Ricci flow with cutoff

4.1 Suppose we are given a collection of smooth solutions $g_{ij}(t)$ to the Ricci flow, defined on $M_k\times[t_k^-,t_k^+),$ which go singular as $t\to t_k^+.$ Let $(\Omega _k, \bar{g}_{ij}^k)$ be the limits of the corresponding solutions as $t\to t_k^+,$ as in the previous section. Suppose also that for each $k$ we have $t_k^- =t_{k-1}^+,$ and $(\Omega _{k-1},\bar{g}_{ij}^{k-1})$ and $(M_k,g_{ij}^k(t_k^-))$ contain compact (possibly disconnected) three-dimensional submanifolds with smooth boundary, which are isometric. Then we can identify these isometric submanifolds and talk about the solution to the Ricci flow with surgery on the union of all $[t_k^-,t_k^+).$

Fix a small number $\epsilon >0$ which is admissible in sections 1,2. In this section we consider only solutions to the Ricci flow with surgery, which satisfy the following a priori assumptions:

(pinching) There exists a function $\phi,$ decreasing to zero at infinity, such that $Rm\ge -\phi(R)R,$

(canonical neighborhood) There exists $r>0,$ such that every point where scalar curvature is at least $r^{-2}$ has a neighborhood, satisfying the conclusions of 1.5. (In particular, this means that if in case (a) the neighborhood in question is $B(x_0,t_0,\epsilon ^{-1}r_0),$ then the solution is required to be defined in the whole $P(x_0,t_0,\epsilon ^{-1}r_0,-r_0^2);$ however, this does not rule out a surgery in the time interval $(t_0-r_0^2,t_0),$ that occurs sufficiently far from $x_0$.)

Recall that from the pinching estimate of Ivey and Hamilton, and Theorem I.12.1, we know that the a priori assumptions above hold for a smooth solution on any finite time interval. For Ricci flow with surgery they will be justified in the next section.

4.2 Claim 1. Suppose we have a solution to the Ricci flow with surgery, satisfying the canonical neighborhood assumption, and let $Q=R(x_0,t_0)+r^{-2}.$ Then we have estimate $R(x,t)\le 8Q$ for those $(x,t)\in P(x_0,t_0,\frac{1}{2}\eta^{-1}Q^{-\frac{1}{2}},-\frac{1}{8}\eta^{-1}Q^{-1}),$ for which the solution is defined.

Indeed, this follows from estimates (1.3).

Claim 2. For any $A<\infty$ one can find $Q=Q(A)<\infty$ and $\xi=\xi(A)>0$ with the following property. Suppose we have a solution to the Ricci flow with surgery, satisfying the pinching and the canonical neighborhood assumptions. Let $\gamma$ be a shortest geodesic in $g_{ij}(t_0)$ with endpoints $x_0$ and $x,$ such that $R(y,t_0)>r^{-2}$ for each $y\in\gamma ,$ and $Q_0=R(x_0,t_0)$ is so large that $\phi(Q_0)<\xi.$ Finally, let $z\in \gamma$ be any point satisfying $R(z,t_0)>10C_2R(x_0,t_0).$ Then $\mathrm{dist}_{t_0}(x_0,z)\ge AQ_0^{-\frac{1}{2}}$ whenever $R(x,t_0)>QQ_0.$

The proof is exactly the same as for Claim 2 in Theorem I.12.1; in the very end of it, when we get a piece of a non-flat metric cone as a blow-up limit, we get a contradiction to the canonical neighborhood assumption, because the canonical neighborhoods of types other than (a) are not close to a piece of metric cone, and type (a) is ruled out by the strong maximum principle, since the $\epsilon$-neck in question is strong.

4.3 Suppose we have a solution to the Ricci flow with surgery, satisfying our a priori assumptions, defined on $[0,T),$ and going singular at time $T.$ Choose a small $\delta >0$ and let $\rho=\delta r.$ As in the previous section, consider the limit $(\Omega ,\bar{g}_{ij})$ of our solution as $t\to T,$ and the corresponding compact set $\Omega _{\rho}.$

Lemma. There exists a radius $h, 0<h<\delta \rho,$ depending only on $\delta ,\rho$ and the pinching function $\phi,$ such that for each point $x$ with $h(x)=\bar{R}^{-\frac{1}{2}}(x)\le h$ in an $\epsilon$-horn of $(\Omega ,\bar{g}_{ij})$ with boundary in $\Omega _{\rho} ,$ the neighborhood $P(x,T,\delta ^{-1}h(x),-h^2(x))$ is a strong $\delta$-neck.

