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 '19 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 '19 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 $V_2 \otimes V_1^{\vee}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.