Deformations of Nakajima quiver varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:07:05Z http://mathoverflow.net/feeds/question/32069 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties Deformations of Nakajima quiver varieties Richard Thomas 2010-07-15T21:30:43Z 2011-01-19T15:42:00Z <p>Are deformations of Nakajima quiver varieties also Nakajima quiver varieties ?</p> <p>In case the answer to this is (don't k)no(w), here are some simpler things to ask for.</p> <ol> <li><p>(If you're a differential geometer) Is any hyperkahler rotation / twistor deformation of a Nakajima quiver variety also a Nakajima quiver variety ?</p></li> <li><p>(An example, for algebraic geometers) Consider the Hilbert scheme $H$ of $k$ points on the minimal resolution of $\mathbb C^2/(\mathbb Z/n)$. Or restrict to $n=2=k$ and consider $Hilb^2T^*\mathbb P^1$. Its exceptional divisor over the symmetric product defines a class in $H^1(\Omega_H)$ (despite the noncompactness). Using the holomorphic symplectic form, we get a deformation class in $H^1(T_H)$. (The corresponding deformation is not so far from the twistor deformation, and can be realised as a composition of a deformation of the ALE space followed by a twistor deformation.) Is there a Nakajima quiver variety in the direction of this deformation ?</p></li> </ol> <p>For instance if I take the quiver $\bullet^{\ \rightrightarrows}_{\ \leftleftarrows}\bullet$, dimension vector (1,1), and an appropriate stability condition (or value of the real moment map) then I get $T^*\mathbb P^1$ as the moduli space over the value $0$ of the complex moment map, and the smoothing of the surface ordinary double point over nonzero values.</p> <p>Now if I take dimension vector (2,2) I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. However, as I vary the value of the complex moment map I simply vary the surface that I take $Hilb^2$ of, rather than getting the deformation I'm after. (The hyperkahler rotation is not a $Hilb^2$, since the exceptional divisor disappears in this deformation.)</p> <p>But is there another quivery way of producing this deformation ?</p> http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties/32095#32095 Answer by Ben Webster for Deformations of Nakajima quiver varieties Ben Webster 2010-07-16T00:39:26Z 2010-07-16T00:39:26Z <p>There is one sense in which all deformations of Nakajima quiver varieties are at least morally again quiver varieties. Let me not try to be precise since I will screw up if I do that.</p> <p>First, note that any quiver variety <code>$X=\mu^{-1}_{\mathbb {C}}(0)//_{\alpha}G$</code> written as a hyperkaehler reduction of a vector space <code>$T^*E$</code> by a group $G$ has a natural deformation over $\mathfrak{g}^*$ given by $T^*E//_{\alpha}G$; the fibers are the reductions at different complex moment map levels.</p> <p>There is a theorem of <a href="http://arxiv.org/abs/math/0005007" rel="nofollow">Kaledin and Verbitsky</a> which implies that in any smooth symplectic family $\tilde X\to S$, (one equipped with a non-degenerate element of $\Omega^2(\tilde{X}/S)$) analytically locally at any closed point $s$ where the fiber is a Nakajima quiver variety "looks like varying the moment map parameter." Formally, this means there's a coarse moduli space for formal deformations of the Nakajima quiver variety given by $H^2(X;\mathbb{C})$ (the map is just taking the cohomology class of the symplectic form). Taking the symplectic form of the different reductions makes $H^2(X,\mathbb{C})$ a quotient of $\mathfrak{g}^*$ by Kirwan surjectivity (proven by <a href="http://arxiv.org/abs/0807.4734" rel="nofollow">Harada and Wilkin</a>) so all formal deformations come from this one (in a coarse, not a fine sense, I think).</p> <p>I'm not really sure how "global" this can be made. Perhaps someone who understands the deformation theory of non-compact varieties can say more.</p> http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties/32153#32153 Answer by Richard Thomas for Deformations of Nakajima quiver varieties Richard Thomas 2010-07-16T11:25:07Z 2010-07-16T11:25:07Z <p>Thanks for this Ben (and for the title upgrade). In fact Ivan Smith pointed the Kaledin-Verbitsky paper out to me some time ago and I forgot about it. But it sounds like the Harada-Wilkin paper might be enough to resolve this puzzle.</p> <p>To make things simple and concrete, I want to concentrate on the $Hilb^2T^*\mathbb P^1$ example, i.e. the quiver $\bullet^{\ \rightrightarrows}_{\ \leftleftarrows}\bullet$ with dimension vector $(2,2)$.</p> <p>If I understand correctly, I think there is only one complex moment map deformation: the dual of the lie algebra of the centre $\mathbb C^\ast\times\mathbb C^\ast$ of $GL(2)\times GL(2)$ divided by the diagonal $\mathbb C^\ast$.</p> <p>But I think that $Hilb^2T^*\mathbb P^1$ has 2-dimensional $H^2$.</p> <p>Does this contradict Harada-Wilkin ? I looked at their paper, but one has to know their entire paper to know how to combine their results on Nakajima quiver varieties with their results on quiver varieties (the latter having no complex moment map condition). </p> <p>2 potential solutions:</p> <p>(a) Harada-Wilkin seems to relate cohomology to the dual of the whole lie algebra, not just the centre. So perhaps one cannot interpret their results in terms of values of the moment map ?</p> <p>(b) Their condition on the dimension vector containing a 1 in Proposition 7.6 (p42) might be a problem ? If so, would my problem go away (i.e. would I get more moment map deformations) by expressing $Hilb^2T^*\mathbb P^1$ as a Nakajima quiver variety in a different way, using a construction with a 1 in the dimension vector ? Does anyone know such a construction ?</p> <p>Many thanks for your time, I'm very impressed by the generosity available on this website.</p> http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties/32165#32165 Answer by Hiraku Nakajima for Deformations of Nakajima quiver varieties Hiraku Nakajima 2010-07-16T13:13:22Z 2010-07-16T13:35:54Z <p>I do not know a general statement. I just want to give a comment:</p> <blockquote> <p>Now if I take dimension vector $(2,2)$ I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. </p> </blockquote> <p>No. You only get the symmetric product of $T^*P^1$ if you work on quiver varieties with the dimension vector $(2,2)$.</p> <p>To get a $Hilb^2$ of the surface, one need to put the one-dimensional vector space $W$at the vertex 0, and take a suitable stability condition. (I hope you are familiar with convention for quiver varieties.)</p> <p>Then we have two dimensional family of quiver varieties from the complex moment map deformation. Thus we get one more dimension from the deformation of the underlying surface.</p> http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties/32168#32168 Answer by Hiraku Nakajima for Deformations of Nakajima quiver varieties Hiraku Nakajima 2010-07-16T13:30:37Z 2010-07-16T13:30:37Z <p>One more example:</p> <p>Take the quiver of type $A_1$, and vector spaces $V$, $W$ with $\dim V > \dim W$. Then the quiver variety is empty regardless of (generic) stability parematers nor complex moment map parameters. But we have one dimensional deformation space for the moment map equation.</p> <p>Does someone know the deformation theory of the empty set ?</p> http://mathoverflow.net/questions/32069/deformations-of-nakajima-quiver-varieties/32367#32367 Answer by Alexander Cruz for Deformations of Nakajima quiver varieties Alexander Cruz 2010-07-18T14:47:17Z 2010-07-18T14:47:17Z <p>It's not an answer, it's just to say I agree with Richard Thomas' proposal. That book would be very relevant.</p>