Let $F\rightarrow E\rightarrow B$ be a fiber bundle. Let $\pi_1$ be the fundamental group of $B$ with base point say $b_0$. In the following we are considering cohomology with coefficients in $\mathbb{Z}$ (but I would not mind to consider coefficients in $\mathbb{Q}$ if torsion elements turn out to be troublesome regarding the question). Each (equivalence class) of loop $[\gamma]\in \pi_1$ induces an automorphism on $F$ which provides an action of $\pi_1$ on $H^*(F)$.

Not being an algebraic topologist, my naive intuition tells me that it should sometimes happen that \begin{align} (\star) \hspace{1cm} H^*(E)\simeq H^*(B)\otimes H^*(F)^{\pi_1} \end{align} At least we know that this is indeed the case when $\pi_1$ acts trivially on $H^*(F)$ and that the spectral sequence degenerates at the second page (i.e. all differentials $d^k$ are trivial for $k\geq 2$). If I were fluent regarding spectral sequences I would not be asking these two questions but since I'm not here they are:

Q1: Do we know other (general) situations for which $(\star)$ is known to be an isomorphism (I don't really care if it is canonical or not) ?

Analogues of this question could as well be asked in the setting of discrete groups. For example say that $1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$ is a split short exact sequence of discrete groups.

Q2 Are there (general) situations for which $H^*(E)\simeq H^*(G)\otimes H^*(A)^G$ ?

added: Here is one interesting example (if my calculations are correct) where ($\star$) is an isomorphism. Let $S^1$ be the circle and set $B=(S^1)^{b}$ and $F=(S^1)^a$. Consider a group homomorphism $\rho:\pi_1=\pi_1(B)\simeq \mathbb{Z}^b\rightarrow GL_a(\mathbb{Z})\simeq Aut(F)$. Then it seems to me that ($\star$) holds true. Let me be more explicit about an example coming from number theory. Take for example $L=\mathbb{Z}[\sqrt{2}]$ and $\Lambda=\epsilon^{\mathbb{Z}}$ where $\epsilon=3+2\sqrt{2}$. Then $\Lambda$ acts naturally on $L$ and one may consider the arithmetic group $\Gamma=L\rtimes \Lambda$. Let $\mathfrak{h}$ be the Poincare upper half-plane. Then an element of the group $(v,\lambda)\in\Gamma$ acts naturally on $\mathfrak{h}^2$ by the usual rule \begin{align*} (v,\lambda)*(z_1,z_2)=(\lambda^{(1)}z_1+v^{(1)},\lambda^{(2)}z_2+v^{(2)}) \end{align*} This gives rise to the following bundle \begin{align*} F=\mathfrak{h}^2/V\rightarrow E=\mathfrak{h}^2/\Gamma\rightarrow B=\mathbb{R}_{>0}^2/\Lambda \end{align*} The first term is homotopy equivalent to $(S^1)^2$ and the second to $S^1$. Now if one considers the de Rham cohomology group $H^*(E)$ then it seems to me that all the closed differential forms on $E$ which are $\Gamma$-invariant are obtained from the following tensor product: \begin{align} \{\mathbb{R}1+\mathbb{R}dx_1\wedge dx_2 \} \otimes \{\mathbb{R}1+\mathbb{R}\frac{dy_1}{y_1}\}, \end{align} where $z_j=x_j+i y_j\in \mathfrak{h}$ for $j=1,2$. Note that $\frac{dy_1}{y_1}$ represents the same cohomology class as $-\frac{dy_2}{y_2}$ in $H^*(B)$ since the function $\log y_1y_2$ is $\Gamma$-invariant.

