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So, today I started learning the definition of a quiver variety, and wanted to make sure I'm understanding things right, so first, my setup:

I've been looking at the simplest case that didn't look completely trivial: two vertices with one directed edge. Now, my understanding is that then we have two vector spaces $V$ and $W$ of dimensions $n$ and $m$, and the quiver variety is $\hom(V,W)\oplus \hom(W,V)/GL(V)\times GL(W)$, with the action of each group on the domain or codomain, as applicable.

Now, a naive dimension count (this is one place that I may be very, very wrong) is that this is a $2nm-n^2-m^2=-(m-n)^2$ dimensional space. So presumably if $m=n$, the GIT quotient is a single point.

Now, what if $n\neq m$? Presumably, we don't really get anything for the GIT quotient, either a point or the empty set (I don't know GIT very well, so perhaps no stable points?)

Finally, does this object have interesting geometry as a stack? It seems obvious that this should be a smooth Artin stack (if my intuition for them is even vaguely accurate) which just happens to be negative dimensional, but what kinds of properties does it have?

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Do you mean two directed edges in opposite directions? One directed edges would just give you hom(V,W) and not the other one. –  Greg Kuperberg Dec 10 '09 at 0:00
The definition I'd been given was that you take a digraph, get $\hom(V_i,V_f)$ where $V_i$ is the initial space and $V_f$ the final one, then you take the cotangent bundle (which in this case is the dual, and thus gives a $\hom(V_f,V_i)$ and then mod out. Is that not the standard definition? Perhaps I was supposed to use symplectic reduction rather than GIT quotient with this? –  Charles Siegel Dec 10 '09 at 0:05
My misconcption, I guess. I suppose that "quiver variety" usually means a Nakajima quiver variety, which is more complicated than just a moduli space of quiver representations as in Gabriel's theorem. –  Greg Kuperberg Dec 10 '09 at 0:16
...ok...I knew about half those words. VERY new to the subject, and am trying to figure out the simplest case. I take it that either this is a nonstandard definition, or else I'm understanding things wrong? –  Charles Siegel Dec 10 '09 at 0:36
No, I think that you are right and I am wrong. Some of it is new to me too. The Nakajima quiver variety is defined as you said, according to several references. There is also a representation variety of a quiver, but that is a simpler object. It so happens in this one case that the Nakajima quiver variety is the representation variety of another relatively tractable quiver. –  Greg Kuperberg Dec 10 '09 at 0:54

4 Answers 4

up vote 6 down vote accepted

First, one important point: people who study quiver varieties seem not to usually take stack quotients, but rather GIT quotients (though there ARE very good reasons to do this if you like geometric representation theory) which leads them to come up different dimension formulae from you. The reason is that you are think of a fine moduli space, whose stack dimension is the geometric dimension of the underlying variety minus the dimension of the automorphism group of the object, and modules over an algebra ALWAYS have automorphisms (if nothing else, they have multiplication by constants).

Now, the answer proper: You seem to have mixed up two of the more popular notions of a quiver variety (leading your confusing dialogue with Greg in the comments), and Kevin seems to have mixed in a third, which may or may not actually be the one you had in mind (all of which are, of course, closely related). If you have two vertices and one arrow, then there are two things you can do.

You can take the moduli space of quiver representations of the path algebra of that quiver, which is given by $\mathrm{Hom}(V,W)/GL(V)\times GL(W)$. This has a very small dimension $(nm-n^2-m^2)$ and should probably be thought of as finitely many points (indeed, this quiver only has finitely many representations of any given dimension. The indecomposibles look like $k\to 0$, $0\to k$ and $k\to k$), all of which have a bunch of automorphisms.

Now it sounds like what you intended to do was take the "hyperkählerization" of this quiver variety. What you should do for this is take the cotangent bundle of $\mathrm{Hom}(V,W)$, but before you mod out, you have to impose a moment map condition. The reason this is a good idea is that you want a resulting variety which is a holomorphic symplectic result (just like the cotangent bundle), which you can also think of as hyperkähler by picking a hermitian metric on $\mathrm{Hom}(V,W)$. This moment map condition is basically that both possible compositions of maps along your arrows are 0 (note: I think this is not a flat map, so I think if you want to really think properly about this story, you should probably take the derived fiber). Then take the quotient of that.

What Kevin is referring to is probably the most popular definition of quiver varieties for geometric representation theory. This is yet another definition, where he interpreted one of your vertices as a shadow vertex. This extra layer of confusion results from some (IMHO poor) notational choices of Nakajima.

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Thanks Ben, that's very helpful. I recall it coming up that there was some hyperkahler stuff going on, and that the quotient should be symplectic. You've cleared a lot of stuff up for me. –  Charles Siegel Dec 10 '09 at 4:41
Ben, what's misguided about Nakajima's shadow vertices? In a sense they were the crucial new ingredient in the definition of quiver varieties. –  Kevin McGerty Dec 11 '09 at 16:00
So, I've edited the answer to better reflect my personal opinion. I certainly have nothing against the varieties with shadow vertices. I just think describing them using shadow vertices is a bad idea. They're the actual moduli space of representations of the preprojective algebra for a different quiver, and I think it makes a lot more sense to think of them that way. I started ranting about this in my answer, and then realized it had nothing to do with Charles's question. –  Ben Webster Dec 11 '09 at 17:50
So, I guess what I mean is, no, they're not crucial at all, but nobody seems to have noticed this. –  Ben Webster Dec 11 '09 at 17:51

I can just tell you what the space looks like up to the action. The orbit classification of $\text{hom}(V,W) \oplus \text{hom}(W,V)$ looks a lot like the orbit classification of $\text{hom}(V,V)$, which as you know from linear algebra is given by Jordan canonical form. In fact, if you compose the two homs, the classification is clearly at least as complicated as JCF. There is just a modest amount more structure because the nilpotent part is more complicated. Although the formal dimension of the quotient is indeed negative or 0, its geometric dimension is strictly positive when $V$ and $W$ are non-trivial.

