Let $\Sigma_g$ be a Riemannian surface of genus $g$. Let $M^4$ be a surface bundle over surface: $\Sigma_g \to M^4 \to \Sigma_h$. $\Sigma_g$ is the fiber and $\Sigma_h$ is the base space.

My question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying (1) $\int_{M^4} c^2 =\pm 1$, and (2) $\int_{\Sigma_g} c =0$

Also: How to construct surface bundles with known odd intersection form? This may help to answer the first question.

See a related question: Oddness of intersection form of surface bundle

A further question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying (1) $\int_{M^4} c^2 =\pm 1$, (2) $\int_{\Sigma_g} c =0$, and (3) $c = w_2$ mod 2. What is the signature and $g$ for such surface bundle?

Note that the condition (1) implies that $M^4$ is not spin and $w_2$ is non-trivial.

  • 2
    $\begingroup$ If $c \in H^2(M; \mathbb{Z})$, what do you mean by $\int_{\Sigma_g}c$? $\endgroup$ Jul 20, 2020 at 19:49
  • 1
    $\begingroup$ Integration along fibres. $\endgroup$ Jul 20, 2020 at 20:04

1 Answer 1


Yes, such a thing exists, but I don't know an explicit example.

To see that it exists, it is clearest to me to consider the universal situation. For any $k \in \mathbb{Z}$ there is a space $\mathcal{S}_g(k)$ which classifies oriented surface bundles $$\Sigma_g \to E \overset{\pi}\to B$$ equipped with a class $c \in H^2(E; \mathbb{Z})$ such that $\int_{\Sigma_g} c = k$. Associated to such a family there are characteristic classes $$\kappa_{i,j} = \int_\pi e(T_\pi E)^{i+1} \cdot c^j \in H^{2(i+j)}(B;\mathbb{Z}),$$ where $T_\pi E$ denotes the tangent bundle of $E$ along the fibres of $\pi$, and $e(T_\pi E)$ denotes its Euler class. (The classes $\kappa_{i,0}$ are the usual Miller--Morita--Mumford classes $\kappa_i$.)


J. Ebert and O. Randal-Williams, Stable cohomology of the universal Picard varieties and the extended mapping class group. Doc. Math. 17 (2012), 417–450.

Johannes Ebert and I studied, among other things, the low-dimensional integral cohomology of $\mathcal{S}_g(k)$, and showed that as long as $g$ is large enough (I think $g \geq 6$ will do) one has $$H^1(\mathcal{S}_g(k);\mathbb{Z})=0 \quad\quad H^2(\mathcal{S}_g(k);\mathbb{Z})\cong\mathbb{Z}^3$$ where the isomorphism in the second case is given by a basis of cohomology classes $\lambda, \kappa_{0,1}, \zeta$, where the outer two are related to the $\kappa_{i,j}$ by the identities $$12 \lambda = \kappa_{1,0} \quad\quad 2\zeta = \kappa_{0,1} - \kappa_{-1,2}.$$

In particular, applying this with $k=0$ and using that every second homology class is represented by a map from an oriented surface, it follows that there is a surface bundle $$\Sigma_g \to E \overset{\pi}\to \Sigma_h$$ for some $h$ (which is uncontrollable using this method) with a class $c \in H^2(E;\mathbb{Z})$ satisfying $\int_{\Sigma_g}c = 0$, and having $$\int_{\Sigma_h}\lambda=\text{whatever you like} \quad\quad \int_{\Sigma_h}\kappa_{0,1}=1 \quad\quad \int_{\Sigma_h}\zeta = 0$$ and hence having $$\int_E c^2 = \int_{\Sigma_h} \int_\pi c^2 = \int_{\Sigma_h} \kappa_{-1,2} = \int_{\Sigma_h} \kappa_{0,1} = 1.$$

  • $\begingroup$ Thank you very much for the answer, so soon for such a specific question. The surface bundle you described has a fiber $\Sigma_g$ with $g\geq 6$. I wonder if a stronger result with $g\geq 2$ exits. I know $g\leq 1$ does not work. $\endgroup$ Jul 20, 2020 at 21:50
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    $\begingroup$ It is also possible for $g=0$: you can take the Hirzebruch surface $S^2 \to H_1^4 \to S^2$. In the basis of homology given by the fibre and the section at infinity, its intersection form is $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$. $\endgroup$ Jul 21, 2020 at 8:33
  • $\begingroup$ Sorry, the bottom-right entry should be $-1$. $\endgroup$ Jul 21, 2020 at 8:40
  • $\begingroup$ For $S^2_f\to H_1^4\to S^2_b$, are two conditions (1) $\int_{H_1^4} c^2 =\pm 1$, and (2) $\int_{S^2_f} c =0$ satisfied? I think (2) is not satisfied. $\endgroup$ Jul 21, 2020 at 15:29
  • $\begingroup$ Also, I like to know if the $c$ satisfying the above two conditions exists for each $g\geq 6$? $\endgroup$ Jul 21, 2020 at 15:48

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