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To define a Nakajima quiver variety associated to a quiver $Q = (Q_0,Q_1)$ (vertices and arrows), one first doubles it to $Q^\heartsuit$ by attaching an extra vertex to every old vertex in $Q_0$. Then a weight space of a representation should arise as the top homology of a moduli space ${\mathcal M}(v,w)$ of representations of $Q^\heartsuit$, where $v$ is the dimension vector on the original vertices and $w$ on the new vertices. (I am following Ginzburg's lectures on this topic.)

  1. Am I right in thinking that the relevant weight space is the $\sum_i w_i \omega_i - \sum_i v_i \alpha_i$ of the $\sum_i w_i \omega_i$ representation, where the $(\omega_i)$ are the fundamental weights and the $\alpha_i$ are the simple roots? Edit: I tracked down this formula in Proposition 6.9 of this.

  2. If so, can I Can I therefore think of the second set of vertices as Langlands dual to the first set, in any useful way?

  3. In particular, has anyone looked into attaching the second set of vertices to one another, to give a second copy of $Q$? (I really want to say "the Langlands dual of $Q$" but I guess we're automatically in the simply-laced case.)

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    $\begingroup$ These types of questions are intriguing, but at the same time Langlands duality is most visible in settings where there are two root lengths. Still, I don't know what to expect here. $\endgroup$
    – Jim Humphreys
    Commented Nov 29, 2010 at 23:24
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    $\begingroup$ A correction: $M(v,w)$ is not exactly the moduli space of reps of $Q^\heartsuit$. Namely, its points are representations of $Q^\heartsuit$ but the notion of isomorphism is different: points correspond to orbits of $GL(v)$ action, not $GL(v)\times GL(w)$ action. $\endgroup$
    – Sasha Kirillov
    Commented Dec 1, 2010 at 3:22
  • $\begingroup$ There is a new graph that the Nakajima quiver variety is the moduli of representations of, but one gets it by adding one vertex, and connecting each old vertex to the new one by w_i edges (and then doubling everything). $\endgroup$
    – Ben Webster ♦
    Commented Dec 1, 2010 at 4:27
  • $\begingroup$ I would say I'm not very optimistic about this idea; I suppose it's worth a little thinking, but it doesn't match my understanding of this story at all. $\endgroup$
    – Ben Webster ♦
    Commented Dec 1, 2010 at 4:28
  • $\begingroup$ @Sasha: I was probably being too glib in hoping to capture that with the phrase "a moduli space". $\endgroup$
    – Allen Knutson
    Commented Dec 1, 2010 at 5:31

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