User hiraku nakajima - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:44:23Z http://mathoverflow.net/feeds/user/3837 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107621/status-of-a-conjectural-definition-of-h-nakajima/124956#124956 Answer by Hiraku Nakajima for Status of a conjectural definition of H. Nakajima Hiraku Nakajima 2013-03-19T12:53:45Z 2013-03-19T12:53:45Z <p>See <a href="http://arxiv.org/abs/1004.2321" rel="nofollow">http://arxiv.org/abs/1004.2321</a>.</p> http://mathoverflow.net/questions/114557/cotangent-to-flags-as-a-quiver-variety/124953#124953 Answer by Hiraku Nakajima for cotangent to flags as a quiver variety Hiraku Nakajima 2013-03-19T12:43:26Z 2013-03-19T12:43:26Z <p>Consider the stability parameter lives in the Cartan subalgebra. The corresponding variety is smooth if it lies in a chamber, i.e., outside of any root hyperplanes. It is clear that the variety remains the same if the parameter stay in the same chamber. If one cross the wall, the variety is changed.</p> <p>On the other hand, one can change the stability condition by what I call reflection functors. They are quiver varieties analog of reflection functors for quiver reprsentations, but behave much nicer than the original ones.</p> <p>They are compatible with the Weyl group action on the space of stability parameters, i.e., the Cartan subalgebra. The dimension vector $v$ is also changed compatible with the Weyl group action, if we think $w - Cv$ as a weight.</p> <p>The reflection functors give an isomorphism between quiver varieties. Therefore, for a finite type quiver varieties, as in this question. One can take any stability parameter. It can be moved to the standard one by successive applications of reflection functors.</p> http://mathoverflow.net/questions/122683/fundamental-groups-of-symplectic-leaves/124766#124766 Answer by Hiraku Nakajima for Fundamental groups of symplectic leaves Hiraku Nakajima 2013-03-17T09:55:55Z 2013-03-17T09:55:55Z <p>I do not know the answer. I believe that it is so. See <a href="http://arxiv.org/abs/1301.1008" rel="nofollow">http://arxiv.org/abs/1301.1008</a> for algebraic fundamental groups.</p> <p>If you consider quiver varieties of affine types, they are moduli spaces of instantons on ALE spaces. If you replace the base space by $\mathbb R^4$ and assume rank is $2$, then the finiteness follows from Atiyah-Jones conjecture for rank $2$. The Atiyah-Jones conjecture actually says much stronger: there is a homotopy equivalence between moduli spaces and the infinite dimensional space of all connections up to dimension explicitly given by the instanton number. The conjecture was proved by Boyer et al.</p> <p>There are lots of generalizations of the Atiyah-Jones conjecture. So it probably applies to quiver varieties of affine types.</p> http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology/99747#99747 Answer by Hiraku Nakajima for Do fixed point sets in equivariant crepant resolutions have the same cohomology? How about for the specific case of Nakajima quiver varieties? Hiraku Nakajima 2012-06-15T21:17:02Z 2012-06-15T21:17:02Z <p>In the situation when Proposition 5.7 in my paper in <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.dmj/1077286968" rel="nofollow">Duke 1994</a> is applicable, the answer is YES: Apply the Poincare duality to $F_\alpha$ and use the formula for the Morse index $m_\alpha$. Then $\dim F_\alpha + m_\alpha = \frac12 \dim M$ is independent of $\alpha$, so the sum of $(\frac12 \dim M-i)$-th Betti numbers of $F_\alpha$ computes the $i$-th Betti number of $M$, which is independent of the choice of the crepant resolution.</p> <p>For the $\mathbb C^*$-action $t(x,y) = (tx,y)$, the symplectic form is multiplied with weight $1$, but the condition that the orientation $\Omega$ contains no cycles is violated. So I must be more careful. (First of all, Poincare duality changes ordinary cohomology to cohomology with compact support.) I never considered this situation before, but may have a chance to say something using the same argument with care.