Expanding solitons with non-negative curvature operator coming out of cones

Abstract

We consider Ricci flow of complete Riemannian manifolds which have bounded non-negative curvature operator, non-zero asymptotic volume ratio and no boundary. We prove scale invariant estimates for these solutions. Using these estimates, we show that there is a limit solution, obtained by scaling down this solution at a fixed point in space. This limit solution is an expanding soliton coming out of the asymptotic cone at infinity.

Appendices

Appendix A: de Rham splitting

In this appendix, we explain some known splitting results, which follow from the de Rham Splitting Theorem. We give proofs for the reader’s convenience. We follow essentially the argument given in the Proof of Theorem 2.1 of [5] which follows closely that of Lemma 8.2/Theorem 8.3 of [19].

Theorem 7.1

Let ${h(t)}_{t \in [0,T)}$ be a smooth (in space and time) bounded family of symmetric two tensors defined on a simply connected complete manifold $M^n$ without boundary, satisfying the evolution equation

$$\begin{aligned} {\partial \over {\partial t}} h_{ij}= {}^{g(t)}\Delta h_{ij} + \phi _{ij} \end{aligned}$$

(7.1)

where $h(x,t) \ge 0$ for all $(x,t) \in M \times [0,T)$ and $\phi (x,t)(v,v) \ge 0$ for all $(x,t) \in M \times [0,T)$ and for all $v \in T_x M$ which satisfy $h(x,t)(v,v) =0$. We also assume $g$ is a smooth family of metrics (in space and time) satisfying $^{g_0}|D^{(i,j)} g| + {}^{g_0}|D^{(i,j)} h| \le k(i,j)< \infty$ everywhere, where $i,j \in \mathbb{N }$ and $D^{(i,j)}$ refers to taking $i$ time derivatives and $j$ covariant derivatives with respect to $g_0$, and $k(i,j) \in \mathbb{R }$ are constants. Then for all $x \in M, t>0$, the null space of $h(x,t)$ is invariant under parallel translation and constant in time. There is a splitting, $(M,g(t))= (N \times P, r(t) \oplus l(t))$, where $r,l$ are smooth families of Riemannian metrics such that $h>0$ on $N$ (as a two tensor), and $h = 0$ on $P$.

Proof

Let $0 \le \sigma _1(x,t) \le \sigma _2(x,t) \le \ldots \le \sigma _n(x,t)$ be the eigenvalues of $h(x,t)$. Assume that $\sigma _1(x_0,t_0) + \sigma _2(x_0,t_0) + \ldots + \sigma _k(x_0,t_0) >0$ at some point $x_0$ and some time $t_0$. Define a smooth function $\eta _{t_0}:M \rightarrow \mathbb{R }^+_0$ which is positive at $x_0$ and zero outside of $B_1(x_0,0)$ (the ball in $M$ of radius one with respect to $g_0$), and satisfies $\sigma _1(\cdot ,t_0) + \sigma _2(\cdot ,t_0) + \ldots + \sigma _k(\cdot ,t_0) > \eta _{t_0}(\cdot ).$ Solve the Dirichlet problem:

$$\begin{aligned} {\partial \over {\partial t}} \eta _i&= {} ^{g(t)} \Delta \eta _i\nonumber \\ \eta _i(\cdot ,t) |_{\partial B_i(x_0,0)}&= 0 \ \ \forall \ t \in [t_0,T)\nonumber \\ \eta _i(\cdot ,t_0)&= \eta _{t_0}(\cdot ) \end{aligned}$$

(7.2)

Using the estimates for $g$ we see that the solutions exist for all time and all satisfy interior estimates independent of $i$ (see for example Theorem 10.1, chapter IV, §10 in [24]), and we may take a subsequence to obtain a smooth solution $\eta : M \times [t_0,T) \rightarrow \mathbb{R }$ of the equation

$$\begin{aligned} {\partial \over {\partial t}} \eta&= {}^{g(t)} \Delta \eta \nonumber \\ \eta (\cdot ,t_0)&= \eta _{t_0}(\cdot ) \end{aligned}$$

(7.3)

