Sewing Riemannian Manifolds with Positive Scalar Curvature

Abstract

We explore to what extent one may hope to preserve geometric properties of three-dimensional manifolds with lower scalar curvature bounds under Gromov–Hausdorff and Intrinsic Flat limits. We introduce a new construction, called sewing, of three-dimensional manifolds that preserves positive scalar curvature. We then use sewing to produce sequences of such manifolds which converge to spaces that fail to have nonnegative scalar curvature in a standard generalized sense. Since the notion of nonnegative scalar curvature is not strong enough to persist alone, we propose that one pair a lower scalar curvature bound with a lower bound on the area of a closed minimal surface when taking sequences as this will exclude the possibility of sewing of manifolds.

Appendix: Short Tunnels with Positive Scalar Curvature by Jorge Basilio and Józef Dodziuk

There is a deep connection between the geometry of Riemannian manifolds $M^n$ with positive scalar curvature and surgery theory. The subject began with the surprising discovery by Gromov and Lawson [12] (for $n \ge 3$) and Schoen and Yau [32] that a manifold obtained via a surgery of codimension 3 from a manifold $M^n$ with a metric of positive scalar curvature may also be given a metric with positive scalar curvature. The key to the tunnel construction of [12] is defining a curve $\gamma$ which begins along the vertical axis then bends upwards as it moves to the right and ends with a horizontal line segment, cf. Fig. 6 below. The tunnel then is the surface of revolution determined by $\gamma$. We note that the “bending argument” has attracted some attention (See [24]).

As the goals of the surgery theory were topological in nature, Gromov and Lawson did not estimate with diameters or volumes of these tunnels. Indeed, the tunnels they constructed may be thin but long (See [11]). To build sewn manifolds, we need tunnels with diameters shrinking to zero as the size of the original balls decreases to zero (see (7), (8) (9)). Therefore, we prove Lemma 2.1 to obtain a refinement of the Gromov and Lawson construction showing the existence of tiny (in sense of (10)) and arbitrarily short tunnels with a metric of positive scalar curvature.

Proof of Lemma 2.1

To aid the reader, we provide a summary of our proof and introduce additional notation.

1.1 Outline of Proof of Lemma 2.1

To aid the reader, we provide a summary of our proof and introduce additional notation.

Step 1: Setup and notation

Let $\epsilon >0$ be given. We shall specify $0<\delta _0<\delta /2$ below.

Given that $B_1=B(p_1,\delta /2) \subset M^3$ has constant sectional curvature $K>0$, we may choose coordinates so that it is realized as a hypersurface of revolution. This is also true for $B(p_1,\delta _0) \subset B_1$ for $0<\delta _0<\delta /2$ centered at the same $p_1$. Thus, $B(p_1,\delta _0)$ is a hypersurface of revolution $U_{\gamma _0}'$ with the induced metric in ${\mathbb {R}}^4$ determined by revolving a segment of the circle $\gamma _0$ in the $(x_0,x_1)$-plane about the $x_0$-axis. We set things up so that the vertical $x_1$-axis corresponds to boundary points of $B(p_1,\delta _0)$. We then proceed as Gromov and Lawson to deform $\gamma _0$ away from vertical axis bending it upwards as we move to the right and ending with an arbitrarily short horizontal line segment. We call this curve $\gamma$, cf. Fig. 6. The curve $\gamma$ begins exactly as $\gamma _0$ so that we may attach the corresponding hypersurface onto the larger $B(p_1,\delta /2)$ in a natural way. We do exactly the same for $B_2 \subset M^3$ and identify the two hypersurfaces along their common boundary, i.e., the “tiny neck,” forming $2U_{\gamma }'=U_{\gamma }' \sqcup U_{\gamma }'$. We then define the tunnel $U=U_\delta$ by

$$\begin{aligned} U=U_\delta = ((B(p_1,\delta /2) \setminus B(p_1,\delta _0)) \sqcup (2U_{\delta _0,\gamma }') \sqcup ((B(p_2,\delta /2) \setminus B(p_2,\delta _0)), \end{aligned}$$

(224)

where $0<\delta _0<\delta /2$ and $U_{\gamma }'=U_{\delta _0,\gamma }'$ is a modified Gromov–Lawson tunnel, see Fig. 1.

The boundary of $2U_{\gamma }'$ is isometric to a collar of $B(p_1,\delta _0) \sqcup B(p_2,\delta _0)$, so we may smoothly attach it to form (224).

