This is an old revision of the document!
Table of Contents
2023-07-24
I should know what is expected to be true, and what is the story.
I should also know what is known to be true, i.e. proven. And I should know how the proof goes.
I should also know what is known to be false, and counter examples.
Chriss-Ginzburg and Bez-Finkelberg-Mirkovic
Q1: from $G$ to $T$
For our flag variety, why is it true that, $T$-equivariant homology and $G$-equivariant homology only differ by Weyl covering.
(1) First of all, why is that for any Borel $B$ of $G$, the abstract Cartan $B/[B,B]$ for different $B$ are identified? Well, we may consider conjugation action of $G$ on the set of $B$. Suppose $B_1 = g B_2 g^{-1}$ for some $g$, the ambiguity of $g$: is right multiplication by $B_2$, left multiplication by $B_1$. For any $g' \in B_1 g B_2$, maybe consider the induced map on.
What is conjugation action on $B$? Fix a $b_0 \in B$, we can send $b \mapsto b_0 b b_0^{-1} = b b^{-1} b_0 b b_0^{-1}$. So, thing in the same conjugacy class in $B$ are also in the same unipotent orbit.
A conjugacy class of $B$ is contained in the unipotent orbit. But not the other way around. For example, identity matrix is its own conjugacy class (even under $G$ conjugation). However, multiplying by $[B,B]$ gives a lot.
(2) On the Lie algebra and Lie group level. Taking $W$ invariant.
Ex: can you show that $\C[x_1,x_2]$ is a $\C[x_1+x_2,x_1 x_2]$ free module? Well,let's choose some generator in $ \C[x_1,x_2] / (x_1+x_2, x_1 x_2)$, say $e_1 = 1, e_2 = x_1$. Well, that is a bit arbitrary, how about $e_1 = 1, e_2 =x_1 - x_2$, so under $W=S_2$ action, they are eigenvectors. Then, for any polynomial $f(x_1, x_2)$, we can do $$ A = f(x_1, x_2) + f(x_2, x_1) = g (x_1, x_2), \quad B = f(x_1, x_2) - f(x_2, x_1) = h(x_1, x_2) (x_1 - x_2) $$ Well, one can find $g,h$ as symmetric functions.
How about $\C[x_1, x_2, x_3]$?
Well, we need a theorem of Pittie-Steinberg theorem. In the case for type $A$ group, we should be able to check explicitly.
You have a coordinate ring, of the torus, or of the plane. And you have the $W$ action. You pass to the $W$ invariant subring. Then, the original ring turns out to be a free module of finite rank over the W-invariant ring.
How do you see that explicitly?