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.



Resolution of conical singularities have even-only cohomology?
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.