From the strong maximum principle, $\eta (\cdot ,t) >0$ for all $t> t_0$. Also, the construction and the estimates on $g$ guarantee that $\sup _{(M \backslash B_i(x_0,0)) \times [t_0,S]} |\eta (\cdot ,t)| \rightarrow 0$ as $i \rightarrow \infty$. for all $S < T$. We claim that $\sigma _1(\cdot ,t) + \cdots + \sigma _k(\cdot ,t) \ge \eta (\cdot ,t)$ for all $t\ge t_0$. One proves first, that $\sigma _1(\cdot ,t) +\cdots + \sigma _k(\cdot ,t) -\eta (\cdot ,t) + \varepsilon e^{\rho ^2(\cdot ,t)(1+at) + at} \ge 0$ for arbitrary small $\varepsilon >0$ and an appropriately chosen constant $a$, where here $\rho (x,t)= \text{ dist }(x,x_0,t)$ ($a>0$ does not depend on $\varepsilon$: $a$ depends on the constants in the statement of the Theorem). This is done by using the maximum principle. See for example the argument in the Proof of Lemma 5.1 in [31] for details. Now let $\varepsilon$ go to zero. This implies $\sigma _1(\cdot ,t) + \cdots + \sigma _k(\cdot ,t) \ge \eta (\cdot ,t)$ for all $t \ge t_0$ and hence $\sigma _1(\cdot ,t) + \cdots + \sigma _k(\cdot ,t) >0$ for all $t>t_0$. Hence

$$\begin{aligned} dim (null (h(x,t)))= \min \{ i \in \{0, \ldots , n\} | \sigma _1(x,t) + \cdots + \sigma _i(x,t)= 0 \} \end{aligned}$$

is constant on some short time interval $t_0 < t < t_0 + \delta$ for any $t_0 \in [0,T)$. Hence $rank(h(x,t))$ is constant in space and time for some short time interval $t_0 < t < t_0 + \delta$ for any $t_0 \in [0,T)$. Now we let $v$ be a smooth vector field in space and time lying in the null space of $h$ (at each point in space and time). We can always construct such sections which have length one in a small neighbourhood, by defining it locally smoothly, and then multiplying by a cut-off function. We follow closely the Proof of Lemma 8.2 of Hamilton ([19]) and Theorem 2.1 of [5]. In the following we use the notation $\nabla$ and $\Delta$ to refer to ${}^{g(t)}\nabla$ and ${}^{g(t)}\Delta$. Using $h(v,v) \equiv 0$ we get

$$\begin{aligned} 0= {\partial \over {\partial t}} (h(v,v))&= \left( {\partial \over {\partial t}} h_{ij}\right) v^iv^j + h_{ij}\left( \left( {\partial \over {\partial t}} v^i\right) v^j + v^i\left( {\partial \over {\partial t}} v^j\right) \right) \nonumber \\&= \left( {\partial \over {\partial t}} h_{ij}\right) v^iv^j, \end{aligned}$$

(7.4)

since $h_{ij}v^i= 0$ and $h_{ij}v^j = 0$ (since $v$ is in the null space of $h$). Furthermore, since $h_{ij} v^i v^j \equiv 0$ we get

$$\begin{aligned} 0&= \Delta (h_{ij} v^i v^j)\nonumber \\&= (\Delta h)_{ij} v^i v^j + 2 \Delta (v)^i h_{ij} v^j \nonumber \\&+4 g^{kl} v^j\nabla _k h_{ij} \nabla _l v^i + 2 g^{kl} h_{ij} \nabla _k v^j \nabla _l v^i \end{aligned}$$

(7.5)

The term $2 \Delta (v)^i h_{ij} v^j$ is once again zero, since $h_{ij} v^j= 0$. Using this, (7.5), (7.4) and the evolution equation for $h$ we get

$$\begin{aligned} 0&= \left( {\partial \over {\partial t}} (h_{ij}) - (\Delta h)_{ij} - \phi _{ij}\right) (v^iv^j)\nonumber \\&= 4 g^{kl} v^j \nabla _k h_{ij} \nabla _l v^i + 2 g^{kl} h_{ij} \nabla _k v^j \nabla _l v^i - \phi _{ij}v^iv^j \end{aligned}$$

(7.6)

Now use

$$\begin{aligned} v^j\nabla _k h_{ij}= \nabla _k( v^j h_{ij}) - h_{ij} \nabla _k v^j = - h_{ij} \nabla _k v^j \end{aligned}$$

to conclude

$$\begin{aligned} 2 g^{kl} h_{ij} \nabla _k v^j \nabla _l v^i + \phi _{ij}v^i v^j= 0 \end{aligned}$$

