Nonpositively Curved Surfaces are Loewner

Abstract

We show that every closed nonpositively curved surface satisfies Loewner’s systolic inequality. The proof relies on a combination of the Gauss–Bonnet formula with an averaging argument using the invariance of the Liouville measure under the geodesic flow. This enables us to find a disk with large total curvature around its center yielding a large area.

1 Introduction

The systole of a closed nonsimply connected surface M endowed with a Riemannian metric, denoted by ${{\,\mathrm{\textrm{sys}}\,}}(M)$, is defined as the length of the shortest noncontractible loop of M. It is attained by the length of a noncontractible closed geodesic. The systolic area of M is defined as

$$\begin{aligned} \sigma (M) = \frac{{{\,\mathrm{\textrm{area}}\,}}(M)}{{{\,\mathrm{\textrm{sys}}\,}}(M)^2}. \end{aligned}$$

We will say that M is Loewner if its systolic area satisfies

$$\begin{aligned} \sigma (M) \ge \frac{\sqrt{3}}{2} \approx 0.866. \end{aligned}$$

The first systolic inequality, due to Loewner, asserts that every metric on the torus ${\mathbb {T}}^2$ is Loewner; see [13]. By [8], all surfaces with a Riemannian metric in a hyperelliptic conformal class are Loewner. In particular, every genus two surface is Loewner since every conformal class in genus two is hyperelliptic. By [7], surfaces of genus at least 20 are Loewner. Pushing this technique further, Li and Su announce that this still holds true for surfaces of genus 18 and 19, and even for surfaces of genus $\ge 11$ when one restricts to nonpositively curved metrics; see [12]. For the other cases the problem is still open.

In this article, we resolve this problem for nonpositively curved surfaces of any genus (as well as for nonorientable surfaces) relying on a new approach.

Theorem 1.1

Every closed nonpositively curved surface is Loewner.

Actually, we prove a stronger statement; see Proposition 3.3. Let M be a closed nonpositively curved surface. For every $r \in [0,\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)]$, there is a ball $B(r) \subseteq M$ of radius r with

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B(r) \ge \pi r^2 - \frac{2\pi \chi (M)}{{{\,\mathrm{\textrm{area}}\,}}(M)} \cdot \frac{\pi }{12} r^4. \end{aligned}$$

In particular, for $r=\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)$, we have

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B\left( \tfrac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)\right) \ge \left( \frac{\pi }{4} - \frac{\pi ^2 \chi (M)}{96} \cdot \frac{{{\,\mathrm{\textrm{sys}}\,}}(M)^2}{{{\,\mathrm{\textrm{area}}\,}}(M)} \right) {{\,\mathrm{\textrm{sys}}\,}}(M)^2. \end{aligned}$$

Now, if M is not a torus, we can show (see Corollary 4.2) that its systolic area is at least

$$\begin{aligned} \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{7}{3}} \right) \approx 0.992 \end{aligned}$$

which is enough to conclude.

Our strategy differs from previous works. Here is the rough idea. Suppose that M is a genus g surface which is not Loewner. Normalize the metric so that ${{\,\mathrm{\textrm{sys}}\,}}(M)=1$. Since the metric is nonpositively curved, every disk of radius $\frac{1}{4}$ has area at least $\frac{\pi }{16}$, which represents at least $\frac{\pi }{16} /\frac{\sqrt{3}}{2} \approx 22.6\%$ of the total area. Now, by an averaging argument using the invariance of the Liouville measure under the geodesic flow, one should be able to find a disk D of radius $\frac{1}{4}$ with at least $22.6\%$ of the total curvature. The disk $D_+$ of radius $\frac{1}{2}$ centered at the same point has a lot of (negative) curvature around its center, namely in D. This should force the disk $D_+$ to have a lot of area contradicting the $\frac{\sqrt{3}}{2}$-bound on the surface area. At implementation level, the existence of a curvature-rich disk relies on an integral-geometric formula relating the weighted average of the curvature K on tangent disks in TM with the Euler characteristic and a comparison result between this weighted average and another weighted average of K on metric disks in M. The area lower bound on this curvature-rich disk follows from an expression relating the area of this disk with the previous weighted average of K on the same disk.

