Ricci flow as a gradient flow

1.1   Consider the functional $\mathcal {F}=\int_M{(R+\vert\nabla f\vert^2)e^{-f}dV}$ for a riemannian metric $g_{ij}$ and a function $f$ on a closed manifold $M$. Its first variation can be expressed as follows:

$$\delta \mathcal {F}(v_{ij},h)=\int_M e^{-f}[-\triangle v+\nabla_i\nabla_jv_{ij}-R_{ij}v_{ij}$$

$$-v_{ij}\nabla_i f\nabla_j f+2<\nabla f,\nabla h>+(R+\vert\nabla f\vert^2)(v/2-h)]$$

$$=\int_M{e^{-f}[-v_{ij}(R_{ij}+\nabla_i\nabla_j f)+(v/2-h)(2\triangle f-\vert\nabla f\vert^2+R)]},$$

where $\delta g_{ij}=v_{ij}$, $\delta f=h$, $v=g^{ij}v_{ij}$. Notice that $v/2-h$ vanishes identically iff the measure $dm=e^{-f}dV$ is kept fixed. Therefore, the symmetric tensor $-(R_{ij}+\nabla_i\nabla_j f)$ is the $L^2$ gradient of the functional $\mathcal {F}^m =\int_M{(R+\vert\nabla f\vert^2)dm}$, where now $f$ denotes $\log(dV/dm)$. Thus given a measure $m$ , we may consider the gradient flow $(g_{ij})_t=-2(R_{ij}+\nabla_i\nabla_j f)$ for $\mathcal {F}^m$. For general $m$ this flow may not exist even for short time; however, when it exists, it is just the Ricci flow, modified by a diffeomorphism. The remarkable fact here is that different choices of $m$ lead to the same flow, up to a diffeomorphism; that is, the choice of $m$ is analogous to the choice of gauge.

Proposition 1.2   Suppose that the gradient flow for $\mathcal {F}^m$ exists for $t\in[0,T].$ Then at $t=0$ we have $\mathcal {F}^m\le \frac{n}{2T}\int_M{dm}.$

Proof. We may assume $\int_M{dm}=1.$ The evolution equations for the gradient flow of $\mathcal {F}^m$ are

$$(g_{ij})_t=-2(R_{ij}+\nabla_i\nabla_j f) ,\ \ f_t=-R-\triangle f ,$$ (1.1)

and $\mathcal {F}^m$ satisfies

$$\mathcal {F}^m_t=2\int{\vert R_{ij}+\nabla_i\nabla_j f\vert^2 dm}$$ (1.2)

Modifying by an appropriate diffeomorphism, we get evolution equations

$$(g_{ij})_t=-2R_{ij} , f_t=-\triangle f + \vert\nabla f\vert^2 - R ,$$ (1.3)

and retain (1.2) in the form

$$\mathcal {F}_t=2\int{\vert R_{ij}+\nabla_i\nabla_j f\vert^2e^{-f}dV}$$ (1.4)

Now we compute

$$\mathcal {F}_t\ge\frac{2}{n}\int{(R+\triangle f)^2e^{-f}dV}\ge\frac{2}{n}(\int{(R+\triangle f)e^{-f}dV})^2=\frac{2}{n}\mathcal {F}^2,$$

and the proposition follows.

1.3 Remark. The functional $\mathcal {F}^m$ has a natural interpretation in terms of Bochner-Lichnerovicz formulas. The classical formulas of Bochner (for one-forms) and Lichnerovicz (for spinors) are $\nabla^*\nabla u_i=(d^*d+dd^*)u_i-R_{ij}u_j$ and $\nabla^*\nabla \psi=\delta^2\psi-1/4 R\psi.$ Here the operators $\nabla^*$ , $d^*$ are defined using the riemannian volume form; this volume form is also implicitly used in the definition of the Dirac operator $\delta$ via the requirement $\delta^*=\delta.$ A routine computation shows that if we substitute $dm=e^{-f}dV$ for $dV$ , we get modified Bochner-Lichnerovicz formulas $\nabla^{*m}\nabla u_i=(d^{*m}d+dd^{*m})u_i-R_{ij}^m u_j$ and $\nabla^{*m}\nabla\psi=(\delta^m)^2\psi-1/4R^m\psi,$ where $\delta^m\psi=\delta\psi-1/2(\nabla f)\cdot\psi$ , $R_{ij}^m=R_{ij}+\nabla_i\nabla_j f$ , $R^m=2\triangle f-\vert\nabla f\vert^2 +R .$ Note that $g^{ij}R_{ij}^m= R + \triangle f \ne R^m .$ However, we do have the Bianchi identity $\nabla_i^{*m}R_{ij}^m=\nabla_iR_{ij}^m-R_{ij}\nabla_i f=1/2\nabla_jR^m .$ Now $\mathcal {F}^m=\int_M{{R^m}dm}=\int_M{g^{ij}R_{ij}^m dm}.$

1.4* The Ricci flow modified by a diffeomorphism was considered by DeTurck, who observed that by an appropriate choice of diffeomorphism one can turn the equation from weakly parabolic into strongly parabolic, thus considerably simplifying the proof of short time existence and uniqueness; a nice version of DeTurck trick can be found in [H 4,$\S 6$].

The functional $\mathcal {F}$ and its first variation formula can be found in the literature on the string theory, where it describes the low energy effective action; the function $f$ is called dilaton field; see [D,$\S 6$] for instance.

The Ricci tensor $R_{ij}^m$ for a riemannian manifold with a smooth measure has been used by Bakry and Emery [B-Em]. See also a very recent paper [Lott].