In this old blog post I showed that Perelman’s modified Ricci flow is a gradient flow for closed manifolds. I’m copying some of what I wrote there, edited a bit.
I want to interpret what Perelman might mean by Ricci flow being
the gradient flow for the first eigenvalue of $-4\Delta +R$.
First, what do we mean by gradient flow? Given a smooth manifold,
the space of smooth Riemannian metrics is denoted $Riem(M)$, and
is an open cone in the space of smooth symmetric bilinear forms
$S^2M$ on $M$, so that for $g\in Riem(M)$, $T_g(Riem(M))\cong
S^2M$. There is a natural inner product on $T_g(Riem(M))$, given
by $(h,h')_g=\int_M h_{ik}h'_{lm}g^{il}g^{km} dv_g$ for $h,h'\in
S^2M$ where we use the Einstein summation convention ($h_{ik}$ is
the tensor written in local coordinates, and $g^{il}$ is the
inverse tensor so that $g^{il}g_{lj}=\delta^i_j$). (Note: I think
the convention is to usually assume that $g_{ij}=\delta_{ij}$ is
diagonalized, in which case we can write $(h,h')_g=\int_M
h_{ik}h'_{ik} dv_g$, summing over repeated indices). Thus, a flow
$g_t=V(g)$, $V:Riem(M)\to S^2M$, is a gradient flow with respect
to this $L^2$ inner product if there is a functional
$\mathcal{F}:Riem(M)\to \mathbb{R}$ such that
$\frac{\partial\mathcal{F}(g)}{\partial h}=(h,V(g))_g$, in which
case $\frac{\partial\mathcal{F}(g_t)}{\partial
t}=(V(g),V(g))_g\geq 0$, since $(,)_g$ is positive definite. This
is of interest, since then the functional $\mathcal{F}$ is
monotonic along the flow, and one might be able to use it to
analyze the evolution of $g_t$ under the flow. It turns out
Perelman has shown that flows related to the Ricci flow are
gradient flows, but with respect to modified $L^2$ metrics on
$Riem(M)$.
We'll use $R$ to denote the scalar curvature (suppressing the
Riemannian structure $(M,g)$). The operator $-4\Delta+R$ is a
Schrodinger operator. It turns out to have a unique normalized
positive eigenvector, that is there is a unique function
$\Phi:M\to \mathbb{R}$ such that $\Phi(x)>0$ for all $x\in M$, $-4\Delta
\Phi+R\Phi=\lambda \Phi$, and $\|\Phi\|_2^2=\int_M \Phi^2 dv_g=1$. In fact, $\Phi$ has minimal
eigenvalue for $-4\Delta+R$. For example, if $R$ is constant (as
for a Yamabe metric), then $\Phi$ is harmonic, so it must be the
constant positive function such that $\|\Phi\|_2=1$. Kleiner and Lott
claim that the eigenvalue (and presumably the eigenfunction)
depend smoothly on $g$ (see after Definition 6.4). We will denote the eigenvalue $\lambda$
and the eigenfunction $\Phi=e^{-f/2}$, following Perelman. Now,
define an $L^2$ metric on $T_g(Riem(M))$ by
$[h,h']_g=\frac12\int_M h_{ik}h'_{lm}g^{il}g^{km} e^{-f} dv_g$. I
claim that the flow $g_t=-2(Ric+\nabla^2f)$, or written locally as
$(g_{ij})_t=-2(R_{ij}+\nabla_i\nabla_j f)$, is the gradient flow
for the functional $\mathcal{F}(g)=\int_M (R+|\nabla f|^2)e^{-f}
dv_g=\int_M \Phi(-4\Delta \Phi+R\Phi) dv_g=\lambda$ with respect
to the metric $[,]$.
To see this, note that the eigenvalue equation
$-4\Delta\Phi+R\Phi=\lambda\Phi$ is equivalent to the equation
$2\Delta f-|\nabla f|^2+R=\lambda$ (this is claimed in 2.4, and follows easily from the computations in section
4 of Kleiner-Lott). The Rayleigh Ritz quotient method for obtaining
the lowest eigenvalue of an operator $L=-4\Delta+R$ is given by
$$\lambda(g)=\inf_{\Phi} \frac{(\Phi,L\Phi)_2}{\|\Phi\|_2^2},$$
or normalizing $\lambda=\inf \{ (\Phi,L\Phi)_2 | \|\Phi\|_2=1 \}$
(where $(,)_2$ is the inner product on $L^2(M)$), in which case
the eigenvalue equation can be seen as a generalization of
Lagrange multipliers). On p. 5, we see the formula
for the variation of $\mathcal{F}(g)$ with $\delta g_{ij}=v_{ij}$
and $\delta f=h$ as
$$\delta \mathcal{F}(v_{ij},h)=\int_M
e^{-f}[-v_{ij}(R_{ij}+\nabla_i \nabla_j f)+(v/2-h)(2\Delta
f-|\nabla f|^2+R)]dv_g,$$ so that if we let $f$ satisfy
$\|\Phi\|_2^2=\int_M e^{-f}dv_{g_t}=1$ and $2\Delta f-|\nabla
f|^2+R=\lambda(g_t)$, then $\delta \|\Phi\|_2^2 = \int_M (v/2-h)
dv_g=0$ (from Proposition 4.3 of Kleiner-Lott), and we have $\delta
\mathcal{F}= [v_{ij},-2 (R_{ij}+\nabla_i\nabla_j f)]_{g} +\lambda
\int_M (v/2-h)e^{-f} dv_g = [v_{ij},-2 (R_{ij}+\nabla_i\nabla_j
f)]_{g} $. Now, the flow $g_t=-2(Ric+\nabla^2 f)$ is equivalent to
Ricci flow $g_t=-2Ric$ up to a diffeomorphism, since the term
$\nabla^2 f$ only modifies the metric by the Lie derivative of the
vector field $\nabla f$.