Isomorphism of cohomology related to Kunneth formula

Question

Let $S$ be a smooth complex (rational) algebraic surface, and $\mathcal{F}$ be a quasi-coherent sheaf such that $H^2(S, \mathcal{F}) = 0$.

Then, by Kunneth formula, we have $H^2(S \times S, \mathcal{F} \boxtimes \mathcal{F} ) \simeq H^1(S, \mathcal{F}) \otimes H^1(S, \mathcal{F})$.

Consider the action of $\mathcal{S}_2$ on $S \times S$ by commuting coordinates.

Is the map $(H^1(S, \mathcal{F}) \otimes H^1(S, \mathcal{F}))^{\mathcal{S}_2} =Sym^2 (H^1(S, \mathcal{F})) \rightarrow H^2(S, \mathcal{F}^{\otimes 2})$ isomorphism ?

(In the first place, how $\mathcal{S}_2$ acts on $H^2(S \times S, \mathcal{F} \boxtimes \mathcal{F} ) \simeq H^1(S, \mathcal{F}) \otimes H^1(S, \mathcal{F})$ ?)

Thanks in advance.

Answer 1

No, it is not true. Take for instance $S = \mathbb{P}^2$ and $F = \mathcal{O}(-2)$. Then $H^2(F) = H^1(F) = 0$, but $H^2(F \otimes F) = H^2(\mathcal{O}(-4)) \ne 0$.