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A crepant resolution $f:Y\to X$ is a resolution of singularities with $f^*(K_X)=K_Y$. Crepant resolutions do not always exist, and when they exist they may not be unique. However, different crepant resolutions $Y_1$ and $Y_2$ share many properties. In particular, Kontsevich introduced motivic integration to prove that the $Y_i$ have the same Betti numbers.

Suppose that $X$ has a $\mathbb{C}^*$ action, and that $f_i:Y_i\to X; i=1,2$ are equivariant crepant resolutions -- that is, the $Y_i$ have $\mathbb{C}^*$ actions and the $f_i$ are equivariant maps. Let $F_i\subset Y_i$ be the fixed point sets. Do the $F_i$ have the same Betti numbers?

How about for Nakajima quiver varieties?

This may be too much to expect in general, so let me mention that the case of particular interest to me is when $Y$ and $X$ are Nakajima quiver varieties (see Ginzburg).

More specifically, $X$ is the space $(\mathbb{C}^2/G)^n/S_n$, where a generator $g\in G=\mathbb{Z}/r\mathbb{Z}$ acts on $\mathbb{C}^2$ by $g(x,y)=(\omega x, \omega^{-1} y)$ for $\omega$ a primitive $r$th root of unity, and the $Y_i$ are certain other quiver varieties. Nakajima considers the $\mathbb{C}^*$ action given by $t(x,y)=(tx,y)$. I care about a more general action $t(x,y)=(t^a x,t^b y), a,b>0$. This $X$ is a Nakajima quiver variety, and other Nakajima quiver varieties provide natural equivariant crepant resolutions.

Note that in this case it is known that the $Y_i$ are in fact diffeomorphic; however they are not equivariantly diffeomorphic for my torus action. Also note that using the action of the larger torus $(\mathbb{C}^*)^2$, we can see that $\chi(F_1)=\chi(Y_1)=\chi(F_2)$, so they at least have the same euler characteristic.

For these Nakajima quiver varieties you can compute the $H^*(F_i)$ on a case by case basis using the obvious $(\mathbb{C}^*)^2$ action, and the question seems to have an affirmative answer. However, actually proving it in general using this method boils down to a difficult combinatorial question about partitions that is what I was originally considering. I only recently came up with the current formulation of the question, which seems like quite a natural question, and I was hoping that I would find the answer to my question already in the literature (for instance, in Kaledin's work on the symplectic McKay correspondence or symplectic resolutions more generally), but have had no luck so far, and so I turn to MO.

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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. –  Hiraku Nakajima Jun 9 '12 at 3:22
    
Hiraku, I could have been clearer. We certainly have $\chi(F_1)=\chi(X)=\chi(F_2)$ by localization. But it appears that more is true -- namely, the betti numbers of the $F_i$ are equal. I've edited the question to make this a bit clearer. –  Paul Johnson Jun 15 '12 at 11:11

2 Answers 2

For a Nakajima quiver variety satisfying Kirwan surjectivity on $H^2$ (which may well be all of them; I don't know a counterexample), all crepant resolutions are diffeomorphic, since they are quiver varieties themselves (you can check that you get every class in the Picard group from GIT reductions). If the $\mathbb{C}^*$ of interest to you comes from a linear action of the cotangent bundle you reduced to get the quiver variety, then you can make the diffeomorphism $S^1$-equivariant, and thus induce a diffeomorphism on fixed point sets (look at the proof of 3.4 in Proudfoot's thesis; that's for the abelian case, but the proof is the same.

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Thanks for the reference -- unfortunately, it appears that with my group action they aren't in fact equivariantly diffeomorphic. I'm really looking at just the ordinary hilbert scheme of points on C^2. Z_r acting antidiagonal on C^2 induces an action on the hilbert scheme, and the components of the fixed point set are the quiver varieties. The C^* action I have is the induced action from a C^$ action on C^2. With the whole (C^*)^2, you know the fixed point sets agree -- they are finite sets. But for different resolutions the normal bundles of the fixed point sets are different. –  Paul Johnson May 17 '12 at 17:46

In the situation when Proposition 5.7 in my paper in Duke 1994 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.

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.

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.

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It seems that arxiv.org/abs/1206.5640 gives a positive answer when $a+b=r$. –  Hiraku Nakajima Jun 26 '12 at 9:21
    
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. –  Hiraku Nakajima Jun 26 '12 at 11:21

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