Consider :
The quasi affine variety $V$ composed of triples $(B_1,B_2,v)\in \mbox{Mat}_n(\mathbb C)\oplus\mbox{Mat}_n(\mathbb C)\oplus\mathbb C^n$ such that the map $\mathbb C[X,Y]\rightarrow\mathbb C^n, P\mapsto P(B_1,B_2)v$ is surjective.
The affine subvariety of $W$ of $\mbox{Mat}_n(\mathbb C)$ composed of trace $0$-matrices.
The morphism $f:V\rightarrow W$, given by $(B_1,B_2,v)\mapsto[B_1,B_2]$.
I have heard that this is not a submersion, i.e. there exists a point of $V$ in which the differential of $f$ is not surjective.
Any idea of how to show it?
By the way, this is related to the smoothness of a Hilbert´s Scheme of points.
Thank you for reading, Lisa
Well, $L=d_{B_1,B_2,v}f$ is $E,H \longmapsto [B_1,H]+[E,B_2]$. In particular, take $B_1=B_2$ a diagonal matrix with pairwise distinct coefficients. Then the commutant of $B_1$ has dimension $n$, and thus the kernel of $[\cdot,B_i]$ has dimension $n$, and thus its image has dimension $n^2-n$, and thus the image of $L$ has dimension $n^2-n$. So if $n>1$, then the image of $L$ can’t be all of $W$.
Note finally that if $v$ is the vector made with only ones, and $B$ is diagonal with pairwise distinct diagonal coefficients, then $(B,B,v) \in V$.