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Suppose that we are given a smooth projective variety $X$ with a full exceptional collection of vector bundles $(F_1, F_2, \ldots, F_k)$ in $D^b(X)$ and two vector bundles $E_1$, $E_2$ on $X$. Consider the following statement: $$ \text{If }H^i(X, E_1\otimes F_j) = H^i(X, E_2\otimes F_j)\text{ for all }i, j\text{, then } E_1=E_2. $$

Question 1. Is the above true?

If no,

Question 2. What if we assume that the collection is strong?

Question 3. What if we consider all twists by the ample sheaf $\mathcal{O}(1)$, i. e. $H^i(X, E_1\otimes F_j(t)) = H^i(X, E_2\otimes F_j(t))$ for $i, j$ and all $t$?

Question 4. Are there any known cohomological criteria for isomorphism?

If yes,

Question 5. Does it suffice to assume that $(F_1, F_2, \ldots, F_k)$ generate the derived category, without them forming an exceptional collection?

Remark. Take $X=\mathbb{P}^n$ and $F_i = \mathcal{O}(i)$ or $F_i = \Omega^i(i)$. Then by the Beilinson spectral sequence, the cohomology groups $H^i(X, E\otimes F_i)$ would determine $E$ unambiguously if we knew the differentials.

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up vote 2 down vote accepted

It is hard to think of a general way to obtain an isomorphism of vector bundles from several isomorphisms of cohomology spaces. In particular, there can be moduli of vector bundles, I presume, even over projective spaces. But an isomorphism of cohomology only means an equality of dimensions, and dimensions are discrete invariants, taking values in the integers.

It would be an entirely different matter if the question were, is a given morphism of vector bundles an isomorphism provided that it induces isomorphisms of certain cohomology spaces. For this version of your question to have a positive answer, it suffices to require that (Fi) generate the derived category (due to Serre duality).

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You convinced me, except for my Question 3. Actually, in my case the bundle $E_1$ is the Frobenius push-forward of $\mathcal{O}_X$. There are easy cohomological criteria for $E_1 = \mathcal{O}_X$. What about $E_1 = F^s_\ast \mathcal{O}_X$? – Piotr Achinger Feb 26 '10 at 20:17
With Question 3, you get an infinite sequence of discrete invariants, indexed by the integers and taking value in the integers. If there are actually moduli of vector bundles over your variety, and if the cardinality of the ground field is more than continuum, such invariants cannot distinguish vector bundles, just for this simple set-theoretical reason. I cannot tell anything about Frobenius push-forwards, unfortunately. – Leonid Positselski Feb 26 '10 at 22:27

If the isomorphism above is as dg-modules over the Ext dg-algebra of $\bigoplus F_i$, then the answer to question 1 is yes. This is a general statement about generating sets in derived categories. The reason is that we can get $E_i^*$ by tensoring $\bigoplus F_i$ with the cohomologies over the dg-Ext algebra.

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