A Proof of Milnor conjecture in dimension 3

TL;DR

This paper proves Milnor's conjecture in three dimensions using Cheeger-Colding theory, offering an alternative approach to previous minimal surface methods.

Contribution

It provides a novel proof of Milnor's conjecture in dimension 3 based on Ricci curvature and limit space analysis, differing from prior minimal surface techniques.

Findings

Milnor conjecture is proven in dimension 3.

Cheeger-Colding theory is effectively applied to this problem.

The proof offers a new perspective distinct from minimal surface approaches.

Abstract

We present a proof of Milnor conjecture in dimension 3 based on Cheeger-Colding theory on limit spaces of manifolds with Ricci curvature bounded below. It is different from [Liu] that relies on minimal surface theory.

Equations

$$({\tilde{M}}_i,{\tilde{p}}_i,{\Gamma }_i)\stackrel{GH}{⟶}(\tilde{X},\tilde{p},G),$$

$$\#{\Gamma }_i(1)\le N$$

$${\Gamma }_i(1)=\{\gamma \in {\Gamma }_i|d(\gamma {\tilde{p}}_i,{\tilde{p}}_i)\le 1\}.$$

$$\#{\Gamma }_i(1)\le \#G(2)<\infty$$

$$(s_i^{-1}\tilde{M},p,\Gamma )\stackrel{GH}{⟶}(C_{\infty }\tilde{M},\tilde{o},G),$$

$$(r_i^{-1}\tilde{M},p,\Gamma )\stackrel{GH}{⟶}(\tilde{Y},\tilde{y},G),$$

$$(r_i^{-1}\tilde{M},\tilde{p},H_i,{\gamma }_i)\stackrel{GH}{⟶}(\tilde{Y},\tilde{y},H,g),$$

$$\mathrm{vol}(B_1(\tilde{p},r_i^{-1}\tilde{M}))\ge v$$

$$\underset{r\to \infty }{\lim }\frac{\mathrm{vol}(B_r(\tilde{p}))}{r^n}\ge v.$$

$$(r_i^{-1}M,p)\stackrel{GH}{⟶}(Y,y),$$

$$r_i^{-1}{d}_M(x_i,\alpha (d{r}_i))\to 2d,r_i^{-1}{d}_i\to 0,$$

$$d_i^{-1}{d}_M(c_i(0),x_i)\to \infty ,d_i^{-1}{d}_M(c_i(0),\alpha (d{r}_i))\to \infty ,d_i^{-1}{d}_M(c_i,p)=1,$$

$$(r_{2k+1}^{-1}S,p)\stackrel{GH}{⟶}([-1,\infty ),0)$$

Full text

1. Introduction

Milnor [Mi] in $1968$ conjectured that any open $n$-manifold $M$ with $\mathrm{Ric}_{M}\geq 0$ has a finitely generated fundamental group. This conjecture remains open today. It was proven for manifolds with Euclidean volume growth by Anderson [An] and Li [Li] independently, and manifolds with small diameter growth by Sormani [Sor]. For background and relevant examples regarding Milnor conjecture, see [SS].

For $3$-manifolds, Schoen and Yau [SY] developed minimal surfaces theory in dimension $3$ and proved that any $3$-manifold of positive Ricci curvature is diffeomorphic to $\mathbb{R}^{3}$. Recently, based on minimal surface theory, Liu [Liu] proved that any $3$-manifold with $\mathrm{Ric}\geq 0$ either is diffeomorphic to $\mathbb{R}^{3}$ or its universal cover splits. In particular, this confirms Milnor conjecture in dimension $3$.

There are some interests to find a proof of Milnor conjecture in dimension $3$ not relying on minimal surface theory. Our main attempt is to accomplish this by using structure results for limits spaces of manifolds with Ricci curvature bounded below [Co, CC1, CC2, CN], equivariant Gromov-Hausdorff convergence [FY] and pole group theorem [Sor].

Theorem 1**.**

Let $M$ be an open $3$-manifold with $\mathrm{Ric}_{M}\geq 0$, then $\pi_{1}(M)$ is finitely generated.

