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 ?



crepant resolution
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!


