{-}}$$bundles {#sending-fiber-bundles-to-principal-ghbox-bundles} ::: {.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 $P { \underset{\scriptscriptstyle {G} }{\times} } F / (pg, f)\sim (p, gf)$ 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: $(U\times G) \overset{\scriptscriptstyle {G} }{\times} F &\xrightarrow{\sim} U\times F \\ (p, g, f) &\mapsto (p, gf) .$ This is well defined since $$(p, gh, f)$$ and $$(p, g, hf)$$ map to $$(p, ghf)$$. The inverse is $$(p, f) \mapsto (p, 1, gf)$$. ::: ::: {.exercise title="?"} Check that this is a fiber bundle with $$G{\hbox{-}}$$structure. ::: # Tuesday, August 31 ::: {.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 *mixing* construction. Since $$G\curvearrowright F$$, we get an identification $$G \subseteq {\operatorname{Homeo}}(F, F)$$. We constructed $P { \underset{\scriptscriptstyle {G} }{\times} } F \coloneqq(P\times F)/(pg, f)\sim (p, gf) .$ 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. $\left\{{\substack{ \text{Fiber bundles with }G{\hbox{-}}\text{structures} \\ \text{ and fiber }F }}\right\} \adjunction{\text{Clutching}}{\text{Mixing}}{}{} \left\{{\substack{ \text{Principal G{\hbox{-}}bundles} }}\right\}$ ::: ::: {.example title="?"} 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. ::: {.exercise title="?"} Show that there is a homeomorphism $\mathop{\mathrm{Frame}}(E) { \underset{\scriptscriptstyle {\operatorname{GL}_n} }{\times} } {\mathbb{R}}^n \xrightarrow{\sim}E .$ ::: 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: $f(( \mathbf{e}_1, \cdots, \mathbf{e}_n) A, \mathbf{b}) = f( (\mathbf{e}_1, \cdots, \mathbf{e}_n), A\mathbf{b}) && \forall A\in \operatorname{GL}_n({\mathbb{R}}) .$ The left hand side is $b_1 (a_{1, 1} \mathbf{e}_1 + \cdots + a_{n, 1} \mathbf{e}_n ) + \cdots + b_n(a_{1, n} \mathbf{e}_1 + \cdots + a_{n, n} \mathbf{e}_n) = \sum_{i=1}^n b_i \qty{ \sum_{j=1}^n a_{j, i} \mathbf{e}_j } .$ The right-hand side is $(a_{1, 1}b_1 + \cdots + a_{1, n}b_n)\mathbf{e}_1 + \cdots + (a_{n, 1} b_1 + \cdots + a_{n,n} b_n)\mathbf{e}_n = \sum_{i=1}^n \qty{\sum_{j=1}^n a_{i, j} b_i } \mathbf{e}_i ,$ and one can check that these sums match term by term. ::: ::: {.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 $${\mathbb{R}}^n$$. ::: ::: {.remark} What are local charts for a principal bundle? For $$P{ \underset{\scriptscriptstyle {G} }{\times} } F$$, pick charts $$(U, \phi)$$ for $$P \xrightarrow{\pi} B$$: $\phi: \pi^{-1}(U) &\to U \times G \\ x & \mapsto (\pi(x), \gamma(x)) .$ Then a local chart for the principal bundle is of the form $\pi^{-1}(U) { \underset{\scriptscriptstyle {G} }{\times} } F &\xrightarrow{\tilde \phi} U \times F \\ (x, f) &\mapsto ( \pi(x), \gamma(x) f) .$ We also have $(U \times G){ \underset{\scriptscriptstyle {G} }{\times} } F &\to U \times F \\ ( (x, g), f) &\mapsto (x, gf) .$ One can invert $$\tilde \phi$$ using $$(a, f) \mapsto (\phi^{-1}(a, 1), f)$$. This yields transition functions: writing $\phi_V: \pi^{-1}(V) &\to V \times G \\ x &\mapsto (\pi(x), \psi(x) ) ,$ then $\phi_{VU} = \phi_V \circ \phi_U^{-1}: (a, f) & \\ &\xrightarrow{\phi_U^{-1}} ( \phi^{-1}_U(a, 1), \,\, f) \\ &\xrightarrow{\phi_V} (\pi \phi_U^{-1}(a, 1), \,\, \psi( \phi_U^{-1}(a, 1))f ) \\ &= (a,\,\, \psi(\phi_U^{-1}(a, 1)) f) .$ This says that $$(a, 1) \mapsto \psi( \phi^{-1}_U(a, 1))$$. ::: ::: {.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 $(U_i \times F) \ni (b, f)\sim (b, \phi_{ij}(b) f) \in (U_j \times F) .$ This is an equivalence relation precisely when the cocycle condition holds. This is referred to as **clutching data**. ::: ::: {.example title="The Mobius band is clutch"} Let $${\mathbb{Z}}/2\curvearrowright{\mathbb{R}}$$ by $$t\mapsto -t$$ with $$U, V$$ defined as follows: {=tex} \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}  Labeling the intersections as 1, 2, we set $\phi_{VU}: (U \cap V) = (U \cap V)_1 {\textstyle\coprod}(U \cap V)_2 &\to {\mathbb{Z}}/2 && \subseteq {\operatorname{Homeo}}({\mathbb{R}}) \\ x {\textstyle\coprod}y &\mapsto x {\textstyle\coprod}-y .$ This yields the open Mobius band. ::: ::: {.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. 1. Given clutching data $$\left\{{\phi_{ij}}\right\}$$, when is the resulting fiber bundle trivial? 2. 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? 3. Given two sets of clutching data $$\left\{{\phi_{ij}}\right\}$$ and $$\left\{{\psi_{ij}}\right\}$$ with the *different* open cover $$\left\{{U_i}\right\} \rightrightarrows X$$ and $$\left\{{V_i}\right\} \rightrightarrows X$$, when are the corresponding bundles $$G{\hbox{-}}$$isomorphic? ::: ::: {.lemma title="?"} The fiber bundle obtained from $$\phi_{ij}$$ is trivial iff there exists a map $$\gamma_i: U_i \to G$$ such that $$\phi_{ij} = \gamma_i \gamma_j^{-1}$$. ::: ::: {.proof title="?"} The trivial bundle is $$B\times F\to B$$, so if we have $$E\to B$$, we can take a map $U_i \times F &\to U_i \times F \\ (b, f) &\mapsto (b, \gamma_i(b) f) .$ Use that $$B\times F$$ is a trivial bundle, so it is its own trivialization. > To be continued next time. ::: # Thursday, September 02 ::: {.remark} Recall that we have a correspondence $\left\{{\substack{ \text{Vector bundles }E }}\right\} &\adjunction{\text{clutching}}{\text{mixing}}{}{} \left\{{\substack{ \text{Principal \operatorname{GL}_n{\hbox{-}}bundles \mathop{\mathrm{Frame}}(E)} }}\right\}$ 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? ::: ::: {.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 $(b, f) \in (U_i \cap U_j )\times F \subseteq U_i\times F ,$ we send this to $(b, \phi_{ji}(b) f ) \in (U_i \cap U_j)\times F \subseteq U_j \times F .$ 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. ::: ::: {.question} How can we tell when two fiber bundles constructed via clutching are isomorphic? ::: ::: {.lemma title="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 $$\phi_{ji} = \gamma_j \circ \gamma_i^{-1}$$. ::: ::: {.remark} For principal bundles, these $$\gamma_i$$ will give sections assembling to a global section obtained from clutching data: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJVX2kgXFx0aW1lcyBHIl0sWzIsMiwiVV9pIl0sWzAsMiwiQiJdLFswLDAsIlAiXSxbMywyXSxbMCwxXSxbMSwwLCJzKGIpID0gKGIsIFxcZ2FtbWFfaShiKSkiLDIseyJjdXJ2ZSI6NH1dLFsyLDEsIlxccHNpX0IiLDJdLFszLDAsIlxccHNpIiwyXV0=) The map on $$U_i \to \displaystyle\bigcup_i U_i \times F$$ will be $$(b, \gamma_i(b))$$, and we can use that $(b, \gamma_i(b)) \sim (b, \phi_{ji}(b)\gamma_i(b) ) \sim (b, \gamma_j(b)) ,$ so these agree on overlaps. ::: ::: {.lemma title="?"} 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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwxLCJQIl0sWzIsMywiQiJdLFs0LDEsIkJcXHRpbWVzIEciXSxbMiw0LCJiIl0sWzAsMCwiKGIsIHMoYilnKSJdLFs0LDAsIihiLCBnKSJdLFsyLDEsInBfMSIsMl0sWzAsMSwiXFxwaSJdLFswLDIsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDAsInMiLDAseyJjdXJ2ZSI6LTR9XV0=) ::: ::: {.proof title="of lemma about when clutched bundles are trivial"} $$\implies$$: If $$E$$ is trivial, we have an isomorphism {=tex} \begin{tikzcd} E \ar[rd, "f"] \ar[rr, "\pi"] & & P\times G \ar[ld, "p_1"] \\ & B & \end{tikzcd}  We have a $$G{\hbox{-}}$$isomorphism $$E_1 \xrightarrow{f} E_2$$, and so a composition {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIoVVxcaW50ZXJzZWN0IFYpIFxcdGltZXMgRiJdLFsyLDAsIlxccGlcXGludihVIFxcaW50ZXJzZWN0IFYpIl0sWzQsMCwiXFxwaVxcaW52KFUgXFxpbnRlcnNlY3QgVikiXSxbNiwwLCIoVVxcaW50ZXJzZWN0IFYpIFxcdGltZXMgRiJdLFsxLDAsIlxccGhpX1UiXSxbMSwyLCJmIiwyXSxbMiwzLCJcXHBoaV9WIiwyXSxbMCwzLCJGIiwwLHsiY3VydmUiOi01fV1d) Here we've used that $$f$$ commutes with the projection maps. We want to show $$\operatorname{im}(F) \subseteq G$$. We have a composite {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJVXFx0aW1lcyBGIl0sWzQsMCwiVVxcdGltZXMgRiJdLFsyLDAsIlxccGlcXGludihVX2kpIl0sWzIsMCwiXFxwaGlfaSJdLFsyLDEsImYiLDJdLFswLDEsIlxcZ2FtbWFfaSA9IFxccGhpX2lcXGludiBcXGNpcmMgZjogVVxcdG8gRyIsMCx7ImN1cnZlIjotNX1dXQ==) We can fill this in to a commutative diagram: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) The converse direction proceeds similarly! ::: ::: {.lemma title="?"} 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 $\gamma_j \phi_{ji} \gamma_i^{-1}= \psi_{ji} .$ ::: ::: {.proof title="?"} We can form the composite {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJVX2lcXHRpbWVzIEYiXSxbMiwwLCJcXHBpXzFcXGludihVX2kpIl0sWzQsMCwiXFxwaV8yXFxpbnYoVV9pKSJdLFs2LDAsIlVfaSBcXHRpbWVzIEYiXSxbMSwwLCJcXHBoaV9pIl0sWzEsMiwiZiIsMl0sWzIsMywiXFxwc2lfaiIsMl0sWzAsMywiXFxnYW1tYV9pOiBVX2lcXHRvIEciLDAseyJjdXJ2ZSI6NX1dXQ==) And then assemble a commuting diagram: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) ::: ## Nonabelian Čech Cohomology ::: {.definition title="Čech complex"} Let $${\mathcal{U}}\coloneqq\left\{{U_i}\right\}_{i\in I}\rightrightarrows B$$ an open cover, and define ${\check{C}}^0({\mathcal{U}}; G) \coloneqq\left\{{ \left\{{\gamma_i: U_i\to G}\right\}_{i\in I} }\right\} ,$ which is a group under pointwise multiplication. Define ${\check{C}}^2({\mathcal{U}}; G) &\coloneqq\left\{{ \left\{{\phi_{ij}: U_i \cap U_j \to G}\right\}_{i, j\in I} }\right\}\\ {\check{C}}^3({\mathcal{U}}; G) &\coloneqq\left\{{ \left\{{\phi_{ijk}: U_i \cap U_j \cap U_k \to G}\right\}_{i,j,k \in I} }\right\} ,$ and boundary maps $\delta^{0}: {\check{C}}^0({\mathcal{U}}; G) &\to {\check{C}}^1({\mathcal{U}}; G)\\ \left\{{\gamma_i: U_i\to G}\right\} &\mapsto \left\{{\phi_{ji} \coloneqq\gamma_j\gamma_i^{-1}: U_i \cap U_j \to G }\right\} ,$ $\delta^{1}: {\check{C}}^1({\mathcal{U}}; G) &\to {\check{C}}^2({\mathcal{U}}; G)\\ \left\{{\phi_{ij}: U_i \cap U_j \to G}\right\} &\mapsto \left\{{\eta{ijk} \coloneqq\phi_{ij} \phi_{jk} \phi_{ik}^{-1}: U_i \cap U_j \cap U_k \to G }\right\} .$ ::: ::: {.remark} One can check that $$\delta^1 \circ \delta^0 = 0$$ is trivial. And 1-cocycle will yield a fiber bundle. ::: ::: {.lemma title="1"} A bundle is trivial iff it is a 1-coboundary, where we take $$Z^1({\mathcal{U}}; G) \coloneqq\ker \delta^1$$, $$B^1({\mathcal{U}}; G) \coloneqq\operatorname{im}\delta^0$$. ::: ::: {.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. ::: ::: {.definition title="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 ${\check{H}}^1({\mathcal{U}}; G) \coloneqq Z^1({\mathcal{U}}; G) / \sim .$ ::: ::: {.lemma title="2"} Two bundles are isomorphic iff they yield the same element in $${\check{H}}^1({\mathcal{U}}; G)$$. ::: ::: {.remark} This works when bundles have the same open cover, and if not, we can take a common refinement. ::: # Tuesday, September 07 ::: {.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: ${\check{C}}^0({\mathcal{U}}; G) &= \left\{{\left\{{ \gamma_i: U_i \to G}\right\}_{i\in I}}\right\}\\ {\check{C}}^1({\mathcal{U}}; G) &= \left\{{\left\{{ \phi_{ij}: U_i \cap U_j \to G }\right\}_{i, j \in I}}\right\}\\ {\check{C}}^2({\mathcal{U}}; G) &= \left\{{\left\{{ \eta_{ijk}: U_i \cap U_j \cap U_k \to G }\right\}_{i,j,k\in I}}\right\}$ with boundary maps $$\delta_i: {\check{C}}^{i-1} \to {\check{C}}^i$$: $(\delta_1 \gamma)_{ij} &= \gamma_i \gamma_j^{-1}\\ (\delta_2 \phi)_{ijk} &= \phi_{ij} \phi_{jk} \phi_{ik}^{-1} .$ Note that - $$\delta_2 \circ \delta_1 = 0$$ - $$\ker \delta_2 = Z^1({\mathcal{U}}; G)$$ yields clutching data, i.e. a fiber bundle with fiber $$F$$, - $$\operatorname{im}\delta_1$$ yields trivial bundles, - $${\check{H}}^1({\mathcal{U}}; G) \coloneqq Z^1({\mathcal{U}}; G) / \operatorname{im}({\check{C}}^0({\mathcal{U}}; G) \to Z^1({\mathcal{U}}; G))$$. 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. ::: ::: {.definition title="Refinement of covers"} We say a cover $${\mathcal{V}}\coloneqq\left\{{V_j}\right\}_{j\in J}$$ is a **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 $$j$$. > DZG: I'll write $${\mathcal{V}}\leq {\mathcal{U}}$$ if $${\mathcal{V}}$$ is a refinement of $${\mathcal{U}}$$. ::: ::: {.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 ${\check{H}}^1(B; G) \coloneqq\colim_{{\mathcal{U}}\rightrightarrows B} {\check{H}}^1({\mathcal{U}}; G) .$ ::: ::: {.lemma title="?"} There is a bijection $\left\{{\substack{ \text{Fiber bundles with fiber }F \\ \text{and structure group }G }}\right\}_{/ {\sim}} &\rightleftharpoons {\check{H}}^1(B; G)$ In particular, these classes are independent of $$F$$. ::: ::: {.corollary title="?"} There is an equivalence of categories $\left\{{\substack{ \text{Fiber bundles with fiber F} \\ \text{and structure group }G \\ \text{over }B }}\right\}_{/ {\sim}} &\rightleftharpoons \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G) _{/ {B}} ,$ where the right-hand side are principal $$G{\hbox{-}}$$bundles. ::: ::: {.definition title="$G\\dash$structures"} Given a map $$G \to {\operatorname{Homeo}}(F)$$, a **$$G{\hbox{-}}$$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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwyLCJVX2lcXGludGVyc2VjdCBVX2oiXSxbMiwwLCJHIl0sWzIsMiwiXFxIb21lbyhGKSJdLFswLDIsIlxccGhpX3tpan0iLDJdLFsxLDJdLFswLDEsIlxcdGlsZGUgXFxwaGlfe2lqfSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) ::: ::: {.remark} Note that we need to impose the cocycle condition, since lifts may not be unique and some choices may not glue correctly! ::: ::: {.example title="$\\Spin_n\\dash$structures"} Using the known $${\operatorname{Spin}}$$ double covers, we can form the composition ${\operatorname{Spin}}_n({\mathbb{R}}) \xrightarrow{\times 2} {\operatorname{SO}}_n({\mathbb{R}}) \hookrightarrow{\operatorname{Homeo}}({\mathbb{R}}^n) .$ 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. ::: ::: {.definition title="Fiber products"} We can fill in a commutative square in the following way: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJYIl0sWzIsMiwiQiJdLFsyLDAsIloiXSxbMCwwLCJYXFxmaWJlcnByb2R7Qn1aIixbMjQwLDYwLDYwLDFdXSxbMiwxLCJcXHBpIl0sWzAsMSwiZiIsMl0sWzMsMCwiIiwyLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsMiwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsMSwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d) Here we can construct the fiber product as $X{ \underset{\scriptscriptstyle {B} }{\times} } Z = \left\{{(x, e) {~\mathrel{\Big\vert}~}\pi(e) = f(x)}\right\} .$ It satisfies the following universal property: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiw0LCJYIl0sWzQsNCwiQiJdLFs0LDIsIloiXSxbMiwyLCJYXFxmaWJlcnByb2R7Qn1aIl0sWzAsMCwiVyIsWzAsNjAsNjAsMV1dLFsyLDEsIlxccGkiXSxbMCwxLCJmIiwyXSxbMywwXSxbMywyXSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwyLCIiLDAseyJjb2xvdXIiOlswLDYwLDYwXX1dLFs0LDAsIiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdfV0sWzQsMywiXFxleGlzdHMhIiwwLHsiY29sb3VyIjpbMCw2MCw2MF0sInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX0sWzAsNjAsNjAsMV1dXQ==) ::: ::: {.lemma title="?"} 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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJmXiogcCBcXGRhIFlcXGZpYmVycHJvZHtYfSBQIixbMjQwLDYwLDYwLDFdXSxbMCwyLCJZIixbMjQwLDYwLDYwLDFdXSxbMiwyLCJYIl0sWzIsMCwiUCJdLFswLDEsIlxccHJfMSIsMCx7ImNvbG91ciI6WzI0MCw2MCw2MF19LFsyNDAsNjAsNjAsMV1dLFszLDIsIlxccGkiXSxbMSwyLCJmIiwyXSxbMCwzLCJcXGV4aXN0cyBcXHRpbGRlIGYiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwyLCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) 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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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. ::: ::: {.remark} So the functor $$X\mapsto \mathop{\mathrm{Prin}}_G(X)$$ is contravariant functor. ::: ::: {.theorem title="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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJQXzAiXSxbMiwwLCJcXHJve1B9ezBcXHRpbWVzIEJ9Il0sWzAsMiwiQiJdLFsyLDIsIjBcXHRpbWVzQiJdLFswLDEsIlxcaWQiXSxbMCwyLCJcXHBpIiwyXSxbMiwzLCJcXGlkIiwyXSxbMSwzXV0=) This extends to an isomorphism $$I\times P_0 \to I\times B$$ and $$P\to I\times B$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) ::: ::: {.corollary title="?"} $$P_1 = { \left.{{P}} \right|_{{1\times B}} } \cong P_0$$. ::: ::: {.corollary title="?"} If $$f_0 \sim f_1: Y\to X$$ are homotopic and $$P\to X$$, then $$f_0^*P \cong f_1^* P$$. ::: ::: {.proof title="?"