Differences

This shows you the differences between two versions of the page.

--- blog:2023-07-24 [2023/07/24 22:43] – pzhou
+++ blog:2023-07-24 [2023/07/25 06:32] (current) – pzhou
@@ Line 1: / Line 1: @@
 ====== 2023-07-24 ======
+  * reading Chriss-Ginzburg the whole morning
+  * watched a youtube video [[https://www.youtube.com/watch?v=NOl0v54DaXo | 8 traits of successful people]]
+  * try to fix the annoying keyboard on a macbook pro, which turns out to be [[https://support.apple.com/keyboard-service-program-for-mac-notebooks| not my (or my wife's) fault.]]
+  * found an interesting lecture note of Teleman on rep theory. Never really understood what is character formula.
+===== Chriss-Ginzburg and Bez-Finkelberg-Mirkovic =====
 I should know what is expected to be true, and what is the story.
@@ Line 8: / Line 17: @@
-===== Chriss-Ginzburg and Bez-Finkelberg-Mirkovic =====
+==== Q1: from $G$ to $T$ ====
-===== 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.
+(1) (what is T? no, not the diagonal, but a quotient) 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.
-Naturally, $B/[B,B]$ is abelian group, orbit being the (inner) conjugacy class.
+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) $\C[x_1, \cdots, x_n]$ is a free $\C[x_1, \cdots, x_n]^{S_n}$-mod. So is the multiplicative case.
+Not trivial, Steinberg-Pittie proved this. It turns out not be about permutation representation.
+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.
+{{:blog:pasted:20230724-162938.png}}
+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.
+You have a bunch of fundamental weight, they are element in the weight lattice, that are dual to the coroots. (ok, what are roots and coroots? Just do $gl_n$. you pick a cartan subalgebra, and let the cartan acts on Lie algebra. I guess I am just not used to having a family of commuting operator acting on something. (how about module over a commutative ring?) ok fine. ok, as h-mod, $\frak g$ lives over a bunch of points on $\frak h^*$. So far, these are canonical, we don't have Killing form. Let's assume 'semi-simple', which says, the root vectors span $\frak h^*$. Take a half-plane, and take some primitive roots, call them simple root.  read this note https://math.mit.edu/~dav/roots.pdf for what is coroot)
+So, we have $R(G) = R(T)^W$ and $\C[G]^G = \C[T]^W$. The first one is the 'integral form'. What does a 'coordinate ring over $\Z$' even mean? And, how much information did I lose?
-(2)
-https://pzhou.org/lib/images/smaller.gif