Peng Zhou

stream of notes

User Tools

Site Tools


blog:2023-07-24

2023-07-24

  • reading Chriss-Ginzburg the whole morning
  • watched a youtube video 8 traits of successful people
  • try to fix the annoying keyboard on a macbook pro, which turns out to be 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.

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.

Q1: from GG to TT

For our flag variety, why is it true that, TT-equivariant homology and GG-equivariant homology only differ by Weyl covering.

(1) (what is T? no, not the diagonal, but a quotient) First of all, why is that for any Borel BB of GG, the abstract Cartan B/[B,B]B/[B,B] for different BB are identified? Well, we may consider conjugation action of GG on the set of BB. Suppose B1=gB2g1B_1 = g B_2 g^{-1} for some gg, the ambiguity of gg: is right multiplication by B2B_2, left multiplication by B1B_1. For any gB1gB2g' \in B_1 g B_2, maybe consider the induced map on.

What is conjugation action on BB? Fix a b0Bb_0 \in B, we can send bb0bb01=bb1b0bb01b \mapsto b_0 b b_0^{-1} = b b^{-1} b_0 b b_0^{-1}. So, thing in the same conjugacy class in BB are also in the same unipotent orbit.

A conjugacy class of BB is contained in the unipotent orbit. But not the other way around. For example, identity matrix is its own conjugacy class (even under GG conjugation). However, multiplying by [B,B][B,B] gives a lot.

(2) C[x1,,xn]\C[x_1, \cdots, x_n] is a free C[x1,,xn]Sn\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[x1,x2]\C[x_1,x_2] is a C[x1+x2,x1x2]\C[x_1+x_2,x_1 x_2] free module? Well,let's choose some generator in C[x1,x2]/(x1+x2,x1x2) \C[x_1,x_2] / (x_1+x_2, x_1 x_2), say e1=1,e2=x1e_1 = 1, e_2 = x_1. Well, that is a bit arbitrary, how about e1=1,e2=x1x2e_1 = 1, e_2 =x_1 - x_2, so under W=S2W=S_2 action, they are eigenvectors. Then, for any polynomial f(x1,x2)f(x_1, x_2), we can do A=f(x1,x2)+f(x2,x1)=g(x1,x2),B=f(x1,x2)f(x2,x1)=h(x1,x2)(x1x2) 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,hg,h as symmetric functions.

How about C[x1,x2,x3]\C[x_1, x_2, x_3]?

Well, we need a theorem of Pittie-Steinberg theorem. In the case for type AA group, we should be able to check explicitly.

You have a coordinate ring, of the torus, or of the plane. And you have the WW action. You pass to the WW 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 glngl_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, g\frak g lives over a bunch of points on h\frak h^*. So far, these are canonical, we don't have Killing form. Let's assume 'semi-simple', which says, the root vectors span h\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)WR(G) = R(T)^W and C[G]G=C[T]W\C[G]^G = \C[T]^W. The first one is the 'integral form'. What does a 'coordinate ring over Z\Z' even mean? And, how much information did I lose?

blog/2023-07-24.txt · Last modified: 2023/07/25 06:32 by pzhou