</p> <p>More general action, when the symplectic form is not necessarily multiplied with weight $1$, I do not know a clean formula for $m_\alpha$. So I do not have any idea why the Betti numbers are the same.</p> http://mathoverflow.net/questions/75509/compact-holomorphic-symplectic-manifolds-whats-the-state-of-the-art/75564#75564 Answer by Hiraku Nakajima for Compact holomorphic symplectic manifolds: what's the state of the art? Hiraku Nakajima 2011-09-15T23:43:42Z 2011-09-15T23:43:42Z <p>Hypertoric varieties (and quiver varieties) are deformation equivalent to affine algebraic varieties. Therefore they are not compact unless they are 0-dimensional or empty sets.</p> http://mathoverflow.net/questions/10409/introduction-to-w-algebras-why-w-algebras/74154#74154 Answer by Hiraku Nakajima for Introduction to W-Algebras/Why W-algebras? Hiraku Nakajima 2011-08-31T11:49:11Z 2011-08-31T11:49:11Z <p>AGT conjecture predicts that the direct sum of intersection cohomology groups of moduli spaces of $G$-instantons on $\mathbf R^4$ is a representation of the $W$-algebra attached to $G$. See <a href="http://arxiv.org/abs/1108.5632" rel="nofollow">http://arxiv.org/abs/1108.5632</a>.</p> http://mathoverflow.net/questions/48407/derived-category-of-equivariant-coherent-sheaves-and-fixed-points derived category of equivariant coherent sheaves and fixed points Hiraku Nakajima 2010-12-06T00:09:13Z 2010-12-06T00:09:13Z <p>The K-group $K^T(X)$ of $T$(torus)-equivariant coherent sheaves on a variety $X$ is isomorphic to $K^T(X^T)$, that of the fixed point locus via the inclusion homomorphism, when we tensor the quotient field of the representation ring $R(T)$ of $T$. </p> <p>Is there a similar result for $D_T(Coh X)$ and $D_T(Coh X^T)$, derived categories of $T$-equivariant coherent sheaves ? I do not know even how to formulate the quotient field of $R(T)$' in the derived category case.</p> http://mathoverflow.net/questions/40161/why-do-we-care-about-the-hilbert-scheme-of-points/40459#40459 Answer by Hiraku Nakajima for Why do we care about the Hilbert scheme of points? Hiraku Nakajima 2010-09-29T12:05:42Z 2010-09-29T12:05:42Z <blockquote> <p>What can someone who knows a lot about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects?</p> </blockquote> <p>I know a little about $X^{[n]}$. And I have no contribution to mathematics nor other areas of algebraic geometry. But I find study of Hilbert schemes is very interesting. Isn't it enough to motivate to study Hilbert schemes ?</p> http://mathoverflow.net/questions/23982/where-have-you-used-computer-programming-in-your-career-as-an-applied-pure-math/37013#37013 Answer by Hiraku Nakajima for Where have you used computer programming in your career as an (applied/pure) mathematician? Hiraku Nakajima 2010-08-29T01:31:48Z 2010-08-29T01:31:48Z <p>I am a pure mathematician interested in representation theory.</p> <p>I computed $q$-characters of $\ell$-fundamental representations for the quantum affine $E_8$ by a SUPERCOMPUTER. See</p> <p><a href="http://arxiv.org/abs/math/0606637" rel="nofollow">http://arxiv.org/abs/math/0606637</a></p> <p>There is more famous project on $E_8$:</p> <p><a href="http://www.aimath.org/E8/" rel="nofollow">http://www.aimath.org/E8/</a></p> <p>I believe there are lots of other computation in the representation theory of exceptional groups, which require lots of memory. </p> <p>They are usually based on recursive algorithms, and one cannot use the parallel computing.</p> <p>When I computed $q$-characters, I could not find any guides explaining how to code a program for such a problem. I appreciate very much if an expert could give me any references.</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/14841/are-the-strata-of-nakajima-quiver-varieties-simply-connected-do-they-have-odd-c/14896#14896 Answer by Hiraku Nakajima for Are the strata of Nakajima quiver varieties simply-connected? Do they have odd cohomology? Hiraku Nakajima 2010-02-10T12:20:37Z 2010-02-10T12:20:37Z <p>As Ben wrote, there are examples of non-simply connected strata.