Step 2: Construction of the curve$\gamma$, Part 1:$C^{1}$

In this step, we construct a $C^1$, and piecewise $C^\infty$, curve $\gamma$. The construction is based on the bending argument of Gromov and Lawson and uses the fundamental theorem of plane curves, i.e., the fact that a smooth curve parametrized by arclength is uniquely determined by its curvature, the initial point, and the initial tangent vector. Care must be taken to ensure that the induced metric on $U_{\gamma }'$ maintains positive scalar curvature and that the length of $\gamma$ is controlled to yield diameter and volume estimates of Lemma 2.1. This step is quite technical and forms the heart of the proof.

Step 3: Construction of the curve$\gamma$, Part 2: from$C^{1}$to$C^{\infty }$

In this step, we show how to modify the curve constructed in Step 2 to obtain a smooth curve ${\bar{\gamma }}$ while maintaining all the required features. The modification is elementary and, once it is completed, we rename ${\bar{\gamma }}$ back to $\gamma$.

Step 4: Diameter estimates (7), (9) and volume estimates (10), (11)

This is very straightforward since the previous steps give an estimate of the length of the tunnel.

Fig. 6

The curve $\gamma$

We remark here that the choice of $\delta _0$ is used only to insure that the tunnel $U'$ (see Fig. 1) has sufficiently small volume.

1.2 Step 1 of the Proof

We now set up our notation further, describe U explicitly in terms of a special curve $\gamma$, and state the important curvature formulas needed in later steps. The construction of $\gamma$ is done in the next two sub-sections (Steps 2 and 3).

As mentioned in Sect. 1, because we assume that $B_1$ and $B_2$ have constant sectional curvature K, we may work directly in Euclidean space ${\mathbb {R}}^4$ with coordinates $(x_0,x_1,x_2,x_3)$ and its standard metric. Let $\gamma (s)$ be a curve in the $(x_0,x_1)$-plane, parametrized by arc-length, written as $\gamma (s)=(x_0(s),x_1(s))$. This curve specifies a hypersurface in ${\mathbb {R}}^4$ (by rotating $\gamma$ about the $x_0$-axis),

$$\begin{aligned} U' = U_{\gamma }' = \{\, (x_0,x_1,x_2,x_3 \in {\mathbb {R}}^4 \mid x_0=x_0(s),\,x_1^2+x_2^2+x_3^2 = x_1(s)^2\, \}, \end{aligned}$$

(225)

which we endow with the induced metric. Our curve $\gamma$ will always lie in the first quadrant of $(x_0,x_1)$-plane and will be parametrized so that $x_0(s)$ will be increasing. We denote by $\theta (s)$ the angle between the horizontal direction and the upward normal vector, and by $\varphi (s)$ the angle between the horizontal direction and the tangent vector to $\gamma$.

We remark that the two angle functions are related by

$$\begin{aligned} \theta (s)=\varphi (s) + \frac{\pi }{2}, \end{aligned}$$

(226)

See Fig. 6. In particular, $\varphi \in (-\pi /2,0]$.

Denote by k(s) the geodesic curvature of $\gamma$. It is a signed quantity so that $\gamma$ bends away from the horizontal axis if $k(s)>0$ and towards the $x_0$-axis when $k(s)<0$. If $\gamma (s_0)=(c,d)$ and $\varphi _0= \varphi (s_0)$ then (cf. Theorem 6.7, [13]) the function k(s) determines $\gamma$ by the formulae

$$\begin{aligned} \varphi (s)=\varphi _0+\int _{s_0}^s k(u)\, \mathrm{{d}}u \end{aligned}$$

(227)

and

$$\begin{aligned} \gamma (s) = \left( c+\int _{s_0}^s \cos (\varphi (u))\, \mathrm{{d}}u,\, d+\int _{s_0}^s \sin (\varphi (u))\, \mathrm{{d}}u \right) . \end{aligned}$$

(228)

Our aim is to define a function k(s) so that the resulting threefold of revolution $U'$ has positive scalar curvature. The formula on page 226 of [12] for $n=3$ gives a relation between the two curvatures. Namely

$$\begin{aligned} \mathrm{Scal}_{U'}(s) = \frac{2\, \sin \theta (s)}{x_1(s)} \left[ \frac{\sin \theta (s)}{x_1(s)} -2 k(s) \right] , \end{aligned}$$

(229)

where $\mathrm{Scal}_{U'}(s)$ is the scalar curvature of the induced metric on $U'$ and k is the geodesic curvature of $\gamma$. In particular, the formula holds if $\gamma$ is the intersection of the 3-sphere around the origin with the $(x_0,x_1)$-plane in which case k is a negative constant.

