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 blowup $C(Gr)$, we always obtain something smooth, but my question is:
Is the blowup a crepant resolution of the singularities of $C(Gr)$?
In fact the answer is no when $n=1$ or $n=m1$, but what happens in the other cases ?



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 $1m$ 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 $n = m1$ when it is smooth. However, it has a noncommutative crepant resolution! 


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 $$\omega_G\simeq \mathcal O_{\mathbb P^N}(1)_G.$$ 

