It is easy to realize cotangent space to the flag variety $Fl=SL_n/B$ as a Nakajima quiver variety: consider the finite quiver of type A, the dimension vectors v=(1,2,...,n-1), w=(0,...,0,n); an appropriate stability condition (polarization) amounts to the condition that the arrow from the i-dimensional space to the (i+1)-dimensional one is injective, and we end up with a complete flag in the n-space, the arrows in the opposite direction giving a cotangent vector.

Now, if I understand correctly, the other stability conditions (of which there is n!) should produce quiver varieties which are also isomorphic to $T^*(Fl)$. How to see this, preferably using equally explicit linear algebra? Is it explained in the literature?

  • $\begingroup$ I'm not sure this is correct - can you elaborate on what your notion of 'stability' is? If you take the trivial character and form the GIT quotient of the fibre over 0 of the moment map used to define Nakajima quiver varieties then we obtain the categorical affine quotient which is Spec of the invariant polynomial functions on this fibre (see Ginzburg's notes 'Lectures on Nakajima Quiver Varieties', Thm. 4.5.6). This is how one obtains the 'quiver variety analog' of the Springer resolution. Perhaps I am misunderstanding your stability condition - where are the n! possibilities coming from? $\endgroup$ – George Melvin Nov 26 '12 at 18:44
  • $\begingroup$ I just mean the usual $\theta$ stability, as defined for example in Ginzburg's notes cited $\endgroup$ – Roman Nov 27 '12 at 2:13
  • $\begingroup$ Can you give a little more information on how 'the other stability conditions' arise? In particular, why are there n! different stability conditions? The $\theta=0$ stability condition (as mentioned in my previous comment) gives a singular variety so cannot be the same as the cotangent bundle. $\endgroup$ – George Melvin Nov 27 '12 at 19:30
  • $\begingroup$ Of course I am considering a generic $\theta$ in order to get cotangent bundle which is nonsingular variety. For example $\theta=(1,...,1)$. $\endgroup$ – Roman Nov 28 '12 at 0:50

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.

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.

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.

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.

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  • $\begingroup$ Do they give actual isomorphisms, not "just" hyperkahler isomorphisms? I want to be sure that the complex structures haven't gotten switched around. $\endgroup$ – Allen Knutson Sep 4 '17 at 13:39
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    $\begingroup$ Do you mean a Riemannian isometry ? Reflection functors are hyper-Kaehler isomorphisms, meaning a Riemannian isometry respecting all complex structures, I, J, K. Much nicer than just isomorphisms of algebraic varieties. $\endgroup$ – Hiraku Nakajima Sep 4 '17 at 22:54
  • $\begingroup$ Is this comment: ''One can take any stability parameter. It can be moved to the standard one by successive applications of reflection functors'' true for any quiver? $\endgroup$ – Filip92 Feb 12 '19 at 0:09
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    $\begingroup$ No. As I wrote above, the space of the stability parameter is the Cartan subalgebra. For general quiver, the Cartan subalgebra is the one for Kac-Moody Lie algebra associated with the quiver. It is well-known that Weyl group action is not transitive on the set of chambers in general. $\endgroup$ – Hiraku Nakajima Feb 13 '19 at 2:27

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