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?



Derivation of yamabe flow
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.




---


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.