We begin defining our curve $\gamma (s)$ so that $\gamma (0)$ corresponds to a point on $\partial B(p_1,\delta _0)$ and $\gamma (s)$, for small values of $s\in [0,s_0]$, parametrizes the intersection of $B(p_1,\delta _0)$ with the $(x_0,x_1)$-plane. In particular, for small s, $k(s)\equiv -\sqrt{K}$. We choose $s_0 =\delta _0/2$ and then extend (in Step 2, Sect. 1) the function k(s) to a suitable step function on a longer interval [0, L] so that the resulting curve $\gamma (s)$ has the following properties.

• (I) The graph of $\gamma$ lies strictly in the first quadrant, beginning at $p_I=\gamma (0)=(0,\cos (-\pi /2+\delta _0)/\sqrt{K})$ and ending at $p_F=\gamma (L)$ with $x_0(L)>0$, $x_1(L)>0$, where L is the length of the curve. Moreover, a point of $\gamma$ moves to the right when s increases.

• (II) Let $\theta (s)$ be the angle between the upward pointing normal to $\gamma$ and the $x_0$-axis. The curve $\gamma$ ends at $p_F$ with $\theta (L)=\pi /2$ and has $\theta =\pi /2$ (so that it is a horizontal line segment) for an arbitrarily small interval $(L',L]$ (where $L'<L$).

• (III) The curve $\gamma$ has constant curvature $-\sqrt{K}$ near 0 so that the boundary of U has a neighborhood that is isometric to a collar of $B_1 \cup B_2$.

• (IV) The curvature function k(s) satisfies

  $$\begin{aligned} k(s) < \frac{\sin (\theta (s))}{2x_{1}(s)} \qquad s \in [0,L], \end{aligned}$$

  (230)

  so that the expression on the right-hand side of (229) is positive for all $s \in [0,L]$. We remark here that in certain stages of the construction k(s) will have discontinuities so that $\mathrm{Scal}_{U'}(s)$ is not defined but this will cause no difficulties.

• (V) The length of $\gamma$, L, is $O(\delta _{0})$.

Due to properties (I) and (II) of $\gamma$ above, we may smoothly attach two copies of $U'$ along their common boundary at $s=L$ to define $2U'=U_{\gamma }' \sqcup U_{\gamma }'$ and then, using property (III), attach $2U'$ to form U as in (224).

In the next step, we construct a piecewise $C^1$ curve $\gamma$ in the $(x_0,x_1)$-plane which satisfies properties (I) through (V). Then, in Step 3, we modify the construction once more to produce a smooth curve, ${\bar{\gamma }}$, with these same properties.

1.3 Step 2 of the Proof: Construction of $\gamma$, Part 1: $C^{1}$

As above, let $s_0=\delta _0/2$ and let $q_{0}=(a_{0},b_{0})$ be the coordinates of the point $\gamma (s_{0})$ that is already defined. By choosing $\delta _0$ sufficiently small, we can assume that the tangent vector to $\gamma$ at $s=s_{0}$ is nearly vertical and is pointing downward at $s=s_{0}$. We also have $k(s)\equiv -\sqrt{K}$ on $[0,s_0]$.

We will use a finite induction to define a sequence of extensions of $\gamma$ over intervals $[s_{i},s_{i+1}]$, with $s_{i}<s_{i+1}$ for a finite number of steps $0 \le i \le n$, where $n=n(\delta _{0})$ is the number of steps required such that properties (I),  (III),  (IV), and (V) all hold at each extension. We denote by $(a_{i},b_{i})$ the coordinates of the point $\gamma (s_{i})$ for $0 \le i \le n$.

Let us first choose the curvature function k(s) of $\gamma (s)$ on the first extended interval $[s_{0},s_{1}]$. Observe that equation (230) limits the amount of positive curvature allowed for k(s). In fact, we choose k(s) to be the constant $k_{1}>0$ over the interval $[s_{0},s_{1}]$ based only the initial data at $s_{0}$

$$\begin{aligned} k_{1} = \frac{\sin (\theta (s_{0}))}{4b_{0}}>0, \end{aligned}$$

(231)

where $\theta (s_{0})=\frac{\pi }{2}+\varphi (s_{0}) = \delta _{0}-\sqrt{K}s_{0}>0$ and $b_{0}=x_{1}(s_{0})$. Note that constant positive curvature means that $\gamma (s)$ moves along the arc of a circle of curvature $1/\sqrt{k_{1}}$ bending away from the origin.

