# Thursday, September 09

## Corollaries of the homotopy bundle lemma

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

\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 $\ro{f}{0\times P_0}$ is the identity.
:::

:::{.corollary title="?"}
If $P\to I\times X \in \Prin\Bun(G)$ then $P_0 \cong P_1$ where $P_i \da \ro{P}{i\times X}$.
:::

:::{.corollary title="?"}
If $f_0, f_1: Y\to X$ with $P \mapsvia{\pi} X$, then $f_0^* P \cong f_1^* P$ are isomorphic bundles.
:::

:::{.corollary title="?"}
If $X$ is contractible, then any $P\in \Prin\Bun(G)\slice X$ is trivial.
:::

:::{.proof title="?"}
Consider the two maps

\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 $f_0 \homotopic f_1$, and conclude by noting that 
\[
f_0^*P = X\fiberprod{x_0} P = X\times \pi\inv(x_0) = X\times G
\]
and $f_1^* P = P$. 
:::

##  Existence/Uniqueness of Metrics

:::{.definition title="Riemannian metrics"}
A **Riemannian metric** on a vector bundle $E \mapsvia{\pi} X$ is a continuous map $E\fiberprod{X} E\to \RR$ 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 $\GL_n(\RR)$ to $\Orth_n(\RR)$.
:::

:::{.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[^loc_finite]
refinement $\mcu = \ts{U_i}_{i\in I}$ and pick a partition of unity $\ts{\chi_i}_{i\in I}$ subordinate to $\mcu$.

[^loc_finite]: 
Here *locally finite* means every point is covered by finitely many opens in the cover.

Define an inner product $g_i$ on $\pi\inv(U_i)$ where $\phi_i: \pi\inv(U_i)\to U_i \times \RR^n$ by pulling back the inner product on $\RR^n$, i.e. taking $e_1\mapsvia{\phi_i} (p_1, \vector v_1)$ and $e_2 \mapsvia{\phi}(p_2, \vector v_2)$ and setting 
\[
g_i(e_1, e_2) \da \inner{\vector v_1}{\vector v_2}_{\RR^n}
.\]
Then define 
\[
g_p(\wait, \wait) \da \sum_i \chi_i(p) g_i(\wait, \wait)
.\]

**Uniqueness:**
Consider two inner products $g_0(\wait, \wait), g_1(\wait, \wait)$ on the bundle $E \mapsvia{\pi} X$, then define
\[
g_t(\wait, \wait) = tg_0(\wait, \wait)+ (1-t)g_1(\wait, \wait)
.\]
Then $I\times E \mapsvia{\id, \pi} I\times X$ is a bundle, and $g_t$ is a Riemannian metric on $I\times E$.
Consider its corresponding principal bundle 
\[
P\to I\times X \in \Prin\Bun(\Orth_n(\RR))
\]
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 $\Orth_n$, this shows the two bundles with metric are isomorphic.
:::

:::{.definition title="Universal $G\dash$bundles"}
A **universal $G\dash$bundle** is a principal $G\dash$bundle $\pi: EG\to \B G$ such that $\pi_i EG = 0$ for all $i$ (so $EG$ is *weakly contractible*).
:::

:::{.example title="?"}
\envlist

- $\qty{\RR \to S^1}\in \Prin \Bun(\ZZ)\slice{S^1}$ since all of the regular covers are principal bundles.
  Since $\RR$ is contractible, this is the universal $\ZZ\dash$bundle, so $S^1 \homotopic \B \ZZ$.

- $\qty{S^\infty \to \RP^\infty} \in \Prin\Bun(C_2)$ is a universal $C_2\dash$bundle, so $\RP^\infty \homotopic \B C_2$
- $\qty{S^\infty \to \CP^\infty}$ is a universal $S^1 = U_1$ bundle, so $\CP^\infty \homotopic \B U_1 \homotopic \B S^1 \homotopic \B \CC\units$:

  \begin{tikzcd}
    {F_z = \ts{\lambda z \st \lambda \neq 0}} &&& {\tv{z_1, \cdots, z_n}} \\
    \CC\units && {\CC^n} \\
    \\
    && {\CP^{n-1}} & {z \da \tv{z_1: \cdots : z_n}}
    \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 $X\in \CW \subseteq \Top$ and $EG \mapsvia{\pi} \B G \in \Bun_G$ is universal, then there is a bijection
\[
\Prin\Bun(G)\slice X &\mapstofrom [X, \B G] \\
f*EG &\mapsfrom f
.\]

:::

:::{.lemma title="?"}
If $E \mapsvia{\pi} X$ 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 $n\dash$cell boundary $\bd e^n$.
- Write $\ro{E}{e_n} = e^n\times F$, which is contractible since $e_n$ is contractible.
- Then $s:\bd e^n = S^{n-1} \to F$ with $\pi_n(F) = 0$, so the section extends:

\begin{tikzcd}
	&& {\ro{E}{e^n}} && {e^n\times F} \\
	&& {} \\
	{\bd e^n} && {e^n}
	\arrow[from=1-3, to=3-3]
	\arrow["\cong", from=1-3, to=1-5]
	\arrow["{\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 $G\dash$action defined by $g\cdot e \da eg\inv$.
Recall that we can use the mixing construction:

\begin{tikzcd}
	F && P && EG && {P\fiberprod{G} 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 $G\dash$equivariant maps $P\to EG$.
Writing $s(x) = [P, e] \sim [Pg, g\cdot e] \da [Pg, g\inv e]$, so given $p\in \pi\inv(x)$ we can send $p\mapsto e\in EG$ such that $[p, e]\in s(x)$.
This is essentially currying an argument.
Conversely, given a $G\dash$equivariant map 
\[
P\to EG\\
p\mapsto e
,\]
we can define $s(x) \da [p, e]$ 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 $EG\to P\fiberprod{G} EG \to X$ has a section $s: X\to P\fiberprod{G}EG$ and this we get a $G\dash$equivariant map $P\to EG$ which induces a map $P/G \mapsvia{h} EG/G$, where $P/G = X$ and $EG/G = \B G$.

\begin{tikzcd}
	P &&&& {h^* EG} && EG \\
	&&&&& {} \\
	&&&& X && {\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 \mapsvia{\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.

:::

:::