Recent progress in systolic geometry includes [6] and [11].

2 Disks and Curvature

Let M be a closed surface of nonpositive Euler characteristic $\chi (M)$ endowed with a Riemannian metric. Let $\pi :UM \rightarrow M$ be the canonical projection defined on the unit tangent bundle UM of M. We will sometimes denote a unit tangent vector $u \in UM$ by $u_x \in U_xM$ when we want to emphasize its basepoint $x =\pi (u)$ in M.

The Liouville measure on UM decomposes as

$$\begin{aligned} du = dx \, du_x \end{aligned}$$

(2.1)

where dx is the area measure of M and $du_x$ is the canonical length measure of $U_xM$; see [2, §1.M]. Note that the Liouville measure is invariant under the geodesic flow $\varphi _t:UM \rightarrow UM$ of M and that

$$\begin{aligned} {{\,\mathrm{\textrm{vol}}\,}}(UM) = 2 \pi {{\,\mathrm{\textrm{area}}\,}}(M). \end{aligned}$$

Let $K$ be the Gaussian curvature of M. The Gauss–Bonnet formula for a domain D of M with piecewise smooth boundary $\partial D$ can be written

$$\begin{aligned} \int _D K(x) \, dx + \int _{\partial D} \kappa (s) \, ds + \tau _{\partial D} = 2\pi \chi (D) \end{aligned}$$

(2.2)

where $\chi (D)$ is the Euler characteristic of D, $\kappa$ is the geodesic curvature of $\partial D$ and $\tau _{\partial D}$ is the sum of the angular differences of the tangent vectors at the corner points of $\partial D$. When $D=M$, the Gauss–Bonnet formula for M takes the form

$$\begin{aligned} \int _M K(x) \, dx = 2\pi \chi (M). \end{aligned}$$

(2.3)

It will be convenient to introduce the function

$$\begin{aligned} \overline{K}:UM \rightarrow {\mathbb {R}}\end{aligned}$$

defined as $\overline{K}=K \circ \pi$.

The results of this section can be summarized as follows. First, there is an integral-geometric formula relating a weighted average of $\overline{K}$ of certain horizontal disks of radii at most r in UM in terms of the Euler characteristic and r. Furthermore, the curvature condition yields an inequality between this weighted average and the average of K of certain disks in M. The inequality between the two averages entails the existence of a curvature-rich disk.

The following function will play a key role in our approach.

Definition 2.1

Let $r>0$. Define $F_r:M \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} F_r(x) = \int _0^r (r-\rho ) \int _{U_xM} \int _0^\rho \overline{K}(\varphi _t(u_x))) \, t \, dt \, du_x \, d\rho . \end{aligned}$$

Let us compute the integral of $F_r$.

Lemma 2.2

We have

$$\begin{aligned} \int _{M} F_r(x) \, dx = 2\pi \chi (M) \cdot \frac{\pi r^4}{12}. \end{aligned}$$

Proof

By the Liouville measure decomposition Eq. (2.1), we derive

$$\begin{aligned} \int _{M} F_r(x) \, dx&= \int _{M} \int _0^r (r-\rho ) \int _{U_xM} \int _0^\rho \overline{K}(\varphi _t(u_x))) \, t \, dt \, du_x \, d\rho \, dx \\&= \int _0^r (r-\rho ) \int _0^\rho \int _{UM} \overline{K}(\varphi _t(u))) \, du \, t \, dt \, d\rho . \end{aligned}$$

Now, by the invariance of the Liouville measure under the geodesic flow, we obtain

$$\begin{aligned} \int _{M} F_r(x) \, dx&= \int _0^r (r-\rho ) \int _0^\rho \int _{UM} \overline{K}(u)) \, du \, t \, dt \, d\rho \\&= 2 \pi \, \int _{M} K(x) \, dx \, \int _0^r (r-\rho ) \int _0^\rho t \, dt \, d\rho \\&= 2\pi \chi (M) \cdot \frac{\pi r^4}{12} \end{aligned}$$

where the last equality follows from the Gauss–Bonnet formula Eq. (2.3). $\square$

Let us introduce the following function in connection with the area lower bound in Proposition 3.1.

