Almost nonnegative curvature in dimension three

12.1 Let $\phi$ be a decreasing function of one variable, tending to zero at infinity. A solution to the Ricci flow is said to have $\phi$-almost nonnegative curvature if it satisfies $Rm(x,t)\ge -\phi (R(x,t))R(x,t)$ for each $(x,t).$

Theorem. Given $\epsilon >0,\kappa>0$ and a function $\phi$ as above, one can find $r_0>0$ with the following property. If $g_{ij}(t), 0\le t\le T$ is a solution to the Ricci flow on a closed three-manifold $M,$ which has $\phi$-almost nonnegative curvature and is $\kappa$-noncollapsed on scales $<r_0,$ then for any point $(x_0,t_0)$ with $t_0\ge 1$ and $Q=R(x_0,t_0)\ge r_0^{-2},$ the solution in $\{(x,t):$dist$^2_{t_0}(x,x_0)<(\epsilon Q)^{-1}, t_0-(\epsilon Q)^{-1}\le t\le t_0\}$ is , after scaling by the factor $Q,$ $\epsilon$-close to the corresponding subset of some ancient solution, satisfying the assumptions in 11.1.

Proof. An argument by contradiction. Take a sequence of $r_0$ converging to zero, and consider the solutions $g_{ij}(t),$ such that the conclusion does not hold for some $(x_0,t_0);$ moreover, by tampering with the condition $t_0\ge 1$ a little bit, choose among all such $(x_0,t_0),$ in the solution under consideration, the one with nearly the smallest curvature $Q.$ (More precisely, we can choose $(x_0,t_0)$ in such a way that the conclusion of the theorem holds for all $(x,t),$ satisfying $R(x,t)>2Q, t_0-HQ^{-1}\le t\le t_0,$ where $H\to\infty$ as $r_0\to 0)$ Our goal is to show that the sequence of blow-ups of such solutions at such points with factors $Q$ would converge, along some subsequence of $r_0\to 0,$ to an ancient solution, satisfying 11.1.

Claim 1. For each $(\bar{x},\bar{t})$ with $t_0-HQ^{-1}\le \bar{t}\le t_0$ we have $R(x,t)\le 4\bar{Q}$ whenever $\bar{t}-c\bar{Q}^{-1}\le t\le \bar{t}$ and dist$_{\bar{t}}(x,\bar{x})\le c\bar{Q}^{-\frac{1}{2}},$ where $\bar{Q}=Q+R(\bar{x},\bar{t})$ and $c=c(\kappa)>0$ is a small constant.

Proof of Claim 1. Use the fact ( following from the choice of $(x_0,t_0)$ and the description of the ancient solutions) that for each $(x,t)$ with $R(x,t)>2Q$ and $t_0-HQ^{-1}\le t\le t_0$ we have the estimates $\vert R_t(x,t)\vert\le CR^2(x,t)$, $\vert\nabla R\vert(x,t)\le CR^{\frac{3}{2}}(x,t).$

Claim 2. There exists $c=c(\kappa)>0$ and for any $A>0$ there exist $D=D(A)<\infty , \rho_0=\rho_0(A)>0,$ with the following property. Suppose that $r_0<\rho_0,$ and let $\gamma$ be a shortest geodesic with endpoints $\bar{x},x$ in $g_{ij}(\bar{t}),$ for some $\bar{t}\in[t_0-HQ^{-1},t_0],$ such that $R(y,\bar{t})>2Q$ for each $y\in\gamma.$ Let $z\in\gamma$ satisfy $cR(z,\bar{t})>R(\bar{x},\bar{t})=\bar{Q}.$ Then dist$_{\bar{t}}(\bar{x},z)\ge A\bar{Q}^{-\frac{1}{2}}$ whenever $R(x,\bar{t})\ge D\bar{Q}.$

Proof of Claim 2. Note that from the choice of $(x_0,t_0)$ and the description of the ancient solutions it follows that an appropriate parabolic (backward in time) neighborhood of a point $y\in\gamma$ at $t=\bar{t}$ is $\epsilon$-close to the evolving round cylinder, provided $c^{-1}\bar{Q}\le R(y,\bar{t})\le cR(x,\bar{t})$ for an appropriate $c=c(\kappa).$ Now assume that the conclusion of the claim does not hold, take $r_0$ to zero, $R(x,\bar{t})$ - to infinity, and consider the scalings around $(\bar{x},\bar{t})$ with factors $\bar{Q}.$ We can imagine two possibilities for the behavior of the curvature along $\gamma$ in the scaled metric: either it stays bounded at bounded distances from $\bar{x},$ or not. In the first case we can take a limit (for a subsequence) of the scaled metrics along $\gamma$ and get a nonnegatively curved almost cylindrical metric, with $\gamma$ going to infinity. Clearly, in this case the curvature at any point of the limit does not exceed $c^{-1};$ therefore, the point $z$ must have escaped to infinity, and the conclusion of the claim stands.

