Let $X=M(v,w)$ be a Nakajima quiver variety for a quiver $Q$. Can one calculate the second singular cohomology groups $H^2(X,\mathbb Z)$ or $H^2(X,\mathbb C)$ explicitly, and if not, are there some particular cohomology classes in these groups which one can identify, or a class of quivers where one can calculate these groups?

I know that in type A the cotangent bundle to the flag variety is a Nakajima quiver variety for the corresponding Dynkin quiver of type A, and the second cohomology group for it is calculated in M. McBreen's talk, which can be found here: http://www.northeastern.edu/iloseu/McBreenTalk.pdf but, if possible, I would like to see a different explanation for the answer in this case from the explanation in this talk.


There's a very natural set of cohomology classes in $H^2(X,\mathbb{Z})$; for each node $i$ in your quiver, there's a tautological bundle $\mathcal{V}_i$. The Chern classes $c_1(\mathcal{V}_i)$ give natural elements of $H^2(X,\mathbb{Z})$. These are usually linearly independent, but not always (for example, sometimes $X$ is a point). Attached to $v,w$, there's a weight space of a highest weight representation for the associated Kac-Moody algebra, and these classes will be linearly independent if the weight is in the interior of the weight polytope for the representation.

It seems very likely that these classes span, and this is proven for type ADE quivers. This is actually a special case of a much more general problem, called hyperkähler Kirwan surjectivity (see, for example, this paper).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.