====== 2022-11-08, Cherednik on Hecke ======

===== 1: Hecke algebra in rep theory =====
very funny comment about real math and imaginary math, one can see he is an concrete guy with feet on ground (i.e. real axis). 
  - What is a zonal spherical function? From wiki, it is a //function on a locally compact group G with compact subgroup K (often a maximal compact subgroup) that arises as the matrix coefficient of a K-invariant vector in an irreducible representation of G.// But why we study it?
  - Why we study characters of any Lie algebra? Well, you may want to spectral decompose a huge thing, and character formula is just a way to decompose. A little mind trying to understand a huge-connected thing as many unrelated things (me?)
  - Tensor multiplicities, now out of fashion since we are dealing with infinite dimensional gadgets now.
  - $[M_\mu, L_\lambda]$, the Kazhdan-Lusztig polynomial. (why it is not counted as 'real' math in the beginning?)

Then, we have the updates. 
  - Hypergeometric functions. hmm, why do we care such functions? just a differential equation solution, very generic type, and with many many parameters. 
  - rational and elliptic KZ equation? 
  - Verlinde algebra? Fusion not tensor? aha, I see. previously we have simple module tensor and decompose, now we have 'primary field' in a vertex algebra, fuse and decompose.
  - Now, I am really confused and intrigued. These simple-verma coefficients are related to modular representation. 

Most of the remaining discussion is too high for me now. Skip it.

===== 2: affine KZ =====
It is a first order differential equation, taking value in an infinite dimensional algebra called ** degenerate affine hecke algebra **

Here is the equation
$$ \frac{\d \Phi}{\d u} = \left( k \frac{s}{e^u - 1} + x \right) \Phi(u) $$
  * There is $e^u$, so the solution is like 1-periodic, or if we use $U = e^u$, then it lives on $\C^*$. But we prefer this way, since $\Phi(U)$ will be multivalued. Wait, around $u=0$, will it also have interesting monodromy? 
  * $k$ is a parameter $s,x$ are linear operators, such that $$ s^2=1, xs + sx = k. $$ Why we have thse? 
  * What?  The above relation among $s,x,k$ generate the $H_{A_1}'$, degenerate affine Hecke of type $A_1$. 
  * Now, if we change variable, let $z = e^{-u}$ (as one should), and repackage the equation, we see poles at $z=0, \infty, 1$. But still, this is a complicated equation. However, if we don't solve for the most general curve in the Hecke algebra (indeed, we are looking at a pair-of-pants in the Hecke algebra parameterized by $k$), but rather we take representation of $H_{A_1}'$, then it become much manageable. 

{{:notes:pasted:20221108-082040.png}}

OK, too lazy to copy down the formula. A few comments
  * $s$ should be a permutation matrix, but here we diagonlized it. Not sure what $x$ should be. 
  * When you solve the equation for $\Phi(u)$, or $\Phi(z)$ for $z = e^{-u}$, the first component $\Phi_1(z)$ will be the hypergeometric equation. There are the Poch-hammer symbol $(x)_n$, and series summation. (We should now the series only converge for some range $z$, no?)

The AKZ equation for $GL_n$. OK, we have 
$$ A_i = \sum_{j \neq i} \frac{\Omega_{ij}}{z_i - z_j} $$
where $z_i \in \C^*$. So affine means working with $z_i \in \C^*$? why such a big deal? 
One need to make them satisfy 
$$ \d_j A_i - \d_i A_j - [A_i, A_j] = 0 $$
Nothing but the flat connection property. 

But, then, what is $\Omega_{ij}$? Is it independent of $z_i$? I guess so, they need to satisfy useful commutation relations. 

So, indeed