Definition 2.3

Let $r \in (0,\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M))$. Define $G_r:M \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} G_r(x) = \int _0^r (r-\rho ) \int _{B_x(\rho )} K(y) \, dy \, d\rho \end{aligned}$$

where $B_x(\rho )$ is the ball of radius $\rho$ centered at x.

The functions $F_r$ and $G_r$ are related through the following comparison result.

Lemma 2.4

Suppose M is nonpositively curved. Let $r\in (0,\frac{1}{2}{{\,\mathrm{\textrm{sys}}\,}}(M))$. Then, for every $x \in M$,

$$\begin{aligned} F_r(x) \ge G_r(x). \end{aligned}$$

Proof

By Gauss’ Lemma and since $r < \frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)$, the exponential map at x in polar coordinates induces a diffeomorphism

$$\begin{aligned} \begin{aligned} U_xM \times (0,r]&\longrightarrow B_x(r) \setminus \{x\} \\ (u_x,t)&\longmapsto y=\pi (\varphi _t(u_x)) \end{aligned} \end{aligned}$$

Since M is nonpositively curved, this map is distance nondecreasing by Rauch’s comparison theorem, see [4, §1.11]. It follows that

$$\begin{aligned} t \, dt \, du_x \le dy. \end{aligned}$$

After integration and since $K\le 0$, this implies

$$\begin{aligned} \int _0^r (r-\rho ) \int _{U_xM} \int _0^\rho \overline{K}(\varphi _t(u_x))) \, t \, dt \, du_x \, d\rho \ge \int _0^r (r-\rho ) \int _{B_x(\rho )} K(y) \, dy \, d\rho . \end{aligned}$$

That is, $F_r(x) \ge G_r(x)$. $\square$

We can now derive our key estimate.

Proposition 2.5

Assume M is nonpositively curved. Let $r \in (0,\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M))$. Then there exists $x_0 \in M$ such that

$$\begin{aligned} G_r(x_0) \le \frac{2\pi \chi (M)}{{{\,\mathrm{\textrm{area}}\,}}(M)} \cdot \frac{\pi r^4}{12}. \end{aligned}$$

Proof

Taking the average integral over M in Lemma 2.4 leads to

By Lemma 2.2, this yields

and the result immediately follows. $\square$

3 Disks and Area

In this section, we show that disks with large (negative) curvature have a large area, and proceed to the proof of the main theorem.

We will need the following area lower bound.

Proposition 3.1

Let $r \in (0,\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M))$. Then

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B_x(r) \ge \pi r^2 - G_r(x). \end{aligned}$$

(3.1)

Proof

First, approximate the metric on M by a real analytic metric. For this new metric, the component $\mathcal {C}_x(s)$ of the circle of radius $s \le r$ centered at x surrounding its center is a piecewise smooth curve. Moreover, the length function $s \mapsto L(\mathcal {C}_x(s))$ is differentiable except for a finite number of values of s, and its derivative is given by the first variation formula. Specifically, as long as $B_x(s)$ is nonempty, we have

$$\begin{aligned} L'(\mathcal {C}_x(s)) = \int _{\mathcal {C}_x(s)} \kappa (t) \, dt + \tau _s \end{aligned}$$

for almost every s, where $\kappa$ is the geodesic curvature of the curve $\mathcal {C}_x(s)$ and $\tau _s$ is the sum of the angular difference of the tangent vectors at the corner points of $\mathcal {C}_x(s)$. By the Gauss–Bonnet formula Eq. (2.2) for domains with boundary and since $\mathcal {C}_x(s)$ bounds a topological disk $\mathcal {D}_x(s)$, we derive

$$\begin{aligned} L'(\mathcal {C}_x(s)) = 2 \pi - \int _{\mathcal {D}_x(s)} K(y) \, dy. \end{aligned}$$

