[Math] Near Trivial Quiver Varieties

ag.algebraic-geometrygeometric-invariant-theoryquiversstacks

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?

Best Answer

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.