Author: Eiko
Time: 2024-10-08 10:44:24 - 2024-10-08 10:44:24 (UTC)
(seminar notes)
Cohomological Integrality For Symmetric Quotient Stacks
arXiv: 2406.09218 2408.15786
M.Reineke Donaldson-Thomas (DT) Invariants
Bettei Title: DT invariants of symmetric representations of reductive groups
We work over .
Situation
, more generally, reductive, i.e. unipotent radical is trivial)
Which is equivalent to linearly reductive (all finite dimension representation are semisimple).
A non-example is
is a nontrivial extension of by itself.
There is a natural pairing
where .
Consider the action of , the maximal torus acts
thich means are weight spaces of weight .
Symmetric representation
Example
Weil Groups
By definition, where is the normalizer.
For example , is the symmetric group.
acts on .
Cohomological Integrality
Setup: is a finite dimensional representation.
to compute it find with a free action , then send .
Example
, , so .
For we have .
,
For example ,
In general, is always a polynomial ring .
Goal of Cohomological Integrality
is to extract a finite dimensional subspace depending on (while is infinite dimensional and does not depend on ), that generates in the sense of parabolic induction
\[P_0 = \text{unipotent cohomology of }V/G =
\begin{cases}
\text{character sheaves} & Lusztig
\text{Hecke eigensheaves & Geometric Langlands
\end{cases}
\]
Context and Motivation
Topology of , also .
The ring is finitely generated by Hilbert’s theorem. classifies closed -orbits in .
by rescaling, then everything contracts to the origin, so there is only one closed orbit (the origin). i.e. .
In terms of invariants, there are no invariants under scaling except constants, which match our expectation since .
by , classical example, whose closed orbits are hyperbolas . and the origin.
The invariants are , which is a polynomial ring in 1 variable. This action gives
where .
Computing generators of is very difficult, even for , which are polynomial invariants of binary forms.
But we can try to understand the topology of . Here comes the cohomological integrality.
where stands for intersection cohomology.
Here
The topology of a smooth artin stack a good moduli space.

Stacks can fully encode equivariant cohomologies.
Introduce and new enumerative invariants fo .
Operations
a representation of
(a three block matrix)
, a representation of
$ G^{} = {g | _{t} (t)g(t)^{-1} }G$
is the parabolic subgroup of corresponding to .
$ G^{}V^{} = {v | _{t} (t)v }$
Induction Diagram

The induction map is
i.e.
where is the Weyl group of .
Explicit Formula
Let
where .
A few more things
On we define a equivalence relation:
is a finite set.
For , .
is characterized by
Main Theorem
Let be a symmetric represetnation of reductive .
Then there exists finite dimensional subspaces , graded and stable under action, such that
The left side is isotypic component for -action.