For any open $3$-manifold $M$ of $\mathrm{Ric}_{M}\geq 0$ and any sequence $r_{i}\to\infty$, by Gromov’s precompactness theorem [Gro2], we can pass to some subsequences and consider tangent cones at infinity of $M$ and its Riemannian universal cover $\widetilde{M}$ coming from the sequence $r_{i}^{-1}\to 0$:

\begin{CD}(r^{-1}_{i}\widetilde{M},\tilde{p})@>{GH}>{}>(C_{\infty}\widetilde{M},\tilde{o})\\ @V{}V{\pi}V@V{}V{\pi}V\\ (r^{-1}_{i}M,p)@>{GH}>{}>(C_{\infty}M,o).\end{CD}

We roughly illustrate our approach to prove Theorem Theorem 1. If $\pi_{1}(M,p)$ is not finitely generated, then we draw a contradiction by choosing some sequence $r_{i}\to\infty$ and eliminating all the possibilities regarding the dimension of $C_{\infty}\widetilde{M}$ and $C_{\infty}M$ above in the Colding-Naber sense [CN], which are integers $1,2$ or $3$.

We also make use of some reduction results by Wilking [Wi] and Evans-Moser [EM]. The first reduces any non-finitely generated fundamental groups to abelian ones in any dimension, while the latter further reduces abelian non-finitely generated ones to some subgroup of the additive group of rationals in dimension $3$. In particular, we can assume that $\pi_{1}(M)$ is torsion free if it is not finitely generated. One observation is that, if $\pi_{1}(M,p)$ is torsion free, then in the space $(C_{\infty}\widetilde{M},\tilde{v},G)$ above, the orbit $G\cdot\tilde{v}$ is not discrete (See Corollary 2.4). This observation plays a key role in the proof.

The author would like to thank Professor Xiaochun Rong and Professor Jeff Cheeger for suggestions during the preparation of this note.

2. Proof of Theorem Theorem 1

We start with the following reductions by Wilking and Evan-Moser.

Theorem 2.1**.**

[Wi]** Let $M$ be an open manifold with $\mathrm{Ric}_{M}\geq 0$. If $\pi_{1}(M)$ is not finitely generated, then it contains a non-finitely generated abelian subgroup.

Theorem 2.2**.**

[EM]** Let $M$ be a $3$-manifold. If $\pi_{1}(M)$ is abelian and not finitely generated, then $\pi_{1}(M)$ is torsion free.

Evans-Moser [EM] actually showed that $\pi_{1}(M)$ is a subgroup of the additive group of rationals. Being torsion free is sufficient for us to prove Theorem Theorem 1.

Gromov [Gro1] introduced the notion of short generators of $\pi_{1}(M,p)$. By path lifting, $\pi_{1}(M,p)$ acts on $\widetilde{M}$ isometrically. We say that $\{\gamma_{1},...,\gamma_{i},...\}$ is a set of short generators of $\pi_{1}(M,p)$, if

$d(\gamma_{1}\tilde{p},\tilde{p})\leq d(\gamma\tilde{p},\tilde{p})$ for all $\gamma\in\pi_{1}(M,p)$,

and for each $i$,

$d(\gamma_{i}\tilde{p},\tilde{p})\leq d(\gamma\tilde{p},\tilde{p})$ for all $\gamma\in\pi_{1}(M,p)-\langle\gamma_{1},...,\gamma_{i-1}\rangle$,

where $\langle\gamma_{1},...,\gamma_{i-1}\rangle$ is the subgroup generated by $\gamma_{1},...,\gamma_{i-1}$.

Let $M$ be an open $3$-manifold with $\mathrm{Ric}_{M}\geq 0$. We always denote $\pi_{1}(M,p)$ by $\Gamma$. Suppose that $\Gamma$ is not finitely generated, then by Theorems 2.1 and 2.2, we can assume that $\Gamma$ is torsion free. Let $\{\gamma_{1},...,\gamma_{i},...\}$ be an infinite set of short generators at $p$. Since $\Gamma$ is a discrete group acting freely on $\widetilde{M}$, we have $r_{i}=d(\tilde{p},\gamma_{i}\tilde{p})\to\infty$. When considering a tangent cone at infinity of $\widetilde{M}$ coming from the sequence $r_{i}^{-1}\to 0$, we also take $\Gamma$-action into account. Passing to some subsequences if necessary, we assume the following sequences converge in equivariant Gromov-Hausdorff topology [FY]:

\begin{CD}(r^{-1}_{i}\widetilde{M},\tilde{p},\Gamma)@>{GH}>{}>(\widetilde{Y},\tilde{y},G)\\ @V{}V{\pi}V@V{}V{\pi}V\\ (r^{-1}_{i}M,p)@>{GH}>{}>(Y=\widetilde{Y}/G,y).\end{CD} $(\star)$

Colding-Naber [CN] showed that the isometry group of any Ricci limit space is a Lie group. In particular, $G$ above, as a closed subgroup of $\mathrm{Isom}(\widetilde{Y})$, is a Lie group.

