No local collapsing theorem II

8.1 Let us first formalize the notion of local collapsing, that was used in 7.3.

Definition. A solution to the Ricci flow $(g_{ij})_t=-2R_{ij}$ is said to be $\kappa$-collapsed at $(x_0,t_0)$ on the scale $r>0$ if $\vert Rm\vert(x,t)\le r^{-2}$ for all $(x,t)$ satisfying dist$_{t_0}(x,x_0)<r$ and $t_0-r^2\le t\le t_0,$ and the volume of the metric ball $B(x_0,r^2)$ at time $t_0$ is less than $\kappa r^n.$

8.2 Theorem. For any $A>0$ there exists $\kappa=\kappa(A)>0$ with the following property. If $g_{ij}(t)$ is a smooth solution to the Ricci flow $(g_{ij})_t=-2R_{ij}, 0\le t\le r_0^2,$ which has $\vert Rm\vert(x,t)\le r_0^{-2}$ for all $(x,t),$ satisfying dist$_0(x,x_0)<r_0,$ and the volume of the metric ball $B(x_0,r_0)$ at time zero is at least $A^{-1}r_0^n,$ then $g_{ij}(t)$ can not be $\kappa$-collapsed on the scales less than $r_0$ at a point $(x,r_0^2)$ with dist$_{r_0^2}(x,x_0)\le Ar_0.$

Proof. By scaling we may assume $r_0=1;$ we may also assume dist$_1(x,x_0)=A.$ Let us apply the constructions of 7.1 choosing $p=x, \tau(t)=1-t.$ Arguing as in 7.3, we see that if our solution is collapsed at $x$ on the scale $r\le 1,$ then the reduced volume $\tilde{V}(r^2)$ must be very small; on the other hand, $\tilde{V}(1)$ can not be small unless min$\ l(x,\frac{1}{2})$ over $x$ satisfying dist$_{\frac{1}{2}}(x,x_0)\le\frac{1}{10}$ is large. Thus all we need is to estimate $l,$ or equivalently $\bar{L},$ in that ball. Recall that $\bar{L}$ satisfies the differential inequality (7.15). In order to use it efficiently in a maximum principle argument, we need first to check the following simple assertion.

8.3 Lemma. Suppose we have a solution to the Ricci flow $(g_{ij})_t=-2R_{ij}.$

(a) Suppose Ric$(x,t_0)\le (n-1)K$ when    dist$_{t_0}(x,x_0)<r_0.$ Then the distance function $d(x,t)=$dist$_t(x,x_0)$ satisfies at $t=t_0$ outside $B(x_0,r_0)$ the differential inequality

$$d_t-\triangle d\ge -(n-1)(\frac{2}{3}Kr_0+r_0^{-1})$$

(the inequality must be understood in the barrier sense, when necessary)

(b) (cf. [H 4,$\S 17$]) Suppose Ric$(x,t_0)\le (n-1)K$ when dist$_{t_0}(x,x_0)<r_0,$ or dist$_{t_0}(x,x_1)<r_0.$ Then

$$\frac{d}{dt}$$dist$$_t(x_0,x_1)\ge -2(n-1)(\frac{2}{3}Kr_0+r_0^{-1}) \$$   at$$\ \ \ t=t_0$$

Proof of Lemma. (a) Clearly, $d_t(x)=\int_{\gamma}{-\mbox{Ric}(X,X)},$ where $\gamma$ is the shortest geodesic between $x$ and $x_0$ and $X$ is its unit tangent vector, On the other hand, $\triangle d\le \sum_{k=1}^{n-1}{s_{Y_k}''(\gamma)},$ where $Y_k$ are vector fields along $\gamma,$ vanishing at $x_0$ and forming an orthonormal basis at $x$ when complemented by $X,$ and $s_{Y_k}''(\gamma)$ denotes the second variation along $Y_k$ of the length of $\gamma.$ Take $Y_k$ to be parallel between $x$ and $x_1,$ and linear between $x_1$ and $x_0,$ where $d(x_1,t_0)=r_0.$ Then

$$\triangle d\le\sum_{k=1}^{n-1}s_{Y_k}''(\gamma)=\int_{r_0}^{d(x,... ...)ds}+\int_0^{r_0} {(\frac{s^2}{r_0^2}(-\mbox{Ric}(X,X))+\frac{n-1}{r_0^2})ds}$$

$$=\int_{\gamma}{-\mbox{Ric}(X,X)} +\int_0^{r_0}{(\mbox{Ric}(X,X)(1-\frac{s^2}{r_0^2})+\frac{n-1}{r_0^2})ds}\le d_t+(n-1)(\frac{2}{3}Kr_0+r_0^{-1})$$

The proof of (b) is similar.

Continuing the proof of theorem, apply the maximum principle to the function $h(y,t)=\phi(d(y,t)-A(2t-1))(\bar{L}(y,1-t)+2n+1),$ where $d(y,t)=$dist$_t(x,x_0),$ and $\phi$ is a function of one variable, equal $1$ on $(-\infty,\frac{1}{20}),$ and rapidly increasing to infinity on $(\frac{1}{20},\frac{1}{10}),$ in such a way that

$$2(\phi ')^2/\phi-\phi ''\ge (2A+100n)\phi '-C(A)\phi,$$ (8.1)

for some constant $C(A)<\infty.$ Note that $\bar{L}+2n+1\ge 1$ for $t\ge\frac{1}{2}$ by the remark in the very end of 7.1. Clearly, min$\ h(y,1)\le h(x,1)=2n+1.$ On the other hand, min$\ h(y,\frac{1}{2})$ is achieved for some $y$ satisfying $d(y,\frac{1}{2})\le \frac{1}{10}.$ Now we compute

$$\Box h=(\bar{L}+2n+1)(-\phi ''+(d_t-\triangle d-2A)\phi ')-2<\nabla\phi\nabla\bar{L}>+(\bar{L}_t-\triangle\bar{L})\phi$$ (8.2)

$$\nabla h=(\bar{L}+2n+1)\nabla\phi+\phi\nabla\bar{L}$$ (8.3)

At a minimum point of $h$ we have $\nabla h=0,$ so (8.2) becomes

$$\Box h=(\bar{L}+2n+1)(-\phi ''+(d_t-\triangle d-2A)\phi '+2(\phi ')^2/\phi)+(\bar{L}_t-\triangle\bar{L})\phi$$ (8.4)

Now since $d(y,t)\ge\frac{1}{20}$ whenever $\phi '\neq 0,$ and since Ric$\le n-1$ in $B(x_0,\frac{1}{20}),$ we can apply our lemma (a) to get $d_t-\triangle d\ge-100(n-1)$ on the set where $\phi '\neq 0.$ Thus, using (8.1) and (7.15), we get

$$\Box h\ge-(\bar{L}+2n+1)C(A)\phi-2n\phi\ge-(2n+C(A))h$$

This implies that min$h$ can not decrease too fast, and we get the required estimate.