# When is the cohomology of a fiber bundle a tensor product?

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.

• That is not what degenerate means. A spectral sequence degenerates at $E^k$ if all $d^i = 0$ for $i \geq k$.
– mme
Aug 21 '19 at 15:17
• Yes you are right, sorry about that. I'll change it. Aug 21 '19 at 15:32
• 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.
– mme
Aug 21 '19 at 15:35
• 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 $H^*(B)\otimes H^*(F)$. Aug 21 '19 at 16:56
• 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 $char k=2$. 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 $$H^*(\mathbb RP^2; \mathbb Q) = H^*(\mathbb RP^{\infty};\mathbb Q) \otimes H^*(S^2;\mathbb Q)^{C_2}.$$

• Include some finite type hypotheses so that the cohomology Kunneth theorem holds. Aug 21 '19 at 18:10
• This is right, I misread the equation in the question. Aug 21 '19 at 18:39
• 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). Aug 21 '19 at 18:47
• Thanks a lot Nicholas for the nice generic construction. Aug 21 '19 at 21:59