In the second case, we can also take a limit along $\gamma;$ it is a smooth nonnegatively curved manifold near $\bar{x}$ and has cylindrical shape where curvature is large; the radius of the cylinder goes to zero as we approach the (first) singular point, which is located at finite distance from $\bar{x};$ the region beyond the first singular point will be ignored. Thus, at $t=\bar{t}$ we have a metric, which is a smooth metric of nonnegative curvature away from a single singular point $o$. Since the metric is cylindrical at points close to $o,$ and the radius of the cylinder is at most $\epsilon$ times the distance from $o,$ the curvature at $o$ is nonnegative in Aleksandrov sense. Thus, the metric near $o$ must be cone-like. In other words, the scalings of our metric at points $x_i\to o$ with factors $R(x_i,\bar{t})$ converge to a piece of nonnegatively curved non-flat metric cone. Moreover, using claim 1, we see that we actually have the convergence of the solutions to the Ricci flow on some time interval, and not just metrics at $t=\bar{t}.$ Therefore, we get a contradiction with the strong maximum principle of Hamilton [H 2].

Now continue the proof of theorem, and recall that we are considering scalings at $(x_0,t_0)$ with factor $Q.$ It follows from claim 2 that at $t=t_0$ the curvature of the scaled metric is bounded at bounded distances from $x_0.$ This allows us to extract a smooth limit at $t=t_0$ (of course, we use the $\kappa$-noncollapsing assumption here). The limit has bounded nonnegative curvature (if the curvatures were unbounded, we would have a sequence of cylindrical necks with radii going to zero in a complete manifold of nonnegative curvature). Therefore, by claim 1, we have a limit not only at $t=t_0,$ but also in some interval of times smaller than $t_0.$

We want to show that the limit actually exists for all $t<t_0.$ Assume that this is not the case, and let $t'$ be the smallest value of time, such that the blow-up limit can be taken on $(t',t_0].$ From the differential Harnack inequality of Hamilton [H 3] we have an estimate $R_t(x,t)\ge -R(x,t)(t-t')^{-1},$ therefore, if $\tilde{Q}$ denotes the maximum of scalar curvature at $t=t_0,$ then $R(x,t)\le \tilde{Q}\frac{t_0-t'}{t-t'}.$ Hence by lemma 8.3(b) dist$_t(x,y)\le$   dist$_{t_0}(x,y)+C$ for all $t,$ where $C=10n(t_0-t')\sqrt{\tilde{Q}}.$

The next step is needed only if our limit is noncompact. In this case there exists $D>0,$ such that for any $y$ satisfying $d=$dist$_{t_0}(x_0,y)>D,$ one can find $x$ satisfying dist$_{t_0}(x,y)=d,$   dist$_{t_0}(x,x_0)>\frac{3}{2}d.$ We claim that the scalar curvature $R(y,t)$ is uniformly bounded for all such $y$ and all $t\in (t',t_0].$ Indeed, if $R(y,t)$ is large, then the neighborhood of $(y,t)$ is like in an ancient solution; therefore, (long) shortest geodesics $\gamma$ and $\gamma_0,$ connecting at time $t$ the point $y$ to $x$ and $x_0$ respectively, make the angle close to 0 or $\pi$ at $y;$ the former case is ruled out by the assumptions on distances, if $D>10C;$ in the latter case, $x$ and $x_0$ are separated at time $t$ by a small neighborhood of $y,$ with diameter of order $R(y,t)^{-\frac{1}{2}},$ hence the same must be true at time $t_0,$ which is impossible if $R(y,t)$ is too large.

Thus we have a uniform bound on curvature outside a certain compact set, which has uniformly bounded diameter for all $t\in (t',t_0].$ Then claim 2 gives a uniform bound on curvature everywhere. Hence, by claim 1, we can extend our blow-up limit past $t'$ - a contradiction.

12.2 Theorem. Given a function $\phi$ as above, for any $A>0$ there exists $K=K(A)<\infty$ 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 with $r_0=1.$ Then $R(x,1)\le K$ whenever dist$_1(x,x_0)<A.$

Proof. In the first step of the proof we check the following

Claim. There exists $K=K(A)<\infty ,$ such that a point $(x,1)$ satisfies the conclusion of the previous theorem 12.1 (for some fixed small $\epsilon >0$), whenever $R(x,1)>K$ and dist$_1(x,x_0)<A.$

