Positive scalar curvature on the total space of a circle bundle

Question

Let $(\Sigma_\gamma,g)$ be a closed and orientable Riemannian surface of genus $\gamma \geq 1$, $(M^3,\tilde{g})$ be a closed, connected and orientable Riemannian $3$-manifold, and $\pi : M \to \Sigma_\gamma$ be a Riemannian fibre bundle whose fibers are minimal circles. Is it known whether the scalar curvature of $(M,\tilde{g})$ can be strictly positive?

Answer 1

It is a theorem of Gromov and Lawson, also Schoen and Yau, that no closed orientable three-manifold which contains an aspherical factor in its prime decomposition can admit a metric of positive scalar curvature, see Theorem IV.6.18 of Spin Geometry by Lawson and Michelsohn. In particular, as the three-manifold you're interested in is aspherical, it does not admit a metric of positive scalar curvature (irrespective of the nature of the fibers).

In fact, thanks to the solution of the elliptisation conjecture, we now know that a closed orientable three-manifold admits a metric of positive scalar curvature if and only if its prime decomposition contains no aspherical factors.

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A natural question to ask is whether your manifold can admit a metric of non-negative scalar curvature. In the absence of positive scalar curvature metrics, such a metric is Ricci-flat, and on a three-manifold, a Ricci-flat metric is flat. If $M$ admits a flat metric, it is finitely covered by $T^3$ and hence $b_1(M) \leq 3$. As $M$ is an orientable circle-bundle over $\Sigma_{\gamma}$, the Gysin sequence tells us that $\pi^* : H^1(\Sigma_{\gamma}) \to H^1(M)$ is injective, and hence $\gamma = 1$. If the Euler class of the bundle is non-zero, then $M$ is finitely covered by the Heisenberg manifold $H(3, \mathbb{R})/H(3, \mathbb{Z})$ which does not admit a flat metric. On the other hand, if the Euler class is zero, then $M = T^3$ which certainly does.