I'm reading Aubin's work on Yamabe problem(the book Some Nonlinear Problems in Riemannian Geometry page150)

He wrote that by a homothetic change of metric we can set the volume equal to one. So henceforth, without loss of generality, we suppose the volume equal to one.

What does he mean, does he mean that we can set the volume to be 1 for any problems? or there is some condition? It seems that this can simplify many estimates.


2 Answers 2


A homothetic change to the metric is simply a rescaling by a positive constant, i.e. replacing $g$ by $cg$ where $c \in (0, \infty)$. Now note that $d\mu_{cg} = c^{n/2}d\mu_g$. This can be seen in local coordinates as follows:

\begin{align*} d\mu_{cg} &= \sqrt{\det(cg)}dx^1\wedge\dots\wedge dx^n\\ &= \sqrt{c^n\det(g)}dx^1\wedge\dots\wedge dx^n\\ &= c^{n/2}\sqrt{\det(g)}dx^1\wedge\dots\wedge dx^n\\ &= c^{n/2}d\mu_g. \end{align*}

Therefore we have

$$\operatorname{Vol}(M, cg) = \int_M d\mu_{cg} = \int_M c^{n/2}d\mu_g = c^{n/2}\int_M d\mu_g = c^{n/2}\operatorname{Vol}(M, g).$$

Choosing $c = \operatorname{Vol}(M, g)^{-2/n}$, we obtain $\operatorname{Vol}(M, cg) = 1$.


The book is using homethetic in the sense described for example here. Essentially, you just scale the entire manifold up or down without changing its shape and after doing this you can assume the volume is equal to 1.


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