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? – Francesco Polizzi May 10 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.) – Willie Wong May 10 at 14:50
• @WillieWong: oh right, the volume form varies with $f$. Thanks :) – Francesco Polizzi May 10 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! – L.F. Cavenaghi May 10 at 15:53
• Nice! I found my mistake. Pretty cool answer, by the way. Thank you! – L.F. Cavenaghi May 10 at 16:43