} Use the homotopy lifting property to get a map $$h$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJoXipQIl0sWzAsMiwiSVxcdGltZXMgWSJdLFsyLDAsIlAiXSxbMiwyLCJZIl0sWzEsMywiaCJdLFswLDFdLFsyLDMsIlxccGkiLDJdLFswLDJdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Then $${ \left.{{h^* P}} \right|_{{0\times Y}} } \simeq{ \left.{{h^*P}} \right|_{{1\times Y}} } \cong f_1^*P$$. ::: # Thursday, September 09 ## Corollaries of the homotopy bundle lemma ::: {.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 {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJJXFx0aW1lcyBQXzAiXSxbMCwyLCJJXFx0aW1lcyBYIl0sWzMsMCwiUCJdLFszLDIsIklcXHRpbWVzIFgiXSxbMCwxXSxbMiwzXSxbNCw1LCJmXFxjb25nIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==) where $${ \left.{{f}} \right|_{{0\times P_0}} }$$ is the identity. ::: ::: {.corollary title="?"} If $$P\to I\times X \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)$$ then $$P_0 \cong P_1$$ where $$P_i \coloneqq{ \left.{{P}} \right|_{{i\times X}} }$$. ::: ::: {.corollary title="?"} If $$f_0, f_1: Y\to X$$ with $$P \xrightarrow{\pi} X$$, then $$f_0^* P \cong f_1^* P$$ are isomorphic bundles. ::: ::: {.corollary title="?"} If $$X$$ is contractible, then any $$P\in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)_{/ {X}}$$ is trivial. ::: ::: {.proof title="?"} Consider the two maps {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwxLCJYIl0sWzIsMSwiWCJdLFswLDAsIngiXSxbMiwwLCJ4XzAiXSxbMCwyLCJ4Il0sWzIsMiwieCJdLFswLDEsImZfMSIsMix7Im9mZnNldCI6MiwiY3VydmUiOjF9XSxbMCwxLCJmXzAiLDAseyJvZmZzZXQiOi0zLCJjdXJ2ZSI6LTF9XSxbNCw1LCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV0sWzIsMywiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==) Then $$f_0 \simeq f_1$$, and conclude by noting that $f_0^*P = X{ \underset{\scriptscriptstyle {x_0} }{\times} } P = X\times \pi^{-1}(x_0) = X\times G$ and $$f_1^* P = P$$. ::: ## Existence/Uniqueness of Metrics ::: {.definition title="Riemannian metrics"} A **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. ::: ::: {.proposition title="?"} A Riemannian metric on $$E$$ corresponds to a restriction of the structure group from $$\operatorname{GL}_n({\mathbb{R}})$$ to $${\operatorname{O}}_n({\mathbb{R}})$$. ::: ::: {.proposition title="?"} Every vector bundle over a paracompact $$X$$ has a unique Riemannian metric. ::: ::: {.proof title="?"} **Existence:** Cover $$X$$ by charts and choose a locally finite[^1] 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 $g_i(e_1, e_2) \coloneqq{\left\langle {\mathbf{v}_1},~{\mathbf{v}_2} \right\rangle}_{{\mathbb{R}}^n} .$ Then define $g_p({-}, {-}) \coloneqq\sum_i \chi_i(p) g_i({-}, {-}) .$ **Uniqueness:** Consider two inner products $$g_0({-}, {-}), g_1({-}, {-})$$ on the bundle $$E \xrightarrow{\pi} X$$, then define $g_t({-}, {-}) = tg_0({-}, {-})+ (1-t)g_1({-}, {-}) .$ 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 $P\to I\times X \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n({\mathbb{R}}))$ 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. ::: ::: {.definition title="Universal $G\\dash$bundles"} A **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 *weakly contractible*). ::: ::: {.example title="?"} {=tex} \envlist  - $$\qty{{\mathbb{R}}\to S^1}\in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\mathbb{Z}})_{/ {S^1}}$$ since all of the regular covers are principal bundles. Since $${\mathbb{R}}$$ is contractible, this is the universal $${\mathbb{Z}}{\hbox{-}}$$bundle, so $$S^1 \simeq{\mathbf{B}}{\mathbb{Z}}$$. - $$\qty{S^\infty \to {\mathbb{RP}}^\infty} \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}(C_2)$$ is a universal $$C_2{\hbox{-}}$$bundle, so $${\mathbb{RP}}^\infty \simeq{\mathbf{B}}C_2$$ - $$\qty{S^\infty \to {\mathbb{CP}}^\infty}$$ is a universal $$S^1 = U_1$$ bundle, so $${\mathbb{CP}}^\infty \simeq{\mathbf{B}}U_1 \simeq{\mathbf{B}}S^1 \simeq{\mathbf{B}}{\mathbb{C}}^{\times}$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwxLCJcXENDXm4iXSxbMiwzLCJcXENQXntuLTF9Il0sWzAsMSwiXFxDQ1xcdW5pdHMiXSxbMywwLCJcXHR2e3pfMSwgXFxjZG90cywgel9ufSJdLFszLDMsInogXFxkYSBcXHR2e3pfMTogXFxjZG90cyA6IHpfbn0iXSxbMCwwLCJGX3ogPSBcXHRze1xcbGFtYmRhIHogXFxzdCBcXGxhbWJkYSBcXG5lcSAwfSJdLFsyLDBdLFswLDFdLFszLDQsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbNSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d) ::: ::: {.theorem title="?"} 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 $\mathop{\mathrm{Prin}}{\mathsf{Bun}}(G)_{/ {X}} &\rightleftharpoons[X, {\mathbf{B}}G] \\ f*EG &\mapsfrom f .$ ::: ::: {.lemma title="?"} If $$E \xrightarrow{\pi} X$$ is a fiber bundle with fiber $$F$$ and $$X$$ is weakly contractible then 1. $$\pi$$ admits a section, and 2. Any two sections are homotopic (through other sections). ::: ::: {.proof title="of lemma, part 1"} **Step 1:** build a section inductively. - Define a section over the 0-skeleton arbitrarily. - Inductively, suppose the section is defined on the $$n-1$$ skeleton, so it's defined over every $$n{\hbox{-}}$$cell boundary $${{\partial}}e^n$$. - Write $${ \left.{{E}} \right|_{{e_n}} } = e^n\times F$$, which is contractible since $$e_n$$ is contractible. - Then $$s:{{\partial}}e^n = S^{n-1} \to F$$ with $$\pi_n(F) = 0$$, so the section extends: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwwLCJcXHJve0V9e2Vebn0iXSxbNCwwLCJlXm5cXHRpbWVzIEYiXSxbMiwxXSxbMiwyLCJlXm4iXSxbMCwyLCJcXGJkIGVebiJdLFswLDNdLFswLDEsIlxcY29uZyJdLFsxLDMsIlxccHJfMSJdLFs0LDNdLFs0LDAsInMiXSxbMywwLCJcXGV4aXN0cyBcXHRpbGRlIHMiLDAseyJjdXJ2ZSI6LTIsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) **Step 2:** Build the homotopy between sections inductively cell-by-cell as in part (1). ::: ::: {.proof title="of theorem"} We want to show that the assignment $$f\mapsto f^* EG$$ is bijective. **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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCJQIl0sWzIsMiwiWCJdLFs2LDAsIlBcXGZpYmVycHJvZHtHfSBFRyJdLFs2LDIsIlgiXSxbNCwwLCJFRyJdLFszLDFdLFs0LDFdLFs1LDFdLFswLDAsIkYiXSxbMCwxLCJcXHBpIl0sWzIsM10sWzIsM10sWzQsMl0sWzUsNywiXFx0ZXh0e21peGluZ30iLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJzcXVpZ2dseSJ9fX1dLFs4LDBdXQ==) 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 $P\to EG\\ p\mapsto e ,$ 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 $$EG/G = {\mathbf{B}}G$$. {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbNiwwLCJFRyJdLFs2LDIsIlxcQiBHIl0sWzQsMiwiWCJdLFswLDAsIlAiXSxbNCwwLCJoXiogRUciXSxbNSwxXSxbMiwxLCJoIl0sWzAsMV0sWzQsMF0sWzQsMl0sWzMsMCwiZiIsMCx7ImN1cnZlIjotNX1dLFszLDQsIlxcZXhpc3RzIHAgXFxtYXBzdmlhe1xcc2ltfSAoXFxwaShwKSwgZihwKSkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMywyLCJcXHBpIiwyXV0=) ::: {.exercise title="?"} Show that this map is an isomorphism. ::: ::: # Universal Bundles (Thursday, September 16) ::: {.definition title="Universal $G\\dash$bundles"} A **universal $$G{\hbox{-}}$$bundle** is a principal $$G{\hbox{-}}$$bundle $$EG \xrightarrow{\pi} {\mathbf{B}}G$$ such that $$EG$$ is weakly contractible, i.e. $$\pi_*(EG) = 0$$. ::: ::: {.remark} We looked at a theorem stating the correspondence $\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}} \rightleftharpoons[X, {\mathbf{B}}G] .$ ::: ::: {.proof title="of surjectivity in theorem, continued"} We showed surjectivity of the following map: $[X, {\mathbf{B}}G] &\twoheadrightarrow\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }_{/ {X}} \\ (f \in [X, {\mathbf{B}}G]) &\mapsto f^* EG .$ 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 $$h^* EG \cong P$$. ::: ::: {.proof title="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 $$s_0$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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 {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwyLCJoXzBeKiBFRyJdLFsyLDIsImhfMV4qIEVHIl0sWzQsMiwiRUciXSxbMCwwLCIoeCwgeSkiXSxbMiwwLCIoeCwgXFxwaGkoeSkpIl0sWzQsMCwiXFxwaGkoeSkiXSxbMiw0LCJYIl0sWzQsNCwiXFxCIEciXSxbMSwyLCJmXzEiXSxbMCwyLCJmXzFcXGNpcmMgXFxwaGkiLDAseyJjdXJ2ZSI6LTV9XSxbMCwxLCJcXHBoaSJdLFsyLDddLFsxLDZdLFs2LDddLFswLDZdXQ==) So we have $s_0(x) &\coloneqq[x,y,y] \\ s_1(x) &\coloneqq[x,y,\phi(y)] .$ By the lemma, $$s_0\simeq s_1$$ through sections, so there is a homotopy $s: I\times X &\to I \times h_0^* EG\overset{\scriptscriptstyle {G} }{\times} EG \\ (t, x) &\mapsto (t, x, y, z) .$ But this is a section of a principle $$EG{\hbox{-}}$$bundle over a CW complex, which yields a $$G{\hbox{-}}$$equivariant map $f: I \times h_0^* EG &\to EG \\ (t, x, y) &\mapsto z .$ Then - At $$t=0$$, we have $$(0, x)\mapsto (0, x,y,y)$$, so $$f(0, x, y) = y$$, - At $$t=1$$, we have $$(1, x)\mapsto (1, x,y,\phi(y))$$, so $$f(1,x,y) = \phi(y)$$. Since $$f$$ is $$G{\hbox{-}}$$equivariant, we can quotient both sides by $$G$$ to get a map $h: I\times X &\to {\mathbf{B}}G \\ (0, x) &\mapsto h_0(x) \\ (1, x) &\mapsto h_1(x) .$ ::: ::: {.exercise title="?"} Try this proof yourself! ::: ::: {.remark} The same proof shows the following: ::: ::: {.lemma title="?"} 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. ::: ::: {.proposition title="?"} 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 $$h \mapsto h^* P$$. ::: ## Existence of Universal Bundles ::: {.theorem title="Milnor, 1966"} For any group $$G \in {\mathsf{Top}}{\mathsf{Grp}}$$, there is a universal bundle $$EG\to {\mathbf{B}}G$$. ::: ::: {.remark} Uniqueness up to homotopy: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwLCJFIl0sWzEsMiwiQiJdLFszLDAsIkUnIl0sWzMsMiwiQiciXSxbMCwwLCJoXypFJyJdLFs0LDAsImgnXypFIl0sWzAsMiwiZiJdLFsxLDMsImgiLDJdLFswLDFdLFsyLDNdLFsyLDAsImYnIiwyLHsiY3VydmUiOjV9XSxbMywxLCJoJyIsMCx7ImN1cnZlIjotNX1dLFs0LDAsIlxcc2ltIiwyXSxbNSwyLCJcXHNpbSIsMl1d) 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 $$B \simeq B'$$. ::: ::: {.exercise title="?"} Show $$E \simeq E'$$. ::: ::: {.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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJLKEcsIDEpIl0sWzIsMiwiWCJdLFsyLDAsIlxcdGlsZGUgWFxcaG9tb3RvcGljIFxccHQiXSxbMywxLCJcXGluIFxcUHJpblxcQnVuZ1xcc2xpY2UgWCBcXHRleHR7IHVuaXZlcnNhbH0iXSxbMiwxLCJcXHBpIl0sWzAsMV1d) ::: ::: {.definition title="Nerve of a category"} Given a category $$\mathsf{C}$$, the **nerve** $${ \mathcal{N}({\mathsf{C}}) }$$ is the following $$\Delta{\hbox{-}}$$complex: - $$0{\hbox{-}}$$simplices are objects of $$\mathsf{C}$$ - $$n{\hbox{-}}$$simplices for $$n\geq 1$$ are sequences of composable morphisms $x_0 \xrightarrow{f_0} x_1 \xrightarrow{f_1} \cdots \xrightarrow{f_{n-1}} x_n .$ - Gluing data for 1-simplices: for $$x_0 \xrightarrow{f} x_1$$, set $${{\partial}}_1(f) = x_1, {{\partial}}_0(f) = x_0$$. - Gluing data for $$n{\hbox{-}}$$simplices: the $$i$$th boundary maps are given by dropping vertex $$i$$: ${{\partial}}_i(f_1, \cdots, f_n) = \begin{cases} (f_2, f_3, \cdots, f_n) & i=0 \\ (f_1, f_2, \cdots, f_{i+1} \circ f_{i}, \cdots, f_n) & i=0 \\ (f_1, f_2, \cdots, f_{n-1}) & i=n. \end{cases}$ {=tex} \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}  ::: # Thursday, September 16 ::: {.remark} Let $$G\in {\mathsf{Grp}}$$, and consider the following two categories. $${\mathbf{B}}G$$ will be the category: - $${\operatorname{Ob}}({\mathbf{B}}G) = \left\{{{\operatorname{pt}}}\right\}$$. - $$\mathop{\mathrm{Mor}}_{{\mathbf{B}}G}({\operatorname{pt}}, {\operatorname{pt}}) = \left\{{g\in G}\right\}$$, 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. $$E G$$ will be the category: - $${\operatorname{Ob}}(EG) = \left\{{g\in G}\right\}$$, one object for each element of $$G$$, - $$\mathop{\mathrm{Mor}}(g, h)=\left\{{g^{-1}h}\right\}$$, a single (conveniently labeled!) morphism for each ordered pair $$(g, h)$$. Note that $$G\curvearrowright EG$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJnXzAiXSxbMiwwLCJnXzEiXSxbMCwyLCJnZ18wIl0sWzIsMiwiZ2dfMSJdLFswLDEsImdfMFxcaW52IGdfMSJdLFsyLDMsImdcXGNkb3QgZ18wXFxpbnYgZ18xIl0sWzQsNSwiIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==) 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 $g_0 \xrightarrow{g_0^{-1}g_1} g_1 \xrightarrow{g_1^{-1}g_2} g_2 \to \cdots \to g_n .$ Acting on this by $$G$$ yields $gg_0 \xrightarrow{g_0^{-1}g_1} gg_1 \xrightarrow{g_1^{-1}g_2} gg_2 \to \cdots \to gg_n ,$ noting we leave the morphism labeling unchanged, and that uniqueness of morphisms makes the simplicial boundary map behave nicely. ::: ::: {.exercise title="?"} Show that ${ \mathcal{N}({EG}) }/G = { \mathcal{N}({{\mathbf{B}}G}) } .$ ::: ::: {.remark} Note that ${ \mathcal{N}({{\mathbf{B}}G}) } &= \Delta^0 {\textstyle\coprod}\Delta^1 \times G {\textstyle\coprod}\Delta^2 \times G{ {}^{ \scriptscriptstyle\times^{2} } } {\textstyle\coprod}\Delta^3 \times G{ {}^{ \scriptscriptstyle\times^{3} } }\cdots \\ { \mathcal{N}({E G}) } &= \Delta^0 \times G {\textstyle\coprod}\Delta^1 \times G{ {}^{ \scriptscriptstyle\times^{2} } } {\textstyle\coprod}\Delta^2 \times G{ {}^{ \scriptscriptstyle\times^{3} } }\cdots ,$ where the gluing data for $${ \mathcal{N}({{\mathbf{B}}G}) }$$ is given by ${{\partial}}_n: \Delta^n \times G{ {}^{ \scriptscriptstyle\times^{n} } } &\to \Delta^{n-1} \times G{ {}^{ \scriptscriptstyle\times^{n-1} } } \\ (\mathbf{x}, \mathbf{g}) &\mapsto (\mathbf{x}\setminus\left\{{x_n}\right\}, \mathbf{g} \setminus\left\{{g_n}\right\})$ and for $${ \mathcal{N}({EG}) }$$ is ${{\partial}}_n: \Delta^n \times G{ {}^{ \scriptscriptstyle\times^{n+1} } } &\to \Delta \times G{ {}^{ \scriptscriptstyle\times^{n} } } \\ (\mathbf{x}, \mathbf{g}) &\mapsto (\mathbf{x}\setminus\left\{{x_n}\right\}, \mathbf{g} \setminus\left\{{g_n}\right\}) .$ The action $$G\curvearrowright EG$$ is the following: $g\cdot (\mathbf{x}, \mathbf{g}) \mapsto (\mathbf{x}, {\left[ {gg_0, gg_1,\cdots, gg_n} \right]} ) .$ ::: ::: {.example title="?"} 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 $[(0, x_1, x_2), (a, a^2)] &\sim [(x_1, x_2), (a^2)] \\ [(x_0, 0, x_2), (a, a^2)] &\sim [(x_0, x_2), (a)] \\ [(x_0, x_1, 0), (a, a^2)] &\sim [(x_0, x_1), (a)] \\ .$ 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: {=html}  ![](figures/2021-09-16_16-59-53.png) ::: ::: {.lemma title="?"} If $$\mathsf{C}$$ has an initial or terminal object, then $${ \mathcal{N}({\mathsf{C}}) }$$ is contractible. ::: ::: {.remark} This clearly holds for $$E G$$, since every object is initial and terminal. ::: ::: {.proof title="?"} 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: $$x_0 \xrightarrow{f_0} x_1 \cdots \to x_n {\color{green} \xrightarrow{f_{x_n}} y} \leadsto y$$. ::: ::: {.corollary title="?"} $${ \mathcal{N}({EG}) }$$ is contractible, and the quotient $${ \mathcal{N}({EG}) }\to { \mathcal{N}({{\mathbf{B}}G}) }$$ is a universal $$G{\hbox{-}}$$bundle. ::: ::: {.exercise title="?"} Construct $$E G$$ and $${\mathbf{B}}G$$ for $$G = C_4, C_2{ {}^{ \scriptscriptstyle\times^{2} } }$$ and explicitly compare their 3-skeleta. ::: # Tuesday, September 21 ::: {.remark} Today: a short discussion on generalizations of $${\mathbf{B}}G$$ to topological groups. ::: ::: {.definition title="Topological categories"} A **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: - $$\mathrm{source}, \mathrm{target}: \mathop{\mathrm{Mor}}_{\mathsf{C}} \to {\operatorname{Ob}}(\mathsf{C})$$, - $$\operatorname{id}: {\operatorname{Ob}}(\mathsf{C})\to \mathop{\mathrm{Mor}}_{\mathsf{C}}$$, - Composition: $$\mathsf{C}(x, y) \times \mathsf{C}(y,z) \to \mathsf{C}(x, z)$$. I.e. it is a category enriched over topological spaces (plus conditions). ::: ::: {.example title="?"} 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. ::: ::: {.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 $${ \mathcal{N}({{\mathcal{E}}G}) }$$. ::: ::: {.definition title="Absolute Neighborhood Retract"} A space is called an **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 $$Y$$. ::: ::: {.example title="?"} Every CW complex is an ANR. This is also true if every point of $$X$$ has a contractible neighborhood. ::: ::: {.lemma title="?"} If $$G$$ is ANR, then $$EG = { \mathcal{N}({{\mathcal{E}}G}) }\to { \mathcal{N}({{\mathcal{B}}G}) } = {\mathbf{B}}G \in \mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }$$. ::: ::: {.proof title="?"} 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 $\Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } } \times G &\to E G \\ ( (x_0,\cdots, x_i), (g_1, \cdots, g_i), g) &\mapsto (\operatorname{id}(\cdots), (g, gg_1, gg_1g_2, \cdots, gg_1\cdots g_i)) ,$ which corresponds to the sequence of composable morphisms $(g \xrightarrow{g_1} gg_1 \xrightarrow{g_2} g g_1 g_2 \to \cdots \to gg_1\cdots g_i) .