Let $$f:V \to W \qquad g:W \to V$$ be the two linear maps. Then $f$ has a kernel, $g \circ f$ could have a larger kernel, etc. Define the stable kernel $V_0$ and the stable image $V_1$ of $g \circ f$ to be the direct limits of the kernel and the image of $(g \circ f)^n$. As in one proof of Jordan canonical form, $V = V_0 \oplus V_1$. Similarly $W = W_0 \oplus W_1$. The pair $(f,g)$ canonically splits into two pairs, $(f_0,g_0)$ and $(f_1,g_1)$. The pair $(f_0,g_0)$ is nilpotent, while the pair $(f_1,g_1)$ is invertible and establishes an isomorhism $V_1 \cong W_1$.

Because of the isomorphisms between $V_1$ and $W_1$, the invariant information in the pair $(f_1,g_1)$ is the Jordan canonical form of $g_1 \circ f_1$, which is the same as the JCF of the other composition. In other words, either $f_1$ or $g_1$ can be any isomorphism, and then the other one can be chosen to establish a prescribed Jordan canonical form. Any eigenvalue can appear other than 0.

The nilpotent pair $(f_0,g_0)$ is a little more interesting. It looks like an Ouroboros. An indecomposable nilpotent pair is any finite chain $$0 \to k \to k \to \cdots \to k \to 0,$$ rolled up from $\mathbb{Z}$-graded to $\mathbb{Z}/2$-graded. (The connecting maps in the middle are all isomorphisms, not differentials.) The chains can have odd length, so that $V_1$ and $W_1$ don't have to have the same dimension. A chain of any length can also descend in two different ways to $V_1$ and $W_1$.

I gather from Ben's comment that this is a right answer to a wrong question. It is a good description of the representation variety of a cyclic quiver; there is nothing special about cycle length 2 in the analysis. But the $A_2$ quiver variety is something else.

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Ahh, so in particular, my naive dimension count fails. And that's because the adjoint action doesn't have a dense orbit. Got it. Does the stackiness of the quotient give us anything interesting? –  Charles Siegel Dec 10 '09 at 1:12
I hardly understand stacks in the sense of algebraic geometry. But just in the sense of group actions: If $g \circ f$ is diagonalizable and either both annihilate the nilpotent part or one is an isomorphism on the nilpotent part, then the stabilizer is semisimple. Otherwise the stabilizer is a parabolic subgroup. The computation is similar to deriving the centralizer of an endomorphism of a vector space, or the non-uniqueness of a JCF basis. –  Greg Kuperberg Dec 10 '09 at 1:25

As the comments say, one of the problems here is the multiplicity of things that get called "quiver varieties". The example you are looking at would correspond in Nakajima's context to a quiver variety attached to $SL_2$, and maybe that's worth making explicit. Nakajima starts with a quiver Q, and does two things: first he adds a "shadow vertex" for each original vertex, and an edge from that vertex to it's corresponding vertex in the original quiver, then he take the contangent bundle of everything, which in graph terms means adding for each edge you now have another edge in the opposite direction. Thus for $SL_2$ the original quiver is just a single vertex with no edges. The Nakajima procedure adds another vertex and two edges one in each direction.

The quotients that Nakajima would then consider would be with respect to the action of the general linear groups on the original vertices: in our case just one of them, say $V$. For the GIT quotient, you would use some choice of stability, which in this case could be just that the map $x \colon W \to V$ is surjective. Then you need the moment map for the action of the group $\text{GL}(V)$: this is just the map to the composite $xy \in \text{End}(V)$ (where $y\colon V \to W$ is the other linear map). The symplectic quotient is then the quotient by the action of $\text{GL}(V)$ on the stable part of the zero locus of the moment map: so pairs $(x,y)$ where $x$ is surjective, and the image of $y$ lies in the kernel of $x$. The surjectivity of $x$ means that the $\text{GL}(V)$ action is free, and the quotient can then be identified with the cotangent bundle to a Grassmannian of $W$: it is the image of the map $(x,y) \mapsto (\text{ker}(x),yx)$. The affine quotient is the image of the map $(x,y) \mapsto yx$, which gives you the closure of a nilpotent orbit in $\text{End}(V)$.

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This answer is merely in relation to the naive dimension count calculated in the question. When taking a GIT quotient, one cannot merely subtract the dimensions to get the dimension of the quotient. For example if you quotient gln by GLn with the adjoint action, then the GIT quotient is n-dimensional. (the quotient here is actually affine n-space and the map sends a matrix to its characteristic polynomial).

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