We verify that property (IV) holds with our choice of $k_{1}$ in (231). From (227), we see that $\varphi (s)$ is an increasing function with range in the interval $(-\pi /2,0)$, hence $\theta (s)$ is also increasing by (226). Moreover, from (227) and (228), we see that the $x_{1}$-coordinate function is decreasing on the interval $(s_{0}, s_{1})$ since $x_{1}'(s)=\sin (\varphi (s))<0$. Thus, the expression on the right-hand side of (230), $\sin (\theta (s))/(2x_{1}(s))$, is an increasing function on $(s_{0},s_{1})$ so that

$$\begin{aligned} \frac{\sin (\theta (s_{0}))}{2x_{1}(s_{0})} \le \frac{\sin (\theta (s))}{2x_{1}(s)} \qquad s \in [s_{0},s_{1}]. \end{aligned}$$

(232)

Since $k(s)\equiv k_{1}$ is constant it follows that the property (IV) holds for $s \in [s_{0},s_{1}]$.

Next, we choose the length of the extension $\Delta s_{1} = s_{1} -s_{0}$, so that properties (I) and (V) hold. This is achieved by setting

$$\begin{aligned} \Delta s_{1} = \frac{b_{0}}{2}>0 \end{aligned}$$

(233)

Observe that $x_0(s)$ is increasing since $x_{0}'(s)=\cos (\varphi (s))>0$ as $\varphi \in (-\pi /2,0)$.

Clearly we have

$$\begin{aligned} b_{0} < \delta _{0} \end{aligned}$$

(234)

since $b_0$ is the vertical distance of $\gamma (s_0)$ to the $x_0$-axis which is less than the distance along the sphere.

Of course, we do not achieve a final angle of $\pi /2$ of the normal at $s_{1}$ and gain only a small but definite increase in the angle. The change in angle of the normal with the $x_{0}$-axis is

$$\begin{aligned} \Delta \theta _{1} =\theta (s_{1}) - \theta (s_{0}) = \int _{s_{0}}^{s_{1}} k(s)\, \mathrm{{d}}s = k_{1} \cdot \Delta s_{1} = \frac{\sin (\theta (s_{0}))}{8} >0 \end{aligned}$$

by (231) and (233).

With $\gamma$ extended over the first interval $[s_{0},s_{1}]$, we now inductively define further extensions. Assume that $\Delta s_{j}$, $s_{j}$ and $k_{j}$ have been chosen for $j=1,2,\ldots ,(i-1)$, and $\gamma$ extended on the intervals $[s_{j},s_{j+1}]$, we then define

$$\begin{aligned} \Delta s_{i} = \frac{b_{i-1}}{2}, \qquad s_{i} = s_{i-1} + \Delta s_{i} \qquad \text {and} \qquad k_{i} = \frac{\sin (\theta (s_{i-1}))}{4b_{i-1}}, \end{aligned}$$

(235)

where $\gamma (s_{i})=(a_{i},b_{i})$. In what follows we will also write $\theta _j$ and $\varphi _j$ for $\theta (s_j)$ and $\varphi (s_j)$, respectively. We remark that $b_{i+1} < b_i$ by (228) since the angle $\varphi$ is negative and that $k_{i+1} > k_i$ since the ratio $\frac{\sin (\theta (s))}{x_{1}(s)}$ is increasing. Observe that properties (I), (IV), and (V) of $\gamma$ hold on $[s_{i-1},s_{i}]$ for all i by our choices in (235) by arguments analogous to those given for the first extension of $\gamma$ on $[s_{0},s_{1}]$.

We observe that we gain a definite amount of angle $\theta$ with each extension since, by (235), for each $j \in \{1,2,\ldots , i\}$,

$$\begin{aligned} \Delta \theta _{j} =\theta (s_{j}) - \theta (s_{j-1})&= \int _{s_{j-1}}^{s_{i}} k(s)\, \mathrm{{d}}s = k_{j} \cdot \Delta s_{j} = \frac{\sin (\theta (s_{j-1}))}{8} \qquad \nonumber \\&\ge \frac{\sin (\theta (s_{0}))}{8}, \end{aligned}$$

(236)

because $\theta (s_{j-1}) \ge \theta (s_{0})$ and the values of $\theta$ are in the range $(0,\pi /2)$ so that the sine is an increasing function. We stop the construction when $\theta (s)$ reaches the value $\pi /2$. Thus the total change in the angle $\theta$ over the interval $[0,s_{i}]$ is bounded from below by

$$\begin{aligned} \Delta \theta = \sum _{j=1}^{i} \Delta \theta _{j} \ge i \cdot \frac{\sin (\theta _{0})}{8}. \end{aligned}$$

(237)

To prove property (V), that the length of $\gamma$ is on the order of $\delta _{0}$, we need the sequence of $b_{i}$’s to be summable and will want to compare it to the geometric progression. The difficulty here is that, since our curve is bending more and more upwards, the ratios $b_i/b_{i-1}$ increase. For this reason, we stop our induction when $\theta$ reaches the value of $\pi /4$. It will turn out that once this value is reached, we can complete the construction of k(s) by a single extension albeit with $\Delta s$ not given by (235).

