Let $X$ be a projective variety over a field $k$ equipped with a very ample line bundle $\mathcal{O}_X(1)$. Suppose that $E, F$ are locally free sheaves of finite rank on $X$ and $c\in \mathrm{Ext}^i(E, F)$ is a non-zero class.

Question: Do there always exist integers $n, d$ such that the map $H^n(X, E\otimes\mathcal{O}_X(d))\to H^{n+i}(X, F\otimes \mathcal{O}_X(d))$ induced by $c$ is non-zero?

As Will Sawin notes in the comments below, the answer to the question as stated is no. I would still be very interested in a positive answer to a modified question:

Question 2: Do there exist a degree $n\in\mathbb{N}$ and a line bundle $L$ such that the induced map $H^n(X, E\otimes L)\to H^{n+i}(X, F\otimes L)$ is non-zero?

I've tried to construct a class in some $H^n(X, E\otimes \mathcal{O}_X(d))$ that is not killed by $c$ by resolving $E$ and $F$ by direct sums of powers of $\mathcal{O}_X(1)$ but that didn't seem to help. Another thing to note is that, for $i>0$, there is only a finite range of degrees $d$ where both groups $H^n(X, E(d))$ and $H^{n+i}(X,F(d))$ are non-zero for some $n$, because of Serre duality and Serre vanishing.

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    $\begingroup$ If $\mathcal O_X(1)$ is a line bundle of high enough degree on a curve, there could be no degrees where both those groups are nonvanishing (for $i=1$ in this case). But there are plenty of nontrivial Ext classes. For example $X = \mathbb P^1$, $E$ a line bundle of degree $1$, $F$ a line bundle of degree $-1$, and $\mathcal O_X(1)$ a line bundle of degree at least $2$. $\endgroup$ – Will Sawin Apr 13 at 18:12
  • $\begingroup$ @WillSawin Ah, good point, do you think that this could be salvaged if in place of $\mathcal{O}_X(d)$ we allow an arbitrary line bundle? (I've edited the question accordingly) $\endgroup$ – SashaP Apr 13 at 18:44

This fails even on curves, where one has more line bundles to play with.

Let $X$ be a curve of genus $>1$. Let $i=1$. Take $E$ a stable vector bundle of rank $2$ and degree $0$ and $F = E \otimes K_X $.

We have a $Ext^1(E, F) = H^1( X, K_X \otimes E \otimes E^\vee)$ which admits $H^1(X, K_X) \neq 0$ as a summand.

We have $H^0 (X ,E \otimes L) =0 $ for all $L$ of degree $\leq 0$ by stability of $E$. But $H^1 (X, F \otimes L) = H^1(X, E \otimes K_X \otimes L) = H^0( X, E^\vee \otimes L^{-1}) =0$ for all $L$ of degree $\geq 0$, again by stability.

So this map will always vanish.

  • $\begingroup$ Thanks! This was too good to be true. $\endgroup$ – SashaP Apr 13 at 19:31

The answer is no. For instance, let $X = \mathrm{Gr}(2,4)$ and consider the tautological exact sequence $$ 0 \to S \to \mathcal{O}^{\oplus 4} \to Q \to 0 $$ of vector bundles. It represents a nontrivial extension class $$ \epsilon \in \mathrm{Ext}^1(Q,S). $$ However, $\mathrm{Pic}(X) = \mathbb{Z} \cdot\mathcal{O}_X(1)$ and for any $d \in \mathbb{Z}$ one has $$ H^i(X,S \otimes \mathcal{O}_X(d)) = H^i(X,Q \otimes \mathcal{O}_X(d)) = 0 $$ unless $i = 0$ or $i = 4$. In particular, all the maps induced by $\epsilon$ are zero.


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