The length of the shortest closed geodesic on positively curved 2-spheres

Abstract

We show that the shortest closed geodesic on a 2-sphere with non-negative curvature has length bounded above by three times the diameter. We prove a new isoperimetric inequality for 2-spheres with pinched curvature; this allows us to improve our bound on the length of the shortest closed geodesic in the pinched curvature setting.

Appendix

Using more advanced techniques from Sturm–Liouville theory we can improve the isoperimetric inequality in Theorem 4.1 when the sphere is not round. This improved version of the inequality is not needed for the proof of Theorem 1.2 that was presented in Sect. 4.

Theorem 5.1

Let $(S^2,g)$ be a $\delta$-pinched metric and denote by $\eta =D\sqrt{K_{min}}$. Then

$$\begin{aligned} \pi A \le \frac{4D^2}{\delta (2-\sin (\eta /2))}. \end{aligned}$$

First note that $2- \sin (\eta /2) \ge 1$ so that this isoperimetric inequality is indeed an improvement on that of Theorem 4.1. Moreover, because $2- \sin (\eta /2) > 1$ when $\eta \ne \pi$ we note that Theorem 5.1 together with Cheng’s [6] rigidity result when $D=\pi /\sqrt{K_{min}}$ provide an alternate proof of the equality case of Theorem 1.2 that does not depend on the Blaschke ideas from Sect. 4.

This new inequality follows as before by combining the inequality due to Hirsch with an improved lower bound on the first eigenvalue. We therefore set out to calculate the linear approximation of $\lambda _1(d,1)$ at $d=\pi$. The main idea is that because the coefficients of the SLP are constant while the domain varies, we can restrict and extend eigenfunctions to compare the values $\lambda _1(d,1)$ as $d\rightarrow \pi$. Let us then define by $\mu (d)$ the SLP eigenvalue $\lambda _1(d,1)$. We then have

Lemma 5.2

For $\pi \ge d >0$

$$\begin{aligned}2 + 2(1-\sin (d/2)) \le \mu (d). \end{aligned}$$

Proof

Define the change of variable $\tanh (u)=\sin (x)$ on the SLP, helpfully suggested to us by Ben Andrews. This becomes

$$\begin{aligned} y'' + \mu .{{\,\mathrm{sech}\,}}^2(u)y = 0,\quad y'(\pm T)=0,\, \quad \tanh (T)=\sin (d/2) \end{aligned}$$

(3)

In particular, for $d=1, T=+\infty$, the eigenvalue $\mu =2$ is realized by $y_{+\infty }=\tanh (u)$.

Recall that the first eigenvalue for a SLP can be written as a Rayleigh quotient

$$\begin{aligned} \mu (d) = \inf _{y\ne 0, y'(\pm T)=0} R[y,T] \end{aligned}$$

where

$$\begin{aligned}R[y,t]= \frac{\int _{-T}^T -y.y''du}{\int _{-T}^T y^2{{\,\mathrm{sech}\,}}^2(u)du} \end{aligned}$$

Denote by $\hat{y}_T$ the eigenfunction of the SLP (3) normalized so $\int _{-T}^T \hat{y}_T^2{{\,\mathrm{sech}\,}}^2(u)du=1$. Extended $\hat{y}_T$ as a constant to the entire real line so (while keeping the same notation):

$$\begin{aligned} \hat{y}_T(u) = {\left\{ \begin{array}{ll} \hat{y}_T(u) &{} \hbox { if}\ |u|\le T \\ \hat{y}_T(T) &{} \hbox { if}\ u>T\\ \hat{y}_T(-T) &{} \hbox { if}\ u<-T \end{array}\right. } \end{aligned}$$

In order to use $\mu (1)\le R[\hat{y}_T,+\infty ]$ we will need to bound $\hat{y}_T(\pm T)$. Observe first that given the symmetries of the SLP (3) the function $\hat{y}_T$ is an odd function. Since $\mu (d)$ is the first eigenvalue, we know that $\hat{y}'_T$ does not vanish in the open interval $(-T,T)$. Hence $|\hat{y}_T(\pm T)| = \max _{-T\le u \le T} |\hat{y}_T(u)|$, so then

$$\begin{aligned} 1=\int _{-T}^T \hat{y}^2_T{{\,\mathrm{sech}\,}}^2(u)du \le |\hat{y}_T(\pm T)|^2 \int _{-T}^T {{\,\mathrm{sech}\,}}^2(u)du \le 2|\hat{y}_T(\pm T)|^2 \end{aligned}$$

