A comparison geometry approach to the Ricci flow

7.1 In this section we consider an evolving metric $(g_{ij})_{\tau}=2R_{ij}$ on a manifold $M;$ we assume that either $M$ is closed, or $g_{ij}(\tau)$ are complete and have uniformly bounded curvatures. To each curve $\gamma(\tau), 0<\tau_1\le\tau\le\tau_2,$ we associate its $\mathcal{L}$-length

$$\mathcal{L}(\gamma)=\int_{\tau_1}^{\tau_2}{\sqrt{\tau}(R(\gamma(\tau))+\vert\dot{\gamma}(\tau)\vert^2)d\tau}$$

(of course, $R(\gamma(\tau))$ and $\vert\dot{\gamma}(\tau)\vert^2$ are computed using $g_{ij}(\tau)$)

Let $X(\tau)=\dot{\gamma}(\tau),$ and let $Y(\tau)$ be any vector field along $\gamma(\tau).$ Then the first variation formula can be derived as follows:

$$\delta_Y(\mathcal{L}) =$$

$$\int_{\tau_1}^{\tau_2}{\sqrt{\tau}(<Y,\nabla R>+2<\nabla_Y X,X>)d\tau} =\int_{\tau_1}^{\tau_2}{\sqrt{\tau}(<Y,\nabla R>+2<\nabla_X Y,X>)d\tau}$$

$$=\int_{\tau_1}^{\tau_2}{\sqrt{\tau}(<Y,\nabla R>+2\frac{d}{d\tau}<Y,X>-2<Y,\nabla_X X>-4\mbox{Ric}(Y,X))d\tau}$$

$$=\left.2\sqrt{\tau}<X,Y>\right\vert _{\tau_1}^{\tau_2}+\int_{\tau... ...{\sqrt{\tau}<Y,\nabla R-2\nabla_X X-4\mbox{Ric}(X,\cdot)-\frac{1}{\tau}X>d\tau}$$ (7.1)

Thus $\mathcal{L}$-geodesics must satisfy

$$\nabla_X X-\frac{1}{2}\nabla R+ \frac{1}{2\tau}X+2 (X,\cdot)=0$$ (7.2)

Given two points $p,q$ and $\tau_2>\tau_1>0,$ we can always find an $\mathcal{L}$-shortest curve $\gamma(\tau), \tau\in[\tau_1,\tau_2]$ between them, and every such $\mathcal{L}$-shortest curve is $\mathcal{L}$-geodesic. It is easy to extend this to the case $\tau_1=0;$ in this case $\sqrt{\tau}X(\tau)$ has a limit as $\tau\to 0.$ From now on we fix $p$ and $\tau_1=0$ and denote by $L(q,\bar{\tau})$ the $\mathcal{L}$-length of the $\mathcal{L}$-shortest curve $\gamma(\tau), 0\le\tau\le\bar{\tau},$ connecting $p$ and $q.$ In the computations below we pretend that shortest $\mathcal{L}$-geodesics between $p$ and $q$ are unique for all pairs $(q,\bar{\tau});$ if this is not the case, the inequalities that we obtain are still valid when understood in the barrier sense, or in the sense of distributions.

The first variation formula (7.1) implies that $\nabla L(q,\bar{\tau})=2\sqrt{\bar{\tau}}X(\bar{\tau}),$ so that $\vert\nabla L\vert^2=4\bar{\tau}\vert X\vert^2=-4\bar{\tau}R+4\bar{\tau}(R+\vert X\vert^2).$ We can also compute

$$L_{\bar{\tau}}(q,\bar{\tau})=\sqrt{\bar{\tau}}(R+\vert X\vert^2)-<X,\nabla L>=2\sqrt{\bar{\tau}}R-\sqrt{\bar{\tau}}(R+\vert X\vert^2)$$

To evaluate $R+\vert X\vert^2$ we compute (using (7.2))

$$\frac{d}{d\tau}(R(\gamma(\tau))+\vert X(\tau)\vert^2)=R_{\tau}+<\nabla R,X>+2<\nabla_X X,X>+2$$Ric$$(X,X)$$

$$=R_{\tau}+\frac{1}{\tau}R+2<\nabla R,X>-2$$Ric$$(X,X)-\frac{1}{\tau}(R+\vert X\vert^2)$$

$$=-H(X)-\frac{1}{\tau}(R+\vert X\vert^2),$$ (7.3)

where $H(X)$ is the Hamilton's expression for the trace Harnack inequality (with $t=-\tau$). Hence,

