Derivation of yamabe flow

Question

I am reading papers about yamabe flow. I have a problem about how people derive it as a gradient flow.

Suppose we have $(M,g_0)$, $g(t)=u^{\frac{4}{n-2}}(t)g_0$ is another conformal metric. Let $R=R(t)$ be the scalar curvature and $s=\frac{\int_M Rd\mu}{\int_M d\mu}$ be the average scalar curvature at time $t$. I know many people use $$u_t=(s-R)u$$ as yamabe flow. I want to view it as the gradient flow the following functional $$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-2}}d\mu_0\right)^{\frac{n-2}{n}}}$$ where $V(t)$ is the volume.

By some lengthy calculation, the frechet derivative of $E$ is $$\langle E'(u),v\rangle=\frac{2}{V(t)^{\frac{n-2}{n}}}\int_{M}(R-s)u^{-1}v\,d\mu$$ It seems that $E'(u)=\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$ in the $L^2$ sense. So the $L^2$ gradient flow of this functional is $$u_t=-\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$$ This is totally different from the yamabe flow as I mentioned. I expect we should have $u$ instead of $u^{-1}$ on the right hand side.

So what is the problem? how should I accommodate them?

Answer 1

The Yamabe flow is (up to a constant) the gradient flow of the Yamabe functional on the unit volume conformal class, as you expected. The comment by @Mark Peletier hints at your error: you aren't using the correct "inner product."

We briefly discuss the Ebin metric on the space of all metrics $Met$. Recall that $T_gMet = Sym^2 T^*M$. Then the Ebin/$L^2$ metric at $g$ is defined by $$ g_E(h_1,h_2)|_g = \int_M tr(g^{-1}h_1g^{-1}h_2). $$ See e.g. http://arxiv.org/pdf/0904.0177v1.pdf for more information on the $L^2$ metric. Here, we are interested in the restriction of the Ebin/$L^2$ metric to the unit volume conformal class $[g]_1$. Now, the correct statement is

The Yamabe flow is (up to a constant multiple) the (negative) Ebin/$L^2$-gradient flow of the Yamabe functional on $[g]_1$.

First, note that the tangent space to $[g]_1$ at $g$ is $$ T_g[g]_1 = \left\{w g : \int_M w dV_g = 0\right\}. $$ and the Ebin, or $L^2$ metric restricted to $[g]_1$ is given by \begin{align*} g_E(w_1g,w_2g)|_g & = \int_M tr(g^{-1} w_1 g g^{-1} w_2g)\\ & = \sum_{i,j=1}^n \int_M w_1 w_2 g^{ij}g_{j}^{k} g_{kl}g^l_i \, dV_g\\ & = n \int_M w_1 w_2 dV_g. \end{align*} So, up to a constant (which we'll ignore), $g_E|_{[g]_1}$ at $g$ is the $L^2$ inner product of the conformal factor. \begin{align*} \frac{d}{dt}\Big|_{t=0} Y((1+tw)^{N-2}g) & = c \int_M (R_{g}-r_{g})wdV_g\\ & = g_E((R_g-r_g)g,wg)|_g \end{align*} This shows that (up to a constant), $$ \nabla_{[g]_1} Y|_g = c_n(R_g-r_g)g, $$ which is the Yamabe flow.

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If you would rather think of the flow as a flow on the level of conformal factors, you may be a bit dissatisfied with the previous computation. So, lets do it again, where we imagine that $g$ is fixed, and the Yamabe flow at time $t$ is given by $v^{N-2}g$ (recall that $N = \frac{2n}{n-2}$). Then, $$ T_{v^{N-2}g}[g]_1 = \left\{w v^{N-3} g : \int_M w v^{N-1} dV_g = 0\right\}, $$ so \begin{align*} g_E(w_1 v^{N-3} g,w_2 v^{N-3} g)|_{v^{N-2}g} & = \int_M tr( v^{2-N} g^{-1} w_1 v^{N-3} g v^{2-N} g^{-1}w_2 v^{N-3} g ) v^N dV_g\\ & = n \int_M w_1 w_2 v^{4-2N+2N-6+N} dV_g\\ & = n \int w_1 w_2 v^{N-2} dV_g. \end{align*} Moreover, \begin{align*} \frac{d}{dt}\Big|_{t=0} Y((v+tw)^{N-2}g) & = c\int_M (R_{v^{N-2}g} -r_{v^{N-2}g})v^{N-1}w dV_g\\ & = c\int_M (R_{v^{N-2}g} -r_{v^{N-2}g})v w v^{N-2} dV_g\\ & =c g_E((R_{v^{N-2}g} -r_{v^{N-2}g})v^{N-2}g,w v^{N-3}g)|_{v^{N-2}g} \end{align*} So, $$ \nabla_{[g]_1} Y|_{v^{N-2}g} = (R_{v^{N-2}g} -r_{v^{N-2}g})v^{N-2}g, $$ which of course is what we expected.