# Multiplicative structure for sheaf cohomology of flag varieties

Let $$F,F'$$ be two locally free sheaves on a smooth complex algebraic variety. There is a cup-product $$H^i(X, F) \otimes H^j(X,F') \to H^{i+j}(X,F \otimes F')$$. In particular if $$F$$ is the sheaf of section of $$\wedge^{\bullet}E$$ for a vector bundle $$E$$, we get a natural algebra structure on $$H^{\bullet}(X, F)$$.

For example (I hope the notation is clear), if $$X = G/B$$ is the flag variety, and $$E = G \times^B \mathfrak n$$, by Hodge theory $$H^{\bullet}(X, F) \cong H_{dR}^{\bullet}(X, \Bbb C)$$. Apparently it is possible way to compute $$H^{\bullet}(X,F)$$ using the filtration of $$E$$ (where each composition factor is a direct sum of line bundles) and apply Borel-Weil-Bott.

Question : Using the filtration method, is it possible to get the product structure as well ?

For example says $$G = \text{SL}_3$$, then $$H^1(X,\mathscr L_{-\alpha_1}) \cong \Bbb C$$, say generated by $$\{x_1\}$$ and similarly for $$x_2$$. There is a isomorphism $$H^1_{dR}(X, \Bbb C) \cong \Bbb C \{x_1\} \oplus \Bbb C\{x_2\}$$. However, my naive hope was that $$x_1x_2 \in H^2_{dR}(X, \Bbb C)$$ could be computed using the map $$\mathscr L_{-\alpha_1} \otimes \mathscr L_{-\alpha_2} \to \mathscr L_{-\alpha_1 - \alpha_2}$$ (for global sections it works like that). However $$\mathscr L_{-\rho}$$ is acyclic and my naive hope doesn't hold (as we know from de Rham cohomology that $$x_1x_2 \neq 0$$).

In practice, I have more complicated sheaves on $$G/B$$ but I understand which line bundles contribute to the cohomology. I would like to know if there is a way to compute the product (other than say write down the Cech complex which is a bit tedious). I guess it boils down to a spectral sequence computation, but I'm not very familiar with it. Even computing the product $$H^1(X, \Omega) \otimes H^1(X, \Omega) \to H^2(X, \Omega^2)$$ for $$G = \text{SL}_3$$ from the filtration could be very useful.

I think to obtain the product you need to use Massey products. Let me illustrate this in the case of $$G = \mathrm{SL}_3$$. Consider the element $$x_1 \otimes x_2 \otimes x_1 \in \mathrm{Ext}^1(\mathcal{O},\mathcal{L}_{-\alpha_1}) \otimes \mathrm{Ext}^1(\mathcal{L}_{-\alpha_1},\mathcal{L}_{-\alpha_1-\alpha_2}) \otimes \mathrm{Ext}^1(\mathcal{L}_{-\alpha_1-\alpha_2},\mathcal{L}_{-2\alpha_1-\alpha_2}).$$ If $$m$$ denotes the usual product then $$m(x_1 \otimes x_2) = 0 \qquad\text{and}\qquad m(x_2 \otimes x_1) = 0,$$ this is precisely the situation when the Massey product $$m_3(x_1\otimes x_2 \otimes x_1) \in \mathrm{Ext}^2(\mathcal{O},\mathcal{L}_{-2\alpha_1-\alpha_2})$$ is defined (note that the cohomological degree of $$m_3$$ is $$-1$$, so the result lives in $$\mathrm{Ext}^2$$). In this situation it is not hard to check that the result gives the nontrivial element in $$H^2(G/B,\mathcal{L}_{-2\alpha_1-\alpha_2})$$. Analogously, $$m_3(x_2 \otimes x_1 \otimes x_2) \in H^2(G/B,\mathcal{L}_{-\alpha_1-2\alpha_2})$$ is another nontrivial element (and these two give a basis of $$H^2(G/B,\Omega^2)$$.