# Intersection form of surface bundle over surface

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.

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

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 $$\kappa_i$$.)

In

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 $$\int_E c^2 = \int_{\Sigma_h} \int_\pi c^2 = \int_{\Sigma_h} \kappa_{-1,2} = \int_{\Sigma_h} \kappa_{0,1} = 1.$$

• 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. Jul 20, 2020 at 21:50
• 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 $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$. Jul 21, 2020 at 8:33
• Sorry, the bottom-right entry should be $-1$. Jul 21, 2020 at 8:40
• 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. Jul 21, 2020 at 15:29
• Also, I like to know if the $c$ satisfying the above two conditions exists for each $g\geq 6$? Jul 21, 2020 at 15:48