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Considering a quotient singularity $\mathbb{C}^n/G,$ its crepant resolution $Y$ (i.e. having $c_1(Y)=0$) has rational cohomology supported in even degrees only. This holds for many other resolutions of singularities, in particular for symplectic resolutions of conic symplectic singularities.

I am wondering whether there is a general statement saying that resolution $Y$, satisfying $c_1(Y)=0$, of a conic singular affine variety $X$ with isolated singularitity has $H^{odd}(Y,\mathbb{Q})=0$? Conic means that there is a $\mathbb{C}^*$-action on $X$ that contracts it to a point. Potentially ask that $X$ is a complete intersection.

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    $\begingroup$ If one doesn't demand that $X$ is conic, there will be no reasonable local condition that implies this, as $X$ and $Y$ can have rational cohomology in odd degrees for global reasons, e.g. when $X$ is a smooth affine variety. $\endgroup$
    – Will Sawin
    Commented Nov 8, 2022 at 17:17
  • $\begingroup$ That's a fair point -- just added it, thanks! $\endgroup$
    – Filip
    Commented Nov 8, 2022 at 17:22
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    $\begingroup$ What about the following. Consider a cubic threefold Z (or some Fano manifold with an embedding given by H such that mH=K for some m>0). Let X be the cone over the |mH|-embedding of X. Then X is conic. It has a resolution given by the total space Y=L of the line bundle O(-mH). The canonical bundle of Y is the pull back of K of X times the relative canonical which is -K (so c_1=0). In the case Z is a cubic threefold, Z has odd cohomology. The line bundle Y is homotopy equivalent to Z. So Y has odd cohomology. $\endgroup$
    – Yosemite Stan
    Commented Nov 8, 2022 at 18:00
  • $\begingroup$ This sounds good, thanks! Just to make sure that I got this correct, in the example when $Z\subset \mathbb{C} P^4$ is a cubic threefold, thus $c_1(Z)=2,$ then $X \subset \mathbb{C}^{16}$ can be written as the cone over the image of $Z$ under the morphism $$\mathbb{C} P^4 \rightarrow \mathbb{C} P^{15}, [z_0: z_1 : \dots : z_4]\mapsto [z_0^2:z_0 z_1: \dots :z_4^2].$$ Here I use degree-2 polynomials as $c_1(Z)=2.$ Am I right? $\endgroup$
    – Filip
    Commented Nov 8, 2022 at 19:17
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    $\begingroup$ Yes, that should be right. It would be even easier to look at the cone over the quartic threefold because that will live in A^5 $\endgroup$
    – Yosemite Stan
    Commented Nov 8, 2022 at 19:40

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Thanks to @Yosemite Stan, we have a counterexample, and actually many of them: Pick a projective variety $Z$ with some odd-cohomology such that $$c_1(\omega_Z)=-m c_1(H), \text{ for some } m>0,$$ where $H$ is the bundle by which $Z$ embeds to $\mathbb{P}^n,$ and $\omega_Z$ is the canonical bundle.

In other words, we have $c_1(Z)=-c_1(\omega_Z)=m c_1(i^*\mathcal{O}(1)),$ where $i:Z \hookrightarrow \mathbb{P}^n$ is the embedding given by $H.$ In particular, a degree $d$-hypersurface in $\mathbb{P}^n$ having some odd cohomology works, with $m = n+1-d$ (due to adjunction formula).

Denoting by $\widetilde{Z}$ its image via the Veronese map $\mathbb{P}^n\rightarrow\mathbb{P^{\binom{n+m}{m}-1}}$, define $X\subset \mathbb{C}^{\binom{n+m}{m}}$ to be the cone over $\widetilde{Z}$. There is a natural resolution $Y$ of $X$ given by the $Y=\overline{\pi^{-1}(X\setminus 0)},$ where $\pi$ is the blow-up at ${0}\subset \mathbb{C}^{\binom{n+m}{m}}.$ As the blow-up at $0$ is $\mathcal{O}(-1)\rightarrow \mathbb{P^{\binom{n+m}{m}-1}},$ we have $$Y=\mathcal{O}(-1)|_\widetilde{Z} \cong \mathcal{O}(-m)|_Z,$$ hence it has $$c_1(Y)=c_1(i^*\mathcal{O}(-m))+c_1(Z)=-mc_1(i^*\mathcal{O}(1)+c_1(Z)=0,$$ and cohomology equal to $Z,$ hence not supported in even degrees only.

In particular, choosing the quartic 3-fold $Z_4\subset \mathbb{P}^4$, we have $b_3(Z_4)=60$ and $m=(4+1)-4=1,$ so $X=V(z_0^4+\dots+z_4^4)$ is an isolated singularity and (trivially) a complete intersection as well, and its resolution given by the blow up of $\mathbb{C}^5$ at the origin yields a counterexample.

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