# Extension between vector bundles inducing non-zero map on cohomology

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.

• 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 $2$. – Will Sawin Apr 13 at 18:12
• @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) – 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.

• Thanks! This was too good to be true. – 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.