$ ::: {.exercise title="?"} Show that this is not compatible with the gluing! ::: 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 $$\pi^{-1}(U_{m+1}) \to U_{m+1}$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) This extends the chart to $${{\mathbf{B}}G}^{(m+1)}$$, noting that $$p$$ is a quotient map and thus preserves open sets. ::: ::: {.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 $p^{-1}(U_m) \subseteq \displaystyle\bigcup_{i\leq m} \Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } } \subseteq \displaystyle\bigcup_{i\leq m-1} \Delta^i \times G{ {}^{ \scriptscriptstyle\times^{i} } } .$ 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. ::: ## Building $${{\mathbf{B}}{\operatorname{O}}}_n$$ and $${\mathsf{E} {\operatorname{O}}}_n$$ {#building-mathbfboperatornameo_n-and-mathsfe-operatornameo_n} ::: {.remark} We'll assume all spaces paracompact from this point forward! We have a correspondence $\left\{{\substack{ \text{n{\hbox{-}}dimensional CW complexes } }}\right\} \rightleftharpoons \left\{{\substack{ \text{n{\hbox{-}}dimensional vector bundles } \\ \text{with an {\operatorname{O}}_n{\hbox{-}}structure} }}\right\} \rightleftharpoons \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)_{/ {X}} \rightleftharpoons [X, {\mathbf{B}}{\operatorname{O}}_n]$ 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 $\qty{\mathbf{v}_1, \cdots, \mathbf{v}_n} \cdot A = \qty{\sum_i a_{i, 1} \mathbf{v}_i, \sum_i a_{i, 2} \mathbf{v}_i, \cdots, \sum_i a_{i, n} \mathbf{v}_i} .$ There is a projection {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwwLCJWX24oXFxSUl5rKSJdLFsyLDIsIlZfMShcXFJSXmspIl0sWzMsMCwiKFxcdmVjdG9yIHZfMSwgXFxjZG90cywgXFx2ZWN0b3Igdl8xKSJdLFszLDIsIlxcdmVjdG9yIHZfMSJdLFswLDAsIkZfe1xcdmVjdG9yIHZfMX0gPSBWX3tuLTF9KFxcUlJee2stMX0pIl0sWzQsMF0sWzAsMV0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==) We'll use the fact that $$V_1({\mathbb{R}}^k)$$ is $$(k-2){\hbox{-}}$$connected, since it's homotopy equivalent to $$S^{k-1}$$. ::: ::: {.lemma title="?"} $$V_n({\mathbb{R}}^{k})$$ is $$(k-n-1){\hbox{-}}$$connected. ::: ::: {.proof title="?"} Induct on $$n$$ using the homotopy LES for the fiber bundle: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) ::: ::: {.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 **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 $$k$$. ::: ::: {.lemma title="?"} $\pi_* V_n({\mathbb{R}}^\infty) = 0 .$ ::: ::: {.proof title="?"} 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. ::: ::: {.definition title="?"} $V_n({\mathbb{R}}^\infty) / {\operatorname{O}}_n = {\operatorname{Gr}}_n({\mathbb{R}}^\infty) .$ ::: ::: {.remark} It will turn out that $${\mathsf{E} {\operatorname{O}}}_n = V_n({\mathbb{R}}^\infty)$$, sometimes referred to as the *Stiefel manifold* of $$n{\hbox{-}}$$frames in $${\mathbb{R}}^\infty$$. ::: # Thursday, September 23 ::: {.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 $${\mathsf{E} {\operatorname{O}}}_n \to {{\mathbf{B}}{\operatorname{O}}}_n$$. ::: ::: {.lemma title="?"} $$V_n({\mathbb{R}}^\infty) \xrightarrow{\pi} {\operatorname{Gr}}_n({\mathbb{R}}^\infty) \in \mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)$$. ::: ::: {.proof title="?"} 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 $U_W \coloneqq\left\{{W' \in {\operatorname{Gr}}_n({\mathbb{R}}^\infty) {~\mathrel{\Big\vert}~}W^\perp \cap W' = 0}\right\} .$ For any such $$W'$$, we have a map $$W'\to W$$ given by orthogonal projection, which is an isomorphism since $$W^\perp \cap W' = 0$$. ::: {.claim} $\pi^{-1}(U_W) \cong U_W \times {\operatorname{O}}_n .$ ::: 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 $U_w \times {\operatorname{O}}_n &\to \pi^{-1}(W) \\ (W', A) &\mapsto F_{W'}( \alpha) \cdot A .$ The claim is that this trivializes the bundle, since this constructs a local section using $${\operatorname{O}}_n$$ translations: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwxLCJcXHBpXFxpbnYoVV9XKSJdLFsyLDEsIlVfdyBcXHRpbWVzIFxcT3J0aF9uIl0sWzIsMCwiKFcnLCBBKSJdLFswLDAsInMoVycpXFxjZG90IEEiXSxbMCwzLCJVX1ciXSxbMiwzLCJXJyJdLFswLDQsIlxccGkiXSxbNCwwLCJzIiwyLHsiY3VydmUiOjN9XSxbMSwwLCJcXGNvbmciXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) 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 $$W'$$. ::: ::: {.remark} {=tex} \envlist  - Replace $${\operatorname{O}}_n$$ with $$U_n$$ and $${\mathbb{R}}$$ with $${\mathbb{C}}$$ to get $${\operatorname{Gr}}_n({\mathbb{C}}^\infty) = {\mathbf{B}}U_n$$. - $$V_n({\mathbb{R}}^\infty)/ {\operatorname{SO}}_n = {\mathbf{B}}{\operatorname{SO}}_n$$ yields the Grassmannian of *oriented* planes. - For $$H\leq G$$, we have $${\mathbf{E}}H = {\mathbf{E}}G$$ and $${\mathbf{B}}H = {\mathbf{E}}G/H$$. ::: > Question: can you get $${\mathbf{B}}{\operatorname{Spin}}_n$$ this way? ::: {.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 **canonical bundle** $$\gamma$$. Recall that every principal $${\operatorname{O}}_n$$ bundle is a pullback of the following form: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJQID0gZl4qIFxcRU9fbiJdLFswLDIsIlgiXSxbMiwwLCJcXEVPX24gPSBWX24oXFxSUl5cXGluZnR5KSJdLFsyLDIsIlxcQk9fbiJdLFsxLDMsImYiXSxbMCwxXSxbMCwyLCIiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiwzXV0=) 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: $$\gamma_n = V_n({\mathbb{R}}^\infty) \overset{\scriptscriptstyle {{\operatorname{O}}_n} }{\times} {\mathbb{R}}^n \to {\operatorname{Gr}}_n({\mathbb{R}}^\infty)$$. {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwxLCJWX24oXFxSUl5cXGluZnR5KSBcXHRpbWVzIFxcUlJebiJdLFsyLDEsIlxcZ2FtbWFfbiJdLFswLDAsIltcXHZlY3RvciB2XzEsIFxcY2RvdHMsIFxcdmVjdG9yIHZfbl0gXFx0aW1lcyBbdF8xLFxcY2RvdHMsIHRfbl0iXSxbMiwwLCJcXHN1bSB0X2kgXFx2ZWN0b3Igdl9pIl0sWzAsMywiXFxHcl9uKFxcUlJeXFxpbmZ0eSkiXSxbMCw0LCJcXHBpJyJdLFsxLDQsIlxccGkiLDJdLFswLDEsIlxcZXhpc3RzIFxcY29uZyIsMl0sWzIsMywiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==) ::: ::: {.definition title="Subbundles"} $$E' \leq E$$ is a **subbundle** iff there is an embedding $$E' \hookrightarrow E$$ over $$X$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJFJyJdLFsyLDAsIkUiXSxbMSwyLCJYIl0sWzAsMiwiXFxwaSciLDJdLFsxLDIsIlxccGkiXSxbMCwxLCJmIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=) 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 $$\dim_{\mathbb{R}}{ \left.{{E}} \right|_{{x}} } = m$$. ::: ::: {.lemma title="?"} Every vector bundle $$E \xrightarrow{\pi} X$$ with $$X\in {\mathsf{CW}}$$ compact is a subbundle of a trivial bundle. ::: ::: {.proof title="?"} 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 $\psi: E &\to {\mathbb{R}}^{nm} = {\mathbb{R}}^n \times {\mathbb{R}}^n \times \cdots \times {\mathbb{R}}^n \\ v &\mapsto {\left[ {\chi_1(v) \phi_1(v), \chi_2(v) \phi_2(v), \cdots, \chi_m(v) \phi_m(v)} \right]} .$ This exhibits $$(E \to X) \leq (X\times {\mathbb{R}}^{nm} \to X)$$ as a subbundle. ::: ::: {.lemma title="?"} 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 $$f:X\to {{\mathbf{B}}{\operatorname{O}}}_n$$. ::: ::: {.example title="?"} 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 $$f^* \gamma_n \cong E$$. ::: ::: {.lemma title="?"} If $$f^* \gamma_n \cong E$$ and $$g^* \gamma_n \cong E$$, then $$f \simeq g$$. ::: ::: {.proof title="?"} 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 $L_t: {\mathbb{R}}^\infty &\to {\mathbb{R}}^\infty \\ \mathbf{x} &\mapsto (t-1){\left[ {x_1,x_2,\cdots} \right]} + t{\left[ {x_1, 0, x_2, 0, x_3, 0, \cdots} \right]} ,$ 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"). ::: # Vector Bundle Classification Theorem (Tuesday, September 28) > See homework posted on the website! Turn in 2 total problem sets, one by mid-October. ::: {.theorem title="?"} $[X, {{\mathbf{B}}{\operatorname{O}}}_n] \cong [X, {\operatorname{Gr}}_n({\mathbb{R}}^\infty)] &{ { \, \xrightarrow{\sim}\, }}{ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X) \\ f & \mapsto f^* \gamma_n .$ ::: ::: {.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. ::: ::: {.lemma title="?"} 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 $$h^* \gamma_n \cong E$$. {=tex} \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}  ::: ::: {.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 $$f_n^* \gamma_i = { \left.{{E}} \right|_{{B_n}} }$$. ::: ::: {.proof title="?"} 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 $${ \left.{{h}} \right|_{{B}} } = f$$. ::: ## Characteristic Classes ::: {.definition title="Characteristic classes and representability"} Let $$F, G$$ be two contravariant functors with source $$\mathsf{C}$$. A **characteristic class of $$F$$ valued in $$\mathsf{G}$$** is a natural transformation $$c: F\to G$$. $$F$$ is **representable** if there exists an object $${\mathbf{B}}F$$ such that $$F(X) = [X, {\mathbf{B}}F]$$ for every $$X\in {\operatorname{Ob}}(\mathsf{C})$$. > Note: we aren't requiring the target categories to coincide! ::: ::: {.example title="?"} {=tex} \envlist  - $$F(X) \coloneqq\mathop{\mathrm{Prin}}{\mathsf{Bun}}({\operatorname{O}}_n)_{/ {X}} = { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} } {}_{/ {X}} = [X {{\mathbf{B}}{\operatorname{O}}}_n]$$ is a contravariant functor $${\mathsf{ho}}{\mathsf{CW}}^{\operatorname{op}}\to {\mathsf{Set}}$$, where contravariance is due to pullbacks. - $$G(X) \coloneqq H^j_{\mathrm{sing}}(X)$$, 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 $$f^* \alpha$$. ::: ::: {.lemma title="?"} If $$F = [{-}, {\mathbf{B}}F]$$ is representable, then characteristic classes of $$F$$ valued in a functor $$G$$ biject with $$G(B)$$ ::: ::: {.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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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 $$c(f) = f^*\alpha \in H^j(X)$$. ::: ::: {.example title="?"} 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 $$f\in [X, {\operatorname{BU}}_n]$$. ::: ::: {.example title="?"} 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 $H^n(X; M) &\to H^m(X; N) \\ (f: X\to K(M, n)) &\mapsto f^* \alpha ,$ i.e. a cohomology operation. If for example $\phi \in \mathop{\mathrm{Hom}}_{\mathsf{Grp}}(M, N) = \mathop{\mathrm{Hom}}_{\mathsf{Grp}}( H_n(K(M, n); {\mathbb{Z}}), N) = H^n( K(M, n); N) ,$ using that $$H_n(K(M, n); {\mathbb{Z}})\cong M$$. This yields a change of coefficient morphism $H^n(X; M) \to H^n(X; N) ,$ 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 $$H^n(X; M)\to H^n(X; N)$$. ::: # Thursday, September 30 ## Line Bundles: Chern and Stiefel-Whitney Classes ::: {.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 $$i=n$$. ::: ::: {.example title="?"} {=tex} \envlist  - $${ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}}, X) { { \, \xrightarrow{\sim}\, }}[X, {{\mathbf{B}}{\operatorname{O}}}_n]$$ 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 $$f^*(\alpha) \in H^j(X; {\mathbb{Z}})$$. - $${ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}}, X) { { \, \xrightarrow{\sim}\, }}[X, {\operatorname{BU}}_n]$$, where $${\operatorname{BU}}_n \simeq{\operatorname{Gr}}_n({\mathbb{C}}^\infty)$$ and we can again pullback cohomology classes. ::: ::: {.example title="?"} 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 **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 $[X, {\mathbb{CP}}^\infty] &{ { \, \xrightarrow{\sim}\, }}H^2(X; {\mathbb{Z}}) \\ f &\mapsto f^*(\alpha),\quad \left\langle{\alpha}\right\rangle = H^2({\mathbb{CP}}^\infty) .$ So the set of bundles is an affine space over $$H^2(X)$$. ::: ::: {.corollary title="?"} There is a bijection $c_1: { {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{R}}, X) { { \, \xrightarrow{\sim}\, }}H^2(X) ,$ ::: ::: {.example title="?"} 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 **first Stiefel-Whitney class** $$w_1$$ and a bijection $w_1: [X, {\mathbb{RP}}^\infty] &{ { \, \xrightarrow{\sim}\, }}H^1(X; {\mathbb{Z}}/2) \\ f &\mapsto f^*(\alpha) .$ ::: ::: {.remark} We can define $$c_1$$ for vector bundles of any dimension by taking a top exterior power to get a line bundle: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCJcXFZlY3RCdW5kbGVfbihcXENDLCBYKSJdLFsyLDAsIlxcVmVjdEJ1bmRsZV8xKFxcQ0MsIFgpIl0sWzAsMSwiRSJdLFswLDMsIlgiXSxbMiwxLCJcXEV4dGFsZ15uIEUiXSxbMiwzLCJYIl0sWzQsMCwiSF4yKFg7IFxcWlopIl0sWzAsMSwiXFxFeHRhbGdebihcXHdhaXQpIl0sWzIsM10sWzQsNV0sWzEsNiwiY18xIl0sWzQsNiwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFsyLDQsIlxcRXh0YWxnXm4oXFx3YWl0KSIsMl0sWzIsNiwiXFxkYSBjXzEgRSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDUsIlxcaWQiLDJdXQ==) So $c_1(E) \coloneqq c_1(\bigwedge\nolimits^n E) .$ ::: ::: {.remark} There is a natural isomorphism $$c(f^*(E)) \cong f^*(c(E))$$, since we can take iterated pullbacks: {=tex} \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}  So we can identify $$c(f^*(E)) = ( g\circ f)^* \alpha$$ and $$f^*(c(E)) = f^*(g^* \alpha) = (g\circ f)^* \alpha$$. ::: ## Euler Classes and the Thom Isomorphism ::: {.remark} Note that any vector bundle with a Riemannian metric admits a unit disc bundle. ::: ::: {.definition title="Oriented disc bundles"} A unit disc bundle $$D \xrightarrow{\pi} B$$ is **oriented** if there is a locally coherent choice of a generator of $$H^n(D_b \coloneqq\pi^{-1}(b), S^b \coloneqq{{\partial}}D_b; {\mathbb{Z}})$$. ::: ::: {.example title="?"} 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. ::: ::: {.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. ::: ::: {.theorem title="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 1. $$H^{i< n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) = 0$$ 2. There exists a generator, the **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 $({-}) \smile u_E : H^j( {\mathbb{D}}E) &{ { \, \xrightarrow{\sim}\, }}H^{j+n}({\mathbb{D}}E, {\mathbb{S}}E) && \forall j\geq 0 \\ \eta &\mapsto \eta \smile u_E ,$ ::: ::: {.proof title="of theorem"} {=tex} \envlist  - **Step 1**: Look locally to see why we might expect this result! If $${\mathbb{D}}E\to X$$ is trivial, then the claims hold. - **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 $$X$$. - **Step 3**: Prove claims for $$H^*({-}; k)$$ for $$k\in \mathsf{Field}$$ - **Step 4**: Prove claims for $$H^*({-}; {\mathbb{Z}})$$. ::: ::: {.proof title="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 $H^k({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) &= H^k({\mathbb{D}}^n\times X, S^{n-1} \times X; {\mathbb{Z}}) \\ &\cong \bigoplus_{0\leq i\leq k} H^i(X; {\mathbb{Z}}) \otimes_{\mathbb{Z}}H^{k-i}({\mathbb{D}}^n, S^{n-1}; {\mathbb{Z}}) \\ &\cong \bigoplus_{0\leq i\leq k} H^i(X; {\mathbb{Z}}) \otimes_{\mathbb{Z}}\qty{{\mathbb{Z}}[i+n ]} \\ &= H^{k-n}(X; {\mathbb{Z}}) & \quad \text{ when } k>n, \text{ else } 0 .$ 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). ::: ::: {.proof title="step 2"} Use Mayer-Vietoris: {=html}  ![](figures/2021-10-03_21-35-21.png) > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwyLCJIXmsoXFxERCBFXFxtaWRfe1UgXFx1bmlvbiBWfSwgXFxTUyBFXFxtaWRfe1UgXFx1bmlvbiBWfSApIl0sWzIsMiwiSF5rKFxcREQgRVxcbWlkX3tVfSwgXFxTUyBFXFxtaWRfe1V9ICkgXFxvcGx1cyBIXmsoXFxERCBFXFxtaWRfe1Z9LCBcXFNTIEVcXG1pZF97Vn0gKSJdLFs0LDIsIkheayhcXEREIEVcXG1pZF97VSBcXGludGVyc2VjdCBWfSwgXFxTUyBFXFxtaWRfe1UgXFxpbnRlcnNlY3QgVn0gKSJdLFs0LDAsIkhee2stMX0oXFxERCBFXFxtaWRfe1UgXFxpbnRlcnNlY3QgVn0sIFxcU1MgRVxcbWlkX3tVIFxcaW50ZXJzZWN0IFZ9ICkiXSxbMCw0LCJcXGNkb3RzIl0sWzAsMV0sWzEsMl0sWzMsMF0sWzIsNF1d) - If $$k< n$$, the union terms vanish in degree $$k$$, since they're surrounded by zeros. - 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 $${ \left.