Long time behavior I

6.1 Let us summarize what we have achieved so far. We have shown the existence of decreasing (piecewise constant) positive functions $r(t)$ and $\bar{\delta }(t)$ (which we may assume converging to zero at infinity), such that if $(M,g_{ij})$ is a normalized manifold, and $0<\delta (t)<\bar{\delta }(t),$ then there exists a solution to the Ricci flow with $\delta (t)$-cutoff on the time interval $[0,+\infty],$ starting from $(M,g_{ij})$ and satisfying on each subinterval $[0,t]$ the canonical neighborhood assumption with parameter $r(t),$ as well as the pinching estimate (5.1).

In particular, if the initial data has positive scalar curvature, say $R\ge a>0,$ then the solution becomes extinct in time at most $\frac{3}{2a},$ and it follows that $M$ in this case is diffeomorphic to a connected sum of several copies of $\mathbb{S}^2\times\mathbb{S}^1$ and metric quotients of round $\mathbb{S}^3.$ ( The topological description of 3-manifolds with positive scalar curvature modulo quotients of homotopy spheres was obtained by Schoen-Yau and Gromov-Lawson more than 20 years ago, see [G-L] for instance; in particular, it is well known and easy to check that every manifold that can be decomposed in a connected sum above admits a metric of positive scalar curvature.) Moreover, if the scalar curvature is only nonnegative, then by the strong maximum principle it instantly becomes positive unless the metric is (Ricci-)flat; thus in this case, we need to add to our list the flat manifolds.

However, if the scalar curvature is negative somewhere, then we need to work more in order to understand the long tome behavior of the solution. To achieve this we need first to prove versions of Theorems I.12.2 and I.12.3 for solutions with cutoff.

6.2 Correction to Theorem I.12.2. Unfortunately, the statement of Theorem I.12.2 was incorrect. The assertion I had in mind is as follows:

Given a function $\phi$ as above, for any $A<\infty$ there exist $K=K(A)<\infty$ and $\rho=\rho(A)>0$ with the following property. Suppose in dimension three we have a solution to the Ricci flow with $\phi$-almost nonnegative curvature, which satisfies the assumptions of theorem 8.2 for some $x_0,r_0$ with $\phi(r_0^{-2})<\rho.$ Then $R(x,r_0^2)\le Kr_0^{-2}$ whenever $\mathrm{dist}_{r_0^2}(x,x_0)<Ar_0.$

It is this assertion that was used in the proof of Theorem I.12.3 and Corollary I.12.4.

6.3 Proposition. For any $A<\infty$ one can find $\kappa =\kappa (A)>0, K_1=K_1(A)<\infty ,K_2=K_2(A)<\infty , \bar{r}=\bar{r}(A)>0,$ such that for any $t_0<\infty$ there exists $\bar{\delta }=\bar{\delta }_A(t_0)>0,$ decreasing in $t_0,$ with the following property. Suppose we have a solution to the Ricci flow with $\delta (t)$-cutoff on time interval $[0,T],\ \ \delta (t)<\bar{\delta }(t)$ on $[0,T],\ \delta (t)<\bar{\delta }$ on $[t_0/2,t_0] ,$ with normalized initial data; assume that the solution is defined in the whole parabolic neighborhood $P(x_0,t_0,r_0,-r_0^2), \ 2r_0^2<t_0,$ and satisfies $\vert Rm\vert\le r_0^{-2}$ there, and that the volume of the ball $B(x_0,t_0,r_0)$ is at least $A^{-1}r_0^3.$ Then

(a) The solution is $\kappa$-noncollapsed on the scales less than $r_0$ in the ball $B(x_0,t_0,Ar_0).$

(b) Every point $x\in B(x_0,t_0,Ar_0)$ with $R(x,t_0)\ge K_1 r_0^{-2}$ has a canonical neighborhood as in 4.1.

(c) If $r_0\le \bar{r}\sqrt{t_0}$ then $R\le K_2r_0^{-2}$ in $B(x_0,t_0,Ar_0).$

Proof. (a) This is an analog of Theorem I.8.2. Clearly we have $\kappa$-noncollapsing on the scales less than $r(t_0),$ so we may assume $r(t_0)\le r_0\le \sqrt{t_0/2}$ , and study the scales $\rho, r(t_0)\le \rho\le r_0.$ In particular, for fixed $t_0$ we are interested in the scales, uniformly equivalent to one.