Proof. An argument by contradiction. Assuming the contrary, take a sequence of solutions with limit metrics $(\Omega ^{\alpha },\bar{g}_{ij}^{\alpha })$ and points $x^{\alpha }$ with $h(x^{\alpha })\to 0.$ Since $x^{\alpha }$ lies deeply inside an $\epsilon$-horn, its canonical neighborhood is a strong $\epsilon$-neck. Now Claim 2 gives the curvature estimate that allows us to take a limit of appropriate scalings of the metrics $g_{ij}^{\alpha }$ on $[T-h^2(x^{\alpha }),T]$ about $x^{\alpha },$ for a subsequence of $\alpha \to\infty.$ By shifting the time parameter we may assume that the limit is defined on $[-1,0].$ Clearly, for each time in this interval, the limit is a complete manifold with nonnegative sectional curvature; moreover, since $x^{\alpha }$ was contained in an $\epsilon$-horn with boundary in $\Omega ^{\alpha }_{\rho},$ and $h(x^{\alpha })/\rho\to 0,$ this manifold has two ends. Thus, by Toponogov, it admits a metric splitting $\mathbb{S}^2\times\mathbb{R}.$ This implies that the canonical neighborhood of the point $(x^{\alpha },T-h^2(x^{\alpha }))$ is also of type (a), that is a strong $\epsilon$-neck, and we can repeat the procedure to get the limit, defined on $[-2,0],$ and so on. This argument works for the limit in any finite time interval $[-A,0],$ because $h(x^{\alpha })/\rho\to 0.$ Therefore, we can construct a limit on $[-\infty,0];$ hence it is the round cylinder, and we get a contradiction.

4.4 Now we can specialize our surgery and define the Ricci flow with $\delta$-cutoff. Fix $\delta >0,$ compute $\rho=\delta r$ and determine $h$ from the lemma above. Given a smooth metric $g_{ij}$ on a closed manifold, run the Ricci flow until it goes singular at some time $t^+;$ form the limit $(\Omega ,\bar{g}_{ij}).$ If $\Omega _{\rho}$ is empty, the procedure stops here, and we say that the solution became extinct. Otherwise we remove the components of $\Omega$ which contain no points of $\Omega _{\rho} ,$ and in every $\epsilon$-horn of each of the remaining components we find a $\delta$-neck of radius $h,$ cut it along the middle two-sphere, remove the horn-shaped end, and glue in an almost standard cap in such a way that the curvature pinching is preserved and a metric ball of radius $(\delta ')^{-1}h$ centered near the center of the cap is, after scaling with factor $h^{-2},$ $\ \ \delta '$-close to the corresponding ball in the standard capped infinite cylinder, considered in section 2. (Here $\ \ \delta '$ is a function of $\delta$ alone, which tends to zero with $\delta .$)

The possibility of capping a $\delta$-neck preserving a certain pinching condition in dimension four was proved by Hamilton [H 5,§4]; his argument works in our case too (and the estimates are much easier to verify). The point is that we can change our $\delta$-neck metric near the middle of the neck by a conformal factor $e^{-f},$ where $f=f(z)$ is positive on the part of the neck we want to remove, and zero on the part we want to preserve, and $z$ is the coordinate along $\mathbb{I}$ in our parametrization $\mathbb{S}^2\times\mathbb{I}$ of the neck. Then, in the region near the middle of the neck, where $f$ is small, the dominating terms in the formulas for the change of curvature are just positive constant multiples of $f'',$ so the pinching improves, and all the curvatures become positive on the set where $f>\delta '.$

Now we can continue our solution until it becomes singular for the next time. Note that after the surgery the manifold may become disconnected; in this case, each component should be dealt with separately. Furthermore, let us agree to declare extinct every component which is $\epsilon$-close to a metric quotient of the round sphere; that allows to exclude such components from the list of canonical neighborhoods. Now since every surgery reduces the volume by at least $h^3,$ the sequence of surgery times is discrete, and, taking for granted the a priori assumptions, we can continue our solution indefinitely, not ruling out the possibility that it may become extinct at some finite time.

4.5 In order to justify the canonical neighborhood assumption in the next section, we need to check several assertions.

Lemma. For any $A<\infty ,0< \theta<1,$ one can find $\bar{\delta }=\bar{\delta }(A,\theta)$ with the following property. Suppose we have a solution to the Ricci flow with $\delta$-cutoff, satisfying the a priori assumptions on $[0,T],$ with $\delta <\bar{\delta }.$ Suppose we have a surgery at time $T_0\in (0,T),$ let $p$ correspond to the center of the standard cap, and let $T_1=\mathrm{min}(T,T_0+\theta h^2).$ Then either

(a) The solution is defined on $P(p,T_0,Ah,T_1-T_0),$ and is, after scaling with factor $h^{-2}$ and shifting time $T_0$ to zero, $A^{-1}$-close to the corresponding subset on the standard solution from section 2, or

(b) The assertion (a) holds with $T_1$ replaced by some time $t^+\in [T_0,T_1),$ where $t^+$ is a surgery time; moreover, for each point in $B(p,T_0,Ah),$ the solution is defined for $t\in [T_0,t^+)$ and is not defined past $t^+.$

