# Friday, November 12

:::{.example title="Projective space"}
Let $G\actson X \in \mods{\CC}^{\dim = n}$ be a linear algebraic group acting on a $\CC\dash$module of dimension $n$, then there is a morphism $G\to \GL_n$ and we'll regard $G \subseteq \GL_n$.
Then $G\actson \PP^n$:

- $\PP(V) = \leftquotient{\CC\units}{V\smz}$, and $G$ acts linearly and commutes with scalar multiplication.
- $\PP(V) = \GL_n / P$ and the $G\dash$action descends since the projection $\GL_n\to \GL_n/P$ is $\GL_n\dash$equivariant.

Note that $G$ also acts on the tautological bundle $\OO(-1)$, since these are lines.
We can write $\OO(-1) = \GL_n \mix{P} \CC_{\tv{1,0,\cdots, 0}}$, using the identification $X^*(T) = \ZZ\cartpower{n}$ and taking the character associated to $\tv{t_1,\cdots, t_n}\mapsto t_1$.
Note that $\OO(-1) \to\GL_n/P$ is $\GL_n$ equivariant.
Write $\zeta \da c_1^G(\OO(1)) \in H_G^2(\PP(V))$ for the equivariant Chern classes.

Recall that if $\GL_n \mix{P} \CC_{\lambda} \to \GL_n/P$ for $\lambda \in X^*(T)$ is a $G\dash$equivariant bundle, we can construct $E\mix{G} \GL_n\mix{P} \CC_{\lambda} \to E\mix{G} \GL_n/P \cong E\mix{G} \times \PP^{n-1}$, and the base here corresponds to $H^*_G(\PP^{n-1})$.
This induces $E\mix{G} \PP^{n-1} \to E/G = \BG$, where now the base corresponds to $H^*_G(\pt)$.
:::

:::{.proposition title="?"}
\[
H^*_G(\PP(V)) \iso H^*_G(\pt)[\zeta] / \gens{\sum_{k=0}^n c_k \zeta^{n-k}}
.\]
:::

:::{.proof title="?"}
Given $\mce\to X$, we know $H^*(\PP(\mce))$ in terms of $H^*(\pt)$.
We have

\begin{tikzcd}
	{E\mix{G} \GL_n/P} && {\PP(E\mix{G} \GL_n \mix{P} \CC_\lambda)} \\
	\\
	{E/G = \BG} && \BG
	\arrow["{=}", from=1-1, to=1-3]
	\arrow[from=1-1, to=3-1]
	\arrow["{=}"', from=3-1, to=3-3]
	\arrow[from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJFXFxtaXh7R30gXFxHTF9uL1AiXSxbMCwyLCJFL0cgPSBcXEJHIl0sWzIsMCwiXFxQUChFXFxtaXh7R30gXFxHTF9uIFxcbWl4e1B9IFxcQ0NfXFxsYW1iZGEpIl0sWzIsMiwiXFxCRyJdLFswLDIsIj0iXSxbMCwxXSxbMSwzLCI9IiwyXSxbMiwzXV0=)

So $\xi$ is a hyperplane class for a projective bundle, and thus $c_i^G(V) = c_i(E\mix{G} V)$.
:::

:::{.example}
For $G = \GL_n$, we have $H^*_G(\pt) = \ZZ[c_1,\cdots, c_n] \subseteq \ZZ[t_1,\cdots, t_n] = H^*_T(\pt)$.
So $H^*_G \PP(V) = \ZZ[c_1,\cdots, c_n][\zeta]/\gens{\zeta^n + c_1\zeta^{n-1} + \cdots + c_n}$.
:::

:::{.example title="?"}
For $G=T$, $H^*_T(\PP(V)) = \ZZ[t_1,\cdots, t_n][\zeta] / \gens{\prod_{1\leq i \leq n} \zeta + t_i}$ where $c_i^T(V) = e_2(t_1,\cdots, t_n)$.
:::

:::{.theorem title="Localization in equivariant cohomology"}
Let $X$ be an $n\dash$dimensional smooth algebraic variety with finitely many $T\dash$fixed points.
Write $X^T$ for the fixed point locus, write $c \da \prod_{p\in X^T} c_n^T(\T_p X) \in H^*_T(\pt)$, noting that since $X$ is smooth these are all the same dimension.
Let $S \subseteq H^*_T(\pt)$ be a multiplicative set containing $c$, which is nonzero since the fixed points are isolated.
Assume there are $m\leq \size X^T$ classes in $H_T^*(X)$ restricting to a basis of $H^*(X)$.
Then there are isomorphisms induced by
\[
H^*_T(X)\localize{S} &\mapsvia{S\inv i^*} H^*_T(X^T)\localize{S} \\
H^*_T(X^T)\localize{S} &\mapsvia{S\inv i_*} H^*_T(X)\localize{S}
.\]
Note that $X^T \mapsvia{i} X$ is $T\dash$equivariant, so $i^*$ on $H^*$ descends to $H_T^*$.
By Poincaré duality, we get $\bar{H}(X^T) \to \bar{H}(X)$.
Without the localization, there is still an injection:
\[
H_T^*(X) \mapsvia{\iota^*} H^*_T(X^T) = \bigoplus H^*_T(\pt)
.\]
:::

:::{.remark}
Note that $H^*_T(\pt; \ZZ) = \ZZ[t_1,\cdots, t_n]$ and $H^*(\pt; \CC) = \CC[t_1,\cdots, t_n] = S(\lieh\dual)$, the symmetric algebra on the Cartan.
This comes up when looking at Soergel bimodules.
Compare to $\K^T(\pt) = R(T)$, the representation ring.
:::


:::{.example title="?"}
For projective space, let $T$ be any torus that acts linearly on a $n\dash$dimensional $\CC\dash$module.
Then $V = \bigoplus_i C_{\lambda_i}$ for some characters $\lambda_i$.
Assume the $\lambda_i$ are distinct, then
\[
H^*_T(\PP^{n-1}) = H^*_T(\pt)[\zeta]/\gens{\prod \zeta + \lambda_i}
,\]
where $\zeta = c_1^T(\OO(1))$.
So write $X^T$ as the set of coordinate lines for $X = \PP^{n-1} = \PP(V)$, i.e. for $p_i\da \tv{0,0,\cdots, 0,1,0,\cdots,0}$, $X^T = \ts{p_1,\cdots, p_n}$.
The tangent spaces are given by $\T_{p_i} \PP^{n-1} = \bigoplus_{j\neq i} \CC_{\lambda_j - \lambda_i} = \T_{p_i} U_i$ where $U_i \cong \CC^{n-1}$ by dividing out by the $i$th coordinate, so
\[
t\tv{x_1: \cdots : 1:\cdots x_n} 
&= \tv{t_1x_1: \cdots : t_i \cdot 1: \cdots t_n x_n } \\
&= \tv{{t_1\over t_i} x_1 : \cdots: 1 : \cdots {t_n \over t_i} x_n}
.\]
Thus $(\lambda_j - \lambda_i)(t) = t_1/t_i$.
Thus $c_{n-1}^T(\T_{p_i} \PP^{n-1}) = \prod_{j\neq i} (\lambda_j - \lambda_i)$.
:::

:::{.proposition}
A self-intersection formula: if $i:Y\injects X$ is a closed embedding of codimension $d$ with normal bundle $N$ of rank $d$, then
\[
i^* i_* ( \alpha) = c_d(N) \alpha
.\]
:::

:::{.exercise title="?"}
Show that the following composite is diagonal:
\[
H_T\sumpower{n} \to H_T(\PP^{n-1}) \to H_T\sumpower{n}
.\]
What is the determinant?
:::