We recall the dimension of Ricci limit spaces in the Colding-Naber sense [CN]. A point $x$ in some Ricci limit space $X$ is $k$-regular, if any tangent cone at $x$ is isometric to $\mathbb{R}^{k}$. Colding-Naber showed that there is a unique $k$ such that $\mathcal{R}_{k}$, the set of $k$-regular points, has full measure in $X$ with respect to any limit renormalized measure (See [CC2, CN]). We regard such $k$ as the dimension of $X$ and denote it by $\dim(X)$. It is unknown whether in general the Hausdorff dimension of $X$ equals to $\dim(X)$. For Ricci limit spaces coming from $3$-manifolds, dimension in the Colding-Naber sense equals to Hausdorff dimension, which follows from Theorem 3.1 in [CC2] and [Hon1].

As indicated in the introduction, we prove Theorem Theorem 1 by eliminating all possibilities regarding the dimension of $Y$ and $\widetilde{Y}$ in ($\star$). There are three possibilities and we rule out each of them, which finishes the proof of Theorem Theorem 1.

Case 1. $\dim(\widetilde{Y})=3$ (Lemma 2.7);

Case 2. $\dim(Y)=\dim(\widetilde{Y})=2$ (Lemma 2.8);

Case 3. $\dim(Y)=1$ (Lemma 2.10).

Lemma 2.3**.**

Let $(M_{i},p_{i})$ be a sequence of complete $n$-manifolds and $(\widetilde{M}_{i},\tilde{p}_{i})$ be their universal covers. Suppose that the following sequence converges

[TABLE]

where $\Gamma_{i}=\pi_{1}(M_{i},p_{i})$ is torsion free for each $i$. If the orbit $G\cdot\tilde{p}$ is discrete in $\widetilde{X}$, then there is $N$ such that

[TABLE]

for all $i$, where $\#\Gamma_{i}(1)$ denotes the number of elements in

[TABLE]

Proof.

We claim that if a sequence $\gamma_{i}\in\Gamma_{i}$ such that $\gamma_{i}\overset{GH}{\to}g\in G$ with $g$ fixing $\tilde{p}$, then $g=e$, the identity element, and $\gamma_{i}=e$ for all $i$ sufficiently large. In fact, suppose that $\gamma_{i}\not=e$ for some subsequence. Since $\gamma_{i}$ is torsion free, we always have $\mathrm{diam}(\langle\gamma_{i}\rangle\cdot\tilde{p}_{i})=\infty$. Together with $d(\gamma_{i}\tilde{p}_{i},\tilde{p}_{i})\to 0$, we see that $G\cdot\tilde{p}$ can not be discrete, which contradicts with the assumption.

Therefore, there exists $i_{0}$ large such that for all $g\in G(2)$ and any two sequences with $\gamma_{i}\overset{GH}{\to}g$ and $\gamma^{\prime}_{i}\overset{GH}{\to}g$, $\gamma_{i}=\gamma^{\prime}_{i}$ holds for all $i\geq i_{0}$. In particular, we conclude that

[TABLE]

for all $i\geq i_{0}$. ∎

Corollary 2.4**.**

Let $(M,p)$ be an open $n$-manifold with $\mathrm{Ric}_{M}\geq 0$ and $(\widetilde{M},\tilde{p})$ be its universal cover. Suppose that $\Gamma=\pi_{1}(M,p)$ is torsion free, then for any $s_{i}\to\infty$ and any convergent sequence

[TABLE]

the orbit $G\cdot\tilde{v}$ is not discrete.

Proof.

The proof follows directly from Lemma 2.3. If $G\cdot\tilde{o}$ is discrete, then there is $N$ such that $\#\Gamma(s_{i})\leq N$ for all $i$. On the other hand, $\#\Gamma(s_{i})\to\infty$ because $\Gamma$ is torsion free. A contradiction. ∎

Lemma 2.5**.**

Let $(M,p)$ be an open $n$-manifold with $\mathrm{Ric}_{M}\geq 0$ and $(\widetilde{M},\tilde{p})$ be its universal cover. Suppose that $\Gamma=\pi_{1}(M,p)$ has infinitely many short generators $\{\gamma_{1},...,\gamma_{i},...\}$. Then in the following tangent cone at infinity of $\widetilde{M}$

[TABLE]

the orbit $G\cdot\tilde{y}$ is not connected, where $r_{i}=d(\gamma_{i}\tilde{p},\tilde{p})\to\infty$.