Thus, define $n=n(\delta _{0})$ to be the first positive integer with

$$\begin{aligned} \frac{\pi }{4} \le \theta _{n} \end{aligned}$$

(238)

which exists by (237). Moreover, if $\theta _{n}>\pi /4$ we re-define $s_{n}$ to be the exact value in $(s_{n-1},\infty )$ such that $\theta (s_{n})=\pi /4$. Thus, for the modified value of $s_n$

$$\begin{aligned} \theta _{n} = \theta (s_n) = \frac{\pi }{4}. \end{aligned}$$

(239)

The following Lemma gives the desired comparison.

Lemma 7.1

There exists a universal constant $C\in (0,1)$, independent of $\delta _{0}$ and K, such that for all $i \le n$

$$\begin{aligned} b_{i} \le C \cdot b_{i-1}, \end{aligned}$$

where $n=n(\delta _0)$ is as above.

The Lemma, to be proven shortly below, implies that the length of the curve $\gamma$ on the entire interval $[0,s_{n}]$ is no larger than a constant (independent of $\delta _{0}$) times $\delta _{0}$. Namely,

$$\begin{aligned} L(\gamma ([0,s_{n}])) = s_{n} = \sum _{j=1}^{n} \Delta s_{j}. \end{aligned}$$

(240)

Thus, from (235) and Lemma (7.1), we have

$$\begin{aligned} \sum _{j=1}^{n} \Delta s_{j} = \sum _{j=1}^{n} \frac{b_{j-1}}{2} \le \frac{b_{0}}{2} \left( \sum _{j=1}^{n-1} C^{j} \right) \le C_1 \delta _0 \end{aligned}$$

(241)

by the lemma and (234). So, $L(\gamma ([0,s_{n}])) \le C_{1} b_{0}$ with $C_{1}= \frac{1}{2-2C}$ which is independent of $\delta _{0}$ since C is. This proves that $L(\gamma ([0,s_{n}])) = O(\delta _0)$.

Proof of Lemma 7.1

Let $1 \le i \le n$. We compute explicitly using (227), (228), and (235),

$$\begin{aligned} \varphi (s_{i}) = \varphi (s_{i-1}) + k_{i}\cdot \Delta s_{i} =\varphi (s_{i-1})+\frac{\sin (\theta _{i-1})}{8} \end{aligned}$$

(242)

and

$$\begin{aligned} b_{i}&= x_{1}(s_{i}) \\&= b_{i-1} + \int _{s_{i-1}}^{s_{i}} \sin (\varphi (s_{i-1})+k_{i}(u-s_{i-1}))\, \mathrm{{d}}u\\&= b_{i-1} - \frac{1}{k_{i}} \left( \cos (\varphi (s_{i})) - \cos (\varphi (s_{i-1})) \right) \\&= b_{i-1} - \frac{4b_{i-1}}{\sin (\theta (s_{i-1}))} \left( \cos \left( \varphi (s_{i-1})+\frac{\sin (\theta _{i-1})}{8} \right) - \cos (\varphi (s_{i-1})) \right) . \end{aligned}$$

Thus,

$$\begin{aligned} \frac{b_{i}}{b_{i-1}}&= 1 - \frac{4}{\sin (\theta (s_{i-1}))} \left( \cos \left( \varphi (s_{i-1})+\frac{\sin (\theta _{i-1})}{8} \right) - \cos (\varphi (s_{i-1})) \right) . \end{aligned}$$

Therefore, by the Mean Value Theorem, there exists $\mu _{i} \in (\varphi (s_{i-1}),\varphi (s_{i-1})+\sin (\theta (s_{i-1}))/8)$ such that

$$\begin{aligned} \frac{b_{i}}{b_{i-1}}&= 1 - \frac{4}{\sin (\theta (s_{i-1}))} (-\sin (\mu _{i})) \cdot \frac{\sin (\theta (s_{i-1}))}{8} = 1 + \frac{\sin (\mu _{i})}{2}. \end{aligned}$$

To complete the proof of the claim, we seek a constant $0<C<1$, independent of $\delta _{0}$, such that

$$\begin{aligned} 1 + \frac{\sin (\mu _{i})}{2}< C < 1. \end{aligned}$$

(243)

Recall that the angle function $\varphi$ takes negative values throughout.

