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I'd like to understand a special case of the following rather general algebraic geometry question:

Given an algebraic group $G$ acting on a variety $V$, can we describe the $G$-invariant subvarieties of $V$?

Since the zero sets of $G$-invariant polynomials are exactly the $G$-invariant hypersurfaces of V, the question seems closely enough related to invariant theory that I imagine that invariant theorists and algebraic geometers would have a sense of when and how this question can be answered. My literature search hasn't turned up anything of use, so I'd appreciate any references or pointers, either regarding the general problem or the specific case I describe below.

The case I'm interested in is this:

Let F be a field. Given a quiver $Q=(\Omega_0,\Omega_1)$ (say, without oriented cycles) and a dimension vector ${\vec d}:\Omega_0\to {\mathbb N}$, let $V$ be the variety of representations of $Q$ with coefficients in F, having dimension vector $\vec d$. That is, $V=\Pi_{(h,t)\in \Omega_1} M_{{\vec d(h)},{\vec d(t)}}$, where $M_{a,b}$ denotes the variety of $a \times b$ matrices with coefficients in F.

Let $G=GL({\vec d})\equiv \Pi_{x \in \Omega_0} GL({\vec d}(x),F)$, and let $G$ act on $V$ in the usual way, so that the isomorphism classes of representations of $Q$ are the orbits of the action of $GL({\vec d})$ on $V$.

Here are my thoughts on the question so far.

For simplicity, let me assume that $\forall x,y \in \Omega_0$, there is at most one path from $x$ to $y$ in $Q$. One way to define a $G$-invariant subvariety of $V$ is this:

For $k>0$, $X=\{x_1,...,x_n\},Y=\{y_1,...,y_m\}$, $X,Y \subset \Omega_0$, let $V_{X,Y,k} \subset V$ be the common zeros of all $k \times k$ minors of $$T_{X,Y}:V \to M_{\Sigma_{i=1}^m {\vec d(y_i)},\Sigma_{i=1}^n {\vec d(x_i)}}$$ where for a representation $R$, $T_{X,Y}(R)$ is the block matrix with $m \times n$ blocks s.t. the $(i,j)^{th}$ block is the 0 matrix if a path from $x_j$ to $y_i$ does not exist in $Q$, and otherwise the $(i,j)^{th}$ block is the product of the matrices in $R$ along the path from $x_j$ to $y_i$.

Since the rank of $T_{X,Y}(R)$ is an invariant of the isomorphism class of $R$, $V_{X,Y,k}$ is $G$-invariant.

I need to think about this more carefully, but my hypothesis (very tentative, as I'm just beginning to study quivers) is that for quivers of finite type, these are all the G-invariant varieties.

In the case of tame quivers, I don't have a good sense of what the G-invariant varieties might be or of how to tackle the question. Are there G-invariant varieties other than the $V_{X,Y,k}$? One relatively simple case for which I don't yet have an answer is:

$\Omega_0=\{ x_1,x_2,x_3,x_4,x_5 \} ;$ $\Omega_1=\{ (x_1,x_5),(x_2,x_5),(x_3,x_5), (x_4,x_5) \} ;$ ${\vec d(x_i)}=1,i=1,2,3,4;$ ${\vec d(x_5)}=2$.

Ultimately, I'd like to find an answer for certain classes of wild quivers, but understanding this example would be a good start.

with thanks, Mike

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Quite a bit is known about representations of quivers, including a description of absolute and relative invariants, which correspond to stable generic orbits, and a complete classification of the isomorphism classes of f.d. representations of quivers of finite type. "Combinatorial commutative algebra" by Miller and Sturmfels has a lot of background material and contains references. –  Victor Protsak Nov 4 '10 at 18:19
    
I can't help but get the feeling from the way you've written the question that you haven't looked very carefully in the quiver literature. This question for finite type quivers is resolved by Gabriel's theorem. For tame quivers, it's a bit more complicated but well-known to people on the field. (These are exposited quite nicely by Crawley-Boevey in amsta.leeds.ac.uk/~pmtwc/quivlecs.pdf). –  Ben Webster Nov 5 '10 at 4:07
    
I also really wish you had included some motivation for the question, as asking for all invariant subvarieties of any moderately complicated space is a bit of a disaster; things will look much cleaner if you ask what the moduli space of indecomposable quiver reps for a given dimension are, and this is essentially the same question. –  Ben Webster Nov 5 '10 at 4:07

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