Proof.

On $r_{i}^{-1}\widetilde{M}$, $\gamma_{i}$ has displacement $1$ at $\tilde{p}$. By basic properties of short generators, $\gamma_{i}\tilde{p}$ has distance $1$ from the orbit $H_{i}\cdot\tilde{p}$, where $H_{i}=\langle\gamma_{1},...,\gamma_{i-1}\rangle$. From equivariant convergence

[TABLE]

we conclude $d(g\tilde{y},H\cdot\tilde{y})=1$. It is obvious that $H$ contains $G_{0}$, the connected component of $G$ containing the identity. Thus $d(g\tilde{y},G_{0}\cdot\tilde{y})\geq 1$ and the orbit $G\cdot\tilde{y}$ is not connected. ∎

We recall cone splitting principle, which follows from splitting theorem for Ricci limit spaces [CC1].

Proposition 2.6**.**

Let $(X,p)$ be the limit of a sequence of complete $n$-manifolds $(M_{i},p_{i})$ of $\mathrm{Ric}_{M_{i}}\geq 0$. Suppose that $X=\mathbb{R}^{k}\times C(Z)$ is a Euclidean cone with vertex $p=(0,z)$. If there is an isometry $g\in\mathrm{Isom}(X)$ with $g(0,z)\not\in\mathbb{R}^{k}\times\{z\}$, then $X$ splits isometrically as $\mathbb{R}^{k+1}\times C(Z^{\prime})$.

Lemma 2.7**.**

Case 1 can not happen.

Proof.

When $\dim(\widetilde{Y})=3$, $\widetilde{Y}$ is a non-collapsing limit space [CC2], that is, there is $v>0$ such that

[TABLE]

for all $i$. By relative volume comparison, this implies that $\widetilde{M}$ has Euclidean volume growth

[TABLE]

By [CC2], $\widetilde{Y}$ is a Euclidean cone $\mathbb{R}^{k}\times C(Z)$ with vertex $\tilde{y}=(0,z)$, where $C(Z)$ does not contain any line and $z$ is the vertex of $C(Z)$. We rule out all the possibilities of $k\in\{0,1,2,3\}$.

If $k=3$, then $\widetilde{Y}=\mathbb{R}^{3}$. Thus $\widetilde{M}$ is isometric to $\mathbb{R}^{3}$ [Co].

If $k=2$, then according to co-dimension $2$ [CC2], actually $\widetilde{Y}=\mathbb{R}^{3}$.

If $k=1$, then $Y=\mathbb{R}\times C(Z)$. By Proposition 2.6, the orbit $G\cdot\tilde{y}$ is contained in $\mathbb{R}\times\{z\}$. Applying Lemma 2.5, we see that $G\cdot\tilde{y}$ is not connected. Note that a non-connected orbit in $\mathbb{R}$ is either a $\mathbb{Z}$-translation orbit, or a $\mathbb{Z}_{2}$-reflection orbit. In particular, the orbit $G\cdot\tilde{y}$ must be discrete. This contradicts with Corollary 2.4.

If $k=0$, then $Y=C(Z)$ with no lines. Again by Proposition 2.6, the orbit $G\cdot\tilde{y}$ must be a single point $\tilde{y}$, which is a contradiction to Lemma 2.5. ∎

Lemma 2.8**.**

Let $(M,p)$ be an open $n$-manifold with $\mathrm{Ric}_{M}\geq 0$ and $(\widetilde{M},\tilde{p})$ be its universal cover. Assume that $\Gamma=\pi_{1}(M,p)$ is torsion free. Then for any $s_{i}\to\infty$ and any convergent sequence

\begin{CD}(s^{-1}_{i}\widetilde{M},\tilde{p},\Gamma)@>{GH}>{}>(C_{\infty}\widetilde{M},\tilde{o},G)\\ @V{}V{\pi}V@V{}V{\pi}V\\ (s^{-1}_{i}M,p)@>{GH}>{}>(C_{\infty}M,o),\end{CD}**

$\dim(C_{\infty}\widetilde{M})=\dim(C_{\infty}M)$* can not happen. In particular, Case 2 can not happen.*

Proof of Lemma 2.8.