We claim that the choice

$$\begin{aligned} C=1+\frac{1}{4} \sin \left( -\frac{\pi }{4}+\frac{\cos (-\frac{\pi }{4})}{8} \right) \approx 0.8395 \end{aligned}$$

(244)

will satisfy our requirement.

This follows from the fact that the sine is an increasing function on the interval $(\varphi (s_{i-1}),\varphi (s_{i-1})+\sin (\theta (s_{i-1}))/8)$ and the fact that both the angles $\varphi _{i}$ and $\theta _{i}$ are increasing, so

$$\begin{aligned} 1+\frac{\sin (\mu _{i})}{2}&\le 1+\frac{1}{2}\sin \left( \varphi (s_{i-1})+\frac{\sin (\theta (s_{i-1}))}{8} \right) \\&\le 1+ \frac{1}{2} \sin \left( \varphi (s_{n})+\frac{\cos (\varphi (s_{n}))}{8} \right) . \end{aligned}$$

By our choice of $s_n$, $\theta (s_{n})=\pi /4$ from (239) and $\varphi (s_{n})=-\pi /4$ so that

$$\begin{aligned} 1+\frac{\sin (\mu _{i})}{2}&\le 1+ \frac{1}{2} \sin \left( -\frac{\pi }{4}+\frac{\cos \left( -\frac{\pi }{4} \right) }{8} \right) \\&< 1+ \frac{1}{4} \sin \left( -\frac{\pi }{4}+\frac{\cos \left( -\frac{\pi }{4} \right) }{8} \right) \\&= C <1. \end{aligned}$$

This finishes the proof of the Lemma. $\square$

At this stage of the construction, $\gamma$ has angle $\theta =\pi /4$ at the endpoint $s_{n}$. We make one additional extension of our step function.

We now define $s_{n+1}>s_{n}$ and $k_{n+1}>0$ as follows.

By (227) $\varphi (s)$ in $[s_{n},s_{n+1}]$ will be given by

$$\begin{aligned} \varphi (s) = \varphi _{n} + \int _{s_{n}}^{s} k(u)\, \mathrm{{d}}u = \varphi _{n} + k_{n+1}(s-s_{n}). \end{aligned}$$

(245)

Let $s_{n+1}$ be determined by $k_{n+1}$ as the first value such that $\varphi (s_{n+1})=0$ (equivalently $\theta (s_{n+1})=\pi /2$). Then

$$\begin{aligned} 0= \varphi (s_{n+1}) = \varphi _{n} + k_{n+1}(s_{n+1}-s_{n}) \end{aligned}$$

(246)

so that

$$\begin{aligned} s_{n+1} = s_{n} - \frac{\varphi _{n}}{k_{n+1}}. \end{aligned}$$

(247)

We require in addition that $b(s_{n+1})>0$ (that is, $\gamma$ remains above the $x_{0}$-axis). Using (247) and (228), we obtain

$$\begin{aligned} b(s_{n+1})&= b_{n} + \int _{s_{n}}^{s_{n+1}} \sin (\varphi (s))\, \mathrm{{d}}s = b_{n} - \frac{\cos (\varphi (s_{n+1})) - \cos (\varphi (s_{n})) }{k_{n+1}} \nonumber \\&= b_{n} - \frac{1 - \cos (\varphi (s_{n})) }{k_{n+1}} \end{aligned}$$

(248)

so that $b(s_{n+1})>0$ is equivalent to

$$\begin{aligned} b_{n} - \frac{ 1 - \cos (\varphi (s_{n})) }{k_{n+1}} > 0 \end{aligned}$$

or

$$\begin{aligned} k_{n+1}\cdot b_{n} > 1 - \cos (\varphi (s_{n})). \end{aligned}$$

(249)

On the other hand, $k_{n+1}$ has to be bounded from above in order to guarantee (230). Therefore, we require that

$$\begin{aligned} k_{n+1} < \frac{\sin (\theta (s_{n}))}{2b_{n}}, \end{aligned}$$

or

$$\begin{aligned} k_{n+1}\cdot b_{n} < \frac{\sin (\theta (s_{n}))}{2}. \end{aligned}$$

(250)

Combining (249) and (250) gives conditions for $k_{n+1}$

$$\begin{aligned} 1 - \cos (\varphi (s_{n}))< k_{n+1}\cdot b_{n} < \frac{\sin (\theta (s_{n}))}{2}. \end{aligned}$$

(251)

