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Let's put $m>n$ two nonnegative integers and $Gr:=Grass(n,k^m)$ the grassmanian of the subspaces of dimension $n$ in $k^m$. We have a natural immersion $Gr \subset P({\Lambda}^{n} k^m)$ and I call $C(Gr) \subset {\Lambda}^{n} k^m$ the affine cone above $Gr$. If we blow-up $C(Gr)$, we always obtain something smooth, but my question is: Is the blow-up a crepant resolution of the singularities of $C(Gr)$?
In fact the answer is no when $n=1$ or $n=m-1$, but what happens in the other cases ?

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I should point out that sometimes a cone can have a crepant resolution even if the cone-vertex-blow-up is not smooth. It might have a small resolution (for example the singularity $xy - uv$ has a small, and thus crepant resolution, even though the blow-up of the origin is not a crepant resolution). – Karl Schwede Jan 31 '11 at 13:46
Bah, and the previous comment should have said "even if the cone-vertex-blow-up is not crepant". – Karl Schwede Jan 31 '11 at 15:22
up vote 2 down vote accepted

The result of the blowup is the total space of line bundle $O(-1)$ on the Grassmannian $Gr$. It follows that its canonical class equals the canonical class of the Grasmannian (i.e. $-m$) plus the relative canonical class of the total space (i.e. $1$). Thus the canonical class is $1-m$ which as you see is always negative. Moreover, it is not a pullback from the cone (since restricts nontrivially to the exceptional divisor of the blowup = zero section of the total space). This shows that the resolution is not crepant.

Moreover, as far as I understand the cone over Grassmannian has NO crepant resolutions unless $n = 1$ or $n = m-1$ when it is smooth. However, it has a noncommutative crepant resolution!

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Dear Sasha, what's a good reference for your last statement? I only know one (Kuznetsov) in the case $n=2$. – Graham Leuschke Feb 1 '11 at 22:01
Yes, this is what I had in mind. The case $n > 2$ is quite similar. If you want to obtain a (weakly) crepant resolution you just need a n exceptional colllection on the Grassmannian containing $O$, $O(1)$, \dots, $O(m-1)$ which is easy to construct. – Sasha Feb 2 '11 at 7:59

Here is how to determine whether the blow up of a cone is a crepant resolution:

Let $X$ be a cone over smooth the projective variety $G\subseteq \mathbb P^N$ and $\phi:Y\to X$ the blow up of the vertex of the cone. Then the exceptional set of $\phi$ is a Cartier divisor $E\subset Y$ such that there is an isomorphism $\alpha: E\to G$. The normal bundle of $E$ in $Y$ can be expressed as $N_{E/Y}\simeq \alpha^*\mathcal O_{\mathbb P^N}(-1)$.

If $\phi$ is crepant, it means that $K_Y\sim \phi^*K_X$ which implies that the normal bundle of $E$ in $Y$ is numerically equivalent to its canonical divisor: $$ K_E \sim (K_Y + E)|_{E} \sim (\phi^*K_X + E)|_E \equiv E|_E. $$ So, you only need to decide whether $G\subseteq \mathbb P^N$ is an anticanonical embedding, that is, whether $$\omega_G\simeq \mathcal O_{\mathbb P^N}(-1)|_G.$$

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