Peng Zhou

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blog:2023-08-05

2023-08-05

  • write up the notes that are useful for myself.
  • why quiver gauge theory has anything to do with Kac-Moody algebra?

The stuff that I typed up below, are so incoherent and dreamy, that I don't know what am I talking about. So they should be either cleaned up or deleted.

I also cleaned up some to read papers.

I don't think I want to write up the example computation of the spaces.

Statements, Examples about Coulomb branches

For $G=GL_n$, you give me an orbit $G(O) t^\lambda G(O)/G(O)$ in the affine Grassmannian $Gr_G = G(K)/G(O)$. Then you want to do some convolution product.

Why not directly work with the $G(O)$ orbit, but rather with its closure, and objects extended from the cell to the closure?

Why you don't do IC homology, but the usual BM homology?

Is (BM) homology well-defined for singular space? i.e. for $xy = z^2$ in $\C^3$.

What do I need smoothness of the space for? Do I need Poincare duality? No, I will just work with the dualizing complex $\omega$ on $Gr_G$.

On one hand, BFM tells me that the pure gauge theory is super-easy.

  • First, one unfolds $H^G(Gr_G)$ to $H^T(Gr_G)$ (enlarge the ring)
  • Then, we study $H^T(Gr_G)$ using localization.

meaning, we look at fixed points of $Gr_G$ under various element $t \in T$, or $X \in Lie(T)$. We get that $H^T(Gr_G)_X = H^T(Gr_{G'})_X$. Then, we learned that $H^T(Gr_{G'})$ is easy. What does $G'$ look like? $G'$ is a Levi subgroup of $G$, simplest non-trivial block-diagonal matries.

Why we can talk about $T$-equivariant homology, using localization?

We always can talk about pushforward of homology, so we always have $$ \iota_*: H^T_*(Gr_T) \to H_*^T(Gr_G) $$ as a $H^T(pt)$ module.

How does that work? On $X=\P^1$, with $T=(\C^*)^2$ action? For equivariant cohomology, we are OK. The equivariant homology, is just $H_i(X) = H^{-i}(X, \omega)$. Since $X$ is smooth orientable, we can choose an orientation and set $\omega_X = \C_X[2]$. Indeed, we should get $H_0(X) = H^2(X)$.

So, when we do $X^T= \{ 0, \infty \}$. What do we get? I think when we do pushforward, it is as homology; when we do pullback, it is as cohomology. They preserve different dimension. So, I guess when you pushforward then pullback, you will get something extra.

In the $K$-theory story, when you have $pt \in \C$, with $\C^*$ action in the standard way, you get $\iota^* \iota_*([0]) = [0] (1 - 1/t)$, where $1/t$ is the representation of the conormal (representation of the linear coordinates).

In the homoloyg story, when we pushforward, we have a $dim=0$ class, so it is $deg=codim=2$ cohomology class, then we pull-back, it is a degree $2$ guy. So, it might be the 'equivarint euler class' of the normal or conormal bundle. I guess it is normal bundle, but I am not sure why. (https://arxiv.org/pdf/1305.4293.pdf) they uses $e_T(\nu_p)$, the normal bundle.

In the example of $\P^1$, at point $[1;0]$, where we use $T_2 / T_1$ as local coordinates. Let me say, the normal bundle is cashed in for $Y_2 - Y_1$. Then, we say $$ [\P^1] = \frac{[0]}{Y_2 - Y_1} + \frac{[\infty]}{Y_1 - Y_2}. $$ If we understand both sides as cohomology, then both are in degree $0$, $deg([0])=2$ but $deg(Y_i)=2$, so they cancel.

Why does localization to fixed point works? Why does it play well with convolution? Isn't convolution very complicated?

blog/2023-08-05.txt · Last modified: 2023/08/06 04:10 by pzhou