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. 



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



Multiplicative structure for sheaf cohomology of flag varieties
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 $H^2(G/B,\Omega^2)$. 


This is related to Conjecture 9.6 in the paper A. Kuznetsov, A. Polishchuk, "Exceptional collections on isotropic Grassmannians", J. Eur. Math. Soc. (JEMS) 18 (2016), no. 3, 507--574.


