2
$\begingroup$

Let $X$ be a projective homogeneous space over $\mathbb{C}$, i.e. $G/P$ where $G$ is a simple, simply connected linear algebraic group and $P$ is a parabolic subgroup. Let $f\colon\mathbb{P}^1\to X$ be a general morphism.

Q. What does $f^*TX$ look like?

By Grothendieck's theorem, $f^*TX$ splits as a sum $\bigoplus_i\mathcal{O}(a_i)$. How to determine the numbers $a_i$?

A general morphism $f$ as above has obviously image an $\mathrm{SL}_2(x)$-orbit, where $x$ is the sum of some root vectors $\sum_ix_{\theta_i}$. It definitely helps to know this, but I don't see the solution clearly.

Presumably, the solution is as follows: Let $d$ be the degree of $f$. Then, $f^*TX=\bigoplus_{\alpha\in R^-\setminus R_P^-}\mathcal{O}(d,\alpha)$. How to prove this?

Here, $R^-\setminus R_P^-$ denotes the roots in $\mathfrak{g}/\mathfrak{p}$.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ Your guess is incorrect. There are many different possibilities for the pullback. One common feature is that the pullback is globally generated. For more about this, see the article of Coskun — Riedl and citations there. $\endgroup$
    – Jason Starr
    Commented Nov 24, 2020 at 11:48
  • 1
    $\begingroup$ I realize now that I may have misunderstood the question. Are you asking about the pullback of the tangent bundle for every morphism from the projective line, or are you asking about the pullback only for the morphism corresponding to the geometric generic point of the Hom scheme? $\endgroup$
    – Jason Starr
    Commented Nov 24, 2020 at 17:42
  • $\begingroup$ Thanks for the reaction and the reference @Jason Starr. I'm asking for a general morphism, i.e. one which is in an open dense subset of the Hom scheme. I guess that it is what you mean with geometric generic point. Please take a look at the related question which popped up in the sidebar after I posted. The Grothendieck decomposition is somehow deformation invariant; it doesn't matter which morphism in the open dense set you choose, the decomposition stays the same. Sorry, if I express myself unclear, I cannot do better. $\endgroup$
    – Christoph Mark
    Commented Nov 25, 2020 at 4:51
  • $\begingroup$ Globally generated in our context means probably that all the numbers in the decomposition are nonnegative. Is that right? $\endgroup$
    – Christoph Mark
    Commented Nov 25, 2020 at 4:54
  • 1
    $\begingroup$ You mention examples of morphisms that are orbit closures (you mention $\textbf{SL}_2$ orbits, but these will also be orbit closures for the action of the $1$-dimensional tori in $\textbf{SL}_2$). For these morphisms, you can compute the integers by equivariant localization. A computation of this kind is the key computation in my (unique) article with Johan de Jong and Xuhua He. I suspect that your description the pullback of the tangent bundle is correct. I will think about this and hopefully reply soon. $\endgroup$
    – Jason Starr
    Commented Nov 26, 2020 at 19:16

0

Reset to default

You must log in to answer this question.