{{u_E}} \right|_{{V}} }$$. Next time: we'll show that $$u_{U \cup V} \smile({-})$$ yields the isomorphism in part 2 of the theorem statement. ::: # Tuesday, October 05 ::: {.remark} Recall that we were proving the Thom isomorphism theorem. Some motivation: ::: ::: {.corollary title="The Gysin Sequence"} Consider an oriented sphere bundle: {=tex} \begin{tikzcd} S^{n-1} \ar[r] & {\mathbb{S}}E \ar[d] \\ & X \end{tikzcd}  Then there is a LES induced by the Euler class $$e\in H^n(X)$$ {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbNCwzXSxbMCwyLCJIXntqLW59KFgpIl0sWzQsMiwiSF57an0oXFxTUyBFKSJdLFsyLDIsIkheaihYKSJdLFs0LDQsIkhee2otMX0oXFxTUyBFKSJdLFswLDAsIlxcY2RvdHMiXSxbMiw0LCJcXGNkb3RzIl0sWzQsMV0sWzEsMywiZVxcY3VwcHJvZChcXHdhaXQpIl0sWzMsMl0sWzIsNV0sWzYsNF1d) ::: ::: {.proof title="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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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. ::: ::: {.proof title="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 $H^i(X; k) &\cong H_i(X; k) {}^{ \vee }\\ &\coloneqq\mathop{\mathrm{Hom}}( H_i(X; k), k) \\ &= \mathop{\mathrm{Hom}}( \colim_j H_i(C_j;k); k) \\ &= \cocolim_j \mathop{\mathrm{Hom}}( H_i(C_j; k), k) \\ &= \cocolim_j H^i(C_j; k) .$ Similarly, $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 [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfTsgaykiXSxbMywyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfSwgXFxyb3tcXFNTIEV9eyBDX2p9IDsgaykiXSxbMCwwLCJIXmkoXFxERCBFOyBrKSJdLFszLDAsIkhebShcXEREIEUsIFxcU1MgRTsgaykiXSxbMCwxLCJcXHJve3VfRX17XFxERCBFfVxcY3VwcHJvZChcXHdhaXQpLCBcXGNvbmciXSxbMiwzLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwwXSxbMywxXV0=) Now use that the vertical maps become isomorphisms after a colimit, so the top map must become an isomorphism as well. ::: ::: {.proof title="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 \[ 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 ,$ since each summand will be zero. For $$i=n$$, the Ext term vanishes, and we have $H^n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) &\cong \mathop{\mathrm{Hom}}( H_n({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}), {\mathbb{Z}}) \\ &\cong \mathop{\mathrm{Hom}}( \colim_j H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}), {\mathbb{Z}}) \\ &\cong \cocolim_j \mathop{\mathrm{Hom}}( H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}), {\mathbb{Z}}) \\ &\cong \cocolim_j H_n( { \left.{{ {\mathbb{D}}E}} \right|_{{C_j}} } , { \left.{{ {\mathbb{S}}E}} \right|_{{C_j}} }; {\mathbb{Z}}) \\ &\cong \left\langle{u_E}\right\rangle \cong {\mathbb{Z}} ,$ using the distinguished generator $$u_E \in H^n({\mathbb{D}}E, {\mathbb{S}}E)$$. So we can define a chain map $u_E \frown({-}): C_{j+n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}}) \to C_j({\mathbb{D}}E) ,$ which shifts degree by $$-n$$ by capping against $$u_E$$. This induces the cup product $$u_E \smile({-}):H^j({\mathbb{D}}E; {\mathbb{Z}}) \to H^{j+n}({\mathbb{D}}E, {\mathbb{S}}E; {\mathbb{Z}})$$. ::: {.lemma title="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 $$R\in \mathsf{Ring}$$. ::: 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. ::: ::: {.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 {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJFIl0sWzIsMiwiWCJdLFswLDIsIlkiXSxbMCwwLCJmXiogRSJdLFszLDBdLFswLDFdLFsyLDFdLFszLDJdLFszLDEsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Then $c(f^* E) = f^*c(E) .$ Note that we can always pull back the canonical: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJmXiogXFxnYW1tYV9uIl0sWzIsMCwiXFxnYW1tYV9uIl0sWzIsMiwiXFxHcl9uKFxcUlJeXFxpbmZ0eSkiXSxbMCwyLCJYIl0sWzEsMl0sWzMsMiwiZiIsMl0sWzAsMV0sWzAsM10sWzAsMiwiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d) 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 $$e(\gamma_n)$$. ::: ::: {.lemma title="?"} The Euler class $$e$$ is natural with respect to pullback and thus a characteristic class. ::: ::: {.proof title="?"} We'll check naturality for the Thom class $$u_E$$. Consider pulling back a disc bundle: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJEJ1xcZGEgZl4qIFxcREQgRSJdLFsyLDAsIkQgXFxkYSBcXEREIEUiXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFszLDIsImYiXSxbMSwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) 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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwzLCJIXm4oXFxERCBFXFxyb3t9e2YoeCl9LCBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzMsMywiSF5uKGZeKiBcXEREIEVcXHJve317Zih4KX0sIGZeKiBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzAsMSwiSF5uKFxcREQgRSwgXFxTUyBFOyBcXFpaKSJdLFszLDEsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFswLDAsInVfRSJdLFszLDAsInUnIl0sWzAsNCwiXFxyb3t1X0V9e3h9Il0sWzMsNCwiXFxyb3t1J317eH0iXSxbMCwxLCJcXGNvbmciXSxbMywxXSxbMiwwXSxbMiwzXV0=) We then get naturality of the Euler class from the following: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMSwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYKSJdLFswLDMsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFszLDMsIkhebihZKSJdLFswLDAsInVfRSJdLFszLDAsImUiXSxbMCw0LCJ1JyJdLFszLDQsImUnIl0sWzEsM10sWzIsM10sWzAsMV0sWzAsMl0sWzQsNV0sWzYsN11d) ::: ::: {.example title="?"} $$e({\mathbb{CP}}^1)$$ is a generator of $$H^2({\mathbb{CP}}^1)$$ Apply the Gysin sequence, taking the canonical line bundle $${\mathbb{S}}E$$ over $${\mathbb{CP}}^1$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbNiwxXSxbMiwwLCJIXntqLTJ9KFxcQ1BeMSkiXSxbNCwwLCJIXntqfShcXENQXjEpIl0sWzYsMCwiSF5qKFxcQ1BeMSkiXSxbMCwwLCJIXntqLTF9KFxcU1MgRSkiXSxbOCwwLCJIXmooXFxTUyBFKSJdLFs0LDFdLFsxLDJdLFsyLDNdLFszLDVdXQ==) The claim is that the total space here is the Hopf fibration: {=tex} \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}  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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) So $$e$$ is a generator of $$H^2({\mathbb{CP}}^1)$$ ::: ::: {.exercise title="?"} Check that $${\mathbb{S}}E\to {\mathbb{CP}}^1$$ with $${\mathbb{S}}E$$ the canonical is the Hopf fibration. ::: ::: {.exercise title="?"} Try to compute $$e({\mathbf{T}}S^2)$$! You may need to add on a bundle to trivialize it. ::: # Thursday, October 07 ## The Euler Class ::: {.remark} Let $$E\to X$$ be an orientable bundle, then {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMCwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYOyBcXFpaKSJdLFswLDEsInVfey1FfSA9IC11X0UiXSxbMywxLCJlKEUpIl0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==) ::: ::: {.lemma title="?"} $$e(-E) = -e(C)$$. ::: ::: {.proof title="?"} Using naturality: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJFID0gZl4qKFxcUlJebikiXSxbMiwwLCJcXFJSXm4iXSxbMiwyLCJcXHB0Il0sWzAsMiwiWCJdLFsxLDIsIlxccGkiXSxbMywyLCJmIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) Then $$e({\mathbb{R}}^n\to {\operatorname{pt}}) = 0$$, and $$0 = f^*(0) = e(f^*({\mathbb{R}}^n\to {\operatorname{pt}})) = e(E)$$. ::: ::: {.lemma title="?"} $$e(E) = 0$$ if $$E$$ is the trivial bundle. ::: ::: {.proof title="?"} Using the exact sequence for the pair $$({\mathbb{D}}E, {\mathbb{S}}E)$$: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) So the second map is the inclusion $$\alpha \mapsto (\alpha, 0)$$, which is thus injective. The first map is 0, so $$e(E) = 0$$. ::: ::: {.example title="?"} Compute $$e( {\mathbf{T}}S^2)$$. ::: ::: {.lemma title="?"} If $$E$$ is an odd dimensional vector bundle, then $$2e(E) = 0$$ ::: ::: {.remark} Check that $$\alpha \smile\beta = (-1)^{ij} \beta \smile\alpha$$. ::: ::: {.proof title="First way"} The antipodal map will give an isomorphism $$E\cong -E$$, so $$e(E) = e(-E) = -e(E)$$ and thus $$2e(E) = 0$$. ::: ::: {.proof title="Second way"} Use the map: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7XFxaWikiXSxbMywwLCJIXnsybn0oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzAsMSwidV9FIl0sWzMsMSwidV9FXFxjdXBwcm9kIHVfRSA9IC0gdV9FIFxcY3VwcHJvZCB1X0UiXSxbMCwxLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d) Then $$2u_E \smile u_E = 0$$, making $$2u_E = 0$$ and thus $$2e(E) = 0$$. ::: ::: {.lemma title="?"} 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 $e(E\times F) = \pi_1^*(e(E)) \smile\pi_2^*(e(F)) = e(E) \otimes e(F) .$ ::: ::: {.exercise title="?"} Prove this using the Kunneth formula. ::: ::: {.corollary title="?"} Let $$E,E'\to X$$ be two vector bundles. Then $$e(E \oplus E') = e(E) \smile e(E')$$. ::: ::: {.proof title="?"} Consider the diagonal $$\Delta: X\to X{ {}^{ \scriptscriptstyle\times^{2} } }$$, then take the pullback: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXERlbHRhXiooRVxcdGltZXMgRScpXFxjb25nIEUgXFxvcGx1cyBFJyJdLFsyLDAsIkVcXHRpbWVzIEUnIl0sWzIsMiwiWCJdLFswLDIsIlhcXGNhcnRwb3dlcnsyfSJdLFszLDIsIlxcRGVsdGEiXSxbMCwzXSxbMSwyXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) Then $e(E \oplus E') &= \Delta^*( e(E\times E'))\\ &= \Delta^*(\pi_1^*(e(E)) \smile\pi_2^*(e(E')) ) \\ &= \Delta^* \pi_1^* (e(E))\smile\Delta^* \pi_2^* e(E') \\ &= e(E) \smile e(E') .$ ::: ::: {.corollary title="?"} If $$E' = L \oplus E$$ for $$L$$ a trivial line bundle, $$e(E') = 0$$. ::: ::: {.proof title="?"} $e(E') = e(L) \smile e(E) = 0 .$ ::: ## Obstruction Theory ::: {.remark} The Euler class is the **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. ::: ::: {.exercise title="?"} Describe the correspondence between line subbundles and nonvanishing global section. ::: ::: {.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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) Thus we have a formula ${\left\langle { \delta \alpha},~{ [\Delta^{k+1}] } \right\rangle} = {\left\langle { \alpha},~{ {{\partial}}[\Delta^{k+1}] } \right\rangle} .$ ::: ::: {.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: {=tex} \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}  > [Link to Diagram](https://q.uiver.app/?q=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) 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 $$[\Delta^n] \xrightarrow{e_s} \deg( {{\partial}}\Delta^n \to S^{n-1})\in {\mathbb{Z}}$$. ::: ::: {.lemma title="?"} $$e_S$$ is independent of the choice of trivialization for the pullback. ::: ::: {.proof title="?"} 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. ::: ::: {.lemma title="?"} If $$s'$$ is a different section over $$X^{(n-1)}$$, then $$e_{s'} - e_s = \delta(\alpha)$$ for some $$\alpha\in C^{n-1}_{{ \mathrm{cell}} }(X)$$. ::: ::: {.proof title="?"} See phone notes. ::: # Tuesday, October 12 ## More Euler Class > Review how to construct the Euler and Thom classes. ::: {.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 [Link to Diagram](https://q.uiver.app/?q=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) We can then define \( e_s(\Delta^n) \coloneqq\deg \psi$$, and we claim that this also equals $$\deg \psi'$$. Look at the LES of a pair: {=tex} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{H_n({\mathbb{D}}^n)=0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_n({\mathbb{D}}^n) = 0} \\ \\ {H_n({\mathbb{D}}^n; S^{n-1})\cong {\mathbb{Z}}} &&& {H_n({\mathbb{D}}^n; S^{n-1})\cong {\mathbb{Z}}} \\ \\ {H_{n-1}(S^{n-1})\cong {\mathbb{Z}}} &&& {H_{n-1}(S^{n-1})\cong {\mathbb{Z}}} \\ \\ \textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}({\mathbb{D}}^n) = 0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}({\mathbb{D}}^n) = 0} \arrow["{\psi^*}", from=3-1, to=3-4] \arrow[from=1-1, to=3-1] \arrow[from=1-4, to=3-4] \arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-1, to=5-1] \arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-4, to=5-4] \arrow["{(\psi')^*}"{description}, from=5-1, to=5-4] \arrow[from=5-1, to=7-1] \arrow[from=5-4, to=7-4] \arrow[from=7-1, to=7-4] \arrow[from=1-1, to=1-4] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=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) We may assume that $$X$$ is $$n{\hbox{-}}$$dimensional since for $$(X, X{ {}^{ (n) } })$$ we have $H^n(X, X{ {}^{ (n) } }) = 0 \to H^n(X) \hookrightarrow H^n(X^n) ,$ so anything equal in $$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)$$ to be the fundamental class (positive generator). Then $e_s(\Delta^n) = {\left\langle {\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* p^* x},~{y} \right\rangle} .$ Consider the map $$(\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) } }$$, we obtain $H^n({\mathbb{D}}E, {\mathbb{S}}E) \xrightarrow{i^*} H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1}) .$ Use that $$\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^*$$, i.e. any two sections of the disc bundle will be homotopic and thus induce equal maps in homology. Now doing exactly what we did for the Euler class, we get a diagram: {=tex} \begin{tikzcd} && {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} \\ & {H^n({\mathbb{D}}E)} & {} & {H^n({\mathbb{D}}E, {\mathbb{S}}E) } && {H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1})} \\ & {H^n(X)} \\ \textcolor{rgb,255:red,92;green,92;blue,214}{e} &&& {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} && {H^n(\Delta^n, {{\partial}}\Delta^n)} \arrow["{i^*}", from=2-4, to=2-6] \arrow["{i_*}", from=4-4, to=4-6] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=2-6, to=4-6] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-4, to=4-4] \arrow["LES"', from=2-4, to=2-2] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-2, to=3-2] \arrow["LES"', from=4-4, to=3-2] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, maps to, from=1-4, to=4-1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=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) More directly, we can map the LESs to get a commutative square: {=tex} \begin{tikzcd} {H^n({\mathbb{D}}E, {\mathbb{S}}E)} && {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} \\ \\ {H^n({\mathbb{D}}E)} && {H^n(X{ {}^{ (n) } })} \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=1-1, to=1-3] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-3, to=3-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEUpIl0sWzAsMiwiSF5uKFxcREQgRSkiXSxbMiwwLCJIXm4oWFxcc2tlbCBuLCBYXFxza2Vse24tMX0pIl0sWzIsMiwiSF5uKFhcXHNrZWwgbikiXSxbMCwyLCJcXGJhcntzfV4qIl0sWzEsMywiXFxiYXJ7c31eKiJdLFswLDFdLFsyLDNdXQ==) In other words, regard $$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) } })$$. Final diagram: {=tex} \begin{tikzcd} && {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} && \textcolor{rgb,255:red,92;green,92;blue,214}{p^*x = i^* u} \\ & {H^n({\mathbb{D}}E)} & {} & {H^n({\mathbb{D}}E, {\mathbb{S}}E) } && {H^n(\Delta^n \times {\mathbb{D}}^n, \Delta^n \times S^{n-1})} \\ \textcolor{rgb,255:red,92;green,92;blue,214}{e} & {H^n(X)} \\ &&& {H^n(X{ {}^{ (n) } }, X{ {}^{ (n-1) } })} && {H^n(\Delta^n, {{\partial}}\Delta^n)} & \textcolor{rgb,255:red,92;green,92;blue,214}{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* p^* x = \mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^* i^* u} \arrow["{i^*}", from=2-4, to=2-6] \arrow["{i_*}", from=4-4, to=4-6] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}", from=2-6, to=4-6] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-4, to=4-4] \arrow["LES"', from=2-4, to=2-2] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mus\mkern-1.5mu}\mkern 1.5mu^*}"', from=2-2, to=3-2] \arrow["LES"', two heads, from=4-4, to=3-2] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=30pt}, dashed, maps to, from=1-4, to=3-1] \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, maps to, from=1-4, to=1-6] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, maps to, from=1-6, to=4-7] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=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) ::: ## Computations on Smooth Manifolds ::: {.remark} Recall that a smooth structure on a manifold $$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$$ we have $\psi_{VU} \coloneqq\phi_V^{-1}\circ \phi_U: \phi_U(U \cap V) \to \phi_V(U \cap V) \in C^\infty ,$ so there are all diffeomorphisms of open subsets of $${\mathbb{R}}^n$$. A smooth atlas is maximal if not contained in any other smooth atlas, and two atlases are compatible if their union is again a smooth atlas. We say two smooth structures are equivalent if they are compatible. ::: ::: {.exercise title="?"} Show that any smooth manifold has a unique maximal smooth atlas. ::: ::: {.remark} Recall that for $$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$$. ::: ::: {.exercise title="?"} Show that every other definition of $${\mathbf{T}}M$$ coincides with this one. ::: # Thursday, October 14 ::: {.remark} Some background from smooth manifolds: a map $$f:M\to N$$ of manifolds is **smooth** if for any smooth charts $$(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$$ is smooth. ::: ::: {.exercise title="?"