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I am trying to understand the solution to the Yamabe problem as presented by Lee and Parker. It seems to me that the constant $\lambda$ which appears in the Yamabe equation $$\square\varphi = \lambda \varphi^{p-1}$$ and the Yamabe invariant $\lambda = \inf_\varphi Q_g(\varphi)$ where $Q_g$ is the functional $$Q_{g}(\varphi) = \frac{\int_{M} \left(a|\nabla \varphi|^{2}+S \varphi^{2}\right) d V_{g}}{\|\varphi\|_{p}^{2}} = \frac{E(\varphi)}{\|\varphi\|_{p}^{2}}$$ are in fact the same thing, that is to say if $\varphi$ is an absolute minimum for the functional $Q_g$ then $\varphi$ is a solution to the Yamabe equation with $\lambda$ as the coefficient of $\varphi^{p-1}$. It seems to me that this is proved explicitely by Aubin in his book "Some nonlinear problems in Riemannian geometry".

However, an explicit calculation of the Euler-Lagrange equation for $Q_g$ (which is shown on page 39 of this paper by Lee and Parker and which can be easily done explicitly)

Lee, John M.; Parker, Thomas H., The Yamabe problem, Bull. Am. Math. Soc., New Ser. 17, 37-91 (1987). ZBL0633.53062.

shows that the Euler equation is $$ a \Delta \varphi+S \varphi-\|\varphi\|_{p}^{-p} E(\varphi) \varphi^{p-1}=0 $$ that is to say the Yamabe equation with $$ \lambda = \frac{E(\varphi)}{\|\varphi\|_{p}^{p}}. $$ For the Yamabe invariant to be the same constant which appears in the Yamabe equation, the exponent at the denominator should be 2 and not $p$. I am therefore a bit confused: is the Yamabe invariant the same constant in the Yamabe equation in presence of a solution $\varphi$? If yes, where am I going wrong?

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  • $\begingroup$ Note that $Q_g(\varphi)$ is invariant under rescaling of $\varphi$ by a constant, but the first version you give of the Yamabe equation is not invariant under this rescaling unless you assume $\lambda$ rescales, too. So it is probably correct only, if you normalize $\varphi$ somehow ($\|\varphi\|_p = 1$ works). $\endgroup$ – Deane Yang Sep 17 at 21:44
  • $\begingroup$ In general a good sanity check when writing equations and estimates is to check that everything rescales correctly when you rescale either the function or metric. $\endgroup$ – Deane Yang Sep 17 at 21:44

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