2
$\begingroup$

Let $X$ be a smooth general ordinary Gushel-Mukai threefold. There is an embedding $X\rightarrow\mathrm{Gr}(2,5):=G$. Consider the normal bundle $\mathcal{N}_{X|G}$, how to compute cohomology of this vector bundle and related vector bundle, i.e, what is $H^k(X,\mathcal{N}_{X|G}\otimes\mathcal{O}_X(2))$, $H^k(X,\mathcal{N}_{X|G})?$

Let $T_X$ be the tangent bundle of $X$($X$ is general OGM threefold or more general, Fano threefold of index 1, Picard rank 1), how to compute $H^0(X,\bigwedge^2T_X)$? I was trying to use normal bundle exact sequence and some other exact sequence involving tangent bundle on $X$, but I could not decide which one I should use?

$\endgroup$

1 Answer 1

6
$\begingroup$

$\mathcal{N}_{X|G}$ is in your case $F|_X$, where $F=O_G(1)^2 \oplus O_G(2))$. In order to compute the first two spaces, you can simply use the Koszul complex for X $$ 0 \to det(F^{\vee}) \to \wedge^2 F^{\vee} \to F^{\vee} \to O_G \to O_X \to 0,$$ twisted with $F$ (or $F(2)$ in the second case).

You can use Borel-Bott-Weil to compute all the terms needed above. Saving you the trouble, the cohomologies you are asking for are 46 dimensional for $H^0(\mathcal{N}_{X|G})$, 313 dimensional for $H^0(\mathcal{N}_{X|G}(2))$, 0 for all other $k$.

Answering your second question. $X$ is a Fano threefold of index 1, therefore $\wedge^2 T_X \cong \Omega^1_X(1)$. You can use the (twisted) cotangent sequence to compute the dimension of the latter. You have $$ 0 \to O_X^2 \oplus O_X(-1) \to \Omega^1_G|_X(1) \to \Omega^1_X(1) \to 0$$

Clearly $H^1( O_X^2 \oplus O_X(-1))=0$. On the other hand you can again use Koszul to compute the cohomology groups of the central object. If I made no mistakes, $H^0(\Omega^1_G|_X(1))= \mathbb C^2$. Therefore $H^0(\wedge^2 T_X)=0$.

You can try to do similar computations for all other Fano 3-folds. Embedding in Grassmannians (or at most weighted projective spaces) are known. Therefore the computations might be longer, but the strategy is the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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