</p> <p>For a quiver variety corresponding to a fundamental representation, it, a priori, only has a small group action. So it is usually far away from homogeneous.</p> <p>So I suspect the answer is NO, though I do not know a concrete example.</p> <p>But, I think the question itself is not right:</p> <p>It is true that perverse sheaves appearing in the push-forward from natural resolutions' are IC sheaves associated with constant sheaves. But it is not because stratum are simply-connected, or equivariantly simply connected. It is from a very different reason:</p> <p>This reason cannot be seen if one treat an individual quiver variety separately. One can see it only when one consider various quiver varieties simultaneously, and relate them to a representation. It is basically because there is a component consisting of a single point, and other components are connected' to it in some sense.</p> <p>(Each component corresponds only to a weight space, and considering several components simultaneously, one get the structure of a representation. The single point corresponds to the highest weight vector.)</p> <p>In summary, quiver varieties are not homogeneous in a conventional sense, but have substitutive property (I do not know how to call it) when one treat several components simultaneously.</p> http://mathoverflow.net/questions/14361/what-do-the-local-systems-in-lusztigs-perverse-sheaves-on-quiver-varieties-look/14366#14366 Answer by Hiraku Nakajima for What do the local systems in Lusztig's perverse sheaves on quiver varieties look like? Hiraku Nakajima 2010-02-06T06:42:31Z 2010-02-06T12:22:34Z <p>For affine quivers, except cyclic ones, there are always perverse sheaves attached to nontrivial local systems.</p> <p>If you just need an example, I recommend you to read McGerty's paper <a href="http://arxiv.org/abs/math/0403279" rel="nofollow">math/0403279</a>, before Lusztig's paper, where the Kronecker quiver case is studied in detail.</p> <p>I also wrote a survey paper </p> <p>Crystal, canonical and PBW bases of quantum affine algebras, in <em>Algebraic Groups and Homogeneous Spaces</em>, Ed. V.B.Mehta, Narosa Publ House. 2007, 389&ndash;421.</p> http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology/99747#99747 Comment by Hiraku Nakajima Hiraku Nakajima 2012-06-26T11:21:09Z 2012-06-26T11:21:09Z I understand the point of the argument. If $r$ is divisible by $a+b$, the action of $a+b$-th roots of $t$ is well-defined. So the symplectic form is of weight $1$, and the above argument works. http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology/99747#99747 Comment by Hiraku Nakajima Hiraku Nakajima 2012-06-26T09:21:49Z 2012-06-26T09:21:49Z It seems that <a href="http://arxiv.org/abs/1206.5640" rel="nofollow">arxiv.org/abs/1206.5640</a> gives a positive answer when $a+b=r$. http://mathoverflow.net/questions/99472/what-else-is-seiberg-witten-theory-equal-to/99479#99479 Comment by Hiraku Nakajima Hiraku Nakajima 2012-06-15T21:44:02Z 2012-06-15T21:44:02Z In <a href="http://arxiv.org/abs/1001.5024" rel="nofollow">arxiv.org/abs/1001.5024</a> I together with Goettsche and Yoshioka proved SW = Donaldson for complex projective surfaces. It is still open for arbitrary $4$-manifolds, but we reduce it to another conjecture involving Nekrasov's partition function. http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology Comment by Hiraku Nakajima Hiraku Nakajima 2012-06-09T03:22:46Z 2012-06-09T03:22:46Z What do you mean by $H^*(F_1) = H^*(F_2)$' ? There is no natural homomorphism between them. If you just compare the dimensions of RHS and LHS, they are the Euler numbers of $Y_1$ and $Y_2$, and hence are equal. http://mathoverflow.net/questions/10409/introduction-to-w-algebras-why-w-algebras/74154#74154 Comment by Hiraku Nakajima Hiraku Nakajima 2011-08-31T12:36:35Z 2011-08-31T12:36:35Z Langlands dual of $G$, a little more