We claim that $G$ is a discrete group when $\dim({C_{\infty}\widetilde{M}})=\dim(C_{\infty}M)=k$. If the claim holds, then the desired contradiction follows from Corollary 2.4.

It remains to verify the claim. Suppose that $G_{0}$ is non-trivial, then we pick $g\not=e$ in $G_{0}$. Note that there is a $k$-regular point $\tilde{q}\in C_{\infty}\widetilde{M}$ such that $d(g\tilde{q},\tilde{q})>0$ and $\tilde{q}$ projects to a $k$-regular point $q\in C_{\infty}M$. In fact, let $\mathcal{R}_{k}(C_{\infty}M)$ be the set of $k$-regular points in $C_{\infty}M$. Since $\mathcal{R}_{k}(C_{\infty}M)$ is dense in $C_{\infty}M$, its pre-image $\pi^{-1}(\mathcal{R}_{k}(C_{\infty}M))$ is also dense in $C_{\infty}\widetilde{M}$. Let $\tilde{q}$ be a point in the pre-image such that $d(g\tilde{q},\tilde{q})>0$. Note that any tangent cone at $\tilde{q}$ splits $\mathbb{R}^{k}$-factor isometrically. By Proposition 3.78 in [Hon2] (also see Corollary 1.10 in [KL]), it follows that any tangent cone at $\tilde{q}$ is isometric to $\mathbb{R}^{k}$. In other words, $\tilde{q}$ is $k$-regular.

Along a one-parameter subgroup of $G_{0}$ containing $g$, we can choose a sequence of elements $g_{j}\in G_{0}$ with $d(g_{j}\tilde{q},\tilde{q})=1/j\to 0$. We consider a tangent cone at $\tilde{y}$ and $y$ respectively coming from the sequence $j\to\infty$. Passing to some subsequences if necessary, we obtain

\begin{CD}(jC_{\infty}\widetilde{M},\tilde{q},G,g_{j})@>{GH}>{}>(C_{\tilde{q}}C_{\infty}\widetilde{M},\tilde{o}^{\prime},H,h)\\ @V{}V{\pi}V@V{}V{\pi}V\\ (jC_{\infty}M,q)@>{GH}>{}>(C_{q}C_{\infty}M,o^{\prime}).\end{CD}

with $C_{\tilde{q}}C_{\infty}\widetilde{M}/H=C_{q}C_{\infty}M$ and $d(h\tilde{o}^{\prime},\tilde{o}^{\prime})=1$. On the other hand, since both $q$ and $\tilde{q}$ are $k$-regular, $C_{\tilde{q}}C_{\infty}\widetilde{M}=C_{q}C_{\infty}M=\mathbb{R}^{k}$. This is a contradiction to $H\not=\{e\}$. Hence the claim holds. ∎

To rule out the last case $\dim(Y)=1$, we recall Sormani’s pole group theorem [Sor]. We say that a length space $X$ has a pole at $x\in X$, if for all $y\not=x$, there is a ray starting from $x$ and going through $y$.

Theorem 2.9**.**

[Sor]** Let $(M,p)$ be an open $n$-manifold with $\mathrm{Ric}_{M}\geq 0$ and $(\widetilde{M},\tilde{p})$ be its universal cover. Suppose that $\Gamma=\pi_{1}(M,p)$ has infinitely many short generators $\{\gamma_{1},...,\gamma_{i},...\}$. Then in the following tangent cone at infinity of $M$

[TABLE]

$Y$* can not have a pole at $y$, where $r_{i}=d(\gamma_{i}\tilde{p},\tilde{p})\to\infty$.*

Lemma 2.10**.**

Case 3 can not happen.

Proof.

By [Hon1] (also see [Chen]), $Y$ is a topological manifold of dimension $1$. Since $Y$ is non-compact, $Y$ is either a line $(-\infty,\infty)$ or a half line $[0,\infty)$. By Theorem 2.9, $Y$ can not have a pole at $y$. Thus there is only one possibility left: $Y=[0,\infty)$ but $y$ is not the endpoint $0\in[0,\infty)$. Put $d=d_{Y}(0,y)>0$. We rule out this case by a rescaling argument and Lemmas 2.7, 2.8 above. (In general, it is possible for an open manifold having a tangent cone at infinity as $[0,\infty)$ with base point not being [math]. See example 2.11.)

