# Thursday, September 23

:::{.remark}
Last time: we were trying to construct $\EO_n$ and $\BO_n$, and we defined $V_n(\RR^\infty) = \directlim_{k} V_n(\RR^k)$, where $V_n(\RR^k)$ was the space of $n$ orthonormal vectors in $\RR^k$.
There is a map $V_n(\RR^\infty)\to \Gr_n(\RR^\infty)$, which will be our candidate for $\EO_n \to \BO_n$.
:::

:::{.lemma title="?"}
$V_n(\RR^\infty) \mapsvia{\pi}  \Gr_n(\RR^\infty) \in \Prin\Bun(\Orth_n)$.
:::

:::{.proof title="?"}
We'll show this directly in charts.
Let $W\in\Gr_n(\RR^\infty)$ be an $n\dash$dimensional plane, the consider an open neighborhood 
\[
U_W \da \ts{W' \in \Gr_n(\RR^\infty) \st W^\perp \intersect W' = 0}
.\]
For any such $W'$, we have a map $W'\to W$ given by orthogonal projection, which is an isomorphism since $W^\perp \intersect W' = 0$.


:::{.claim}
\[
\pi\inv(U_W) \cong U_W \times \Orth_n
.\]
:::

Fix some $\alpha\in \pi\inv(U_W)$ (an orthonormal basis for $W$), apply $f\inv$ to get $f\inv( \alpha)$, then apply Gram-Schmidt to get $\tilde a$, an orthonormal basis for $W'$.
Define $F_{W'}$ to be this composition;
this yields a bijection $\pi\inv(W) \iso \pi\inv(W')$ for all $W'\in U_W$, namely
\[
U_w \times \Orth_n &\to \pi\inv(W) \\
(W', A) &\mapsto F_{W'}( \alpha) \cdot A
.\]
The claim is that this trivializes the bundle, since this constructs a local section using $\Orth_n$ translations:

\begin{tikzcd}
	{s(W')\cdot A} && {(W', A)} \\
	{\pi\inv(U_W)} && {U_w \times \Orth_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 $P\inv$ and the applying Gram-Schmidt to get an orthonormal basis for $W'$.
:::

:::{.remark}
\envlist

- Replace $\Orth_n$ with $U_n$ and $\RR$ with $\CC$ to get $\Gr_n(\CC^\infty) = \B U_n$.
- $V_n(\RR^\infty)/ \SO_n = \B \SO_n$ yields the Grassmannian of *oriented* planes.
- For $H\leq G$, we have $\E H = \E G$ and $\B H = \E G/H$.

:::

> Question: can you get $\B \Spin_n$ this way?

:::{.remark}
An alternative description of $\EO_n$ and $\BO_n$, due to Milnor-Stasheff: write $\BO_n = \Gr_n(\RR^\infty)$ and define the **canonical bundle** $\gamma$.
Recall that every principal $\Orth_n$ bundle is a pullback of the following form:

\begin{tikzcd}
	{P = f^* \EO_n} && {\EO_n = V_n(\RR^\infty)} \\
	\\
	X && {\BO_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, $\Prin(\Orth_n)\slice X = [X, \BO_n] = [X, \Gr_n(\RR^\infty)]$.
Then $\gamma_n \mapsvia{\pi} \Gr_n(\RR^\infty)$ is the $\RR^n\dash$bundle where $\pi\inv(v) = v = V$, regarded as a plane in $\RR^\infty$.
Another description comes from the mixing construction: $\gamma_n = V_n(\RR^\infty) \mix{\Orth_n} \RR^n \to \Gr_n(\RR^\infty)$.

\begin{tikzcd}
	{[\vector v_1, \cdots, \vector v_n] \times [t_1,\cdots, t_n]} && {\sum t_i \vector v_i} \\
	{V_n(\RR^\infty) \times \RR^n} && {\gamma_n} \\
	\\
	{\Gr_n(\RR^\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 $E' \embeds E$ over $X$:

\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 $f_x: \ro{E'}{x} \to \ro{E}{x} \in \Mat(m\times n; \RR)$ is a linear map to an $n\dash$dimensional subspace $\ro{E}{x}$, where $\dim_\RR \ro{E'}{x} = n$ and $\dim_\RR \ro{E}{x} = m$.
:::

:::{.lemma title="?"}
Every vector bundle $E \mapsvia{\pi} X$ with $X\in \CW$ compact is a subbundle of a trivial bundle.
:::

:::{.proof title="?"}
Take $\ts{(U_i, \phi_i)}_{i=1}^m\covers X$ a finite cover by charts, and choose a subordinate partition of unity $\ts{\chi_i}_{i=1}^m$ such that $\supp \chi_i \subseteq U_i$.
Then define
\[
\psi: E &\to \RR^{nm} =
\RR^n \times \RR^n \times \cdots \times \RR^n
\\
v &\mapsto \tv{\chi_1(v) \phi_1(v),  \chi_2(v) \phi_2(v), \cdots, \chi_m(v) \phi_m(v)} 
.\]
This exhibits $(E \to X) \leq (X\times \RR^{nm} \to X)$ as a subbundle.
:::

:::{.lemma title="?"}
Every $(E\to X)\in \VectBunrk{n}\slice X$ for $X\in \CW$ compact is a pullback of the canonical bundle along some map $f:X\to \BO_n$.
:::

:::{.example title="?"}
For $E \mapsvia{\pi} X$ and $f: X\to \BO_n$, \(\psi: E\to \RR^{nm} \subseteq \RR^\infty \) and we can take $f(x) \da \psi(\pi\inv(x)) \in \Gr_n(\RR^\infty)$ to get $f^* \gamma_n \cong E$.
:::

:::{.lemma title="?"}
If $f^* \gamma_n \cong E$ and $g^* \gamma_n \cong E$, then $f \homotopic g$.
:::

:::{.proof title="?"}
Corresponding to $f^* \gamma_n \cong E$ we get a map $\tilde f: E\to \RR^{\infty}$ which restricts to an embedding on all fibers, and similarly $g^*\gamma_n \cong E$ yields $\tilde g: E\to \RR^\infty$.
So take
\[
L_t: \RR^\infty &\to \RR^\infty \\
\vector x &\mapsto (t-1)\tv{x_1,x_2,\cdots} + t\tv{x_1, 0, x_2, 0, x_3, 0, \cdots}
,\]
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").
:::