} {=tex} \envlist  1. Any smooth map $$f:M\to N$$ induces a bundle map $$df: {\mathbf{T}}M\to f^* {\mathbf{T}}N$$. 2. There is a canonical isomorphism $${\mathbf{T}}{\mathbb{R}}^n \to {\mathbb{R}}^n\times {\mathbb{R}}^n$$. 3. Let $$(U, \phi_U)$$ be a chart on $$M$$, then show that there are trivializing charts for $${\mathbf{T}}M$$: $d\phi_U: { \left.{{{\mathbf{T}}M}} \right|_{{U}} } \to { \left.{{ \phi_U^* }} \right|_{{ {\mathbf{T}}{\mathbb{R}}^n }} }{U} \cong U \times {\mathbb{R}}^n .$ In particular, the following diagram commutes: {=tex} \begin{tikzcd} {{ \left.{{{\mathbf{T}}M}} \right|_{{U}} }} && {U\times {\mathbb{R}}^n} \\ \\ U \arrow["\pi", from=1-1, to=3-1] \arrow["{d\phi_U}", from=1-1, to=1-3] \arrow[from=1-3, to=3-1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve1xcVCBNfXtVfSJdLFswLDIsIlUiXSxbMiwwLCJVXFx0aW1lcyBcXFJSXm4iXSxbMCwxLCJcXHBpIl0sWzAsMiwiZFxccGhpX1UiXSxbMiwxXV0=) ::: ::: {.exercise title="?"} Show that there is a canonical isomorphism $T_p M { { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Der}}(M, {\mathbb{R}}) \leq C^\infty(M, {\mathbb{R}}) ,$ where a derivation at $$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)$$. ::: ::: {.remark} $$N \subseteq M$$ is a **smooth submanifold** if for any $$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$$. ::: ::: {.exercise title="?"} Show that this is equivalent to $$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$$. ::: ::: {.definition title="?"} Given a Riemannian metric on $${\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$$. ::: ::: {.exercise title="?"} Show that up to a canonical isomorphism, this is independent of the choice of Riemannian metric. ::: ::: {.definition title="?"} Given a curve $$\gamma: I\to M$$, then $$\gamma'(t) \in {\mathbf{T}}_{\gamma(t)} M$$ is the following derivation: $\gamma'(t) . f \coloneqq\lim_{h\to 0} { f(\gamma(t+h)) - f(\gamma(t)) \over h} .$ A **vector field** on $$M$$ is a section of $${\mathbf{T}}M$$, and a vector field along $$\gamma$$ is a section of $$\gamma^* {\mathbf{T}}M$$: {=tex} \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-15.pdf_tex} }; \end{tikzpicture}  ::: ::: {.example title="?"} $$\gamma'(t)$$ is a vector field along $$\gamma$$. ::: ::: {.remark} A **Lie group** is a group $$G$$ with the structure of a smooth manifold where multiplication and inversion are smooth self-diffeomorphisms. ::: ::: {.definition title="?"} Given a Lie group $$G$$ and smooth principal $$G{\hbox{-}}$$bundle $$P \xrightarrow{\pi} M^n$$, a **connection** on $$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$$ is the horizontal subspace: {=tex} \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-25.pdf_tex} }; \end{tikzpicture}  ::: ::: {.remark} Given a connection, curves $$\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$$: ::: ::: {.definition title="?"} Given a connection on $$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$$. {=tex} \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_13-32.pdf_tex} }; \end{tikzpicture}  ::: ::: {.lemma title="?"} Given a smooth path $$\gamma$$ and $$\tilde \gamma(0)$$, there is a unique horizontal lift $$\tilde \gamma$$ starting at $$\tilde \gamma(0)$$. ::: ::: {.remark} Consider $$\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]$$. ::: ::: {.exercise title="?"} Show that parallel transport yields a well-defined map $${\mathbf{T}}_{\gamma(0)}M \to {\mathbf{T}}_{\gamma(1))}M$$. ::: ::: {.theorem title="?"} Given a Riemannian metric on $${\mathbf{T}}M$$, there is a canonical connection $$\nabla$$, the **Levi-Cevita** connection, which is torsionfree (i.e. $$\nabla_X Y - \nabla_Y X = [X, Y]$$) for $$X, Y$$ vector fields and $$\nabla_X Y$$ denotes parallel transporting $$Y$$ along $$X$$. ::: ::: {.definition title="?"} A curve $$\gamma$$ is a **geodesic** if $$\gamma'(t)$$ is parallel. ::: ::: {.theorem title="?"} For any $$v\in {{\mathbf{T}}M}$$, there is a unique geodesic $$\gamma_v$$ with $$\gamma_v'(0) = v$$. ::: ::: {.definition title="Exponential map"} There is a map $\exp: T_p M &\to M \\ v &\mapsto \gamma_v(1) .$ This is well-defined for $$v$$ of small norm, and for all $$v$$ if $$M$$ is closed. ::: ::: {.theorem title="?"} If $$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$$. {=tex} \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-10-14_14-04.pdf_tex} }; \end{tikzpicture}  ::: # Tuesday, October 19 ::: {.remark} Today: computations involving the Euler class. ::: ::: {.theorem title="?"} Let $$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}})$$, [^2] ${\left\langle { e({\mathbf{T}}M)},~{ [M] } \right\rangle} = \chi_{\mathsf{Top}}(M) ,$ the topological Euler characteristic of $$M$$. ::: ::: {.theorem title="?"} Let $$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$$ transversally. Then - $$z\coloneqq s^{-1}(0)$$ is a submanifold of $$M$$ is a submanifold of $$M$$ - $${ \left.{{ds}} \right|_{{Z}} }: \nu Z { { \, \xrightarrow{\sim}\, }}{ \left.{{E}} \right|_{{Z}} }$$ ::: ::: {.exercise title="?"} Prove this. ::: ::: {.corollary title="?"} $$e(E) = \mathrm{PD}([Z])$$, where $$[Z] \in H_{n-k}(M)$$. ::: ::: {.claim} The second theorem implies the first. ::: ::: {.proof title="?"} Note that a section of $${\mathbf{T}}M$$ is a vector field. Use Morse functions $$f: M\to {\mathbb{R}}$$: - $$\operatorname{crit}(f) = \left\{{p\in M {~\mathrel{\Big\vert}~}df_p = 0}\right\}$$. - $$p\in \operatorname{crit}(f)$$ is **nondegenerate** if the Hessian $$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$$. - $$f$$ is **Morse** if all critical points are nondegenerate. - If $$p\in \operatorname{crit}(f)$$ is nondegenerate, then $$\operatorname{Ind}_p(f)$$ is the number of negative eigenvalues of $$H_p$$. - $$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$$. ::: ::: {.proof} $$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)$$ ::: {.exercise title="?"} Show that the sign of a zero of the gradient vector field is $$(-1)^{\operatorname{Ind}_p(f)}$$. ::: So taking $$Z = \operatorname{crit}(f)$$, we can write $e({\mathbf{T}}M) = \mathrm{PD} { \left[ { \sum_{p\in \operatorname{crit}(f) } (-1)^{\operatorname{Ind}_p(f)} [x] } \right] } ,$ where $$[x]$$ is the dual of a generator of $$H_0(M)$$ Then ${\left\langle { e({\mathbf{T}}M)},~{ [M] } \right\rangle} = \sum_{p\in \operatorname{crit}(f) } (-1)^{\operatorname{Ind}_p(f)} = \# \operatorname{crit}_{\text{even}}(f) - \# \operatorname{crit}_{\text{odd}}(f) = \chi_{\mathsf{Top}}(M) .$ ::: ::: {.remark} Given $$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$$ and apply excision to get the following: {=tex} \begin{tikzcd} {u_{\nu N} \in H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu n)} && {H^{m-n}(M, M\setminus N)} && {H^{m-n}(M)} \\ \\ && {H^{m-n}({\mathbb{D}}\nu N, {\mathbb{D}}\nu N \setminus N)} \arrow["\cong", from=1-1, to=1-3] \arrow["\cong", from=1-1, to=3-3] \arrow["{\text{excision} \cong}", from=1-3, to=3-3] \arrow[from=1-3, to=1-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJ1X3tcXG51IE59IFxcaW4gSF57bS1ufShcXEREIFxcbnUgTiwgXFxTUyBcXG51IG4pIl0sWzIsMCwiSF57bS1ufShNLCBNXFxzbSBOKSJdLFsyLDIsIkhee20tbn0oXFxERCBcXG51IE4sIFxcREQgXFxudSBOIFxcc20gTikiXSxbNCwwLCJIXnttLW59KE0pIl0sWzAsMSwiXFxjb25nIl0sWzAsMiwiXFxjb25nIl0sWzEsMiwiXFx0ZXh0e2V4Y2lzaW9ufSBcXGNvbmciXSxbMSwzXV0=) We can also consider the composition: {=tex} \begin{tikzcd} {[N]\in H_n(N)} && {H_n(M)} && {H^{m-n}(M)} \arrow["{i_*}", from=1-1, to=1-3] \arrow["\mathrm{PD}", from=1-3, to=1-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJbTl1cXGluIEhfbihOKSJdLFsyLDAsIkhfbihNKSJdLFs0LDAsIkhee20tbn0oTSkiXSxbMCwxLCJpXyoiXSxbMSwyLCJcXFBEIl1d) ::: ::: {.claim} These two classes are equal in $$H^{m-n}(M)$$. ::: ::: {.proof title="?"} Consider the following: {=tex} \begin{tikzcd} \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu N}} \\ {\left\langle{u_{\nu N}}\right\rangle = H^{m-n}({\mathbb{D}}\nu n, {\mathbb{S}}\nu n)} && {H_n({\mathbb{D}}\nu N)} \\ \\ {H^{m-n}(M, M\setminus N)} && {H_n(N) = \left\langle{[N]}\right\rangle} & \textcolor{rgb,255:red,92;green,92;blue,214}{\pm [N]} \\ \\ {H^{m-n}(M)} && {H_m(M)} \arrow["{\mathrm{PD}, \cong}", tail reversed, from=2-1, to=2-3] \arrow["{\mathrm{PD}({-}) = [M]\frown({-}), \cong}", tail reversed, from=6-1, to=6-3] \arrow["\cong", from=2-3, to=4-3] \arrow["\cong"', from=2-1, to=4-1] \arrow[from=4-1, to=6-1] \arrow[from=4-3, to=6-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, dashed, from=1-1, to=4-4] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJcXGdlbnN7dV97XFxudSBOfX0gPSBIXnttLW59KFxcREQgXFxudSBuLCBcXFNTIFxcbnUgbikiXSxbMCwzLCJIXnttLW59KE0sIE1cXHNtIE4pIl0sWzAsNSwiSF57bS1ufShNKSJdLFsyLDEsIkhfbihcXEREIFxcbnUgTikiXSxbMiw1LCJIX20oTSkiXSxbMiwzLCJIX24oTikgPSBcXGdlbnN7W05dfSJdLFszLDMsIlxccG0gW05dIixbMjQwLDYwLDYwLDFdXSxbMCwwLCJ1X3tcXG51IE59IixbMjQwLDYwLDYwLDFdXSxbMCwzLCJcXFBELCBcXGNvbmciLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifX19XSxbMiw0LCJcXFBEKFxcd2FpdCkgPSBbTV1cXGNhcHByb2QoXFx3YWl0KSwgXFxjb25nIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn19fV0sWzMsNSwiXFxjb25nIl0sWzAsMSwiXFxjb25nIiwyXSxbMSwyXSxbNSw0XSxbNyw2LCIiLDAseyJjdXJ2ZSI6LTIsImNvbG91ciI6WzI0MCw2MCw2MF0sInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) The claim is that $$u_{\nu N}$$ is mapped to the *positive* generator, so the both paths from $$H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N)\to H^{m-n}(M)$$ agree. ::: ::: {.remark} An aside on cup/cap products. The cap product is a map: $\frown: H_m(X) \otimes_{\mathbb{Z}}H^i(X) &\to H_{m-i}(X) \\ .$ On chains, it's given by $\frown C_m(X) \otimes_{\mathbb{Z}}C^i(X) &\to C_{m-i}(X) \\ (\sigma, \phi) &\mapsto \phi\qty{ { \left.{{\sigma}} \right|_{{{\left[ {v_0,\cdots, v_i} \right]}}} } } { \left.{{\sigma}} \right|_{{v_{i+1}, \cdots, v_m}} } .$ Then $$\mathrm{PD}({-}) \coloneqq[X]\frown({-})$$. The cup product is a map $\smile: H^i(X) \otimes_{\mathbb{Z}}H^j(X) &\to H^{i+j}(X) \\ C^i(X) \otimes_{\mathbb{Z}}C^j(X) &\to C^{i+j}(X) \\ (\phi, \psi) &\mapsto \qty{\sigma \mapsto \phi\qty{{ \left.{{\sigma}} \right|_{{[v_0,\cdots, v_i]}} }} \psi\qty{{ \left.{{\sigma}} \right|_{{[v_{i+1}, \cdots, v_{i+j}}} }} } .$ Then fixing any element yields a map $$\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)$$. ::: # Thursday, October 21 ::: {.remark} Recall that if $$N^n \leq M^n$$ is a submanifold, we have the following diagram: {=tex} \begin{tikzcd} {u_{\nu N}} & {H^{m-n}({\mathbb{D}}\nu N, {\mathbb{S}}\nu N)} && {H^{m-n}(M, M\setminus N)} && {H^{m-n}(M)} \\ \\ & {H_n({\mathbb{D}}\nu N)} && {H_n(N)} && {H_n(M)} \\ &&& {[N]} \arrow["\mathrm{PD}", from=3-6, to=1-6] \arrow["\cong", from=3-2, to=3-4] \arrow["{i_*}", from=3-4, to=3-6] \arrow["{j^*}"', from=1-4, to=1-6] \arrow["{\cong, \text{ excision}}"', from=1-2, to=1-4] \arrow["\mathrm{PD}"', from=3-2, to=1-2] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMSwwLCJIXnttLW59KFxcREQgXFxudSBOLCBcXFNTIFxcbnUgTikiXSxbMywwLCJIXnttLW59KE0sIE1cXHNtIE4pIl0sWzUsMCwiSF57bS1ufShNKSJdLFs1LDIsIkhfbihNKSJdLFsxLDIsIkhfbihcXEREIFxcbnUgTikiXSxbMywyLCJIX24oTikiXSxbMywzLCJbTl0iXSxbMCwwLCJ1X3tcXG51IE59Il0sWzMsMiwiXFxQRCJdLFs0LDUsIlxcY29uZyJdLFs1LDMsImlfKiJdLFsxLDIsImpeKiIsMl0sWzAsMSwiXFxjb25nLCBcXHRleHR7IGV4Y2lzaW9ufSIsMl0sWzQsMCwiXFxQRCIsMl1d) Then the Thom class $$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)$$ is mapped to the generator specified by the orientations on fibers. ::: ::: {.theorem title="?"} For $$A^i \pitchfork B^j \leq X$$ are smooth oriented submanifolds intersecting transversally, then $\mathrm{PD}([A]) \smile\mathrm{PD}([B]) &= \mathrm{PD}([A \frown B]) \\ H^{n-i}(X) \times H^{n-j}(X) &\to H^{2n-i-j}(X) .$ ::: ::: {.remark} Then $[( {\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) \frown N] = [\left\{{x}\right\}] ,$ which is the positive generator. So $\mathrm{PD}[ ({\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) ] \smile\mathrm{PD}[N] = \mathrm{PD}[x] .$ Now we can cap this to obtain $[{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\qty{ \mathrm{PD}[ ({\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x) ] \smile\mathrm{PD}[N]} &= \qty{ [{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\mathrm{PD}[{\mathbb{D}}\nu N_x, {\mathbb{S}}\nu N_x] } \frown\mathrm{PD}[N] \\ &= [{\mathbb{D}}\nu N, {\mathbb{S}}\nu N] \frown\mathrm{PD}[x] \\ &= 1 ,$ where we've used that $${\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$$. ::: ::: {.theorem title="?"} Let $$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$$. Then $e(E) = \mathrm{PD}[Z] && Z \coloneqq\operatorname{im}(s) \cap M = s^{-1}(0) .$ ::: ::: {.remark} Recall that $$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$$ as well. ::: ::: {.proof title="?"} Let $$N$$ be a tubular neighborhood of $$Z$$ in $$M$$, such that $$N \cong {\mathbb{D}}\nu Z$$. By the lemma, we have two maps {=tex} \begin{tikzcd} {H^k(N, N\setminus Z)} && {H^k(M, M\setminus Z)} && {H^k(M)} \\ {u_{\nu Z}} &&&& {\mathrm{PD}[Z]} \arrow["\cong", from=1-1, to=1-3] \arrow["{j^*}", from=1-3, to=1-5] \arrow["{\text{Lemma}}"', curve={height=30pt}, from=1-1, to=1-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJIXmsoTiwgTlxcc20gWikiXSxbMiwwLCJIXmsoTSwgTVxcc20gWikiXSxbNCwwLCJIXmsoTSkiXSxbMCwxLCJ1X3tcXG51IFp9Il0sWzQsMSwiXFxQRFtaXSJdLFswLDEsIlxcY29uZyJdLFsxLDIsImpeKiJdLFswLDIsIlxcdGV4dHtMZW1tYX0iLDIseyJjdXJ2ZSI6NX1dXQ==) on the other hand, since $$Z \xhookrightarrow{i} M$$, we get {=tex} \begin{tikzcd} {{ \left.{{E}} \right|_{{Z}} }} && E \\ \\ & N \arrow["\iota", from=1-1, to=1-3] \arrow["{{ \left.{{ds}} \right|_{{Z}} }}", from=3-2, to=1-1] \arrow["s"', from=3-2, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve0V9e1p9Il0sWzIsMCwiRSJdLFsxLDIsIk4iXSxbMCwxLCJcXGlvdGEiXSxbMiwwLCJcXHJve2RzfXtafSJdLFsyLDEsInMiLDJdXQ==) ::: {.exercise title="?"} Show $${ \left.{{ds}} \right|_{{Z}} } \simeq s$$ are homotopic sections. ::: Then $$({ \left.{{ds}} \right|_{{Z}} })^* \circ i^* (u_E) = u_{\nu Z}$$, so $$u_{\nu Z} = s^* u_E$$, and we have {=tex} \begin{tikzcd} \textcolor{rgb,255:red,92;green,92;blue,214}{u_E} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu Z}} \\ & {H^k(E, E\setminus M)} && {H^k(M, M\setminus Z)} \\ \\ & {H^k(E)} && {H^k(M)} \\ \textcolor{rgb,255:red,92;green,92;blue,214}{?} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\mathrm{PD}[Z]} \\ &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{e(E)} \arrow["{\text{LES}}", from=2-2, to=4-2] \arrow["{s^*}", from=2-2, to=2-4] \arrow["{s^*, \cong}", from=4-2, to=4-4] \arrow[from=2-4, to=4-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=1-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-5, to=5-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=5-1, to=6-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=5-1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMSwxLCJIXmsoRSwgRVxcc20gTSkiXSxbMSwzLCJIXmsoRSkiXSxbMCwwLCJ1X0UiLFsyNDAsNjAsNjAsMV1dLFswLDQsIj8iLFsyNDAsNjAsNjAsMV1dLFszLDEsIkheayhNLCBNXFxzbSBaKSJdLFs0LDAsInVfe1xcbnUgWn0iLFsyNDAsNjAsNjAsMV1dLFszLDMsIkheayhNKSJdLFs0LDQsIlxcUERbWl0iLFsyNDAsNjAsNjAsMV1dLFs0LDUsImUoRSkiLFsyNDAsNjAsNjAsMV1dLFswLDEsIlxcdGV4dHtMRVN9Il0sWzAsNCwic14qIl0sWzEsNiwic14qLCBcXGNvbmciXSxbNCw2XSxbMiw1LCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNSw3LCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMyw4LCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiwzLCIiLDEseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) Since the diagram commutes, we get $$e(E) = \mathrm{PD}[Z]$$. ::: ::: {.remark} The Euler class for a bundle $$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$$ and there exists a nowhere vanishing vector field. ::: # Tuesday, October 26 ::: {.remark} For $$F\to E \xrightarrow{\pi} B$$ a fiber bundle with a $$G{\hbox{-}}$$structure, we have maps: {=tex} \begin{tikzcd} && G \\ \\ {\phi_{ij}(U_i \cap U_j)} && {{\operatorname{Homeo}}(F, F)} \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=3-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMiwwLCJHIl0sWzIsMiwiXFxIb21lbyhGLCBGKSJdLFswLDIsIlxccGhpX3tpan0oVV9pIFxcaW50ZXJzZWN0IFVfaikiXSxbMCwxXSxbMiwxXSxbMiwwXV0=) A vector bundle was an $${\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$$ structure. A **principal** $$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{-}}$$bundles, namely the **clutching** construction: given $${\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}}$$. ::: ::: {.remark} On universal bundles and classifying spaces: for $$X\in {\mathsf{CW}}$$, $\mathop{\mathrm{Prin}}{ {\mathsf{Bun}}\qty{G} }(X) { { \, \xrightarrow{\sim}\, }}[X, {\mathbf{B}}G] \\ f^* EG &\mapsfrom f .