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.

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.

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. Does one have $H^*(F_1)=H^*(F_2)$?

##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 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.

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.