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I recently ran into a 30+ years old literature by Andersen and Jantzen on some calculations on cohomology of flag varieties (Cohomology of Induced Representations for Algebraic Groups). Here is the setting:

$G$ complex simple Lie group, $B = HN$ a Borel subgroup corresponding to the positive roots of $\mathfrak{g}$, and $\mathfrak{n} = Lie(N)$. Theorem 3.6(a) of [Andersen-Jantzen] says that for all $i > 0$ and $n \geq 0$,

$$H^i(G/B, S^n(\mathfrak{n}^*)) = 0.$$

I tried an example for $G = SL(3)$ and $n = 2$, where $\alpha_1 = (1,-1,0)$ and $\alpha_2 = (0,1,-1)$, $\alpha_1 + \alpha_2 = (1,0,-1)$ are the positive roots in $\mathfrak{h}^*$. The weights of $S^2(\mathfrak{n}^*)$ are given by $$S^2(\mathfrak{n}^*) = \mathbb{C}_{(-2,2,0)} \oplus \mathbb{C}_{(0,-2,2)} \oplus \mathbb{C}_{(-2,0,2)} \oplus \mathbb{C}_{(-1,0,1)} \oplus \mathbb{C}_{(-2,1,1)} \oplus \mathbb{C}_{(-1,-1,2)}.$$

Then I tried to apply Bott-Borel-Weil: $$H^i(G/B, \mathbb{C}_{\lambda}^*) = \begin{cases} V_{\mu}^* &\text{if }w(\lambda + \rho) = \mu + \rho\ \text{for some}\ w \in W\ \text{with}\ l(w) = i, \\ 0 & \text{otherwise.} \end{cases} $$ Here $\mu$ must be a dominant weight of $G$.

For instance, if $\mathbb{C}_{(-2,2,0)} = \mathbb{C}_{(2,-2,0)}^*$, then $$(2,-2,0)+\rho = (2,-2,0) + (1,0,-1) = (3,-2,-1).$$

Let $w \in W$ be the transposition of the last two entries with $l(w) = 1$, then $w((2,-2,0)+\rho) = (3,-1,-2) = (2,-1,-1) + \rho$.

Does it imply that $H^1(G/B,S^2(\mathfrak{n}^*))$ contains a copy of $V_{(2,-1,-1)}^* = V_{(1,1,-2)}$, which is non-trivial? Any insight would be appreciated.

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    $\begingroup$ The answer to your question is no, as the vector bundle corresponding to $S^2(\mathfrak{n}^*)$ does not split into line bundles. This is because $S^2(\mathfrak{n}^*)$ does not split into a sum of one-dimensional representations as a $B$-representation ($B$ is not semisimple.) Instead, what you get is a filtration of your vector bundle by line bundles. $\endgroup$
    – dhy
    Commented Jul 30, 2018 at 11:53
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    $\begingroup$ Incidentally, Anderson-Jantzen is mainly about the positive characteristic case. If you are interested only in the characteristic zero case, I think this theorem goes back to Kostant. In fact it also follows from the Grauert-Riemenschneider theorem applied to the Springer resolution (I am not sure who first observed this.) $\endgroup$
    – dhy
    Commented Jul 30, 2018 at 11:55
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    $\begingroup$ Thanks a lot! I suspected the action on $N$ plays a role, and it indeed does. Just another question: I suppose the $B$-action on $S^{\bullet}(\mathfrak{n}^*)$ preserves the grading, so that one has $$H^i(G/B, S^{\bullet}(\mathfrak{n}^*)) = \bigoplus_k H^i(G/B, S^k(\mathfrak{n}^*))?$$ $\endgroup$
    – wky
    Commented Jul 30, 2018 at 12:27
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    $\begingroup$ Yep - that decomposition is fine (and works on the level of vector bundles). $\endgroup$
    – dhy
    Commented Jul 30, 2018 at 12:34
  • $\begingroup$ Here is a more refined question. Given that we know there is no higher cohomology in the above example, then there must be another weight in $S^2(\mathfrak{n}^*)$ that 'cancels' the $H^1$ given by the weight $(-2,2,0)$. And $(-2,1,1)$ seems to play the role that its $H^0$ will cancel out the $H^1$ of $(-2,2,0)$. How can we see this cancellation on the level $B$-modules? Or more generally, given a finite-dimensional $B$ module, can we find its cohomology explicitly? $\endgroup$
    – wky
    Commented Oct 22, 2018 at 8:55

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