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?
 A: 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.
