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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?

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I can't promise this will fix everything, but I think that your expression for $V(t)$ is wrong. The power of $u$ should be $\frac{2n}{n-2}$ rather than $\frac{2n}{n-1}$. (but maybe this was just a typo in your post). – Otis Chodosh Apr 26 '14 at 15:22
Yes. I notice that. It is just a typo. Thank your very much.@OtisChodosh – Slm2004 Apr 26 '14 at 18:30
The computation for $E'$ should not be that disgusting. So maybe you're taking a long route and somewhere there's a mistake? The computation of the gradient of the Yamabe functional can be found here:, but its not so explicit. But it might help you get on the right track. If you can't figure it out, you should post your computation so that someone can help you out, otherwise its impossible to tell what went wrong. – Otis Chodosh Apr 26 '14 at 21:08
As you do your calculation, keep track of how your equations transform if you rescale either $u$ or space by a constant factor. Both sides must scale the same. This will help you figure out where your error is. – Deane Yang Apr 26 '14 at 22:04
Could the difference be whether the gradient flow is with respect to the $L^2(g_0)$ metric or the $L^2(g)$ metric? – Mark Peletier Apr 27 '14 at 9:56
up vote 2 down vote accepted

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. 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.

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.

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Thank you very much. I got impressed by you. @OtisChodosh – Slm2004 May 9 '14 at 19:44
You're welcome! – Otis Chodosh May 10 '14 at 3:03

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