The proof of this statement essentially repeats the proof of the previous theorem (the $\kappa$-noncollapsing assumption is ensured by theorem 8.2). The only difference is in the beginning. So let us argue by contradiction, and suppose we have a sequence of solutions and points $x$ with dist$_1(x,x_0)<A$ and $R(x,1)\to\infty,$ which do not satisfy the conclusion of 12.1. Then an argument, similar to the one proving claims 1,2 in 10.1, delivers points $(\bar{x},\bar{t})$ with $\frac{1}{2}\le \bar{t}\le 1,$   dist$_{\bar{t}}(\bar{x},x_0)<2A,$ with $Q=R(\bar{x},\bar{t})\to\infty ,$ and such that $(x,t)$ satisfies the conclusion of 12.1 whenever $R(x,t)>2Q, \bar{t}-DQ^{-1}\le t\le\bar{t},$   dist$_{\bar{t}}(\bar{x},x)<DQ^{-\frac{1}{2}},$ where $D\to\infty.$ (There is a little subtlety here in the application of lemma 8.3(b); nevertheless, it works, since we need to apply it only when the endpoint other than $x_0$ either satisfies the conclusion of 12.1, or has scalar curvature at most $2Q$) After such $(\bar{x},\bar{t})$ are found, the proof of 12.1 applies.

Now, having checked the claim, we can prove the theorem by applying the claim 2 of the previous theorem to the appropriate segment of the shortest geodesic, connecting $x$ and $x_0.$

12.3 Theorem. For any $w>0$ there exist $\tau=\tau(w)>0, K=K(w)<\infty, \rho=\rho(w)>0$ with the following property. Suppose we have a solution $g_{ij}(t)$ to the Ricci flow, defined on $M\times [0,T),$ where $M$ is a closed three-manifold, and a point $(x_0,t_0),$ such that the ball $B(x_0,r_0)$ at $t=t_0$ has volume $\ge wr_0^n,$ and sectional curvatures $\ge -r_0^{-2}$ at each point. Suppose that $g_{ij}(t)$ is $\phi$-almost nonnegatively curved for some function $\phi$ as above. Then we have an estimate $R(x,t)<Kr_0^{-2}$ whenever $t_0\ge 4\tau r_0^2, t\in [t_0-\tau r_0^2,t_0],$   dist$_t(x,x_0)\le \frac{1}{4}r_0,$ provided that $\phi(r_0^{-2})<\rho.$

Proof. If we knew that sectional curvatures are $\ge -r_0^{-2}$ for all $t,$ then we could just apply corollary 11.6(b) (with the remark after its proof) and take $\tau(w)=\tau_0(w)/2, K(w)=C(w)+2B(w)/\tau_0(w).$ Now fix these values of $\tau ,K,$ consider a $\phi$-almost nonnegatively curved solution $g_{ij}(t),$ a point $(x_0,t_0)$ and a radius $r_0>0,$ such that the assumptions of the theorem do hold whereas the conclusion does not. We may assume that any other point $(x',t')$ and radius $r'>0$ with that property has either $t'>t_0$ or $t'<t_0 -2\tau r_0^2,$ or $2r'>r_0.$ Our goal is to show that $\phi(r_0^{-2})$ is bounded away from zero.

Let $\tau '>0$ be the largest time interval such that $Rm(x,t)\ge -r_0^{-2}$ whenever $t\in[t_0-\tau 'r_0^2,t_0],$   dist$_t(x,x_0)\le r_0.$ If $\tau '\ge 2\tau,$ we are done by corollary 11.6(b). Otherwise, by elementary Aleksandrov space theory, we can find at time $t'=t_0-\tau 'r_0^2$ a ball $B(x',r')\subset B(x_0,r_0)$ with $VolB(x',r')\ge \frac{1}{2}\omega_n(r')^n,$ and with radius $r'\ge cr_0$ for some small constant $c=c(w)>0.$ By the choice of $(x_0,t_0)$ and $r_0,$ the conclusion of our theorem holds for $(x',t'),r'.$ Thus we have an estimate $R(x,t)\le K(r')^{-2}$ whenever $t\in [t'-\tau (r')^2,t'],$   dist$_t(x,x')\le \frac{1}{4}r'.$ Now we can apply the previous theorem (or rather its scaled version) and get an estimate on $R(x,t)$ whenever $t\in [t'-\frac{1}{2}\tau (r')^2,t'],$   dist$_t(x',x)\le 10r_0.$ Therefore, if $r_0>0$ is small enough, we have $Rm(x,t)\ge -r_0^{-2}$ for those $(x,t),$ which is a contradiction to the choice of $\tau '.$

12.4 Corollary (from 12.2 and 12.3) Given a function $\phi$ as above, for any $w>0$ one can find $\rho>0$ such that if $g_{ij}(t)$ is a $\phi$-almost nonnegatively curved solution to the Ricci flow, defined on $M\times [0,T),$ where $M$ is a closed three-manifold, and if $B(x_0,r_0)$ is a metric ball at time $t_0\ge 1,$ with $r_0<\rho,$ and such that $\min Rm(x,t_0)$ over $x\in B(x_0,r_0)$ is equal to $-r_0^{-2},$ then $VolB(x_0,r_0)\le wr_0^n.$