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.


1 Answer 1


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.


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