All Questions
8 questions
6
votes
1
answer
139
views
Quiver variety, generically symplectic
Theorem 11.3.1 (iv) of "Noncommutative Geometry and Quiver algebras" by Crawley-Boevey, Etingof and Ginzburg claims that, for a dimension vector $\mathbf{d}\in\Sigma_0$, the quiver variety $...
- 985
5
votes
0
answers
273
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Intuition for the McGerty-Nevins compactification of quiver varieties
In Section 4 of the paper Kirwan surjectivity for quiver varieties (Inventiones Math. 2018) McGerty and Nevins define a compactification of the moduli space of representations
of the preprojective ...
- 2,964
3
votes
0
answers
111
views
Example of an irreducible component with an open set of infinitely many codimension 2 (codimension 3) orbits
Let $\mathbb{K}$ be an algebraically closed field of characteristics $0$. Let $A$ be a finite dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, ...
- 839
2
votes
0
answers
143
views
Two notions of stability
Let $Q$ be a finite quiver (i.e. an oriented graph). A representation of $Q$ is by definition a module over the path algebra of $Q$. More concretely, a representation associates to every vertex $v \in ...
- 373
2
votes
0
answers
129
views
Getting an equivariant morphism
Let $X\subset\mathbb{A}^n$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristic zero. Suppose we have two linearly reductive algebraic groups $G$, $G'$ ...
- 839
1
vote
0
answers
127
views
How to determine if an invariant rational function is defined at the $\theta$-polystable point
Background:
Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{C}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of ...
- 839
1
vote
0
answers
95
views
Is $U\subseteq X^{s}(L)$?
Let $X$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristics $0$. Let $G$ be a connected, linearly reductive, affine algebraic group acting regularly on $...
- 839
1
vote
0
answers
144
views
Non-empty stable locus of an irreducible component
I have a vague question:
Let $X$ be an algebraic pre-scheme and $G$ be a linear reductive group. Consider the G.I.T. quotient $X{/\!/}G$. Is there any result (maybe in some special case) which tells ...
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