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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?

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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.


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.

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  • $\begingroup$ Thank you for your answer. Why is $M$ aspherical? $\endgroup$ Jun 27, 2020 at 22:51
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    $\begingroup$ It is a circle bundle over an aspherical manifold. In general, if you have a fiber bundle with two of the three spaces aspherical, then it follows for the long exact sequence in homotopy that the third is also aspherical. Alternatively, one can show directly that a circle bundle over $\Sigma_{\gamma}$ for $\gamma \geq 1$ has universal cover $\mathbb{R}^3$. $\endgroup$ Jun 27, 2020 at 22:55

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