Let $\alpha(t)$ be a unit speed ray in $M$ starting from $p$ and converging to the unique ray from $y$ in $Y=[0,\infty)$ with respect to the sequence $(r_{i}^{-1}M,p)\overset{GH}{\longrightarrow}(Y,y)$. Let $x_{i}\in r_{i}^{-1}M_{i}$ be a sequence of points converging to $0\in Y$, then $r_{i}^{-1}d_{M}(p,x_{i})\to d$. For each $i$, let $c_{i}(t)$ be a minimal geodesic from $x_{i}$ to $\alpha(dr_{i})$, and $q_{i}$ be a closest point to $p$ on $c_{i}$. We reparametrize $c_{i}$ so that $c_{i}(0)=q_{i}$. With respect to $(r_{i}^{-1}M,p)\overset{GH}{\longrightarrow}(Y,y)$, $c_{i}$ subconverges to the unique segment between [math] and $2d\in[0,\infty)$. Clearly,

[TABLE]

where $d_{i}=d_{M}(p,c_{i}(0))$.

If $d_{i}\to\infty$, then we rescale $M$ and $\widetilde{M}$ by $d_{i}^{-1}\to 0$. Passing to some subsequences if necessary, we obtain

\begin{CD}(d^{-1}_{i}\widetilde{M},\tilde{p},\Gamma)@>{GH}>{}>(\widetilde{Y}^{\prime},\tilde{y}^{\prime},G^{\prime})\\ @V{}V{\pi}V@V{}V{\pi}V\\ (d^{-1}_{i}M,p)@>{GH}>{}>(Y^{\prime},y^{\prime}).\end{CD}

If $\dim(Y^{\prime})=1$, then we know that $Y^{\prime}=(-\infty,\infty)$ or $[0,\infty)$. On the other hand, since

[TABLE]

$c_{i}$ subconverges to a line $c_{\infty}$ in $Y^{\prime}$ with $d(c_{\infty},y^{\prime})=1$. Clearly this can not happen in $(-\infty,\infty)$ nor $[0,\infty)$. If $\dim(\widetilde{Y}^{\prime})=3$, then $\widetilde{M}$ has Euclidean volume growth and thus $\dim(\widetilde{Y})=3$. This case is already covered in Lemma 2.7. The only situation left is $\dim(\widetilde{Y}^{\prime})=\dim(Y^{\prime})=2$. By Lemma 2.8, this also leads to a contradiction. In conclusion, $d_{i}\to\infty$ can not happen.

If there is some $R>0$ such that $d_{i}\leq R$ for all $i$, then on $M$, $c_{i}$ subconverges to a line $c$ with $c(0)\in B_{2R}(p)$. Consequently, $M$ splits a line isometrically [CG], which contradicts with $Y=[0,\infty)$. This completes the proof. ∎

Example 2.11**.**

We construct a surface $(S,p)$ isometrically embedded in $\mathbb{R}^{3}$ such that $S$ has a tangent cone at infinity as $[0,\infty)$, but $p$ does not correspond to [math]. We first construct a subset of $xy$-plane by gluing intervals. Let $r_{i}\to\infty$ be a positive sequence with $r_{i+1}/r_{i}\to\infty$. Starting with a interval $I_{1}=[-r_{1},r_{2}]$, we attach a second interval $I_{2}=[-r_{3},r_{4}]$ perpendicularly to $I_{1}$ by identifying $r_{2}\in I_{1}$ and $0\in I_{2}$. Repeating this process, suppose that $I_{k}$ is attached, then we attach the next interval $I_{k+1}=[-r_{2k+1},r_{2k+2}]$ perpendicularly to $I_{k}$ by identifying $r_{2k}\in I_{k}$ and $0\in I_{k+1}$. In the end, we get a subset $T$ in the $xy$-plane consisting of segments. We can smooth the $\epsilon$-neighborhood of $T$ in $\mathbb{R}^{3}$ so that it has sectional curvature $\geq-C$ for some $\epsilon,C>0$. We call this surface $S$ and let $p\in S$ be a point closest to $0\in I_{1}$ as base point. If we rescale $(S,p)$ by $r_{2k+1}^{-1}$, then

[TABLE]

because $r_{i+1}/r_{i}\to\infty$. In other words, $S$ has a tangent cone at infinity as the half line, but the base point does not correspond to the end point in this half line.