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Let $(M,g)$ be a compact Riemannian manifold with boundary and assume it has positive scalar curvature.

Question. Is it true that $DM$, the double of $M$, admits a metric of positive scalar curvature?

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The answer is negative at least if you do not add some sort of convexity hypothesis for the boundary, and at least in dimension $2$. Take a round $2$-sphere with $h\ge 2$ round holes. It has positive curvature, but its double has genus $h-1\ge 1$ and cannot have a positively curved metric. Of course, in dimension $2$ scalar curvature is far more rigid than in higher dimension.

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    $\begingroup$ If additionally the boundary of $M$ is mean convex, does it follow that $DM$ admits a metric of positive scalar curvature? $\endgroup$ Commented Apr 28, 2021 at 23:23
  • $\begingroup$ @EduardoLonga: this looks like a reasonnable hypothesis. My guess is yes, by taking the double metric (which is singular along the boundary) and then smoothing it up. The mean convexity of the boundary should be sufficient to ensure the smoothing keeps the scalar curvature positive. This could make a good follow-up question, I would expect some people to be able to provide more details or references. $\endgroup$ Commented Apr 29, 2021 at 17:32
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This is a little expansion on Benoît's answer.

His counterexample actually can be generalized. There is the following conjecture which appears as Conjecture 1.24 in Rosenberg's "Manifolds of positive scalar curvature: A progress report" (http://www2.math.umd.edu/~jmr/psc2006.pdf).

Conjecture A manifold $X$ admits positive scalar curvature (psc), if and only if $X\times S^1$ admits positive scalar curvature.

This conjecture is known to fail in dimension four (a counterexample is $K3\#\overline{\mathbb{CP}}{}^2$), but is verified in many other cases, for example in dimensions 2,3,5,6 or for a great class of Spin-manifolds. This follows for example from the band-width estimates of Räde (https://arxiv.org/abs/2104.10120, Theorem 2.25) and Zeidler (https://arxiv.org/abs/1905.08520, Theorem 1.4).

Now take one of these manifolds, for which the conjecture is verified. By Gromov's h-principle, there is a psc-metric on $X\times \mathbb R$ which can be restricted to give a psc-metric on $M:=X\times [0,1]$. The double of $X$ however is given by $X\times S^1$ and hence admits no psc-metric.

However, the answer is also yes in many cases. Gromov-Lawson have shown in "Spin and scalar curvature in the presence of a fundamental group" (Theorem 5.7) that if M admits a psc-metric with positive mean curvature on the boundary, then the double admits a psc-metric. This has recently been generalized by Bär-Hanke (https://arxiv.org/abs/2012.09127, Corollary 34) to many other boundary condition (positive mean curvature, vanishing mean curvature, non-negative mean curvature, nonnegative second fundamental form, positive second fundamental form, vanishing second fundamental form) in the following way:

Theorem A psc-metric $g$ on $M$ satisfying any of the boundary conditions above can be deformed into a psc-metric $g'$ which can be doubled, i.e. $g'\cup g'$ is a smooth psc-metric on $DM$.

Another very recent result on psc-metrics on manifolds with boundary was obtained by Rosenberg-Weinberger (https://arxiv.org/pdf/2201.01263.pdf), where they give a list of manifold with boundary which admit such a doubling metric.

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