So assume that $x\in B(x_0,t_0,Ar_0)$ and the solution is defined in the whole $P(x,t_0,\rho,-\rho^2)$ and satisfies $\vert Rm\vert\le \rho^{-2}$ there. An inspection of the proof of I.8.2 shows that in order to make the argument work it suffices to check that for any barely admissible curve $\gamma$, parametrized by $t\in [t_{\gamma},t_0], t_0-r_0^2\le t_{\gamma}\le t_0,$ such that $\gamma(t_0)=x,$ we have an estimate

$$2\sqrt{t_0-t_{\gamma}} \int_{t_{\gamma}}^{t_0}{\sqrt{t_0-t}(R(\gamma(t),t)+\vert\dot{\gamma}(t)\vert^2)dt}\ge C(A)r_0^2$$ (6.1)

for a certain function $C(A)$ that can be made explicit. Now we would like to conclude the proof by using Lemma 5.3. However, unlike the situation in Lemma 5.2, here Lemma 5.3 provides the estimate we need only if $t_0-t_{\gamma}$ is bounded away from zero, and otherwise we only get an estimate $\rho^2$ in place of $C(A)r_0^2.$ Therefore we have to return to the proof of I.8.2.

Recall that in that proof we scaled the solution to make $r_0=1$ and worked on the time interval $[1/2,1].$ The maximum principle for the evolution equation of the scalar curvature implies that on this time interval we have $R\ge -3.$ We considered a function of the form $h(y,t)=\phi(\hat{d}(y,t))\hat{L}(y,\tau),$ where $\phi$ is a certain cutoff function, $\tau=1-t, \hat{d}(y,t)=\mathrm{dist}_t(x_0,y)-A(2t-1), \hat{L}(y,\tau)=\bar{L}(y,\tau)+7,$ and $\bar{L}$ was defined in [I,(7.15)]. Now we redefine $\hat{L},$ taking $\hat{L}(y,\tau)=\bar{L}(y,\tau)+2\sqrt{\tau}.$ Clearly, $\hat{L}>0$ because $R\ge -3$ and $2\sqrt{\tau}>4\tau^2$ for $0<\tau\le 1/2.$ Then the computations and estimates of I.8.2 yield

$$\Box h\ge -C(A)h-(6+\frac{1}{\sqrt{\tau}})\phi$$

Now denoting by $h_0(\tau)$ the minimum of $h(y,1-t),$ we can estimate

$$\frac{d}{d\tau}(\mathrm{log}(\frac{h_0(\tau)}{\sqrt{\tau}}))\le C... ...}+1}{2\tau-4\tau^2\sqrt{\tau}}-\frac{1}{2\tau}\le C(A)+ \frac{50}{\sqrt{\tau}},$$ (6.2)

whence

$$h_0(\tau)\le \sqrt{\tau}\ \mathrm{exp}(C(A)\tau +100\sqrt{\tau}),$$ (6.3)

because the left hand side of (6.2) tends to zero as $\tau\to 0+.$

Now we can return to our proof, replace the right hand side of (6.1) by the right hand side of (6.3) times $r_0^2,$ with $\tau=r_0^{-2}(t_0-t_{\gamma}),$ and apply Lemma 5.3.

(b) Assume the contrary, take a sequence $K_1^{\alpha }\to\infty$ and consider the solutions violating the statement. Clearly, $K_1^{\alpha } (r_0^{\alpha })^{-2}<(r(t_0^{\alpha }))^{-2},$ whence $t_0^{\alpha }\to\infty;$

When $K_1$ is large enough, we can, arguing as in the proof of Claim 1 in [I.10.1], find a point $(\bar{x},\bar{t}), x\in B(x_0,\bar{t},2Ar_0), \bar{t}\in [t_0-r_0^2/2,t_0],$ such that $\bar{Q}=R(\bar{x},\bar{t})>K_1r_0^{-2},\ \ (\bar{x},\bar{t})$ does not satisfy the canonical neighborhood assumption, but each point $(x,t)\in \bar{P}$ with $R(x,t)\ge 4\bar{Q}$ does, where $\bar{P}$ is the set of all $(x,t)$ satisfying $\bar{t}-\frac{1}{4}K_1\bar{Q}^{-1}\le t\le \bar{t},\ \ \mathrm{dist}_{t}(x_0... ...\mathrm{dist}_{\bar{t}}(x_0,\bar{x})+K_1^{\frac{1}{2}} \bar{Q}^{-\frac{1}{2}}.$ (Note that $\bar{P}$ is not a parabolic neighborhood.) Clearly we can use (a) with slightly different parameters to ensure $\kappa$-noncollapsing in $\bar{P}.$

