# Natural vector bundle on flag variety coming from the variety of nilpotent matrices of fixed rank?

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.

• It is not a $GL_k$-torsor — what would be the action?? On your flag variety there are two natural rank $k$ vector bundles $E$ (corresponding to the rank $k$ subspace of $V$) and $F$ (quotient of $V$ by the rank $n-k$ subspace); you are looking at the space $\underline{Isom}(E,F)$. It is a torsor both under $\underline{Aut}(E)$ and $\underline{Aut}(F)$. – abx Jul 1 at 17:13
• @abx Thank you. For $k=1$, there is a natural action of $\mathbb G_m$, and it's a $\mathbb G_m$-torsor. I think what you said is correct. – sawdada Jul 2 at 0:44

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 $$V_2 \otimes V_1^{\vee}$$.