  • $\begingroup$ That is not what degenerate means. A spectral sequence degenerates at $E^k$ if all $d^i = 0$ for $i \geq k$. $\endgroup$ – Mike Miller Aug 21 '19 at 15:17
  • $\begingroup$ Yes you are right, sorry about that. I'll change it. $\endgroup$ – Hugo Chapdelaine Aug 21 '19 at 15:32
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    $\begingroup$ If $E, B, F$ all have finite rank cohomology with coefficients in the field $k$, then $H^*(E;k) \cong H^*(B;H^*(F;k))$ if and only if the Serre spectral sequence degenerates on $E^2$, where the latter term should be interpreted as $B$ equipped with a non-trivial local system of (graded) groups; this is how the $\pi_1$-action is encoded. This follows from rank considerations. It is rarely true that $H^*(B;H^*(F;k)^{\pi_1}) \to H^*(B;H^*(F;k))$ is an isomorphism. The failure to be an iso is measured by $H^*(B;H^*(F;k)/H^*(F;k)^{\pi_1});$ you should ask that this is zero. $\endgroup$ – Mike Miller Aug 21 '19 at 15:35
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    $\begingroup$ To develop on @MikeMiller 's comment: the beginning part (cohomology of the total space being isomorphic to cohomology of the base with the coefficient in the cohomology of the fiber) is true whether $k$ is field or not. But when $k$ is not a field, even with the trivial action of $\pi -1(B)$, the RHS may differ from $H^*(B)\otimes H^*(F)$. $\endgroup$ – user43326 Aug 21 '19 at 16:56
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    $\begingroup$ As to the added example, I doubt if it works. Let's take $a=b=1$, and $\rho$ a non-trivial map. Then if I am not mistaken, we have a cofibration $S^1\to E\to Th$ where $Th$ is the Thom space of the Moebius line bundle over $S^1$. Again, if I am not mistaken, the cohomology of $E$ is isomorphic to that of the torus if and only if this bundle is orientable, or $char k=2$. $\endgroup$ – user43326 Aug 21 '19 at 17:04

Lets analyze this when $\pi_1$ is finite, and the bundle is associated to a principle $\pi_1$ bundle, and we are considering rational coefficients.

Very generally, if a finite group $G$ acts freely and properly on a sensible space $\tilde X$, and we let $X = \tilde X/G$, then $H^*(X;\mathbb Q) = H^*(\tilde X;\mathbb Q)^G$. (This is a standard fact proved with transfers.)

Now lets consider the situation of the question, and let $\tilde B$ be the universal cover of $B$. Then $E = (\tilde B \times F)/\pi_1$, and applying the observation above shows that

$$H^*(E;\mathbb Q) = H^*(\tilde B \times F; \mathbb Q)^{\pi_1} = [H^*(\tilde B; \mathbb Q) \otimes H^*(F; \mathbb Q)]^{\pi_1}.$$ Meanwhile

$$H^*(B; \mathbb Q) \otimes H^*(F; \mathbb Q)^{\pi_1}=H^*(\tilde B; \mathbb Q)^{\pi_1} \otimes H^*(F; \mathbb Q)^{\pi_1}= [H^*(\tilde B; \mathbb Q) \otimes H^*(F; \mathbb Q)]^{\pi_1 \times \pi_1}.$$

These two expressions will be equal exactly when $\pi_1$ acts trivially on either $H^*(\tilde B;\mathbb Q)$ or $H^*(F;\mathbb Q)$.

One situation where this first possibility always holds is when $B = B\pi_1$, so that $\tilde B$ is contractible. For example, the fibration $ S^2 \rightarrow \mathbb RP^2 \rightarrow \mathbb RP^{\infty}$ illustrates all of this, as $$H^*(\mathbb RP^2; \mathbb Q) = H^*(\mathbb RP^{\infty};\mathbb Q) \otimes H^*(S^2;\mathbb Q)^{C_2}.$$

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  • $\begingroup$ Include some finite type hypotheses so that the cohomology Kunneth theorem holds. $\endgroup$ – Nicholas Kuhn Aug 21 '19 at 18:10
  • $\begingroup$ This is right, I misread the equation in the question. $\endgroup$ – Will Sawin Aug 21 '19 at 18:39
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    $\begingroup$ Actually this is not completely right in general because we do not always have $E = (\tilde{B} \times F)/\pi_1$, for instance if $B= \tilde{B} = S^2$, $\pi^1$ is trivial, $F= S^1$, but $E= S^3$ (Hopf fibration). $\endgroup$ – Will Sawin Aug 21 '19 at 18:47
  • $\begingroup$ Thanks a lot Nicholas for the nice generic construction. $\endgroup$ – Hugo Chapdelaine Aug 21 '19 at 21:59

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