# Oddness of intersection form of surface bundle

Let $$\Sigma_g$$ be a Riemannian surface of genus $$g$$. Let $$M^4$$ be a surface bundle: $$\Sigma_g \to M^4 \to \Sigma_h$$. When $$g=1$$, $$M^4$$ is called a torus bundle.

My question: is there a torus bundle whose intersection form contains an odd diagonal element (if we choose a basis and view the intersection form as a matrix)?

If $$M^4=\Sigma_1\times \Sigma_h$$, then $$M^4$$ is spin and its intersection form has only even diagonal elements.

More generally: For a given fiber $$\Sigma_g$$, is there a $$\Sigma_g$$-bundle whose intersection form is odd?

• what is the meaning of "odd diagonal element"? Jul 2, 2020 at 22:38
• We choose a basis and view the intersection form as a matrix. I modified my question. What is E(1) 4-manifold? Jul 2, 2020 at 22:44
• $E(1)= CP^2 \# 9 \bar{CP^2}$ which is a torus bundle as you desired. For more details please have a look at the chapter 3 of Gompf and Stipsicz book. In paticular you will be interested in elliptic surfaces. Jul 2, 2020 at 22:50
• @AnubhavMukherjee: It seems to me that the elliptic fibration you are talking about has (generically) 12 nodal fibres, so it is not a torus bundle. Jul 2, 2020 at 22:56
• In fact, its Euler characteristic is $12$, whereas for a torus bundle it is $0$. Jul 2, 2020 at 23:00

Suppose $$\pi\colon M \to \Sigma_g$$ is an oriented smooth torus bundle. If $$w_2(M) = 0$$, then also the second Wu class $$v_2(M) = 0$$ and $$M$$ has even intersection form (the converse holds if $$H_1(M;\mathbb Z)$$ has no $$2$$-torsion, but we do not need this here). I claim that this is always the case in our situation.

Even better, I claim that $$M$$ is always parallelizable: stably, $$TM$$ agrees with the vertical tangent bundle $$T_{\pi}$$, whose classifying map $$E \to B\text{SL}_2(\mathbb R)$$ can be identified with the map $$E \xrightarrow{\pi} \Sigma_g \xrightarrow{(1)} B\text{Diff}^+(T^2) \xrightarrow{(2)} B\text{SL}_2(\mathbb Z) \xrightarrow{(3)} B\text{SL}_2(\mathbb R),$$ where (1) is the classifying map of $$\pi$$, (2) is induced from applying $$\pi_1$$, and (3) is induced from extending coefficients. Since $$H^2(B\text{SL}_2(\mathbb Z);\mathbb Z) = \mathbb Z/12$$ is torsion, the map composition of (1), (2) and (3) is trivial on second cohomology and hence nullhomotopic, as $$B\text{SL}_2(\mathbb R) = K(\mathbb Z,2)$$. Thus, $$T_{\pi}$$ is trivial and $$M$$ is stably parallelizable. Since $$\chi(M) = \chi(\Sigma_g)\chi(T^2) = 0$$, $$M$$ is parallelizable.

If the base is not a surface, I believe that it is possible for torus bundles to be non-spin, see Johannes Ebert's thesis (the last pages of chapter 5), although no concrete examples are constructed there.

For higher genus, note that there are examples of surface bundles over surfaces whose total space has signature $$4$$, in particular, its intersection form cannot possibly be even.

Also, the total space of the (unique!) nontrivial $$S^2$$-bundle over $$S^2$$ is diffeomorphic to $$\mathbb CP^2 \# \overline{\mathbb CP^2}$$, which has odd intersection form.

• Thank you for the answer. I just like to confirm what you said: All torus bundle with base $\Sigma_g$ are spin and have even intersection form. Jul 3, 2020 at 23:46
• For surface bundles over surfaces whose total space has signature 4, the total space is not spin. But can we conclude that the intersection form be odd? Jul 3, 2020 at 23:52
• Yes, I think the above argument shows that the total space of every torus bundle with base a surface is spin (even parallelizable), and therefore its intersection form is even. Jul 4, 2020 at 10:01
• The signature of any even intersection form over Z is divisible by 8. Jul 4, 2020 at 10:07
• You mentioned that the total space of the (unique!) nontrivial $S^2$-bundle over $S^2$ is diffeomorphic to $CP^2\#CP^2$, which has odd intersection form.'' ie there exists a 2-cocycle $c \in H^4(CP^2\#CP^2;Z)$ such that $\int_{CP^2\#CP^2} c^2 =k$ and $k$ is odd. I have two questions here: (1) is $k=\pm 1$? (2) what is the evaluation of the 2-cocycle $c$ on the fiber $S^2_{fiber}$: $\int_{S^2_{fiber}} c =?$. Jul 16, 2020 at 13:37