# A corollary of the non-existence of positive scalar curvature

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.

• 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? May 10, 2021 at 14:47
• @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.) May 10, 2021 at 14:50
• @WillieWong: oh right, the volume form varies with $f$. Thanks :) May 10, 2021 at 15:05

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.

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