Replacing now $R[\hat{y}_T,+\infty ]$

$$\begin{aligned} 2=\mu (\pi )\le R[\hat{y}_T,+\infty ] = \frac{\mu (d)}{1 + 2|\hat{y}_T(T)|^2\int _T^{\infty } {{\,\mathrm{sech}\,}}^2(u)du} \end{aligned}$$

Hence we obtain

$$\begin{aligned} 2 + 2(1-\tanh (T)) \le \mu (d) \end{aligned}$$

Replacing now $\tanh (T)=\sin (d/2)$ and using the Taylor series of sine at $\pi /2$ we finally get

$$\begin{aligned} 2 + 2(1-\sin (d/2)) \le \mu (d). \end{aligned}$$

$\square$

We are now ready to prove Theorem 5.1.

Proof

The proof proceeds exactly as the proof for Theorem 4.1 and we begin by rescaling the metric so that $K_{min}=1$ and by Myers theorem $\eta = D \le \pi$. Lemma 5.2 then gives the bound $2 + 2(1-\sin (d/2)) \le \mu (d) = \lambda _1(d,1)$. Following the ideas of [4] we combine this lower bound on $\lambda _1$ with the upper bound due to Hirsch $\lambda _1 \le \frac{8\pi }{A}$ (see [11, 25]) to yield

$$\begin{aligned} 2 + 2(1-\sin (\eta /2)) \le \frac{8\pi }{A} \end{aligned}$$

which we can manipulate to

$$\begin{aligned} \pi A \le \frac{8\pi ^2}{2 + 2(1-\sin (\eta /2))}. \end{aligned}$$

Klingenberg’s injectivity radius estimate for positive metrics on 2-spheres says that

$$\begin{aligned} D \ge \frac{\pi }{\sqrt{K_{max}}} = \pi \sqrt{\delta }. \end{aligned}$$

We have therefore related both area and diameter to $\pi$, and conclude that

$$\begin{aligned} \pi A \le \frac{8\pi ^2}{2 + 2(1-\sin (\eta /2))} = \frac{4\pi ^2}{K_{min}(2-\sin (\eta /2))} \le \frac{4D^2}{\delta (2-\sin (\eta /2))}. \end{aligned}$$

$\square$

Finally, while not necessary for the proof of Theorem 5.1, the following Lemma of independent interest expands the approach of Lemma 5.2 and gives an idea of the sharpness of its inequality, both in general and at the limiting case of $d=\pi$.

Lemma 5.3

For $\pi \ge d \ge 2\arcsin (\tanh (3/2)) \approx 2.263$

$$\begin{aligned}\mu (d) \le 2 + \cos ^2(d/2) A(d), \end{aligned}$$

where

$$\begin{aligned} A(d) = \frac{6[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]}{\sin ^3(d/2) - 3\cos ^2(d/2)[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]}. \end{aligned}$$

Moreover, at the limiting case $d=\pi$ we have $\mu '(\pi )=0$ and $\frac{1}{4}\le \mu ''(\pi )\le \frac{3}{2}$.

Proof

Define the test function $y_T = \tanh (u) - u{{\,\mathrm{sech}\,}}^2(T)$, which by design satisfies the Neumann initial conditions of (3). By direct calculation

