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.

  • 2
    $\begingroup$ 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 $\underline{Aut}(F)$. $\endgroup$
    – abx
    Jul 1, 2019 at 17:13
  • $\begingroup$ @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. $\endgroup$
    – sawdada
    Jul 2, 2019 at 0:44

1 Answer 1


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


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