$$\bar{\tau}^{\frac{3}{2}}(R+\vert X\vert^2)(\bar{\tau})=-K+\frac{1}{2}L(q,\bar{\tau}),$$ (7.4)

where $K=K(\gamma,\bar{\tau})$ denotes the integral $\int_0^{\bar{\tau}}{\tau^{\frac{3}{2}}H(X)d\tau},$ which we'll encounter a few times below. Thus we get

$$L_{\bar{\tau}}=2\sqrt{\bar{\tau}}R-\frac{1}{2\bar{\tau}}L+\frac{1}{\bar{\tau}}K$$ (7.5)

$$\vert\nabla L\vert^2=-4\bar{\tau}R+\frac{2}{\sqrt{\bar{\tau}}}L-\frac{4}{\sqrt{\bar{\tau}}}K$$ (7.6)

Finally we need to estimate the second variation of $L.$ We compute

$$\delta ^2_Y(\mathcal{L})=\int_0^{\bar{\tau}}{\sqrt{\tau}(Y\cdot Y\cdot R+2<\nabla_Y \nabla_Y X,X>+2\vert\nabla_Y X\vert^2)d\tau}$$

$$=\int_0^{\bar{\tau}}{\sqrt{\tau}(Y\cdot Y\cdot R+2<\nabla_X \nabla_Y Y,X>+2<R(Y,X),Y,X>+2\vert\nabla_X Y\vert^2)d\tau}$$

Now

$$\frac{d}{d\tau}<\nabla_Y Y,X>=<\nabla_X \nabla_Y Y,X>+<\nabla_Y Y,\nabla_X X>+2Y\cdot$$Ric$$(Y,X)-X\cdot$$Ric$$(Y,Y),$$

so, if $Y(0)=0$ then

$$\delta ^2_Y(\mathcal{L})=2<\nabla_Y Y,X>\sqrt{\bar{\tau}}+$$

$$\begin{multline}\int_0^{\bar{\tau}}\sqrt{\tau}(\nabla_Y \nabla_Y R+2<R(Y,X),Y,X>... ...vert^2\\ +2\nabla_X\mbox{Ric}(Y,Y)-4\nabla_Y\mbox{Ric}(Y,X))d\tau,\end{multline}$$

where we discarded the scalar product of $-2\nabla_Y Y$ with the left hand side of (7.2). Now fix the value of $Y$ at $\tau=\bar{\tau}$, assuming $\vert Y(\bar{\tau})\vert=1,$ and construct $Y$ on $[0,\bar{\tau}]$ by solving the ODE

$$\nabla_X Y=- (Y,\cdot)+\frac{1}{2\tau}Y$$ (7.8)

We compute

$$\frac{d}{d\tau}<Y,Y>=2$$Ric$$(Y,Y)+2<\nabla_X Y,Y>=\frac{1}{\tau}<Y,Y>,$$

so $\vert Y(\tau)\vert^2=\frac{\tau}{\bar{\tau}},$ and in particular, $Y(0)=0.$ Making a substitution into (7.7), we get

Hess$$_L(Y,Y)\le$$

$$\int_0^{\bar{\tau}}\sqrt{\tau}(\nabla_Y \nabla_Y R+2<R(Y,X),Y,X>+2\nabla_X$$Ric$$(Y,Y)-4\nabla_Y$$Ric$$(Y,X)$$

$$+2\vert$$Ric$$(Y,\cdot)\vert^2- \frac{2}{\tau}$$Ric$$(Y,Y)+\frac{1}{2\tau\bar{\tau}})d\tau$$

To put this in a more convenient form, observe that

$$\frac{d}{d\tau}$$Ric$$(Y(\tau),Y(\tau))=$$Ric$$_{\tau}(Y,Y)+\nabla_X$$Ric$$(Y,Y)+ 2$$Ric$$(\nabla_X Y,Y)$$

$$=$$Ric$$_{\tau}(Y,Y)+\nabla_X$$Ric$$(Y,Y)+\frac{1}{\tau}$$Ric$$(Y,Y)- 2\vert$$Ric$$(Y,\cdot)\vert^2,$$

so

$$_L(Y,Y)\le\frac{1}{\sqrt{\bar{\tau}}}-2\sqrt{\bar{\tau}} (Y,Y)-\int_0^{\bar{\tau}} {\sqrt{\tau}H(X,Y)d\tau},$$ (7.9)

where

$$H(X,Y)=-\nabla_Y \nabla_Y R-2<R(Y,X)Y,X>-4(\nabla_X$$Ric$$(Y,Y)-\nabla_Y$$Ric$$(Y,X))$$

$$-2$$Ric$$_{\tau}(Y,Y)+ 2\vert$$Ric$$(Y,\cdot)\vert^2-\frac{1}{\tau}$$Ric$$(Y,Y)$$

is the Hamilton's expression for the matrix Harnack inequality (with $t=-\tau$). Thus

$$\triangle L\le-2\sqrt{\tau}R+\frac{n}{\sqrt{\tau}}-\frac{1}{\tau}K$$ (7.10)