(7.7)

Since $\phi (v,v) \ge 0$ (and $h\ge 0$) we see that $\phi _{ij}v^i v^j= 0$. That is, $v$ is also in the null space of $\phi$. But then, (7.7) shows that $X_R(x,t): = \nabla _{R} v(x,t)$ is in the null space of $h$ for any vector $R \in T_x M$ [choose orthonormal coordinates at $x$ at time $t$, so that $\frac{\partial }{\partial x^1} (x):= R/ \Vert R \Vert _{g(x,t)}$ and use this in Eq. (7.7)]. This shows that the null space of $h$ is invariant under parallel transport for each fixed time, as explain in the following for the readers convenience:

• Let $v_1(x), \ldots , v_k(x)$ be a smooth o.n. basis for $null(h(x,t))$ in a small spatial neighbourhood of $x_0$, and extend this to a smooth family $v_1, \ldots , v_n$ of vectors which is an o.n. basis everywhere in a small spatial neighbourhood of $x_0$. Let $X_0\in T_{x_0}M$ satisfy $g(X_0,v_i(x_0)) = 0$ for all $i \in \{k+1, \ldots , n\}$ and let $\gamma :[0,1]\rightarrow M$ be any smooth curve, starting in $x_0$ and whose image is contained in the neighbourhood of $x_0$ in question. Then parallel transport $X_0$ along $\gamma$. Call this vector field $X$. Write $X(\tau ) = \sum _{i = 1}^n X^i(\tau )v_i(\gamma (\tau ))$. We claim $X(\tau ) = \sum _{i = k+1}^n X^i(\tau ) v_i(\gamma (\tau ))$. Let $X^{\top }(\tau ) = \sum _{i = 1}^k X^i(\tau ) v_i(\gamma (\tau ))$, and $X^{\perp }(\tau ) = \sum _{i = k+1}^n X^i(\tau ) v_i(\gamma (\tau ))$. First note that for $i \in \{1, \ldots , k\}$, and $V$ the tangent vector field along $\gamma$:

  $$\begin{aligned} g(\nabla _V (X^{\perp }), v_i) = V(g(X^{\perp }, v_i)) - g(X^{\perp }, \nabla _V v_i) = 0 \end{aligned}$$

  (7.8)

• in view of $\nabla _V v_i \in span\{ v_1, \ldots , v_k\}$ and $X^{\perp } \in span\{ v_{k +1}, \ldots , v_n\}$. Furthermore, for $j \in \{ k+1, \ldots , n\}$ we have

  $$\begin{aligned} g(\nabla _V (X^{\perp }), v_j)&= g(\nabla _V (X- X^{\top }), v_j)\nonumber \\&= - g(\nabla _V (X^{\top }), v_j)\nonumber \\&= -g \left( \sum \limits _{i=1}^k V(X^i)v_i,v_j\right) - g\left( \sum \limits _{i=1}^k X^i \nabla _V v_i,v_j\right) \nonumber \\&= -\sum \limits _{i=1}^k X^i g(\nabla _V v_i,v_j)\nonumber \\&= 0 \end{aligned}$$

  (7.9)

• in view of the fact that $\nabla _V v_i \in span\{ v_1, \ldots , v_k\}$. Hence $X^{\perp }$ is also parallel along $\gamma$. Since $X^\perp (0)=0$ we have $X^\perp \equiv 0$.

We have also shown, that $null(h) \subset null(\phi )$. Let $v(x_0,s)$ for $s \in (t, t+\delta )$ be smoothly dependent on time, and $v(x_0,s) \in null(h(x_0,s))$ for each $s \in (t, t+\delta )$. Extend this vector at each time $s \in (t,t + \delta )$ by parallel transport along geodesics emanating from $x_0$ to obtain a local smooth vector field $v(\cdot ,\cdot )$ which satisfies $v(x,s) \in null(h(x,s))$ for all $x$ (in a small ball) and all $s \in (t,t+ \delta )$. In particular,

$$\begin{aligned} \nabla _iv \in null(h(x_0,s)) \ \ \text{ and }\ \ \Delta v(x,s)= (g^{kl}\nabla _k \nabla _l v)(x,s) \in null(h(x_0,s)). \end{aligned}$$