Now we apply the argument from 5.4. First, by Claim 2 in 4.2, for any $\bar{A}<\infty$ we have an estimate $R\le Q(\bar{A})\bar{Q}$ in $B(\bar{x},\bar{t},\bar{A}\bar{Q}^{-\frac{1}{2}})$ when $K_1$ is large enough; therefore we can take a limit as $\alpha \to\infty$ of scalings with factor $\bar{Q}$ about $(\bar{x},\bar{t}),$ shifting the time $\bar{t}$ to zero; the limit at time zero would be a smooth complete nonnegatively curved manifold. Next we observe that this limit has curvature uniformly bounded, say, by $Q_0,$ and therefore, for each fixed $\bar{A}$ and for sufficiently large $K_1,$ the parabolic neighborhood $P(\bar{x},\bar{t},\bar{A}\bar{Q}^{-\frac{1}{2}},-\epsilon \eta^{-1}Q_0^{-1}\bar{Q}^{-1})$ is contained in $\bar{P}.$ (Here we use the estimate of distance change, given by Lemma I.8.3(a).) Thus we can take a limit on the interval $[-\epsilon \eta^{-1}Q_0^{-1},0].$ (The possibility of surgeries is ruled out as in 5.4) Then we repeat the procedure indefinitely, getting an ancient $\kappa$-solution in the limit, which means a contradiction.

(c) If $x\in B(x_0,t_0,Ar_0)$ has very large curvature, then on the shortest geodesic $\gamma$ at time $t_0,$ that connects $x_0$ and $x,$ we can find a point $y,$ such that $R(y,t_0)=K_1(A)r_0^{-2}$ and the curvature is larger at all points of the segment of $\gamma$ between $x$ and $y.$ Then our statement follows from Claim 2 in 4.2, applied to this segment.

From now on we redefine the function $\bar{\delta }(t)$ to be $\mathrm{min}(\bar{\delta }(t),\bar{\delta }_{2t}(2t)),$ so that the proposition above always holds for $A=t_0.$

6.4 Proposition. There exist $\tau>0, \bar{r}>0, K<\infty$ with the following property. Suppose we have a solution to the Ricci flow with $\delta (t)$-cutoff on the time interval $[0,t_0],$ with normalized initial data. Let $r_0,t_0$ satisfy $2C_1h\le r_0\le \bar{r}\sqrt{t_0},$ where $h$ is the maximal cutoff radius for surgeries in $[t_0/2,t_0] ,$ and assume that the ball $B(x_0,t_0,r_0)$ has sectional curvatures at least $-r_0^{-2}$ at each point, and the volume of any subball $B(x,t_0,r)\subset B(x_0,t_0,r_0)$ with any radius $r>0$ is at least $(1-\epsilon )$ times the volume of the euclidean ball of the same radius. Then the solution is defined in $P(x_0,t_0,r_0/4,-\tau r_0^2)$ and satisfies $R< Kr_0^{-2}$ there.

Proof. Let us first consider the case $r_0\le r(t_0).$ Then clearly $R(x_0,t_0)\le C_1^2r_0^{-2},$ since an $\epsilon$-neck of radius $r$ can not contain an almost euclidean ball of radius $\ge r.$ Thus we can take $K=2C_1^2, \tau=\epsilon \eta^{-1}C_1^{-2}$ in this case, and since $r_0\ge 2C_1h,$ the surgeries do not interfere in $P(x_0,t_0,r_0/4,-\tau r_0^2).$

In order to handle the other case $r(t_0)<r_0\le\bar{r}\sqrt{t_0}$ we need a couple of lemmas.

6.5 Lemma. There exist $\tau_0>0$ and $K_0<\infty,$ such that if we have a smooth solution to the Ricci flow in $P(x_0,0,1,-\tau),\tau\le\tau_0,$ having sectional curvatures at least $-1$, and the volume of the ball $B(x_0,0,1)$ is at least $(1-\epsilon )$ times the volume of the euclidean unit ball, then

(a) $R\le K_0\tau^{-1}$ in $P(x_0,0,1/4,-\tau/2),$ and

(b) the ball $B(x_0,1/4,-\tau)$ has volume at least $\frac{1}{10}$ times the volume of the euclidean ball of the same radius.

