I'm trying to prove that the Yamabe invariant of a compact manifold, that is the sigma constant, is positive iff $M$ admits a metric of positive scalar curvature. I will use $$Y_{[g]} = \underset{g \in [g]}{\inf} \mathcal{E}(g)$$ to be the Yamabe constant with
$$\mathcal{E}(g) = {\frac{{\int_{M}S_g \Omega}}{\left({\int_{M} \Omega}\right)^{{\frac{n-2}{n}}}}}$$
One direction is straightforward. That is, assume $\sigma(M) > 0$, then there exists some conformal class [g] such that $Y_{[g]} > 0$. Then by the Yamabe problem there exists a metric $\tilde{g} \in [g]$ such that $(M,\tilde{g})$ has constant positive scalar curvature.
I'm struggling with the other direction. Suppose $M$ admits a metric of positive scalar curvature $g_0$, say. We want to show that $$\sigma(M) = \underset{[g]\in \mathcal{C}_M }{\sup} Y_{[g]} > 0$$ with $\mathcal{C}_{M}$ being the set of conformal classes on $M$.
So we just need to show there exists a conformal class giving a positive Yamabe constant. Given the information we have I would imagine that this supremum is achieved by $Y_{[g_0]}$. But I'm having trouble showing that this is positive. What's stopping there being some other metric, say $\tilde{g_0} \in [g_0]$ such that $\mathcal{E}(\tilde{g_0}) \leq 0$ thus giving a non-positive value for $Y_{[g_0]}$?
Best Answer
Let $g$ be a metric of positive scalar curvature. We will show that $Y_{[g]}>0$. Note that if $c>0$ is a constant, then $S_{cg}=c^{-1}S_g$, so there is no loss in generality in assuming that $0<\inf S \leq \frac{4(n-1)}{n-2}$.
Let $\hat g\in[g]$. Write $\hat g = u^{\frac{4}{n-2}}g$ for some positive function $u$. Then $$ \tag{$\dagger$} \int S_{\hat g}\Omega_{\hat g} = \int \left( \frac{4(n-1)}{n-2}\lvert\nabla u\rvert_g^2 + S_gu^2 \right)\Omega_g \geq (\inf S_g) \lVert u\rVert_{1,2}^2 , $$ where $\lVert u\rVert_{1,2}^2$ is the $W^{1,2}(M^n)$-norm of $u$. (The first equality follows from the formula for $S_{\hat g}$ in terms of $g$ and $u$.) Note that $\inf S_g>0$ by assumption. Now, the continuity of the embedding $W^{1,2}(M^n)\subset L^{\frac{2n}{n-2}}(M^n)$ implies that there is a constant $C>0$ such that $$ \tag{$\ddagger$} \lVert w\rVert_{\frac{2n}{n-2}} \leq C\lVert w\rVert_{1,2} $$ for all $w\in W^{1,2}(M^n)$. Combining ($\dagger$) and ($\ddagger$) implies that $$ \int S_{\hat g}\Omega_{\hat g} \geq C^\prime \lVert u\rVert_{\frac{2n}{n-2}}^2 = C^\prime\left( \int \Omega_{\hat g} \right)^{\frac{n-2}{n}} , $$ where $C^\prime>0$ is determined by $\inf S$ and $C$; and the last equality is just the formula for $\Omega_{\hat g}$ in terms of $\Omega_g$. This last display proves that $Y_{[g]}>0$.