Since $\sin (\theta (s))=\cos (\varphi (s))$, (251) is equivalent to

$$\begin{aligned} 1 - \cos (\varphi (s_{n}))< k_{n+1}\cdot b_{n} < \frac{\cos (\varphi (s_{n}))}{2}. \end{aligned}$$

(252)

Now, recall that $s_{n}$ was chosen in (239) so that $\varphi (s_{n})=-\pi /4$ so

$$\begin{aligned} 1-\cos (\varphi (s_{n}))=\frac{2-\sqrt{2}}{2} < \frac{\cos (\varphi (s_{n}))}{2} = \frac{\sqrt{2}}{4}. \end{aligned}$$

Now, choose arbitrarily any $\alpha$, satisfying

$$\begin{aligned} \frac{2-\sqrt{2}}{2}< \alpha < \frac{\sqrt{2}}{4} \end{aligned}$$

(253)

and define $k_{n+1}$ by

$$\begin{aligned} k_{n+1}=\alpha /b_{n}. \end{aligned}$$

(254)

With this choice (252), and therefore, (249) and (250) hold.

To ensure property (II), we choose $L>s_{n+1}$ so that $L-s_{n+1}$ is arbitrarily small. We extend $\gamma$ to the interval $[s_{n+1},L]$, where $\gamma$ is a straight horizontal line on $[s_{n+1},L]$ by choosing $k(s)=0$ there. To check that the length of the curve we constructed is $O(\gamma _0)$ we observe that

$$\begin{aligned} s_{n+1} = s_{n} - \varphi _{n}/k_{n+1} = s_{n} + \frac{\pi }{4\alpha }b_{n} \le s_{n} + \frac{\pi }{4\alpha }b_{0} = O(\delta _{0}) \end{aligned}$$

(255)

by (234), (241), and (255).

We note that the choice of L is arbitrary. It will be made explicit in the next step when we construct the curve ${\bar{\gamma }}$, the $C^{\infty }$ version of $\gamma$.

This completes the construction of the continuously differentiable curve $\gamma$ defined on the interval [0, L] satisfying properties (I) through (V).

1.4 Step 3 of the Proof: Construction of $\gamma$, Part 2: From $C^{1}$ to $C^{\infty }$

In this step, barred quantities will refer to the $C^{\infty }$ curve ${\bar{\gamma }}(s)$ to be constructed in this step and all the other quantities related to the construction (for example, ${\bar{\theta }}$, ${\bar{\varphi }}$, ${\bar{k}}(s)$, etc.). Unbarred quantities will refer to the $C^{1}$ curve constructed in the previous step.

The general plan is to replace k(s) as chosen in Step 2 with a smooth version ${\bar{k}}(s)$ as depicted in Fig. 7, which will then define ${\bar{\gamma }}$ by the formulae (227) and (228). Set $k_0=-K^{1/2}$ and modify k(s) on $[s_i,s_{i+1}]$ for $i=0,1,2,\ldots ,n$ so that the graph of ${\bar{k}}(s)$ will connect to the constant function equal to $k_{i}$ smoothly at $s_i$, will rise steeply to the value $k_{i+1}$ in a very short interval $[s_i,s_i+\alpha ]$ and will connect smoothly with constant function equal to $k_{i+1}$ in $[s_i+\alpha , s_{i+1}]$. For each $i=0,1,2,\ldots n$, ${\bar{k}}|[s_i,s_{i+1}]$ can be constructed as follows. Choose and fix a $C^\infty$ function g(s) which is identically 0 for $s<0$, identically 1 for $s>1$, and strictly increasing on [0, 1]. Then ${\bar{k}}| [s_i,s_{i+1}]$ is constructed by appropriate rescaling and translations of the graph of g(s) in both vertical and horizontal directions. The values of $k_i$ and $k_{i+1}$ determine the transformations along the vertical axis but rescaling of the independent variable remains a free parameter $\alpha$ to be set sufficiently small later. We will use the same value of $\alpha$ for every $i=1,2,\ldots n$.