$ We noted - $$G$$ discrete implies $${\mathbf{B}}G \simeq K(G, 1)$$ - $$K(C_2, 1) = {\mathbb{RP}}^{\infty}$$ with $$EC_2 = S^{\infty}$$ - $${\mathbf{B}}U_1 = {\mathbb{CP}}^{\infty}$$ with $$EU_1 = S^{\infty}$$ - $${{\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{-}}$$dimensional frames. - $${\mathbf{B}}{\operatorname{SO}}_n$$ are oriented $$n{\hbox{-}}$$planes in $${\mathbb{R}}^\infty$$. - For $$H\leq G$$, $$EH = EG$$ and $${\mathbf{B}}H = EG/H$$. We had a canonical bundle $$\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)$$. ::: ::: {.example title="?"} A useful application: characteristic classes. For any $$c\in H^d({{\mathbf{B}}{\operatorname{O}}}_n)$$ can be pulled back: $H^d(X) &\to { \mathsf{Vect} }_n(X) \\ f^*c &\mapsfrom f\in [X, {{\mathbf{B}}{\operatorname{O}}}_n] .$ Noting that $${\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)$$, the **first Chern class**. Noting $$H^2(X; {\mathbb{Z}}) \cong [X, K({\mathbb{Z}}, 2)] \cong [X, {\mathbb{CP}}^\infty]$$. ::: ::: {.remark} Note: why $$c_1=0$$ in symplectic settings, related to Maslov index and ensures that the dimension of the relevant moduli space is zero. ::: # Thursday, October 28 ::: {.remark} The Euler class is *natural* in the following sense: for $$E\to X$$ with $$\dim E = n$$ and $$X\in {\mathsf{CW}}$$, we can write {=tex} \begin{tikzcd} {E\cong f^* \gamma_n} && {\gamma_n} \\ \\ X && {{{\mathbf{B}}{\operatorname{O}}}_n} \arrow[from=1-1, to=3-1] \arrow["f", from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJFXFxjb25nIGZeKiBcXGdhbW1hX24iXSxbMCwyLCJYIl0sWzIsMiwiXFxCT19uIl0sWzIsMCwiXFxnYW1tYV9uIl0sWzAsMV0sWzEsMiwiZiJdLFszLDJdLFswLDNdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Then naturality is the following equality: $e(E) = e(f^* \gamma_n) = f^*(e(\gamma_n)) .$ Recall the Thom isomorphism theorem: for an oriented $$E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_r} }^n$$, we have a disk bundle $${\mathbb{D}}E$$ and sphere bundle $${\mathbb{S}}E$$, and $H^j({\mathbb{D}}E; {\mathbb{S}}E) \cong \begin{cases} 0 & j< n \\ {\mathbb{Z}}& j=n \\ H^{j-n}({\mathbb{D}}E) & j> n \end{cases} ,$ noting that we can take $$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$$. We had a claim: $H^n({\mathbb{D}}E, {\mathbb{S}}E) &\to H^n({\mathbb{D}}E) \cong H^n(X) \\ u_E &\mapsto e(E) .$ ::: ::: {.remark} On the **Gysin sequence**: use the bundle $$S^{n-1} \hookrightarrow{\mathbb{S}}E \to X$$ and the LES $\cdots \to H^{j-1}({\mathbb{S}}E) \xrightarrow{\delta} H^{j-n}(X) \xrightarrow{({-}) \smile e(E)} H^j(X) \to H^j({\mathbb{S}}E) \to \cdots .$ The connecting map $$\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)$$. ::: ::: {.proposition title="?"} If $$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$$. ::: ::: {.remark} So $$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) } }$$. We can use the cellular attaching maps to write {=tex} \begin{tikzcd} {\Delta^k \times S^{n-1}} && {i^* {\mathbb{S}}E} && {{\mathbb{S}}E} \\ \\ && {\Delta^k} && X \arrow["i", from=3-3, to=3-5] \arrow[from=1-5, to=3-5] \arrow[from=1-3, to=1-5] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-1, to=3-3] \arrow["{s: {{\partial}}\Delta^k \to S^{n-1} \in \pi_{k}^{S^{n-1}}}", curve={height=-18pt}, dashed, from=3-3, to=1-1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJcXERlbHRhXmsiXSxbNCwyLCJYIl0sWzQsMCwiXFxTUyBFIl0sWzIsMCwiaV4qIFxcU1MgRSJdLFswLDAsIlxcRGVsdGFeayBcXHRpbWVzIFNee24tMX0iXSxbMCwxLCJpIl0sWzIsMV0sWzMsMl0sWzMsMF0sWzQsM10sWzQsMF0sWzAsNCwiczogXFxiZCBcXERlbHRhXmsgXFx0byBTXntuLTF9IFxcaW4gXFxwaV97a31ee1Nee24tMX19IiwwLHsiY3VydmUiOi0zLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) - If $$s \in \pi_k(S^n)$$ and $$k [Link to Diagram](https://q.uiver.app/?q=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) So we can compute \( e(E)$$ for $${\mathbf{T}}M$$ and $$\nu N$$. ::: ::: {.remark} Next topics: Chern and Stiefel-Whitney classes. ::: ::: {.theorem title="?"} For $$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)$$ , the **Stiefel-Whitney classes**, satisfying the following properties: 1. Naturality: $$w_i(f^*(E)) = f^*(w_i(E))$$ 2. $$w(E_1 \bigoplus E_2) = w(E_1 )\smile w(E_2)$$ 3. $$w_i(E) = 0$$ if $$i>\dim_{\mathbb{R}}E$$. 4. If $$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)$$. Moreover, these properties characterize $$w_i(E)$$ uniquely. For complex vector bundles, there are $$c_i(E) \in H^{2i}(X; {\mathbb{Z}})$$ and $$c_(E)$$, the **Chern classes**, which satisfy the same properties with $${\mathbb{C}}$$ instead of $${\mathbb{R}}$$ and $$\left\langle{\alpha}\right\rangle = H^2({\mathbb{CP}}^\infty; {\mathbb{Z}})$$. ::: ::: {.remark} Next time: existence and uniqueness. ::: # Stiefel-Whitney and Chern Classes (Tuesday, November 02) ::: {.remark} The Stiefel-Whitney classes $$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)$$, we mentioned several properties: - $$w_i(f^* E) = f^* w_i(E)$$ - $$w(E_1 \oplus E_2) = w(E_1) \smile w(E_2)$$ - $$w_{> \dim E} E = 0$$ - $$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}}$$. These properties characterize the $$w_i$$ and $$c_i$$ uniquely. ::: ::: {.remark} On why we need $$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$$ the work with, and this we need to pull back something nonzero to get anything interesting. ::: ::: {.corollary title="?"} Some immediate consequences: - $$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$$. - If $$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$$. - If $$E_1 \oplus E_2$$ is trivial, then $$w(E_1) \smile w(E_2) = 1$$. ::: ::: {.example title="?"} If $$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$$ has no nonvanishing sections. ::: ::: {.remark} Consider inverting a formal power series $$\sum_{i\geq 0} a_i$$: $(1 + a_1 + a_2 + a_3 + \cdots)( 1 - a_1 + (a_1^2 + a_2) + (a_1^3 + 2a_1 a_2 - a_3) + \cdots) = 1 .$ So if $$w(E_1)w(E_2) = 1$$, we can solve for $$w(E_1)$$ in terms of $$w(E_2)$$. ::: ::: {.proposition title="?"} Note that $$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}$$ ::: ::: {.proof title="?"} Let $$\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)$$. ::: {.lemma title="?"} ${\mathbf{T}}{\mathbb{RP}}^n = \mathop{\mathrm{Hom}}(\gamma, \gamma^\perp) .$ ::: Recall that $$\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})$$, so ${\mathbf{T}}{\mathbb{RP}}^n = \left\{{ [(x, v), (- \mathbf{x}, - \mathbf{v})] {~\mathrel{\Big\vert}~}v\in {\mathbf{T}}_x S^n }\right\} = \left\{{ [(x, v), (- \mathbf{x}, - \mathbf{v})] {~\mathrel{\Big\vert}~}v\in {\mathbf{T}}_x S^n \iff {\left\langle {x},~{v} \right\rangle} = 0 }\right\} .$ So define a map ${\mathbf{T}}{\mathbb{RP}}^n &\to \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp) \\ [(\mathbf{x}, \mathbf{v}), (-\mathbf{x}, \mathbf{v})] &\mapsto \qty{\lambda \mathbf{x} \xrightarrow{\ell} \lambda \mathbf{v} } ,$ and one can check that this is well-defined. {=tex} \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-11-02_13-45.pdf_tex} }; \end{tikzpicture}  We can thus write ${\mathbf{T}}{\mathbb{RP}}^n \oplus {\mathbb{R}} &= \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp) \oplus \mathop{\mathrm{Hom}}( \gamma, \gamma) \\ &= \mathop{\mathrm{Hom}}( \gamma, \gamma^\perp \oplus \gamma) \\ &= \mathop{\mathrm{Hom}}(\gamma, {\mathbb{R}}^{n+1}) \\ &= \mathop{\mathrm{Hom}}(\gamma, {\mathbb{R}}){ {}^{ \scriptscriptstyle\oplus^{(n+1)} } } .$ Note that $$\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)$$. ::: ::: {.corollary title="?"} $$w({\mathbf{T}}{\mathbb{RP}}^n) = 1$$ iff $$n+1=2^r$$ for some $$r$$. ::: ::: {.proof title="?"} $$\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$$ coefficients. Now write $(1+a)^{2^r} = ( (1+a)^{2^{r-1}} )^2 \overset{IH}{=} (1 + a^{2^{r-1}})^2 = 1 + a^{2^r} .$ Now use that $$\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$$ odd, then $(1 + a)^{ m2^{r}} = ((1+a)^{2^r})^m \\ = (1+a^{2^r})^m \\ = 1 + ma^{2^r} + \cdots ,$ where the first nontrivial term doesn't vanish since $$a^{2^r}\neq 0$$. ::: ::: {.lemma title="?"} If $${\mathbb{RP}}^{2^r}$$ admits an immersion into $${\mathbb{R}}^N$$, then $$N\geq 2^{r+1} - 1$$. ::: ::: {.proof title="?"} Why? $w({\mathbf{T}}{\mathbb{RP}}^{2^r}) = (1+a)^{2^r+1} = (1+a)^{2^r}(1+a) \\ = (1+a^{2^r})(1+a) \\ = 1 +a + a^{2^r} + a^{2^r + 1} \\ = 1 +a + a^{2^r} .$ Try to invert this: let $$n\coloneqq 2^r$$, then $(1+a+a^n)(1 + a + a^2 + \cdots + a^{n-1}) = 1 .$ Then $w(\nu {\mathbb{RP}}^n) = \sum_{0\leq i \leq n-1} a^i ,$ so $$\dim \nu {\mathbb{RP}}^n \geq n-1$$. ::: # Thursday, November 04 ::: {.remark} Recall that $$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}})$$ the positive generator. ::: ::: {.theorem title="Leray-Hirsch"} Fix a commutative ring $$R$$ and a fiber bundle $$F\hookrightarrow E \to B$$ such that 1. $$H^j(F; R) \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is finitely generated and free, 2. For any $$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)$$. Then there is an isomorphism $\Phi: H^*(B; R)\otimes_R H^*(F; R){ { \, \xrightarrow{\sim}\, }}H^*(E; R) \\ b_i \otimes i^*(c_{j_k}) \mapsto \pi^*(b_i) \smile c_{j_k} .$ ::: ::: {.example title="?"} Let $${\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$$. Now take the pullback: {=tex} \begin{tikzcd} && \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{C}}^n} \\ & { \pi^*E = \gamma \oplus E'} && E \\ \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{CP}}^n} \\ & E && X \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-1, to=4-2] \arrow[from=4-2, to=4-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-3, to=2-4] \arrow[from=2-4, to=4-4] \arrow[from=2-2, to=4-2] \arrow[from=2-2, to=2-4] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwxLCIgXFxwaV4qRSA9IFxcZ2FtbWEgXFxvcGx1cyBFJyJdLFszLDEsIkUiXSxbMywzLCJYIl0sWzEsMywiRSJdLFsyLDAsIlxcQ0NebiIsWzI0MCw2MCw2MCwxXV0sWzAsMiwiXFxDUF5uIixbMjQwLDYwLDYwLDFdXSxbNSwzLCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdfV0sWzMsMl0sWzQsMSwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX1dLFsxLDJdLFswLDNdLFswLDFdXQ==) So we have a bundle $${\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}$$: {=tex} \begin{tikzcd} {\gamma_x} && \gamma \\ \\ {{\mathbb{P}}(E)_x = {\mathbb{CP}}^{n-1}} && {{\mathbb{P}}(E)} \arrow[from=1-1, to=3-1] \arrow[from=1-3, to=3-3] \arrow["{\iota_x^*}"{description}, hook, from=3-1, to=3-3] \arrow[hook, from=1-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzIsMiwiXFxQUChFKSJdLFswLDIsIlxcUFAoRSlfeCA9IFxcQ1Bee24tMX0iXSxbMCwwLCJcXGdhbW1hX3giXSxbMywyXSxbMCwxXSxbMiwxLCJpX3giLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d) Then $$\iota_x^* c_1(\gamma) = c_1(\gamma_x) = e(\gamma_x) = u$$, so we have condition 2. ::: {.exercise title="?"} For $$L \in { {\mathsf{Bun}}\qty{\operatorname{GL}_{1}} }({\mathbb{C}})$$, $c_1(L {}^{ \vee }) = c_1(L) .$ ::: So we have classes $$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)$$ such that we can write the former as a linear combination of the latter: $c_1( \gamma {}^{ \vee })^n + c_1(E) c_1( \gamma {}^{ \vee })^{n-1} + \cdots + c_n(E) = 0 ,$ and the $$c_i(E)$$ are called the **Chern classes**. ::: ::: {.remark} Write $$c(E) = 1 + c_1(E) + \cdots + c_n(E)$$, so $\pi^* c(E) &= c(\gamma) c(E') = (1 +c_1(\gamma))(1 + c_1(E') + \cdots + c_{n-1}(E'))\\ &= 1 + (c_1(\gamma) + c_1(E')) + (c_1( \gamma)c_1(E') + c_2(E')) + \cdots + (c_1(\gamma) c_{i-1}(E') + c_1(E') ) .$ Plugging this into the LHS above yields $c_1( \gamma {}^{ \vee })^n + (c_1(\gamma) + c_1(E')) c_1(\gamma {}^{ \vee })^{n-1} + (c_1(\gamma)c_1(E') + c_2(E'))c_1(\gamma {}^{ \vee })^{n-2} + \cdots + c_1(\gamma)c_{n-1}(E') &= (c_1 (\gamma) + c_1(\gamma {}^{ \vee }))( c_1( \gamma {}^{ \vee })^{n-1} + c_1(E') c_1( \gamma {}^{ \vee })^{n-2} + c_2(E') c_1( \gamma {}^{ \vee })^{n-2} + \cdots + c_{n-1}(E') ) .$ Since $$\gamma {}^{ \vee }$$ and $$\gamma$$ are dual, the first term is zero, so this entire expression is zero. ::: ::: {.exercise title="?"} Show that $$c_1(E {}^{ \vee }) = -c_1(E)$$. > Hint: consider an explicit description in terms of transition functions. ::: ::: {.exercise title="?"} Show that given $$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$$. ::: ::: {.remark} Consider the following pullbacks: {=tex} \begin{tikzcd} {f^* E} && {\gamma \oplus E'} && E \\ \\ {{\mathbb{S}}E} && {{\mathbb{P}}(E)} && X \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=3-5] \arrow[from=3-3, to=3-5] \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow["f"{description}, curve={height=30pt}, tail, from=3-1, to=3-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJcXGdhbW1hIFxcb3BsdXMgRSciXSxbNCwwLCJFIl0sWzQsMiwiWCJdLFsyLDIsIlxcUFAoRSkiXSxbMCwyLCJcXFNTIEUiXSxbMCwwLCJmXiogRSJdLFs1LDBdLFswLDFdLFsxLDJdLFszLDJdLFswLDNdLFs0LDNdLFs1LDRdLFs0LDIsImYiLDEseyJjdXJ2ZSI6NSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibW9ubyJ9fX1dXQ==) What is $$f^* E$$? There is a pullback {=tex} \begin{tikzcd} {\varphi^* \gamma} && \gamma \\ \\ {S^{2n-1}} && {{\mathbb{CP}}^{n-1}} \arrow["\varphi", from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJTXnsybi0xfSJdLFsyLDIsIlxcQ1Bee24tMX0iXSxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiXFx2YXJwaGleKiBcXGdhbW1hIl0sWzAsMSwiXFx2YXJwaGkiXSxbMiwxXSxbMywwXSxbMywyXV0=) Here $$\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$$, we have $(\pi \circ q)^* c_i(E) = c_i(E) && i < n-1 ,$ and $$c_n(E) = e(E)$$. To find this we'll need $$\pi \circ q$$ to be injective. Consider {=tex} \begin{tikzcd} S^{2n-1} \ar[r] & {\mathbb{S}}E \ar[d, "\pi \circ q"] \\ & X \end{tikzcd}  By the Gysin sequence, for $$j<2n-1$$ we'll have $$H^j(X) \cong H^j({\mathbb{S}}E)$$ since the red terms vanish: {=tex} \begin{tikzcd} {H^j({\mathbb{S}}E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n+1}(X)} && {H^{j+1}(X)} \\ \\ {H^{j-1}({\mathbb{S}}E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n}(X)} && {H^j(X)} \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=1-1] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwyLCJIXntqLTF9KFxcU1MgRSkiXSxbMCwwLCJIXmooXFxTUyBFKSJdLFsyLDIsIkhee2otMm59KFgpIixbMCw2MCw2MCwxXV0sWzQsMiwiSF5qKFgpIl0sWzIsMCwiSF57ai0ybisxfShYKSIsWzAsNjAsNjAsMV1dLFs0LDAsIkhee2orMX0oWCkiXSxbMCwyXSxbMiwzXSxbMywxXSxbMSw0XSxbNCw1XV0=) This uses that for $$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$$. ::: ::: {.remark} Idea: pull back to split off a line bundle, pull back further to split off another line bundle, and continue. This is why we only need $$c_1$$ to determine a line bundle. ::: # Tuesday, November 09 ::: {.remark} Today: proving/checking the axioms for Chern classes. Recall that given $$E \xrightarrow{\pi} X$$, we can form a pullback: {=tex} \begin{tikzcd} & {\pi^* E \cong \gamma \oplus E'} && E \\ {{\mathbb{CP}}^{n-1}} \\ & {{\mathbb{P}}(E)} && X \arrow["\pi", from=1-4, to=3-4] \arrow["\pi"', from=3-2, to=3-4] \arrow[from=1-2, to=3-2] \arrow[from=1-2, to=1-4] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=3-4] \arrow[hook, from=2-1, to=3-2] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwwLCJcXHBpXiogRSBcXGNvbmcgXFxnYW1tYSBcXG9wbHVzIEUnIl0sWzMsMCwiRSJdLFszLDIsIlgiXSxbMSwyLCJcXFBQKEUpIl0sWzAsMSwiXFxDUF57bi0xfSJdLFsxLDIsIlxccGkiXSxbMywyLCJcXHBpIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwzLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==) Here $$\gamma {}^{ \vee }$$ is the dual of $$\gamma$$ and $$c_1( \gamma {}^{ \vee }) = -_1( \gamma)$$. By Leray-Hirsch, we have $1 + c_1( \gamma {}^{ \vee }) + c_1(\gamma {}^{ \vee })^2 + \cdots + c_1( \gamma {}^{ \vee })^{n-1} \divides (c_1(\gamma {}^{ \vee }))^n$ and for $$c_i(E)$$ the $$i$$th Chern class of $$E$$, $c_1(\gamma {}^{ \vee })^n + c_1(E) c_1(\gamma {}^{ \vee })^{n-1} + \cdots + c_n(E) = 0 .$ ::: ::: {.proposition title="?"} These satisfy some axioms: 1. Naturality: $$c_1(f^* E) = f^* c_1(E)$$ 2. Homomorphism: $$c_(E_1 \oplus E_2) = c(E_1) \smile c(E_2)$$. 