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.



How to compute Galois representations from etale cohomology groups of a generalized flag variety?
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).


