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.



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




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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.