precisely. http://mathoverflow.net/questions/66943/what-is-the-symplectic-duality-between-holomorphic-symplectic-manifolds-where/66967#66967 Comment by Hiraku Nakajima Hiraku Nakajima 2011-06-06T05:44:53Z 2011-06-06T05:44:53Z Well, for example, rather than just a bijection between bases, is there an isomorphism between 1) the cohomology of the core of quiver variety, and 2) the intersection cohomology group of the affine Grassmannian ? I want it to be natural' in some sense, which I do not know how to formulate. And it is not clear to me how this is explained in the framework of Koszul duality, discussed in the earlier part. http://mathoverflow.net/questions/66943/what-is-the-symplectic-duality-between-holomorphic-symplectic-manifolds-where/66967#66967 Comment by Hiraku Nakajima Hiraku Nakajima 2011-06-06T01:25:08Z 2011-06-06T01:25:08Z Hi, Ben. Do you have more precise statements of, for example, duality between quiver varieties and affine Grassmann ? I am not fully satisfied with your evidences in p.38. http://mathoverflow.net/questions/56295/k-uhlenbecks-preprint-a-priori-estimates-for-yang-mills-fields/58202#58202 Comment by Hiraku Nakajima Hiraku Nakajima 2011-03-12T04:42:19Z 2011-03-12T04:42:19Z It is newer than what I have. http://mathoverflow.net/questions/56295/k-uhlenbecks-preprint-a-priori-estimates-for-yang-mills-fields Comment by Hiraku Nakajima Hiraku Nakajima 2011-03-01T08:29:31Z 2011-03-01T08:29:31Z I have a scanned copy of the paper. I do not want to put the file at my webpage, but I am happy to send it to Dick if Karen could not scan. The paper has an overlap with my paper, which is available at <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jmsj/1230129890" rel="nofollow">projecteuclid.org/&hellip;</a> http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients/40556#40556 Comment by Hiraku Nakajima Hiraku Nakajima 2010-10-03T09:49:11Z 2010-10-03T09:49:11Z There is NO issue. Lusztig proved that the base $\{ b_g \}$ coincides with the base given by perverse sheaves (and hence is the canonical base) for type $ADE$. And later Saito proved the same result in general. This result is true for any choice of a reduced expression of the longest element of the Weyl group. Since Saito's proof use the whole of Kashiwara's theory, I am looking for a simpler proof. http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients/40556#40556 Comment by Hiraku Nakajima Hiraku Nakajima 2010-10-01T21:56:53Z 2010-10-01T21:56:53Z I do not know any results on this problem. By the way, could you explain how Leclerc (and you) concludes his base $\{ b_g\}$ coincides with the canonical base ? Easy to show that they coincide up to sign. But how does he remove the sign ambiguity ? The only way I know is to use Saito's result (Publ. RIMS, 1994). I would like to know if there are other methods. http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients/40556#40556 Comment by Hiraku Nakajima Hiraku Nakajima 2010-09-30T23:39:16Z 2010-09-30T23:39:16Z In the last spring Kashiwara gave a talk on KLR algebras and asked us (in particular, to Tsuchioka) what is the base given by simple modules for non-symmetric cases. My comments above were vague, as it was not clear that I can make public Kashiwara's question. Sorry. http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients/40556#40556 Comment by Hiraku Nakajima Hiraku Nakajima 2010-09-30T11:43:15Z 2010-09-30T11:43:15Z Harry - Thank you. After Tsuchioka's answer, it is probably better to change everyone' to nobody'. http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients/40556#40556 Comment by Hiraku Nakajima Hiraku Nakajima 2010-09-30T03:35:05Z 2010-09-30T03:35:05Z Could you exclude Nakajima from everyone' in the 6th line, please ? http://mathoverflow.net/questions/40161/why-do-we-care-about-the-hilbert-scheme-of-points/40459#40459 Comment by Hiraku Nakajima Hiraku Nakajima 2010-09-29T12:43:57Z 2010-09-29T12:43:57Z If I am asked this question by a politician, I will answer differently.