# Wednesday, October 27

:::{.remark}
If $H \leq G \in \Alg\Grp$ is a closed linear subgroup and $Y\in\gspaces{G}$, then there is a commuting diagram

\begin{tikzcd}
	{G\mix{H}Y} && {G/H\times Y} \\
	\\
	& {G/H}
	\arrow["{\pr_1}", from=1-3, to=3-2]
	\arrow["\pi"', from=1-1, to=3-2]
	\arrow["\cong", tail reversed, from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJHXFxtaXh7SH1ZIl0sWzIsMCwiRy9IXFx0aW1lcyBZIl0sWzEsMiwiRy9IIl0sWzEsMiwiXFxwcl8xIl0sWzAsMiwiXFxwaSIsMl0sWzAsMSwiXFxjb25nIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn19fV1d)

The isomorphism $\phi$ is given by 
\[
[g ,y] &\mapsto (\bar g, gy) \\
[g, g\inv y] &\mapsfrom (g, y)
.\]


More generally, if $Y \subseteq X$ one often has $H \leq G$ with $H\actson Y$ and $G\actson X$.
In this case, $\phi: G\mix{H} Y \mapsvia{\phi} G/H \times X$ may be an embedding instead.
:::

:::{.proposition title="?"}
For $G$ connected reductive and $T\leq G$ is a maximal torus,
\[
R_G = 
\K_G(\pt) \cong \K_T(\pt)^W
= R_T^W
.\]
:::

:::{.slogan}
To compute $G\dash$equivariant \(\K\dash\)theory, it suffices to understand $T\dash$equivariant \(\K\dash\)theory and the action of the Weyl group.
:::

:::{.proof title="?"}
Define $\rho R_G\to R_T$ by restriction to $T$, so explicitly $\rho[v] = \sum_{ \lambda} m_{\lambda}e^{\lambda} \in R_T^W$ where the $m_{\lambda}$ are the multiplicities of $e^{\lambda}$ in $V_{ \lambda}$.
Set $G^{sr}$ to be the **semisimple regular** elements in $G$.
Note that a regular element $t\in T$ satisfies $t\not\in \ker \alpha$, and

1. $G^{sr} \embeds G$ is open and dense.
2. Every $g\in G^{rs}$ is conjugate to some $t\in T$.

Let $f\in \CC[G]^G$ be function invariant under $G\dash$conjugation, i.e. a class function, and suppose $\ro{f}{T} = 0$.
By (ii), $\ro{f}{G^{sr}} = 0$, so by (i) $f\equiv 0$ on $G$ since $f$ is continuous and zero on a dense subset.
There is a diagram:

\begin{tikzcd}
	{R(G)} && {R(T)^W} \\
	\\
	{\CC[G]^G} && {\CC[T]^W}
	\arrow["\rho", from=1-1, to=1-3]
	\arrow["{\wait\tensor_\ZZ \CC}"', from=1-1, to=3-1]
	\arrow["{\therefore }"', hook, from=3-1, to=3-3]
	\arrow["{\wait\tensor_\ZZ \CC}", from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJSKEcpIl0sWzIsMCwiUihUKV5XIl0sWzAsMiwiXFxDQ1tHXV5HIl0sWzIsMiwiXFxDQ1tUXV5XIl0sWzAsMSwiXFxyaG8iXSxbMCwyLCJcXHdhaXRcXHRlbnNvcl9cXFpaIFxcQ0MiLDJdLFsyLDMsIlxcdGhlcmVmb3JlICIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMywiXFx3YWl0XFx0ZW5zb3JfXFxaWiBcXENDIl1d)

Here the bottom map is injective by the previous argument.
To prove $\rho$ is surjective, fix $f\in R(T)^W$, then we'll produce an $h\in \CC[G]^G$ such that $\ro{h}{T} = f$.
Choosing $B \supseteq T$ a Borel, then for any such Borel containing $T$ is a canonical isomorphism $T \subseteq B \to B/U$ where we write $B = T\semidirect U$.
So identify $f$ with an element of $R(B/U)^W$.
Let $Z\da G\mix{B} B$, and instead of having the same action of $B$ on both factors (which would be isomorphic to $G$ by mapping to $B/B$ with fiber $G$) let $B\actson G$ by conjugation.
Define a map
\[
\mu: Z\to G \\
[g, b] &\mapsto gbg\inv
,\]
which is a $G\dash$equivariant algebraic morphism.
Then $\mu\inv(g) = \ts{B' \supseteq g}$ are the Borels containing $g$: note the similarity to the Springer resolution with the nilpotent radical.


:::{.exercise title="?"}
Prove this -- a hint is that $G\mix{B} B \mapsvia{\subseteq} G/B\times G$.
:::

Note the two extremal cases:

1. $\mu\inv(1) = G/B$.

2. For $g\in G^{sr}$ regular semisimple, use conditions on dimensions of centralizers and $\dim T \da \dim Z(T)$, how many Borels contain a fixed maximal torus $T$?
  There are at least two, since $T \subseteq B \implies T \subseteq B^-$.
  One can think of the flag variety as parameterizing Borels, so these correspond to $T\dash$fixed points in the flag variety.
  The key is that $W$ acts simply transitively, so $\mu\inv(g) \cong W$.

Define a map
\[
\nu: Z = G\mix{B} B &\to (G\times U)\mix{B} B 
\mapsvia{\wait/U} G\mix{B} B/U 
\iso G/B \cross B/U 
\mapsvia{\pr_2} B/U \\
[g,b] &\mapsto (\bar g, \bar{gb}) \mapsvia{\text{trivial action}} \bar b
,\]
where we've used that relevant actions commute.
Note that this composite map is rare, but allows defining an **abstract Cartan**.
We can then pull back $f$ to a regular function on $Z$, so set $\tilde f \da \nu_* f$, so $\tilde f[g, b] = f(\bar{b})$.


:::{.claim}
$\tilde f\in \CC[Z]^B$.
:::

Next restrict $\tilde f$ to $Z^{sr} = \mu\inv(G^{sr})$, then $W\actson Z^{sr}$ freely and $\nu$ is $W\dash$equivariant.
Since $f$ is $W\dash$invariant, $\ro{ \tilde f}{Z^{sr}}$ to be $W\dash$invariant and $\ro{ \tilde f}{Z^{sr}} \in \CC[Z^{sr}]^W$.

:::{.fact}
If $\xi:X\to Y$ is a quotient by a free action of a finite group, then $\xi$ is **generically Galois**, i.e. $\mu^*: \CC(G^{sr}) \iso \CC(Z^{sr})^W$.
:::

:::{.claim}
$h$ is regular on $G$, i.e. $h\in \CC[G]$.
:::

> See Chriss-Ginzburg 3.1.3.

:::

:::{.remark}
Next time: equivariant cohomology.
:::