Let $Y$ be an orientable, smooth 3-manifold and let $X=Y\times S^{1}$. My question is that: when does $X$ admits a Riemannian metric with positive scalar curvature?

An obvious case is when $Y$ itself admits a psc metric. Are there any other case? It was proved by Gromov and Lawson that any 3-manifold which contains a $K(\pi,1)$ prime factor does not admits a psc metric (another result in the paper implies that $Y$ should not be hyperbolic if $Y\times S^{1}$ admits psc), which, together with the Geometrization conjecture classifies all the 3-manifold admits psc metric. Does their proof works for $Y\times S^{1}$?

(I am most interested in the special case that $Y$ is an irreducible integer homology sphere)


2 Answers 2


It is a theorem of Schoen and Yau (see Jonathan Rosenberg's survey, theorem 1.10) that no closed aspherical 4-manifold is psc. That means that the three-manifold should be non-aspherical, which narrows it down to a very small list (especially if you insist on integer homology spheres).

  • $\begingroup$ Thanks a lot! I checked the theorem of Schoen and Yau. Actually it is slightly stronger: X has no psc metric if it has a non-zero degree map to a compact $K(\pi,1)$. So I think this almost implies that $Y$ is psc it self. $\endgroup$
    – user44651
    Aug 27, 2015 at 22:13
  • $\begingroup$ @user44651 So, what irreducible $Y$ do you get? The Poincare homology sphere? $\endgroup$
    – Igor Rivin
    Aug 27, 2015 at 22:20
  • $\begingroup$ Poincare homology sphere and $S^{3}$ I think. $\endgroup$
    – user44651
    Aug 27, 2015 at 23:37
  • 2
    $\begingroup$ Schoen and Yau only outline a proof of the statement in the answer above. A full proof was recently completed by Chodosh and Li: arxiv.org/pdf/2008.11888.pdf. The stronger claim in user44651's first comment is still open as far as I am aware. $\endgroup$ Sep 7, 2020 at 15:53
  • 1
    $\begingroup$ In case anyone is still interested in this approach, you can find a full proof in these notes web.stanford.edu/~ochodosh/Math258-min-surf.pdf (Corollary 7.34, but you probably have to read a lot of the proceeding material). (@MichaelAlbanese was completely correct when he wrote the above comment, but now the mapping problem has been solved, so the approach works fine). $\endgroup$ Feb 2, 2022 at 14:29

It follows from Theorem 2 of the following paper and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

Schoen, Richard; Yau, Shing-Tung, On the structure of manifolds with positive scalar curvature, Manuscr. Math. 28, 159-183 (1979). ZBL0423.53032.

On p. 9 of the paper, they define $C_3'$ to be the class of closed three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $\geq 1$ (i.e. a Haken 3-manifold). It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$ (a homology class $\alpha\in H_k(M)$ is represented by an oriented $k$-manifold $N$ if there is a map $f:N\to M$ such that $f_\ast([N])=\alpha$, where $[N]\in H_k(N)$ is the fundamental class determined by the orientation of $N$).

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and let $Y'\to Y$ be a finite-sheeted cover. By Theorem 2, there exists an $N\subset Y'\times S^1$ such that $[N]=[Y'\times \ast]\in H_3(Y'\times S^1)$ and $N$ is of class $C_3'$. Moreover, there is a non-zero degree map $N\to Y'$ (by composing the embedding with the projection $Y'\times S^1\to Y'$), hence $Y'\in C_3'$ (the class $C_3'$ is clearly closed under taking non-zero degree maps). Since every finite-sheeted cover of $Y$ is in $C_3'$, we conclude that $Y$ admits a metric of positive scalar curvature.


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