most recent 30 from http://mathoverflow.net 2013-05-26T07:32:14Z http://mathoverflow.net/feeds/question/53865 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53865/crepant-resolution sabrebooth 2011-01-31T08:10:25Z 2011-01-31T13:43:22Z

<p>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 $C(Gr)$?<br> In fact the answer is no when $n=1$ or $n=m-1$, but what happens in the other cases ?</p>

http://mathoverflow.net/questions/53865/crepant-resolution/53874#53874 Sasha 2011-01-31T10:16:35Z 2011-01-31T10:16:35Z

<p>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. </p> <p>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!</p>

http://mathoverflow.net/questions/53865/crepant-resolution/53882#53882 Sándor Kovács 2011-01-31T13:13:17Z 2011-01-31T13:13:17Z

<p>Here is how to determine whether the blow up of a cone is a crepant resolution:</p> <p>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 $N_{E/Y}\simeq \alpha^*\mathcal O_{\mathbb P^N}(-1)$. </p> <p>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 $$\omega_G\simeq \mathcal O_{\mathbb P^N}(-1)|_G.$$ </p>