Let $G$ be a connected reductive group over a number field $K$, $P$ be a parabolic subgroup of $G$ defined over $K$, $X=G/P$ be the generalized flag variety which is a smooth projective variety over $K$ and $p$ be a prime number. For a positive integer $i>0$, consider the etale cohomology $V=H^i(X_{K^{alg}},\mathbb Q_p)$ as a Galois representation of $G_K$.

How to compute such Galois representation? This may be done somewhere but I can't find a reference. Firstly, the dimension may be computed by using Betti numbers and some combination datas from the Lie algebra (the odd dimension shall vanish).

Secondly, is the representation semi-simple? For the projective space it's obviously true as the dimension is no bigger than $1$. If that's true, then must the direct summand be some $\mathbb Q_p(-i/2)$? Some density theorem may reduce this to the finite field case.

In a short word, how to completely decide the Galois representation? As the flag variety has a stratification by affine spaces, this seems reachable.

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    $\begingroup$ The whole cohomology is algebraic, so $H^{2i}(X_{\bar{K}}, \mathbb{Q}_{p}(i))$ is invariant under Galois. Thus $H^{2i}(X_{\bar{K}}, \mathbb{Q}_{p})\cong \mathbb{Q}_{p}(-i)^{b_{2i}}$ (and $H^{2i+1}$ is zero). $\endgroup$
    – abx
    Jan 13, 2019 at 6:55
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    $\begingroup$ @abx: I don't think what you say (about the even cohomology) is correct. If one twists $H^{2i}$ by $\mathbb{Q}_p(i)$ then the Galois action will factor through a finite group and this action can be non-trivial. $\endgroup$
    – naf
    Jan 13, 2019 at 7:40
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    $\begingroup$ @Ulrich: I guess you are right. The cohomology is spanned by the classes of Schubert subvarieties. I was assuming that they can be defined over $K$ but that might be incorrect. $\endgroup$
    – abx
    Jan 13, 2019 at 10:58

1 Answer 1


Over $\overline{K}$ we can decompose $G/P$ into Schubert cells. There is an action of $\operatorname{Gal}(K)$ on this set of Schubert cells. This is not completely obvious, as the Galois conjugate of a Schubert cell may not be a Schubert cell for the same Borel, but we can fix this by noting that it is a translate of one of our original set of Schubert cells, and a unique one because they all have different cohomology classes (there is probably an elementary proof of this). This preserves the dimension of the cells.

Take the permutation representation $\operatorname{Gal}(K)$ on the set of codimension $i$ cells and Tate twist it by $-i/2$ and you will get the Galois representation on cohomology.

The proof is by observing that first, the cycle class map is an isomorphism from the free vector space on the Schubert cell classes to the cohomology is appropriately equivariant and, second, is an isomorphism (because it's an isomorphism over an algebraically closed field).

  • $\begingroup$ Thank you! How to decide the Galois action on codim $i$ Schubert cells? For example, do we know any nontrivial example? $\endgroup$
    – sawdada
    Jan 15, 2019 at 16:19
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    $\begingroup$ @zzy For instance we can take $G$ to be the Weil restriction from a field extension of a split group, and $P$ the Weil restriction of a parabolic subgroup. Over an algebraically closed field, $G$ will look like a power of the split group, so parabolic subgroups will look like tuples of subgroups, one for each embedding of the field extension of the algebraic closure. The Galois group acts by permutation on the embedding, hence by permutation on the tuples. This is nontrivial. $\endgroup$
    – Will Sawin
    Jan 15, 2019 at 17:03
  • $\begingroup$ @zzy The Schubert cells are $B$-orbits for a borel $B$ defined over the algebraic closure. We can take a maximal torus defined over the base field, then the Galois action on the set of Borels factors through the Weyl group. The Galois action on the set of Schubert cells then must factor through the Weyl action on the set of Schubert cells by conjugating the Borel subgroup. $\endgroup$
    – Will Sawin
    Jan 15, 2019 at 17:06
  • $\begingroup$ How do we determine it is $(-i/2)$? Any reference for that? $\endgroup$
    – userabc
    Feb 13, 2019 at 4:31
  • $\begingroup$ @userabc This arises from the construction of the cycle class map in etale cohomology. I first saw it in Milne's etale cohomology book. $\endgroup$
    – Will Sawin
    Feb 13, 2019 at 13:19

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