3. $$c_{>\dim E}(E) = 0$$. 4. For $$\gamma\to {\mathbb{CP}}^\infty$$ the canonical, $$c_1(\gamma)\in H^2({\mathbb{CP}}^\infty)$$ is the positive generator. ::: ::: {.proof title="?"} Here (3) is clear by definition, since we don't even define $$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}})$$ for line bundles. For (1), consider a pullback: {=tex} \begin{tikzcd} \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(f)^* \gamma \oplus {\mathbb{P}}(f)^* E' } &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\gamma \oplus E'} \\ & {f^* E} &&&& E \\ \\ \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(f^* E)} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{P}}(E)} \\ & {X'} &&&& X \arrow["f", from=5-2, to=5-6] \arrow[from=2-6, to=5-6] \arrow[from=2-2, to=2-6] \arrow[from=2-2, to=5-2] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=5-6] \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-5, to=5-6] \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-1, to=5-2] \arrow["{\exists {\mathbb{P}}(f)}", color={rgb,255:red,92;green,92;blue,214}, from=4-1, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=1-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=4-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, color={rgb,255:red,92;green,92;blue,214}, draw=none, from=1-1, to=4-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=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) We then use that on fibers, $$\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)})$$, so $(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n} + c_1(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n-1} + c_2(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(E)} )^{n-2} + \cdots + c_n(E) ,$ and applying $${\mathbb{P}}(f)^*$$ to this entire expression yields zero. Thus $(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n} {\mathbb{P}}(f)^* c_1(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-1} {\mathbb{P}}(f)^* c_2(E) \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-2} + \cdots + + {\mathbb{P}}(f)^* c_n(E) = 0 ,$ and we can note that by definition this equals $(c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n} c_1 f^* E \cdot (c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-1} c_2 f^* E \cdot c_1 \gamma {}^{ \vee }_{{\mathbb{P}}(f)^* E } )^{n-2} + \cdots + c_n f^* E = 0 .$ ::: ::: {.remark} Recall the alternative description of Chern classes: {=tex} \begin{tikzcd} {\pi^* E \cong {\mathbb{R}}\oplus E'} && E \\ \\ {{\mathbb{S}}E} && X \arrow[from=1-3, to=3-3] \arrow["\pi"', from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJcXFNTIEUiXSxbMiwyLCJYIl0sWzIsMCwiRSJdLFswLDAsIlxccGleKiBFIFxcY29uZyBcXFJSIFxcb3BsdXMgRSciXSxbMiwxXSxbMCwxLCJcXHBpIiwyXSxbMywwXSxbMywyXV0=) By the Gysin sequence, we obtained $$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)$$. ::: ## Splitting Principle ::: {.proposition title="?"} Given an $$E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}$$, there exists $$Y\in {\mathsf{Top}}$$ with $$f:Y\to X$$ such that 1. $$f^* E = L_1 \oplus \cdots \oplus L_n$$ with $$L_i$$ line bundles. 2. $$f^*: H^*(X) \hookrightarrow H^*(Y)$$ is injective, so classes in $$H^*(Y)$$ can be uniquely pulled back. ::: ::: {.proof title="?"} By induction, it suffices to find $$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}$$. Equivalently, the following map is injective: $f^* : H^*(X) &\to H^*(Y) \\ \alpha & \mapsto 1\cdot \alpha .$ Here being generated means that $\beta\in H^*(Y) \implies \beta = \sum_{k=0}^{n-1} c_1(\gamma)^k \alpha_k, \quad \alpha_k \in H^*(X) .$ ::: ::: {.corollary title="?"} If a polynomial identity holds for Chern classes in $$E \coloneqq\bigoplus_i L_i$$ for $$L_i$$ line bundles, then it holds for any bundle. ::: ::: {.lemma title="?"} For line bundles $$L_1,\cdots, L_n$$, $c(\bigoplus L_i) = \prod c(L_i) = \prod(1 + c_1(L_i)) = 1 + \qty{\sum c_1(L_i) } + \cdots + \qty{\prod c_1(L_i)} ,$ so for $$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))$$. ::: ::: {.corollary title="?"} The direct sum formula $$c(E \oplus E') = c(E) c(E')$$. Take a pullback {=tex} \begin{tikzcd} {\bigoplus_{i} L_i \oplus \bigoplus_j L_j} && {E \oplus E'} \\ \\ Y && X \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJFIFxcb3BsdXMgRSciXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFswLDAsIlxcYmlnb3BsdXNfe2l9IExfaSBcXG9wbHVzIFxcYmlnb3BsdXNfaiBMX2oiXSxbMywwXSxbMCwxXSxbMiwxXSxbMywyXSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) Then $c( \bigoplus_i L_i \oplus \bigoplus_j L_j) = \qty{\prod_i c(L_i)} \cdot \qty{\prod_j c(L_j)} = c\qty{ \bigoplus_i L_i} \cdot c\qty{\bigoplus L_j} = f^* c(E) \cdot f^* c(E') ,$ so $$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')$$. ::: ::: {.proof title="of lemma"} Write $$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$$. ::: {.claim} $\prod_{i=1}^n \qty{ c_1(\gamma {}^{ \vee }) + c_1( \pi^* L_i) } = 0 .$ ::: Using the claim, we get $0 = c_1( \gamma {}^{ \vee })^n + \qty{ \sum c_1 \pi^* L_i }(c_1(\gamma {}^{ \vee }))^{n-1} + \cdots + \prod c_1(\pi^* L_i) \cdot c_1(\gamma {}^{ \vee })^1 = c_1(\gamma {}^{ \vee })^{n} + \pi^* c_1(E) \cdot (c_1(\gamma {}^{ \vee }))^{n-1} + \cdots + \pi^* c_n(E) \cdot c_1(\gamma {}^{ \vee })^1 ,$ which proves e.g. that $$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, {-})$$: {=tex} \begin{tikzcd} 0 && \gamma && {\bigoplus_i \pi^* L_i} && {\gamma^\perp} && 0 \\ \\ && {\mathop{\mathrm{Hom}}(\gamma, \gamma)} && {\mathop{\mathrm{Hom}}\qty{\gamma, \bigoplus_i \pi^* L_i} \cong \bigoplus_i \mathop{\mathrm{Hom}}(\gamma, \pi^* L_i)} \arrow[from=1-1, to=1-3] \arrow[""{name=0, anchor=center, inner sep=0}, "\iota", from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[""{name=1, anchor=center, inner sep=0}, "{\iota_*}", from=3-3, to=3-5] \arrow["{\mathop{\mathrm{Hom}}(\gamma,{-})}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzIsMCwiXFxnYW1tYSJdLFs0LDAsIlxcYmlnb3BsdXNfaSBcXHBpXiogTF9pIl0sWzYsMCwiXFxnYW1tYV5cXHBlcnAiXSxbOCwwLCIwIl0sWzIsMiwiXFxIb20oXFxnYW1tYSwgXFxnYW1tYSkiXSxbNCwyLCJcXEhvbVxccXR5e1xcZ2FtbWEsIFxcYmlnb3BsdXNfaSBcXHBpXiogTF9pfSBcXGNvbmcgXFxiaWdvcGx1c19pIFxcSG9tKFxcZ2FtbWEsIFxccGleKiBMX2kpIl0sWzAsMV0sWzEsMiwiXFxpb3RhIl0sWzIsM10sWzMsNF0sWzUsNiwiXFxpb3RhXyoiXSxbOCwxMSwiXFxIb20oXFxnYW1tYSxcXHdhaXQpIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==) Now note that we can get a splitting: {=tex} \begin{tikzcd} {\mathop{\mathrm{Hom}}(\gamma, \gamma)} && {\bigoplus_i \mathop{\mathrm{Hom}}(\gamma, \pi^* L_i)} \\ \\ & X \arrow[from=1-3, to=3-2] \arrow[from=1-1, to=3-2] \arrow[from=1-1, to=1-3] \arrow["s"{description}, curve={height=-24pt}, dashed, from=3-2, to=1-1] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXEhvbShcXGdhbW1hLCBcXGdhbW1hKSJdLFsyLDAsIlxcYmlnb3BsdXNfaSBcXEhvbShcXGdhbW1hLCBcXHBpXiogTF9pKSJdLFsxLDIsIlgiXSxbMSwyXSxbMCwyXSxbMCwxXSxbMiwwLCJzIiwxLHsiY3VydmUiOi00LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) So take $$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$$, this concludes the proof. ::: # Chern and Stiefel-Whitney classes (Thursday, November 11) ::: {.remark} Recall that we were proving the splitting principle: given $$E\to X$$ of dimension $$n$$, there is a space $$Y$$ with $$f:Y\to X$$ where - $$f^* E = \bigoplus L_i$$ splits as a direct sum of line bundles. - $$f^*: H^*(X) \to H^*(Y)$$ is injective. ::: ::: {.lemma title="?"} Assuming the exercise from last time that $$c_1(L {}^{ \vee }) = - c_1(L)$$ $c_i(E {}^{ \vee }) = (-1)^i c_i(E) .$ ::: ::: {.proof title="?"} It suffices to show this for $$E = \bigoplus L_i$$. Write $$E {}^{ \vee }= \bigoplus L_i {}^{ \vee }$$, so $c(E {}^{ \vee }) = \prod c(L_i {}^{ \vee }) &= \prod (1- c_1(L_i)) \\ &= 1 - \qty{\sum c_1(L_i)} + \sigma_2(c_1 L_1,\cdots, c_1 L_n) - \cdots \pm \prod c_1(L_i) \\ &= 1 - c_1(E) + c_2(E) \cdots \pm c_1(L_n) .$ ::: ::: {.theorem title="?"} For $$E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}})_{/ {X}}$$, the top Chern class equals the Euler class: $c_n(E) = e(E) .$ Similarly, for $$E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}$$, $w_n(E) \equiv e(E) \operatorname{mod}2 .$ ::: ::: {.proof title="?"} Both classes satisfy the Whitney sum formula, so $E &= \bigoplus L_i \implies c(E) = \prod c(L_i) \\ \implies e(E) &= \prod e(L_i) \\ &= \prod (1 + c_1(L_i)) \\ &= 1 + \sigma_1(c_1 L_1, \cdots, c_1 L_n) + \cdots + \prod c_1(L_i) \\ &\coloneqq 1 + \sigma_2(c_1 L_1, \cdots, c_1 L_n)+ \cdots + c_n(E) .$ Then $$e(E) = \prod e(L_i) = \prod c_1(L_i) = c_n(E)$$. Now the claim follows from the splitting principle and naturality of characteristic classes. ::: ::: {.exercise title="?"} Show that for any $$E\in { {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{C}})_{/ {E}}$$, $c_i(E) \equiv w_{2i} (E) \operatorname{mod}2 .$ ::: ::: {.lemma title="?"} For $$E\in{ {\mathsf{Bun}}\qty{\operatorname{GL}_{n}} }({\mathbb{R}})_{/ {X}}$$,\ Then $c_1(E) = c_1(\bigwedge\nolimits^n E) ,$ noting that $$\bigwedge\nolimits^n E$$ is a line bundle. A similar claim holds for $$c_1$$. ::: ::: {.fact} $c_1(L_1\otimes L_2) = c_1(L_1) + c_1(L_2) .$ ::: :::{.remark} ETS this holds for $$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)$$, we have $c_1(E) = \sigma_1(c_1 L_1,\cdots, c_1 L_n) = \sum c_1 L_i .$ :: ::: {.remark} Recall that the Euler class is the obstruction to finding an obstruction to extending a section over the $$n{\hbox{-}}$$skeleton. ::: ## Obstruction Theory ::: {.lemma title="?"} $$w_1 E$$ is the obstruction to orienting $$E$$ in the following sense: for any $$S^1 \xrightarrow{f} E$$, write $$f_*[S]$$ for the image of the fundamental class, then ${\left\langle {w_1(E)},~{f_*[S^1] } \right\rangle} = \chi_{f^* E \text{ is orientable}} = \begin{cases} 1 & f^*E \text{ is nonorientable} \\ 0 & f^* E \text{ is orientable}. \end{cases} .$ ::: ::: {.remark} Why? For example, consider $$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$$ will be nonorientable. ::: ::: {.proof title="?"} If $$E$$ is a line bundle, define $$\tilde w_1(E) \in H^1(X)$$ by the following ${\left\langle {\tilde w_1(E)},~{ f_* [S^1] } \right\rangle} = \chi_{f^* E \text{ nonorientable}} .$ This is natural under pullback, consider the following diagram: {=tex} \begin{tikzcd} {f^* h^* E} && {h^*E} && E \\ \\ {S^1} && Y && X \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=3-5] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=1-3] \arrow["f", from=3-1, to=3-3] \arrow["h", from=3-3, to=3-5] \arrow[from=1-1, to=3-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJoXipFIl0sWzQsMCwiRSJdLFs0LDIsIlgiXSxbMCwyLCJTXjEiXSxbMiwyLCJZIl0sWzAsMCwiZl4qIGheKiBFIl0sWzAsMV0sWzEsMl0sWzAsNF0sWzUsMF0sWzMsNCwiZiJdLFs0LDIsImgiXSxbNSwzXSxbNSw0LCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbMCwyLCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) We have $&{\left\langle { \tilde w_1 h^* E},~{ f_* [S^1] } \right\rangle} &= {\left\langle { \tilde w_1 h^* E},~{ (h\circ f)_* [S^1] } \right\rangle} \\ &= {\left\langle { h^*\tilde w_1 h^* E},~{ f_* [S^1] } \right\rangle} \\ &= \tilde w_1 (h^* E) \\ &= h^* \tilde w_1(E) .$ We need to show $$\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$$. Take the pullback: {=tex} \begin{tikzcd} {i^*\gamma} && \gamma \\ \\ {{\mathbb{RP}}^1 \cong S^1} && {{\mathbb{RP}}^\infty} \arrow["i", from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiaV4qXFxnYW1tYSJdLFsyLDIsIlxcUlBeXFxpbmZ0eSJdLFswLDIsIlxcUlBeMSBcXGNvbmcgU14xIl0sWzMsMiwiaSJdLFswLDJdLFsxLDNdLFsxLDBdLFsxLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Then $$i^* \gamma$$ is not orientable since locally this looks like a Mobius band, so we have ${\left\langle {\tilde w_1( \gamma)},~{ i_* [{\mathbb{RP}}^1] } \right\rangle} = 1 .$ Now $$w_1(E) = w_1(\bigwedge\nolimits^n E)$$ and we have to bundles: {=tex} \begin{tikzcd} {f^*(\bigwedge\nolimits^n E)} && {f^* E} \\ \\ & {S^1} \arrow[from=1-1, to=3-2] \arrow[from=1-3, to=3-2] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJmXiooXFxFeHRhbGdebiBFKSJdLFsyLDAsImZeKiBFIl0sWzEsMiwiU14xIl0sWzAsMl0sWzEsMl1d) These are either simultaneously orientable or simultaneously nonorientable, by considering $$\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}}$$. ::: ::: {.theorem title="?"} $$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$$. ::: ::: {.remark} For $$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) } }$$. ::: # Pontryagin Classes (Tuesday, November 23) ## Complexification ::: {.question} How to we get integral cohomology classes when we don't have a complex structure? ::: ::: {.definition title="Complexification"} Given a real vector bundle $$E$$, its **complexification** is defined as $E\otimes_{\mathbb{R}}{\mathbb{C}} \coloneqq\displaystyle\coprod_{x\in X} E_x\otimes_{\mathbb{R}}{\mathbb{C}} .$ Then $$\dim_{\mathbb{R}}E\otimes_{\mathbb{R}}{\mathbb{C}}= 2\dim_{\mathbb{R}}E$$. ::: ::: {.remark} In terms of transition functions (i.e. Čech cohomology data), this is the inclusion $$\operatorname{GL}_n({\mathbb{R}}) \hookrightarrow\operatorname{GL}_n({\mathbb{C}})$$. We can now consider the Chern classes $$c_i(E\otimes{\mathbb{C}})$$. ::: ::: {.definition title="Conjugate bundle"} Given a complex vector bundle $$E$$, there is a **conjugate bundle** $$\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$$. ::: ::: {.lemma title="?"} For any real vector bundle $$E$$, $E\otimes{\mathbb{C}}\cong \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{E}}\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu .$ ::: ::: {.proof title="?"} The transition functions for the former are in $$\operatorname{GL}_n({\mathbb{R}})$$, which is fixed under conjugation. ::: ::: {.remark} Note that $$E\otimes{\mathbb{C}}\cong E \oplus E$$, and we can map ${\mathbb{R}}{ {}^{ \scriptscriptstyle\oplus^{2} } }\otimes{\mathbb{C}}&{ { \, \xrightarrow{\sim}\, }}{\mathbb{C}}{ {}^{ \scriptscriptstyle\oplus^{2} } } \\ (x\oplus y) \otimes\left\langle{1, i}\right\rangle &\mapsto (x, ix) \oplus (y, iy) .$ ::: ::: {.exercise title="?"} Show that if $$E$$ is a complex vector bundle, $E\otimes{\mathbb{C}}\cong E \oplus \mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu ,$ and $c_i(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = (-1)^i c_i(E) .$ ::: ## Pontryagin Classes ::: {.lemma title="?"} For $$i$$ odd, $$c_i(E\otimes{\mathbb{C}})$$ is 2-torsion. ::: ::: {.proof title="?"} On one hand, $E\otimes{\mathbb{C}}\cong \mkern 1.5mu\overline{\mkern-1.5muE\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu \implies c_i(E\otimes{\mathbb{C}}) = c_i(\mkern 1.5mu\overline{\mkern-1.5muE\otimes{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu) = (-1)^i c_i(E\otimes{\mathbb{C}}) .$ So if $$i$$ is odd, $$c_i = -c_i$$ and thus $$2c_i = 0$$. ::: ::: {.definition title="Pontryagin classes"} The **$$i$$th Pontryagin class** is $p_i(E) = (-1)^{i} c_{2i} (E\otimes{\mathbb{C}}) \in H^{4i}(X) .$ ::: ::: {.remark} Suppose $$R$$ is a coefficient ring where 2 is invertible (or not a zero divisor). Some properties of the $$p_i$$: a. $$p_i$$ is natural, so $$p_i (f^* E) = f^* p_i(E)$$. b. $$p_{i> \dim(E)/2}$$. c. $$p(E \bigoplus E') = p(E) \smile p(E')$$. Why? Note $$p(E) = 1 - c_2 + c_4 - c_6 + \cdots$$, so multiplying two such things yields $p(E) p(E') = 1 - (c_2 + c_2') + (c_4 + c_2c_2' + c_4') - (c_6 + c_4 c_2' + c_2 c_4' + c_6') - \cdots = p(E \oplus E') .$ d. If $$\dim_{\mathbb{R}}E = 2n$$ then $$p_n(E) = e(E)^2$$. Why? $p_n(E) = (-1)^n c_{2n}(E\otimes{\mathbb{C}}) = (-1)^n e(E\otimes{\mathbb{C}}) = (-1)^n (-1)^{2n(2n+1) \over 2} = (-1)^{n + n(2n+1)} e(E)^2 = (-1)^{n(2n+2)}e(E)^2 = e(E)^2 .$ e. If $$E$$ is a complex line bundle, $c(E \oplus \mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = c(E\otimes{\mathbb{C}}) = 1 - p_1(E) + p_2(E) - \cdots = c(E) c(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu) = (1 + c_1(E) + c_2(E) + \cdots)(1 - c_1(E) + c_2(E) - \cdots) .$ ::: ::: {.example title="?"} We can compute $$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$$. ::: ::: {.corollary title="?"} If $${\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$$. ::: ::: {.theorem title="?"