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Let $X$ be a smooth projective rational variety over $\mathbb{C}$, let $Y$ be another smooth projective variety, both of dimension bigger than 2, and let $\pi : Y \rightarrow X$ be a locally trivial fibration (in the analytic topology) with fibers $F$ isomorphic to a product of Flag varieties.

I claim the following: $H^3(Y,\mathbb{Q})\simeq H^3(X,\mathbb{Q}).$

It is easily seen to be true for trivial bundle $Y \simeq X\times F$ using Kunneth formula and the fact that $X$ being rational, $H^1(X,\mathbb{C})=0$ (since $dim H^{1,0}$ is a birational invariant), and cohomology of $F$ only occurs at even degrees since $F$ is product of Flag varieties.

For the general case, I'm trying to prove it using Leray-Hirsch theorem, but I'm getting stuck at checking the hypothesis of theorem.

Is my claim correct? And can someone please help me how to prove it?

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    $\begingroup$ Since $X$ is rational it is simply connected, hence the local system $R^2\pi _*\mathbb{Q}$ is constant, and therefore has no $H^1$. Thus the only nonzero term in the Leray spectral sequence contributing to $H^3(Y, \mathbb{Q})$ is $E^{3,0}_{\infty}=E^{3,0}_2= H^3(X,\mathbb{Q})$ (using the fact that the Leray spectral sequence degenerates at $E_2$). $\endgroup$
    – abx
    Commented Nov 21, 2019 at 9:05
  • $\begingroup$ @abx : Thanks! Are you using the Deligne's theorem for the above argument? Also, will the isomorphism be exactly given by $\pi ^*$? $\endgroup$
    – Hajime_Saito
    Commented Nov 22, 2019 at 13:57
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    $\begingroup$ Yes (though you could probably avoid Deligne) and yes: the edge-homomorphism $E^{p,0}_{2}\rightarrow H^p(Y)$ in the Leray spectral sequence is given by $\pi ^*$. $\endgroup$
    – abx
    Commented Nov 22, 2019 at 14:15

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