A field $Y(\tau)$ along $\mathcal{L}$-geodesic $\gamma(\tau)$ is called $\mathcal{L}$-Jacobi, if it is the derivative of a variation of $\gamma$ among $\mathcal{L}$-geodesics. For an $\mathcal{L}$-Jacobi field $Y$ with $\vert Y(\bar{\tau})\vert=1$ we have

$$\frac{d}{d\tau}\vert Y\vert^2=2$$Ric$$(Y,Y)+2<\nabla_X Y,Y>=2$$Ric$$(Y,Y)+2<\nabla_Y X,Y>$$

$$=2 (Y,Y)+\frac{1}{\sqrt{\bar{\tau}}} _L(Y,Y)\le\frac{1}{\bar{\tau}}- \frac{1}{\sqrt{\bar{\tau}}}\int_0^{\bar{\tau}}{\tau^{\frac{1}{2}}H(X,\tilde{Y})d\tau},$$ (7.11)

where $\tilde{Y}$ is obtained by solving ODE (7.8) with initial data $\tilde{Y}(\bar{\tau})=Y(\bar{\tau}).$ Moreover, the equality in (7.11) holds only if $\tilde{Y}$ is $\mathcal{L}$-Jacobi and hence $\frac{d}{d\tau}\vert Y\vert^2=2$Ric$(Y,Y)+\frac{1}{\sqrt{\bar{\tau}}}$Hess$_L(Y,Y) =\frac{1}{\bar{\tau}}.$

Now we can deduce an estimate for the jacobian $J$ of the $\mathcal{L}$-exponential map, given by $\mathcal{L}$exp$_X(\bar{\tau})=\gamma(\bar{\tau}),$ where $\gamma(\tau)$ is the $\mathcal{L}$-geodesic, starting at $p$ and having $X$ as the limit of $\sqrt{\tau}\dot{\gamma}(\tau)$ as $\tau\to 0.$ We obtain

$$\frac{d}{d\tau} J(\tau)\le\frac{n}{2\bar{\tau}}-\frac{1}{2}\bar{\tau}^{-\frac{3}{2}}K,$$ (7.12)

with equality only if $2$Ric$+\frac{1}{\sqrt{\bar{\tau}}}$Hess$_L=\frac{1}{\bar{\tau}}g.$ Let $l(q,\tau)=\frac{1}{2\sqrt{\tau}}L(q,\tau)$ be the reduced distance. Then along an $\mathcal{L}$-geodesic $\gamma(\tau)$ we have (by (7.4))

$$\frac{d}{d\tau}l(\tau)=-\frac{1}{2\bar{\tau}}l+\frac{1}{2}(R+\vert X\vert^2) =-\frac{1}{2}\bar{\tau}^{-\frac{3}{2}}K,$$

so (7.12) implies that $\tau^{-\frac{n}{2}}$exp$(-l(\tau))J(\tau)$ is nonincreasing in $\tau$ along $\gamma$, and monotonicity is strict unless we are on a gradient shrinking soliton. Integrating over $M$, we get monotonicity of the reduced volume function $\tilde{V}(\tau)=\int_M{\tau^{-\frac{n}{2}}\mbox{exp}(-l(q,\tau))dq}.$ ( Alternatively, one could obtain the same monotonicity by integrating the differential inequality

$$l_{\bar{\tau}}-\triangle l+\vert\nabla l\vert^2-R+\frac{n}{2\bar{\tau}}\ge0,$$ (7.13)

which follows immediately from (7.5), (7.6) and (7.10). Note also a useful inequality

$$2\triangle l-\vert\nabla l\vert^2 +R+\frac{l-n}{\bar{\tau}}\le 0,$$ (7.14)

which follows from (7.6), (7.10).)

