Let $F$ be a field and $V$ be a $n$-dimensional $F$-vector space, then $\{A \in End(V) | A^2 =0, \operatorname{rank} A=k \}$ gives an algerbraic variety $\mathcal{N}\_{n,k}$ over $F$. There is a natural map $f$ from $\mathcal{N}\_{n,k}$ to a flag variety $Fl\_{(k,n-k)}$ sending $A$ to $0 \subseteq Im A \subseteq Ker A \subseteq V$.


$f$ is surjective (over the algebraic closure of $F$), and the fiber of $f$ can be identified with $GL\_k$, so does $f$ define a $GL\_k$-torsor hence a vecotor bundle on $Fl\_{(k,n-k)}$ ? What is this rank $k$ vector bundle ?


Edit: the comment says it's not a $GL\_k$ torsor in general.


Can we generalize such construction to other linear algebraic groups (using the nilpotent cone)?


Example: if $k=1$ and $n=2$, then this gives $O(-2)$ on $\mathbb P^1$, the canonical bundle. 



Natural vector bundle on flag variety coming from the variety of nilpotent matrices of fixed rank?
If $k=n/2$ exactly, then any rank $k$ matrix with $A^2=0$ has kernel = image. Then you can (non-canonically) identify the fiber with $GL\_k$.


However, there is no natural action of $GL\_k$ on the fiber. (At least no obvious one.) Instead, the fiber is the space of isomorphisms between two vector bundles, those being the $k$-dimensional subspace and the $k$-dimensional quotient space.


The case of $k=1$ is special because $GL\_1$ is abelian, so the $\operatorname{Isom}(V\_1, V\_2)$ can be viewed as a single torsor, for instance as $\operatorname{Isom} (1, V\_2 \otimes V\_1^{-1})$ and so the associated vector bundle is $V\_2 \otimes V\_1^{-1}$. 


But this doesn't work in general, we just have two vector bundles and the tensor product $V\_2 \otimes V\_1^{\vee}$.