Fig. 7

Graph of the smooth curvature ${\bar{k}}(s)$ with “full bend”

Fig. 8

Graph of the curvature, k(s), with “full bend” as a step function

Since

$$\begin{aligned} \Delta {\bar{\theta }}=\int _0^{s_{n+1}}{\bar{k}}\,\mathrm{{d}}s \le \int _0^{s_{n+1}}{k}\,\mathrm{{d}}s = \Delta \theta , \end{aligned}$$

we loose a small amount of “bend” so that ${{\bar{\theta }}}(s_{n+1}) < \frac{\pi }{2}$ by a very small amount controlled by $\alpha$. We compensate for this by one final extension of ${\bar{k}}$ to an interval $[s_{n+1},L]$ with $L=s_{n+1} + 2\beta$. We choose ${\bar{k}}$ so that it connects smoothly with $k_{n+1}$ at $s_{n+1}$, drops smoothly to zero over $[s_{n+1}, s_{n+1}+\beta ]$ and continues identically zero on $[s_{n+1}+\beta , s_{n+1}+2\beta ]$. $\beta$ and ${\bar{k}}$ are chosen so that

$$\begin{aligned} \int _{s_{n+1}}^{s_{n+1}+\beta } {\bar{k}}(s)\, \mathrm{{d}}s = \frac{\pi }{2} - {\bar{\theta }}(s_{n+1}). \end{aligned}$$

This ensures that ${\bar{\theta }}=\frac{\pi }{2}$ in the interval $[s_{n+1}+\beta , s_{n+1}+2\beta ]$. This final extension is constructed as the preceding ones except that we have to use the reflection $s \mapsto -s$ before rescaling and translating the original function g. We note that $\beta =O(\alpha )$ is determined by the choice of $\alpha$ and the requirement that ${\bar{\theta }}(L)=\frac{\pi }{2}$. We also observe that as $\alpha$ tends to zero, the functions ${\bar{\varphi }}$, ${\bar{\theta }}$, $\bar{x_0}$, and $\bar{x_1}$ will converge uniformly on [0, L] to $\varphi$, $\theta$, $x_0$, and $x_1$, respectively, as follows from (227) and (228).

We now check that the properties (I) through (V) on page (I) hold for the curve ${\bar{\gamma }}$ for sufficiently small choice of $\alpha$. Only (IV) and (V) need a verification. (V) follows since $L = s_{n+1} + 2\beta = O(\delta _0) + O(\alpha )$. To prove (IV) we use the uniform convergence on $[0,s_{n+1}]$ as $\alpha$ approaches 0 of $\frac{\sin {\bar{\theta }}(s)}{2{\bar{x}}_1(s)}$ to $\frac{\sin \theta (s)}{2x_1(s)}$. More precisely, on $[s_i,s_{i+1}]$,

$$\begin{aligned} \frac{\sin {\bar{\theta }}(s)}{2{\bar{x}}_1(s)} -{\bar{k}}(s) = \left( \frac{\sin {\bar{\theta }}(s)}{2{\bar{x}}_1(s)} - k_{i+1}\right) +\left( k_{i+1} -{\bar{k}}(s)\right) . \end{aligned}$$

For sufficiently small $\alpha$, the first term on the right becomes positive by the property (IV) for the curve $\gamma$ while the second term is nonnegative by construction (cf. Fig. 8). Finally, in the last interval $[s_{n+1},L]$ the ratio $\frac{\sin {\bar{\theta }} (s)}{2{\bar{x}}_1(s)}$ is nondecreasing so that

$$\begin{aligned} \frac{\sin {\bar{\theta }}(s)}{2{\bar{x}}_1(s)}\ge \frac{\sin {\bar{\theta }}(s_{n+1})}{2{\bar{x}}_1(s_{n+1})} > k_{n+1} \end{aligned}$$

since the last inequality was verified for $s=s_{n+1}$ already. Property (IV) follows since $k_{n+1} > {\bar{k}}(s)$ in $[s_{n+1},L]$. This finishes the construction of ${\bar{\gamma }}$.

1.5 Step 4 of the Proof: Diameter and Volume Estimates of Lemma 2.1

Given the definition of U in (224), the diameter of U is estimated by

$$\begin{aligned} {\text {Diam}}(U) \le \pi \delta +\delta + 2L = O(\delta )+O(\delta _0)=O(\delta ). \end{aligned}$$

To estimate the volume of $U'$, note that the intersection of $U'$ with the hyperplane $x_0=x_0(s)=c$ for $0<s<L$ is a sphere of two dimensions and of radius $x_1(s)< \delta _0$. It follows by Fubini’s theorem that ${\text {Vol}}(U')=O(\delta _0^3)$. To prove (10) recall that U is obtained from the union of two disjoint balls of radius $\delta$ by removing balls of radius $\delta _0$ and attaching $U'$ along the common boundary (cf. Fig. 1). Since the volumes of the removed balls and of the added tunnel are $O(\delta _0^3)$, the estimate (10) follows by choosing $\delta _0$ sufficiently small depending on $\epsilon$. The estimate (11) is proved in the same way. The proof of Lemma 2.1 is now complete. $\square$