} $H^*({{\mathbf{B}}{\operatorname{SO}}}_{2n+1}; R) &{ { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_n] \\ H^*({{\mathbf{B}}{\operatorname{SO}}}_{2n}; R) &{ { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_{n-1}, e] \\ H^*({{\mathbf{B}}{\operatorname{O}}}_{2n+1}; R) &{ { \, \xrightarrow{\sim}\, }}H^*({{\mathbf{B}}{\operatorname{O}}}_{2n}; R) { { \, \xrightarrow{\sim}\, }}R[p_1,\cdots, p_{n}] \\ ,$ where $$p_i$$ are the Pontryagin classes for the respective canonical bundles. ::: # Bordism (Tuesday, November 30) ::: {.definition title="Bordism"} Let $$M_1, M_2\in {\mathsf{sm}}{\mathsf{Mfd}}^n$$ be closed, then $$M_1$$ is **bordant** to $$M_2$$ iff $$\exists W\in {\mathsf{sm}}{\mathsf{Mfd}}^{n+1}$$ with $${{\partial}}W = M_1 {\textstyle\coprod}M_2$$. ::: ::: {.remark} This defines an equivalence relation on $${\mathsf{sm}}{\mathsf{Mfd}}^n$$, and yields a ring $$(\prod_n {\mathcal{N}}^n, {\textstyle\coprod}, \times)$$. We'll add extra structure to get refinements of this ring, for which we'll need a bit about stable normal bundles and the Whitney embedding theorem. ::: ::: {.theorem title="Whitney Embedding"} Any $$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}$$ are isotopic. ::: ::: {.remark} Moreover if $$i, i': M\hookrightarrow{\mathbb{R}}^n$$ with $$n\geq 2n+3$$, then the normal bundles $$\nu M, (\nu M)'$$ are bundle isomorphic. ::: ::: {.corollary title="?"} Every such $$M$$ has a stable normal bundle $$\nu_M$$ which is well-defined up to adding trivial line bundles. ::: ::: {.remark} Note that adding trivial line bundles yields a tower $${{\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}}$$, etc: {=tex} \begin{tikzcd} & \cdots && {X_k} && {X_{k+1}} && \cdots \\ \\ & \cdots && {{{\mathbf{B}}{\operatorname{O}}}_k} && {{{\mathbf{B}}{\operatorname{O}}}_{k+1}} && \cdots \\ M \arrow[from=1-2, to=1-4] \arrow[from=1-4, to=1-6] \arrow[from=1-6, to=1-8] \arrow[from=3-2, to=3-4] \arrow[from=3-4, to=3-6] \arrow[from=3-6, to=3-8] \arrow[from=1-4, to=3-4] \arrow[from=1-6, to=3-6] \arrow["{\nu_M}"{description}, curve={height=6pt}, from=4-1, to=3-4] \arrow[curve={height=24pt}, from=4-1, to=3-6] \arrow[curve={height=12pt}, dashed, from=4-1, to=1-4] \arrow[curve={height=12pt}, dashed, from=4-1, to=1-6] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMSwwLCJcXGNkb3RzIl0sWzMsMCwiWF9rIl0sWzUsMCwiWF97aysxfSJdLFs3LDAsIlxcY2RvdHMiXSxbMywyLCJcXEJPX2siXSxbNSwyLCJcXEJPX3trKzF9Il0sWzcsMiwiXFxjZG90cyJdLFsxLDIsIlxcY2RvdHMiXSxbMCwzLCJNIl0sWzAsMV0sWzEsMl0sWzIsM10sWzcsNF0sWzQsNV0sWzUsNl0sWzEsNF0sWzIsNV0sWzgsNCwiXFxudV9NIiwxLHsiY3VydmUiOjF9XSxbOCw1LCIiLDEseyJjdXJ2ZSI6NH1dLFs4LDEsIiIsMSx7ImN1cnZlIjoyLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbOCwyLCIiLDEseyJjdXJ2ZSI6Miwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) ::: ::: {.definition title="Structures"} An $$X{\hbox{-}}$$structure on $$M$$ is a family of lifts for large $$k$$: {=tex} \begin{tikzcd} && {X_k} \\ \\ M && {{{\mathbf{B}}{\operatorname{O}}}_k} \arrow[from=1-3, to=3-3] \arrow["{\nu_M^k}"', from=3-1, to=3-3] \arrow[dashed, from=3-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMiwwLCJYX2siXSxbMCwyLCJNIl0sWzIsMiwiXFxCT19rIl0sWzAsMl0sWzEsMiwiXFxudV9NXmsiLDJdLFsxLDAsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) We also require these to be compatible with the morphism $${{\mathbf{B}}{\operatorname{O}}}_k\to {{\mathbf{B}}{\operatorname{O}}}_{k+1}$$ and $$X_k\to X_{k+1}$$. ::: ::: {.example title="?"} Some examples: - $${{\mathbf{B}}{\operatorname{SO}}}_k$$ for orientation, - $$V_k$$ (the Stiefel manifold) for framings. ::: ::: {.definition title="$X\\dash$bordism"} Given $$M_i$$ with $$X{\hbox{-}}$$structures, we say $$N$$ is an **$$X{\hbox{-}}$$bordism** if - $${{\partial}}N = M_1 {\textstyle\coprod}M_2$$ - There is an $$X{\hbox{-}}$$structure on $$N$$ that restricts to the $$X{\hbox{-}}$$structures on $$M_1$$ and $$M_2$$. ::: ::: {.remark} For orientations, we get a bordism ring $$\Omega_n$$, and for framings we get $$\Omega_n^{\mathop{\mathrm{Fr}}}$$. ::: ::: {.theorem title="Identification of framed bordism group"} $\Omega_n^{\mathop{\mathrm{Fr}}} \cong \pi_n^s ,$ the stable homotopy groups of spheres $$\colim_k \pi_{n+k} S^n$$ (which stabilize for $$k>n+1$$). ::: ::: {.definition title="Stiefel-Whitney numbers"} Given $$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)$$. We can evaluate this to get $w_I(M) \coloneqq{\left\langle {W_n},~{[M]} \right\rangle} \in C_2 .$ Note that the condition on $$I$$ guarantees that $$W_n\in H^n$$. ::: ::: {.remark} For $$n=3$$, the only possibilities for $$I$$ are - $$(3,0,0) \leadsto W_n = w_1^3({\mathbf{T}}M)$$ - $$(1, 1, 0 \leadsto W_n = w_1({\mathbf{T}}M)w_2({\mathbf{T}}M) )$$ - $$(0,0,1) \leadsto W_n = w_3({\mathbf{T}}M)$$ ::: ::: {.definition title="Pontryagin number"} Given an *oriented* $$M \in {\mathsf{sm}}{\mathsf{Mfd}}^n$$, the **Pontryagin number** of $$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}$$, and evaluating $p_I(M) \coloneqq{\left\langle {P_n},~{[M]} \right\rangle} .$ ::: ::: {.corollary title="?"} If any Pontryagin number is nonzero, then $$M$$ can not admit an orientation reversing diffeomorphism. ::: ::: {.proof title="?"} Use that $P_I(-M) = {\left\langle {\prod p_1 {{\mathbf{T}}M} },~{[-M]} \right\rangle} = -P_I(M) ,$ but also $P_I(M) &= {\left\langle {\prod p_i {\mathbf{T}}M},~{[M]} \right\rangle} \\ &= {\left\langle {f^* \qty{ \prod p_i {\mathbf{T}}M} },~{[M]} \right\rangle} \\ &= {\left\langle {\prod p_i {\mathbf{T}}M},~{[f_* M]} \right\rangle} \\ &= {\left\langle {\prod p_i {\mathbf{T}}M},~{[- M]} \right\rangle} \\ &= P_I(-M) .$ ::: ::: {.example title="?"} Some examples: - $$P_I({\mathbb{CP}}^{2n}) \neq 0$$ by a computation. - $$P_I({\mathbb{CP}}^{2n+1}) = 0$$ since conjugating the complex structure $$J\mapsto -J$$ is an orientation reversing diffeomorphism. ::: ::: {.theorem title="Pontryagin, Thom"} $$M=0$$ in $${\mathcal{N}}_n$$ iff $$w_I(M) = 0$$ for all $$I$$. ::: ::: {.proof title="?"} $$\implies$$: Easy, a simple algebraic topology calculation. $$\impliedby$$: Difficult and omitted! ::: ::: {.corollary title="?"} There is an injective group morphism ${\mathcal{N}}_n ({\mathcal{N}}_{I_1}, \cdots, {\mathcal{N}}_{I_k}) \to C_2{ {}^{ \scriptscriptstyle\times^{k} } } .$ ::: ::: {.theorem title="Pontryagin-Thom"} If $$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$$. ::: ::: {.remark} Milnor and Wall: the only torsion in $$\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$$. For a proof (for at least the first statement), see Milnor-Stasheff. ::: ::: {.remark} We can kill torsion to get an injective map $\Omega_n \otimes{\mathbb{Q}}\xrightarrow{(P_{I_1}, \cdots, P_{i_k} ) } {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{k} } } .$ ::: ::: {.theorem title="Pontryagin-Thom"} The ring $${\mathcal{N}}_n$$ is a polynomial ring over $${\mathbb{F}}_2$$ with one generator in each dimension: ${\mathcal{N}}_n \cong {\mathbb{F}}_2[\left\{{v_i {~\mathrel{\Big\vert}~}i\neq 2^i-1}\right\}] .$ ::: ::: {.theorem title="Pontryagin-Thom"} The ring $$\Omega_* \otimes{\mathbb{Q}}$$ is polynomial ring over $${\mathbb{Q}}$$ generated by $${\mathbb{CP}}^n$$ for $$n=1,2,\cdots$$: $\Omega_* \cong \cong {\mathbb{Q}}[v_1,v_2,\cdots] && {\left\lvert {v_i} \right\rvert} = i .$ ::: # Thursday, December 02 ::: {.remark} Today: the Hirzebruch signature theorem and exotic $$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} } } }$$ can only be torsion. ::: ::: {.definition title="Signature"} For $$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}$$, there is a middle-dimensional pairing $H^{2n}(M; {\mathbb{Z}}){ {}^{ \scriptstyle\otimes_{{\mathbb{Z}}}^{2} } } &\to {\mathbb{Z}}\\ a\otimes b &\mapsto {\left\langle {a \smile b},~{[M]} \right\rangle} ,$ and we write $$\sigma(M)$$ for the signature of this form. ::: ::: {.theorem title="?"} The signature induces a ring morphism $\sigma: (\Omega_*, {\textstyle\coprod}, \times) \to ({\mathbb{Z}}, +, \cdot) ,$ so - $$\sigma(M {\textstyle\coprod}N) = \sigma(M) + \sigma(N)$$, - $$\sigma(M\times N) = \sigma(M) \sigma(N)$$, - $$\sigma(M) = 0$$ if $$M = {{\partial}}M'$$ for some $$M'$$. ::: ::: {.lemma title="?"} If there is a half-dimensional subspace $$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$$. ::: ::: {.proof title="?"} Use Poincare duality to get an isomorphism: {=tex} \begin{tikzcd} {H^{2n}(N; {\mathbb{Q}})} && {H^{2n}(M; {\mathbb{Q}})} && {H^{2n+1}(N, M; {\mathbb{Q}})} \\ \\ && {H_{2n}(M; {\mathbb{Q}})} && {H_{2n}(N; {\mathbb{Q}})} \arrow["{i_*}", from=3-3, to=3-5] \arrow["{\cong, \mathrm{PD}}"', from=1-5, to=3-5] \arrow["\cong", from=1-3, to=3-3] \arrow["{f}", from=1-3, to=1-5] \arrow["{i^*}", from=1-1, to=1-3] \end{tikzcd}  > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbNCwwLCJIXnsybisxfShOLCBNOyBcXFFRKSJdLFs0LDIsIkhfezJufShOOyBcXFFRKSJdLFsyLDAsIkheezJufShNOyBcXFFRKSJdLFsyLDIsIkhfezJufShNOyBcXFFRKSJdLFswLDAsIkheezJufShOOyBcXFFRKSJdLFszLDEsImlfKiJdLFswLDEsIlxcY29uZywgXFxQRCIsMl0sWzIsMywiXFxjb25nIl0sWzIsMF0sWzQsMiwiaV4qIl1d) Note that $$\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^*$$, ${\left\langle { i^*(a\smile b)},~{[M]} \right\rangle} = {\left\langle { \alpha\smile\beta},~{i_*[M]} \right\rangle} =0 ,$ since $$i_*[M] = 0$$. Thus $$\sigma(M) = 0$$. ::: ::: {.corollary title="?"} One can compute the signature in terms of Pontryagin numbers, i.e. there is a formula for $$\sigma(M)$$ in terms of $$\left\{{p_{I_n}(M)}\right\}$$. The explicit formula is the **Hirzebruch signature theorem**. ::: ::: {.example title="?"} Write $$\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}$$, so we have a formula $1 - p_1({\mathbb{CP}}^2) &= c({\mathbb{CP}}^2) \cdot c(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu) \\ &= (1+x)^3(1-x)^3 \\ &= (1-x^2)^3 \\ &= 1 + -3x^2 + 3x^4 - x^6 \\ &= 1 - 3x^2 ,$ where we've used that $$H^{\geq 4} = 0$$. So $$p_1({\mathbb{CP}}^2)= 3x^2$$, which implies ${\left\langle { p_1 {\mathbb{CP}}^2},~{[{\mathbb{CP}}^2]} \right\rangle} = {\left\langle {3x^2},~{[{\mathbb{CP}}^2]} \right\rangle} = 3 .$ Using that $$\sigma({\mathbb{CP}}^2) = 1$$, we can deduce $$\sigma(M) = p_1(M)/3$$ by considering eigenvalues of the intersection pairing. ::: ::: {.example title="?"} Consider $$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$$,then $1 - p_1(v_4) + p_2(v_4) &= c(v_4)c(\mkern 1.5mu\overline{\mkern-1.5muv\mkern-1.5mu}\mkern 1.5mu_4) \\ &= (1+x)^5(1-x)^5 \\ &= (1-x^2)^5 \\ &= 1-5x^2 + 10x^4 ,$ so $$p_1(v_4) = 5x^2$$ and $$p_2(v_4) = 10x^4$$. Then ${\left\langle {p_1^2 v_4},~{[v_4]} \right\rangle} &= {\left\langle {25x^4},~{[v_4]} \right\rangle} = 25 \\ {\left\langle {p_2 v_4},~{[v_4]} \right\rangle} &= {\left\langle {10x^4},~{[v_4]} \right\rangle} = 10 .$ Similarly, $1 + p_1(v_4^2) + p_2(v_4^2) = (1+ 3x^2)(1+3y^2) = 1 + 3x^2 + 3y^2 + 9x^2y^2 ,$ so ${\left\langle {p_1^2 v_2^2},~{[v_2^2]} \right\rangle} = {\left\langle {(3x^2 + 3y^2)^2 },~{[v_2^2]} \right\rangle} = {\left\langle {9x^4 + 9y^4 + 18x^2 y^2 },~{[v_2^2]} \right\rangle} = 18 ,$ since the cohomology ring is $${\mathbb{F}}[x]/\left\langle{x^2}\right\rangle \otimes{\mathbb{F}}[y]/\left\langle{y^2}\right\rangle$$, and a similar calculation shows ${\left\langle {p_2(v_2^2)},~{[v_2^2]} \right\rangle} = 9 .$ Summarizing, where we abuse notation and identity classes with numbers, - $$p_1^2(v_4) = 25$$, - $$p_2(v_4) = 10$$, - $$p_1^2(v_2^2) = 18$$, - $$p_2(v_2^2) = 9$$. Writing $$\sigma = ap_1^2 + b p_2$$, $1 = \sigma(v_4) &= 25 a + 10 b \\ 1 = \sigma(v_2^2) &= 18a + 9b .$ This yields $$b = {1\over 9}-2a$$ and $$1 = 25a + 10\qty{{1\over 9} - 2a}$$ yields $$a=-1/45$$ and $$b=7/45$$. Thus $\sigma = {1\over 45}\qty{2p_2 - p_1^2} .$ ::: ::: {.theorem title="Thom"} $\Omega_7 = 0 .$ ::: ::: {.definition title="?"} Given $$M^7$$ with $$H^3(M) = H^4(M)$$ and suppose $$M = {{\partial}}B^*$$. Let $$i: H^4(B; M) \to H^4(B)$$ and define $\lambda(M) \coloneqq 2 {\left\langle { (i^{-1}p_1 {\mathbf{T}}B)^2 },~{[B]} \right\rangle} - \sigma(B) .$ Here $$\sigma(B)$$ is the signature of $H^4(B, M ; {\mathbb{Q}}){ {}^{ \scriptstyle\otimes_{{\mathbb{Z}}}^{2} } } &\to {\mathbb{Z}}\\ a\otimes b &\mapsto {\left\langle {a \smile b},~{[B]} \right\rangle} .$ ::: ::: {.proposition title="?"} $$\lambda(M)$$ is independent of $$B$$. ::: ::: {.remark} Let $$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)$$. We have $45\sigma(C) + {\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\ \implies -40\sigma(C) + {\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\ \implies \sigma(C) + 2{\left\langle {p_1(C)^2},~{[C]} \right\rangle} &\equiv 0 \operatorname{mod}7 \\ \implies (\sigma(B_1) - \sigma(B_2) ) - 2 \qty{ {\left\langle {(i^{-1}p_1 B_1)^2 },~{[B_1]} \right\rangle} \cdots } &= \lambda_{B_1}(M) - \lambda_{B_2}(M) .$ ::: ::: {.remark} If the $$\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$$ are homeomorphic by finding a Morse function with exactly 2 critical points. ::: # Problem Set 1 ## 1 ::: {.problem title="?"} With the definition of a vector bundle from class, show that the vector space operations define continuous maps: $&+: E { \underset{\scriptscriptstyle {B} }{\times} } E \rightarrow E \\ &\times: \mathbb{R} \times E \rightarrow E$ ::: ::: {.remark} Definition of vector bundle: need charts $$(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$$. What are these maps?? ::: ## 2. ::: {.problem title="?"} Suppose you are given the following data: - Topological spaces $$B$$ and $$F$$ - A set $$E$$ and a map of sets $$\pi: E \rightarrow B$$ - An open cover $$\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$$. Give conditions on the maps $$\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\}$$ as an atlas. ::: ## 3. ::: {.problem title="?"} An *oriented $$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$$, $\left.\phi\right|_{E_{b^{\prime}}}: E_{b^{\prime}} \rightarrow \mathbb{R}^{n}$ is orientation-preserving. Show that given an oriented $$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$$-plane bundle. ::: ## 4. ::: {.problem title="?"} A Riemannian metric on a vector bundle $$\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)$$-bundle there is an induced vector bundle with Riemannian metric. ::: ## 5 . ::: {.problem title="?"} What operation on principal $$O(n)$$-bundles corresponds to dualizing a vector bundle? What about the direct sum of vector bundle? ::: ## 6. ::: {.problem title="?"} For nice spaces $$X$$ (e.g. CW complexes) and abelian groups $$G$$, there is a canonical isomorphism $\check{H}^{i}(X ; G) \cong H^{i}(X ; G)$ between Čech and singular cohomology of $$X$$ with coefficients in $$G$$. > A nice, readable proof can be found in Frank Warner's Foundations of Differential Manifolds and Lie Groups, Chapter 5. In the rest of this problem, cohomology either means Čech cohomology or singular cohomology after applying this isomorphism. (a) Let $$\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}))$$. Show that the sign of the determinant $\operatorname{sgn}\operatorname{det}: \operatorname{GL}_n(\mathbb{R}) \rightarrow\{\pm 1\} \cong \mathbb{Z} / 2 \mathbb{Z}$ induces a map $\check{H}^{1}(B ; G L(n, \mathbb{R})) \rightarrow \check{H}^{1}(B ; \mathbb{Z} / 2 \mathbb{Z}) ,$ and so $$\phi$$ induces an element $$w_{1}(E) \in H^{1}(B ; \mathbb{Z} / 2 \mathbb{Z})$$. (b) Compute $$w_{1}$$ for the trivial line bundle (1-dimensional vector bundle) over the circle and for the Möbius band. (c) Prove that (for nice spaces) a line bundle $$\pi: E \rightarrow B$$ is trivial if and only if $$w_{1}(E)=0 \in$$ $$H^{1}(B ; \mathbb{Z} / 2 \mathbb{Z})$$ ::: ## 7. ::: {.problem title="?"} Show that the exact sequence of abelian topological groups $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{R} \rightarrow S^{1}=G L(1, \mathbb{C}) \rightarrow 0$ induces an exact sequence in Čech cohomology $\check{H}^{1}(B, \mathbb{Z}) \rightarrow \breve{H}^{1}(B, \mathbb{R}) \rightarrow \check{H}^{1}\left(B ; S^{1}\right) \stackrel{\delta}{\rightarrow} \check{H}^{2}(B ; \mathbb{Z})$ Given a complex line bundle (principal $$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}$$. ::: [^1]: Here *locally finite* means every point is covered by finitely many opens in the cover. [^2]: $${\mathbb{Z}}$$ can be replaced with $${\mathbb{Z}}/2$$ if $$M$$ is not oriented.