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Suppose I have a fiber bundle $\pi: E\rightarrow B$, with fiber $F$, such that the Serre spectral sequence on cohomology is immediately degenerate. In other words, $H^*(E)=H^*(B)\otimes H^*(F)$.

I have a subvariety $X\subseteq E$ such that $\pi|_X$ is also a fiber bundle, with image $C$ and fiber $G$. I want to know the class (Poincare dual to the fundamental class) of $X$. Is it always true the $[X]=[C]\otimes [G]$? If so, could someone provide a reference?

(I would be happy with a reference in Chow rather than singular cohomology. If it matters, I'm actually interested in the case where $E$ is a flag variety $G/B$ and $B$ is $G/P$ for some parabolic, but $X$ is NOT a Schubert variety (where the desired fact is also a combinatorial fact about Schubert polynomials).)

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    $\begingroup$ As written, this is essentially never true. The class of $X$ lives in $H^{c+p}(E)$, where $c$ is the codimension of $C$ in $B$ and $p$ the codimension of $G$ in $F$. Your cup-product is in $H^{c+q}(E)$, where $q$ is the codimension of $G$ in $E$. $\endgroup$
    – abx
    Commented Jun 21 at 4:20
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    $\begingroup$ @abx: Oops - will fix. $\endgroup$
    – Alexander Woo
    Commented Jun 21 at 4:25
  • $\begingroup$ Your question still does not make sense. Are you asking for every principal $G$-bundle $E_G$ over $B$, for every space $F$ with a $G$-action such that $H^*(F\times E_G/G)$ equals $H^*(F)\otimes H^*(B)$, and for every $G$-invariant closed subset $C$ of $F$ whether also $H^*(C\times E_G)$ equals $H^*(C)\otimes H^*(B)$? $\endgroup$
    – Jason Starr
    Commented Jun 21 at 11:40
  • $\begingroup$ My formulation of your question above has a negative answer. Similar to user @Sasha, let $B$ be the curve $\mathbb{G}_m$, let $G$ be a cyclic group of order $2$, let $E_G$ be the unique isomorphism class of a non trivial $G$-torsor over $B$, let $F$ be $\mathbb{A}^1$ with the unique nontrivial linear action of $G$, and let $C$ be any orbit with two elements (a faithful orbit). $\endgroup$
    – Jason Starr
    Commented Jun 21 at 11:49

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This is definitely not true. For instance, consider the case where $$ \pi \colon \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1 $$ is just the projection and $X \subset \mathbb{P}^1 \times \mathbb{P}^1$ is a curve of bidegree $(1,n)$, so that $\pi\vert_X$ is an isomorphism (in particular, it is a fiber bundle with fiber a point) and note that the cohomology class of $X$ varies with $n$.

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  • $\begingroup$ Hmm - this is not what I meant, because I want the inclusion of $G$ into $F$ to respect the trivialization of the bundle that identifies the tensor product of the cohomologies of fiber and base with the cohomology of the bundle, and the curves of different bidegrees require different trivializations to make it work. What's the right way to say this precisely? $\endgroup$
    – Alexander Woo
    Commented Jun 21 at 7:09

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