$$\begin{aligned} \begin{aligned} \int _{-T}^T -y_T.y_T''du&= \int _{-T}^T 2\tanh ^2(u){{\,\mathrm{sech}\,}}^2(u)du - {{\,\mathrm{sech}\,}}^2(T)\int _{-T}^T 2u\tanh (u){{\,\mathrm{sech}\,}}^2(u)du\\&= \frac{2}{3}[2\tanh ^3(T)] - {{\,\mathrm{sech}\,}}^2(T)[2\tanh (T) - 2T{{\,\mathrm{sech}\,}}^2(T)] \\ \int _{-T}^T y_T^2{{\,\mathrm{sech}\,}}(u)du&= \int _{-T}^T \left( \tanh ^2(u) -2u\tanh (u)){{\,\mathrm{sech}\,}}^2(T) + u^2{{\,\mathrm{sech}\,}}^4(T)\right) {{\,\mathrm{sech}\,}}^2(u)du\\&= \int _{-T}^T \tanh ^2(u){{\,\mathrm{sech}\,}}^2(u)du -{{\,\mathrm{sech}\,}}^2(T)\int _{-T}^T 2u\tanh (u){{\,\mathrm{sech}\,}}^2(u)du \\&\quad + {{\,\mathrm{sech}\,}}^4(T)\int _{-T}^T u^2{{\,\mathrm{sech}\,}}^2(u)du\\&= \frac{1}{3}[2\tanh ^3(T)] - {{\,\mathrm{sech}\,}}^2(T)[2\tanh (T) - 2T{{\,\mathrm{sech}\,}}^2(T)] \\&\quad + {{\,\mathrm{sech}\,}}^4(T) \int _{-T}^T u^2{{\,\mathrm{sech}\,}}^2(u)du. \end{aligned} \end{aligned}$$

Hence our test function $y_T$ gives us the inequality

$$\begin{aligned} \begin{aligned} \mu (d)&\le \frac{\int _{-T}^T 2\tanh ^2(u){{\,\mathrm{sech}\,}}^2(u)du - {{\,\mathrm{sech}\,}}^2(T)\int _{-T}^T 2u\tanh (u){{\,\mathrm{sech}\,}}^2(u)du}{\int _{-T}^T \tanh ^2(u){{\,\mathrm{sech}\,}}^2(u)du -{{\,\mathrm{sech}\,}}^2(T)\int _{-T}^T 2u\frac{\tanh (u)}{\cosh ^2(u)}du + {{\,\mathrm{sech}\,}}^4(T)\int _{-T}^T u^2{{\,\mathrm{sech}\,}}^2(u)du}\\&\le \frac{\frac{2}{3}[2\tanh ^3(T)] - {{\,\mathrm{sech}\,}}^2(T)[2\tanh (T) - 2T{{\,\mathrm{sech}\,}}^2(T)]}{\frac{1}{3}[2\tanh ^3(T)] - {{\,\mathrm{sech}\,}}^2(T)[2\tanh (T) - 2T{{\,\mathrm{sech}\,}}^2(T)]} \end{aligned} \end{aligned}$$

where the last inequality is valid if $T\ge \frac{3}{2}$, so for $d\ge 2\arcsin (\tanh (3/2))\approx 2.263$.

Using that ${{\,\mathrm{sech}\,}}^2(T)=1-\tanh ^2(T) = 1-\sin ^2(d/2)=\cos ^2(d/2)$ we have the desired inequality:

$$\begin{aligned} \begin{aligned} \mu (d)&\le \frac{2\sin ^3(d/2) - 3\cos ^2(d/2)[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]}{\sin ^3(d/2) - 3\cos ^2(d/2)[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]}\\&= 2 + \cos ^2(d/2) \frac{6[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]}{\sin ^3(d/2) - 3\cos ^2(d/2)[\sin (d/2) - {{\,\mathrm{arctanh}\,}}(\sin (d/2))\cos ^2(d/2)]} \end{aligned} \end{aligned}$$

Turning our attention to the limiting case of $d=\pi$, we Taylor expand cosine at $\pi /2$ to yield

$$\begin{aligned} \mu (d) \le 2 + \frac{3}{2}(\pi -d)^2 + O((\pi -d)^3). \end{aligned}$$

Similarly, we combine Lemma 5.2 and the Taylor expansion of sine at $\pi /2$ to yield

$$\begin{aligned} 2 + \frac{1}{4}(\pi -d)^2 + O((\pi -d)^3) \le \mu (d). \end{aligned}$$

From this pair of Taylor expansions we calculate that

$$\begin{aligned} \mu '(\pi )=0,\quad \frac{1}{4} \le \mu ''(\pi ) \le \frac{3}{2} . \end{aligned}$$

$\square$