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Intersection form of surface bundle over surface

MathOverflow Asked on November 3, 2021

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.

One Answer

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 $$H^1(mathcal{S}_g(k);mathbb{Z})=0 quadquad H^2(mathcal{S}_g(k);mathbb{Z})congmathbb{Z}^3$$ 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 $$12 lambda = kappa_{1,0} quadquad 2zeta = kappa_{0,1} - kappa_{-1,2}.$$

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 $$int_{Sigma_h}lambda=text{whatever you like} quadquad int_{Sigma_h}kappa_{0,1}=1 quadquad int_{Sigma_h}zeta = 0$$ and hence having $$int_E c^2 = int_{Sigma_h} int_pi c^2 = int_{Sigma_h} kappa_{-1,2} = int_{Sigma_h} kappa_{0,1} = 1.$$

Answered by Oscar Randal-Williams on November 3, 2021

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