Proof. Let $Q$ be the maximum of the scalar curvature on the standard solution in the time interval $[0,\theta],$ let $\triangle t=N^{-1}(T_1-T_0)<\epsilon \eta^{-1}Q^{-1}h^2,$ and let $t_k=T_0+k\triangle t, k=0,...,N.$

Assume first that for each point in $B(p,T_0,A_0h),$ where $A_0=\epsilon (\delta ')^{-1},$ the solution is defined on $[t_0,t_1].$ Then by (1.3) and the choice of $\triangle t$ we have a uniform curvature bound on this set for $h^{-2}$-scaled metric. Therefore we can define $A_1,$ depending only on $A_0$ and tending to infinity with $A_0,$ such that the solution in $P(p,T_0,A_1h,t_1-t_0)$ is, after scaling and time shifting, $A_1^{-1}$-close to the corresponding subset in the standard solution. In particular, the scalar curvature on this subset does not exceed $2Qh^{-2}.$ Now if for each point in $B(p,T_0,A_1h)$ the solution is defined on $[t_1,t_2],$ then we can repeat the procedure, defining $A_2$ etc. Continuing this way, we eventually define $A_N,$ and it would remain to choose $\delta$ so small, and correspondingly $A_0$ so large, that $A_N>A.$

Now assume that for some $k, 0\le k<N,$ and for some $x\in B(p,T_0,A_kh)$ the solution is defined on $[t_0,t_k]$ but not on $[t_k,t_{k+1}].$ Then we can find a surgery time $t^+\in[t_k,t_{k+1}],$ such that the solution on $B(p,T_0,A_kh)$ is defined on $[t_0,t^+),$ but for some points of this ball it is not defined past $t^+.$ Clearly, the $A_{k+1}^{-1}$-closeness assertion holds on $P(p,T_0,A_{k+1}h,t^+-T_0).$ On the other hand, the solution on $B(p,T_0,A_kh)$ is at least $\epsilon$-close to the standard one for all $t\in [t_k,t^+),$ hence no point of this set can be the center of a $\delta$-neck neighborhood at time $t^+.$ However, the surgery is always done along the middle two-sphere of such a neck. It follows that for each point of $B(p,T_0,A_kh)$ the solution terminates at $t^+.$

4.6 Corollary. For any $l<\infty$ one can find $A=A(l)<\infty$ and $\theta=\theta(l), 0<\theta<1,$ with the following property. Suppose we are in the situation of the lemma above, with $\delta <\bar{\delta }(A,\theta).$ Consider smooth curves $\gamma$ in the set $B(p,T_0,Ah),$ parametrized by $t\in [T_0,T_{\gamma}],$ such that $\gamma(T_0)\in B(p,T_0,Ah/2)$ and either $T_{\gamma}=T_1<T$, or $T_{\gamma}<T_1$ and $\gamma(T_{\gamma})\in\partial B(p,T_0,Ah).$ Then $\int_{T_0}^{T_{\gamma}}{(R(\gamma(t),t)+\vert\dot{\gamma}(t)\vert^2)dt}>l$.

Proof. Indeed, if $T_{\gamma}=T_1,$ then on the standard solution we would have $\int_{T_0}^{T_{\gamma}}R(\gamma(t),t)dt \ge\mathrm{const}\int_0^{\theta}(1-t)^{-1}dt=-\mathrm{const}\cdot (\mathrm{log}(1-\theta))^{-1},$ so by choosing $\theta$ sufficiently close to one we can handle this case. Then we can choose $A$ so large that on the standard solution $\mathrm{dist}_t(p,\partial B(p,0,A))\ge 3A/4$ for each $t\in [0,\theta].$ Now if $\gamma(T_{\gamma})\in\partial B(p,T_0,Ah)$ then $\int_{T_0}^{T_{\gamma}}\vert\dot{\gamma}(t)\vert^2dt\ge A^2/100,$ so by taking $A$ large enough, we can handle this case as well.

4.7 Corollary. For any $Q<\infty$ there exists $\theta=\theta(Q), 0<\theta<1$ with the following property. Suppose we are in the situation of the lemma above, with $\delta <\bar{\delta }(A,\theta), A>\epsilon ^{-1}.$ Suppose that for some point $x\in B(p,T_0,Ah)$ the solution is defined at $x$ (at least) on $[T_0,T_x], T_x\le T,$ and satisfies $Q^{-1}R(x,t)\le R(x,T_x)\le Q(T_x-T_0)^{-1}$ for all $t\in [T_0,T_x].$ Then $T_x\le T_0 + \theta h^2.$

Proof. Indeed, if $\ \ \ T_x>T_0+\theta h^2,$ then by lemma $\ \ \ R(x,T_0+\theta h^2)\ge \mathrm{const}\cdot (1-\theta)^{-1}h^{-2},$ whence $R(x,T_x)\ge \mathrm{const}\cdot Q^{-1}(1-\theta)^{-1}h^{-2},$ and $T_x-T_0\le \mathrm{const}\cdot Q^2(1-\theta)h^2<\theta h^2$ if $\theta$ is close enough to one.