The proof can be extracted from the proof of Lemma I.11.6.

6.6 Lemma. For any $w>0$ there exists $\theta_0=\theta_0(w)>0,$ such that if $B(x,1)$ is a metric ball of volume at least $w,$ compactly contained in a manifold without boundary with sectional curvatures at least $-1,$ then there exists a ball $B(y,\theta_0)\subset B(x,1),$ such that every subball $B(z,r)\subset B(y,\theta_0)$ of any radius $r$ has volume at least $(1-\epsilon )$ times the volume of the euclidean ball of the same radius.

This is an elementary fact from the theory of Aleksandrov spaces.

6.7 Now we continue the proof of the proposition. We claim that one can take $\tau=\mathrm{min}(\tau_0/2, \epsilon \eta^{-1}C_1^{-2}), K=\mathrm{max}(2K_0\tau^{-1}, 2C_1^2).$ Indeed, assume the contrary, and take a sequence of $\bar{r}^{\alpha }\to 0$ and solutions, violating our assertion for the chosen $\tau,K.$ Let $t_0^{\alpha }$ be the first time when it is violated, and let $B(x_0^{\alpha },t_0^{\alpha },r_0^{\alpha })$ be the counterexample with the smallest radius. Clearly $r_0^{\alpha }>r(t_0^{\alpha })$ and $(r_0^{\alpha })^2 (t_0^{\alpha })^{-1}\to 0$ as $\alpha \to\infty.$

Consider any ball $B(x_1,t_0,r)\subset B(x_0,t_0,r_0), r<r_0.$ Clearly we can apply our proposition to this ball and get the solution in $P(x_1,t_0,r/4,-\tau r^2)$ with the curvature bound $R<Kr^{-2}.$ Now if $r_0^2 t_0^{-1}$ is small enough, then we can apply proposition 6.3(c) to get an estimate $R(x,t)\le K'(A)r^{-2}$ for $(x,t)$ satisfying $t\in [t_0-\tau r^2/2,t_0], \mathrm{dist}_t(x,x_1)<Ar,$ for some function $K'(A)$ that can be made explicit. Let us choose $A=100r_0r^{-1};$ then we get the solution with a curvature estimate in $P(x_0,t_0,r_0,-\triangle t),$ where $\triangle t=K'(A)^{-1}r^2.$ Now the pinching estimate implies $Rm\ge -r_0^{-2}$ on this set, if $r_0^2 t_0^{-1}$ is small enough while $rr_0^{-1}$ is bounded away from zero. Thus we can use lemma 6.5(b) to estimate the volume of the ball $B(x_0,t_0-\triangle t,r_0/4)$ by at least $\frac{1}{10}$ of the volume of the euclidean ball of the same radius, and then by lemma 6.6 we can find a subball $B(x_2,t_0-\triangle t, \theta_0(\frac{1}{10})r_0/4),$ satisfying the assumptions of our proposition. Therefore, if we put $r=\theta_0(\frac{1}{10})r_0/4,$ then we can repeat our procedure as many times as we like, until we reach the time $t_0-\tau_0 r_0^2$, when the lemma 6.5(b) stops working. But once we reach this time, we can apply lemma 6.5(a) and get the required curvature estimate, which is a contradiction.

6.8 Corollary. For any $w>0$ one can find $\tau=\tau(w)>0, K=K(w)<\infty, \bar{r}=\bar{r}(w)>0, \theta=\theta(w)>0$ with the following property. Suppose we have a solution to the Ricci flow with $\delta (t)$-cutoff on the time interval $[0,t_0],$ with normalized initial data. Let $t_0,r_0$ satisfy $\theta^{-1}(w)h\le r_0\le \bar{r}\sqrt{t_0},$ and assume that the ball $B(x_0,t_0,r_0)$ has sectional curvatures at least $-r_0^2$ at each point, and volume at least $wr_0^3.$ Then the solution is defined in $P(x_0,t_0,r_0/4,-\tau r_0^2)$ and satisfies $R< Kr_0^{-2}$ there.

Indeed, we can apply proposition 6.4 to a smaller ball, provided by lemma 6.6, and then use proposition 6.3(c).