Hence

$$\begin{aligned} 0= {\partial \over {\partial t}} (h_{ij} v^i)&= \left( h_{ij} {\partial \over {\partial t}} v^i\right) + (v^i (\Delta h)_{ij}) + v^i \phi _{ij}\nonumber \\&= h_{ij} {\partial \over {\partial t}} v^i + \Delta (v^i h_{ij}) +v^i\phi _{ij}\nonumber \\&= h_{ij} {\partial \over {\partial t}} v^i \end{aligned}$$

(7.10)

where we have used that $v \in null(\phi )$. Hence ${\partial \over {\partial t}} v(x_0,s) \in null(h(x_0,s))$. Assume that at time $s_0$ we have $null(h(x_0,s_0))= \mathbb{R }^k \subset \mathbb{R }^n= T_{x_0} M$ and let $\{ e_1(t), \ldots , e_n(t) \}$ be a smooth (in time) o.n. basis of vectors with $\{ e_1(t), \ldots , e_k(t) \}$ a smooth (in time) o.n. basis of vectors of $null(h(x_0,t))$. Let $e_i^l(t):= \langle e_i(t), e^l(0)\rangle$, where $\{ e^1(0), \ldots , e^n(0) \}$ refer to the standard basis vectors of $\mathbb{R }^n$ and $\langle \cdot , \cdot \rangle$ is the standard inner product on $\mathbb{R }^n$. ${\partial \over {\partial t}} e_i(t) \in null(h(t))$ for all $i \in \{ 1, \ldots , k\}$ implies ${\partial \over {\partial t}} e_i(t) = \sum _{j= 1}^k a_i^j(t)e_j(t)$ for some smooth functions $a_i^j: [0, \infty ) \rightarrow ~\mathbb{R },\,i,j \in \{1, \ldots , k\}$. Then we have a system of ODEs ($l \in \{1, \ldots , n\}, i \in \{ 1, \ldots , k\}$)

$$\begin{aligned} {\partial \over {\partial t}} e_{i}^l (t)&= \sum _{j= 1}^k a_{i}^j(t)e_j^{l} (t) \nonumber \\ e_{i}^l (0)&= \delta _i^l. \end{aligned}$$

(7.11)

By assuming $e_j^{l}(t)= 0$ for all $l \ge k +1$ we still have a solvable system, and hence the solution satisfies (by uniqueness) $e_j^{l}(t)= 0$ for all $l \ge k +1$. That is $\{ e_1(t), \ldots , e_k(t) \}$ remains in $\mathbb{R }^k$.

$null(h(x_0,t))$ is a space which is invariant under parallel transport (from the argument above). Hence the de Rham splitting theorem (see [13]) says, $M$ splits isometrically at time $s$ as $N(s) \oplus P(s)$ where $h(\cdot ,s) = 0$ on $P(s)$ and $h(\cdot ,s)>0$ on $N(s)$. We can do this it every time $s$. But the second part of the argument shows that $N(s)= N(s_0)$ for all $s$ and $P(s)= P(s_0)$ for all $s$.$\square$

Appendix B: An approximation result by V. Kapovitch/G. Perelman

Let $(M_i,d_i,p_0)$ be a non-collapsing sequence of non-negatively curved $n-$dimensional, smooth, complete manifolds without boundary such that $(M^n_i,d_i,p_0) \rightarrow (X,d_X,0)$ as $i \rightarrow ~\infty$ (in the GH sense) where $X =CV$ is an Euclidean cone with non-negative curvature over the metric space $(V,d_V)$ (with sectional curvature not less than 1 in the sense of Alexandrov), and $(M_i,d_i,p_0)$ are smooth with $\sec \ge 0$. This is the situation examined in Sect. 1. It is well known that the space of directions $\Sigma _{0}(X)$ of $(X,d_X)$ at $0$ is $(V,d_V)$: see Theorem 10.9.3 (here we have used that the tangent cone of $X$ at $0$ is equal to $X$, since $X$ is a cone). Now Theorem 5.1 of [23] says that $\Sigma _{0}(X)$ is homeomorphic to $\Sigma _{p_0} M_i$ (for $i$ big enough) which is isometric to the standard sphere $S^{n-1}$ since the $(M_i,g_i)$ are smooth manifolds. That is $V$ is homeomorphic to $S^{n-1}$.