On the other hand, if we denote $\bar{L}(q,\tau)=2\sqrt{\tau}L(q,\tau),$ then from (7.5), (7.10) we obtain

$$\bar{L}_{\bar{\tau}}+\triangle \bar{L}\le2n$$ (7.15)

Therefore, the minimum of $\bar{L}(\cdot,\bar{\tau})-2n\bar{\tau}$ is nonincreasing, so in particular, the minimum of $l(\cdot,\bar{\tau})$ does not exceed $\frac{n}{2}$ for each $\bar{\tau}>0.$ (The lower bound for $l$ is much easier to obtain since the evolution equation $R_{\tau}=-\triangle R-2\vert$Ric$\vert^2$ implies $R(\cdot,\tau)\ge-\frac{n}{2(\tau_0-\tau)},$ whenever the flow exists for $\tau\in[0,\tau_0].$)

7.2 If the metrics $g_{ij}(\tau)$ have nonnegative curvature operator, then Hamilton's differential Harnack inequalities hold, and one can say more about the behavior of $l.$ Indeed, in this case, if the solution is defined for $\tau\in[0,\tau_0],$ then $H(X,Y)\ge-$Ric$(Y,Y)(\frac{1}{\tau}+\frac{1}{\tau_0-\tau})\ge -R(\frac{1}{\tau}+\frac{1}{\tau_0-\tau})\vert Y\vert^2$ and $H(X)\ge-R(\frac{1}{\tau}+\frac{1}{\tau_0-\tau}).$ Therefore, whenever $\tau$ is bounded away from $\tau_0$ (say, $\tau\le(1-c)\tau_0, c>0$), we get (using (7.6), (7.11))

$$\vert\nabla l\vert^2+R\le\frac{Cl}{\tau},$$ (7.16)

and for $\mathcal{L}$-Jacobi fields $Y$

$$\frac{d}{d\tau} \vert Y\vert^2\le\frac{1}{\tau}(Cl+1)$$ (7.17)

7.3 As the first application of the comparison inequalities above, let us give an alternative proof of a weakened version of the no local collapsing theorem 4.1. Namely, rather than assuming $\vert Rm\vert(x,t_k)\le r_k^{-2}$ for $x\in B_k,$ we require $\vert Rm\vert(x,t)\le r_k^{-2}$ whenever $x\in B_k, t_k-r_k^2\le t\le t_k.$ Then the proof can go as follows: let $\tau_k(t)=t_k-t, p=p_k,\epsilon_k=r_k^{-1}Vol(B_k)^{\frac{1}{n}}.$ We claim that $\tilde{V}_k(\epsilon_k r_k^2) < 3\epsilon_k^{\frac{n}{2}}$ when $k$ is large. Indeed, using the $\mathcal{L}$-exponential map we can integrate over $T_pM$ rather than $M;$ the vectors in $T_pM$ of length at most $\frac{1}{2}\epsilon_k^{-\frac{1}{2}}$ give rise to $\mathcal{L}$-geodesics, which can not escape from $B_k$ in time $\epsilon_k r_k^2,$ so their contribution to the reduced volume does not exceed $2\epsilon_k^{\frac{n}{2}};$ on the other hand, the contribution of the longer vectors does not exceed exp$(-\frac{1}{2}\epsilon_k^{-\frac{1}{2}})$ by the jacobian comparison theorem. However, $\tilde{V}_k(t_k)$ (that is, at $t=0$) stays bounded away from zero. Indeed, since min$\ l_k(\cdot,t_k-\frac{1}{2}T)\le\frac{n}{2},$ we can pick a point $q_k,$ where it is attained, and obtain a universal upper bound on $l_k(\cdot, t_k)$ by considering only curves $\gamma$ with $\gamma(t_k-\frac{1}{2}T)=q_k,$ and using the fact that all geometric quantities in $g_{ij}(t)$ are uniformly bounded when $t\in[0,\frac{1}{2}T].$ Since the monotonicity of the reduced volume requires $\tilde{V}_k(t_k)\le\tilde{V}_k(\epsilon_k r_k^2),$ this is a contradiction.

A similar argument shows that the statement of the corollary in 4.2 can be strengthened by adding another property of the ancient solution, obtained as a blow-up limit. Namely, we may claim that if, say, this solution is defined for $t\in(-\infty,0),$ then for any point $p$ and any $t_0>0,$ the reduced volume function $\tilde{V}(\tau),$ constructed using $p$ and $\tau(t)=t_0-t,$ is bounded below by $\kappa.$

7.4* The computations in this section are just natural modifications of those in the classical variational theory of geodesics that can be found in any textbook on Riemannian geometry; an even closer reference is [L-Y], where they use "length", associated to a linear parabolic equation, which is pretty much the same as in our case.