I've been done some work with scalar curvature and managed to give a simple proof for the following result:

Let $M$ be a closed manifold which do not admit a metric of positive scalar curvature. Then for any Riemannian metric $g$ on $M$ it holds that $$\int_M \mathrm{scal}_g \leq 0.$$

In particular, the above condition is necessary and sufficient in order that a manifold does not admit a metric of positive scalar curvature.

Since I am quite new in this area of positive scalar curvature specifically, I would like to know how known is this result and, also, if anyone can provide some references.

EDIT: perhaps the correct statement is under the hypothesis of $M$ being a non-enlargeable spin manifold.

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    $\begingroup$ There is something I do not understand. Assume that $\dim M \geq 3$ and that $M$ does not admit a metric of positive scalar curvature. By Kazhdan-Werner trichotomy theorem, a non-zero function $f$ is the scalar curvature of a Riemannian metric on $M$ if and only if it is negative somewhere. So, you are asserting that, as soon as a smooth function $f$ is negative at some point of $M$, then its integral on $M$ is non-positive. How is this possible? Or am I missing something? $\endgroup$ May 10, 2021 at 14:47
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    $\begingroup$ @FrancescoPolizzi: you forgot that there is a weight (coming from the volume form of the metric that realizes that value of scalar curvature). (Not saying that the claim is correct or not, but that it is plausible.) $\endgroup$ May 10, 2021 at 14:50
  • $\begingroup$ @WillieWong: oh right, the volume form varies with $f$. Thanks :) $\endgroup$ May 10, 2021 at 15:05

1 Answer 1


I am suspicious of your result.

The three torus $\mathbb{T}^3$ is well-known to not admit any metric of positive scalar curvature.

Let $g_0$ be the flat metric on $\mathbb{T}^3$. Given a positive function $u > 0$, consider the metric $g = u^{4} g_0$. Then we have the identity $$ - 8 \Delta_0 u = S_g u^5 \tag{1} $$ where $S_g$ is the scalar curvature of the metric $g$.

Note that the volume form of $g$ is $\mathrm{dvol}_g = u^6~ \mathrm{dvol}_0$. Multiply both sides of (1) by $u ~\mathrm{dvol}_0$ we find $$ - 8 \Delta_0 u \cdot u ~\mathrm{dvol}_0 = S_g ~\mathrm{dvol}_g$$ Integrating both sides, for any non-constant $u$ you have that the left hand side is manifestly positive. But your "result" would imply that the right hand side must be non-positive.

More generally:

Let $g_0$ be any constant scalar curvature ($S_0$) metric, on an $n$-dimensional manifold $M$ with $n > 2$. Let $g_u = u^{4/(n-2)} g_0$, where $u$ is a positive function. Then we have that the scalar curvature $S_u$ of $g_u$ satisfies

$$ - \gamma \Delta_0 u + S_0 u = S_u u^{2n/(n-2)} u^{-1} $$

where $\gamma = 4(n-1)/(n-2)$. Using again that $\mathrm{dvol}_u = u^{2n/(n-2)} ~\mathrm{dvol}_0$ we find that

$$ \int (- \gamma \Delta_0 + S_0)u \cdot u ~\mathrm{dvol}_0 = \int S_u ~\mathrm{dvol}_u $$

Using that $\Delta_0^{-1}$ is compact, the operator $(-\gamma \Delta_0 + S_0)$ has arbitrarily large and positive eigenvalues. This shows that you can always find a metric conformal to $g_0$ with positive Einstein-Hilbert integral.

(Remark: that $g_0$ has constant scalar curvature is inessential; it just makes the description of the spectrum of $-\gamma \Delta_0 + S_0$ easier to state. You can do the same argument with any metric; or you can bring out the big guns and use Yamabe to first transform the metric to one with constant scalar curvature.)

When $n = 2$, your result is true by Gauss-Bonnet.

  • $\begingroup$ good! I am certainly missing something which I will take a look at! $\endgroup$ May 10, 2021 at 15:53
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    $\begingroup$ Nice! I found my mistake. Pretty cool answer, by the way. Thank you! $\endgroup$ May 10, 2021 at 16:43

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