Integrating this relation twice and using the coarea formula lead to

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}\mathcal {D}_x(r)&\ge \pi r^2 - \int _0^r \int _0^\rho \int _{\mathcal {D}_x(s)} K(y) \, dy \, ds \, d\rho \\&\ge \pi r^2 - \int _0^r (r-\rho ) \, \int _{\mathcal {D}_x(\rho )} K(y) \, dy \, d\rho . \nonumber \end{aligned}$$

(3.2)

To conclude, simply observe that, when the real analytic metric approaches the initial metric on M, the domains $\mathcal {D}_x(\rho )$ Hausdorff converge to the balls $B_x(\rho )$ (uniformly in $\rho$). In particular, the area of $\mathcal {D}_x(r)$ converges to the one of $B_x(r)$, and $\int _{\mathcal {D}_x(\rho )} K(y) \, dy$ uniformly converges to $\int _{B_x(\rho )} K(y) \, dy$. Hence,

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B_x(r) \ge \pi r^2 - \int _0^r (r-\rho ) \, \int _{B_x(\rho )} K(y) \, dy \, d\rho . \end{aligned}$$

$\square$

Remark 3.2

When the surface M is nonpositively curved, there is an equality in (3.1). Indeed, in this case, we can approximate the metric by a real analytic metric of negative curvature. In this case, the curve $\mathcal {C}_x(s)$ represents the circle of radius s around x and the domain $\mathcal {D}_x(s)$ it surrounds coincides with the metric ball of radius s around x. Because of the curvature condition, this ball is a topological disk. In this case, the coarea formula leads to an equality in Eq. (3.2) and in the following inequalities.

Putting everything together, we obtain the following area lower bound.

Proposition 3.3

Let $r\in \left( 0,\frac{1}{2}{{\,\mathrm{\textrm{sys}}\,}}(M)\right)$ where M is a nonpositively curved surface. Then there exists a ball B(r) of radius r with

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B(r) \ge \pi r^2 - \frac{2\pi \chi (M)}{{{\,\mathrm{\textrm{area}}\,}}(M)} \cdot \frac{\pi }{12} r^4. \end{aligned}$$

Proof

By Proposition 2.5, there exists $x_0 \in M$ such that

$$\begin{aligned} G_r(x_0) \le \frac{2\pi \chi (M)}{{{\,\mathrm{\textrm{area}}\,}}(M)} \cdot \frac{\pi r^4}{12}. \end{aligned}$$

By Proposition 3.1, it follows that

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}B_{x_0}(r) \ge \pi r^2 - \frac{2\pi \chi (M)}{{{\,\mathrm{\textrm{area}}\,}}(M)} \cdot \frac{\pi }{12} r^4. \end{aligned}$$

$\square$

We can conclude the proof of Theorem 1.1 as follows.

Proof of Theorem 1.1

We apply Proposition 3.3 with $r=\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)$. Since

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}(M) \ge {{\,\mathrm{\textrm{area}}\,}}B\left( \tfrac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)\right) , \end{aligned}$$

(3.3)

we obtain

$$\begin{aligned} {{\,\mathrm{\textrm{area}}\,}}(M) \ge \left( \frac{\pi }{4} - \frac{\pi ^2 \chi (M)}{96} \cdot \frac{{{\,\mathrm{\textrm{sys}}\,}}(M)^2}{{{\,\mathrm{\textrm{area}}\,}}(M)} \right) {{\,\mathrm{\textrm{sys}}\,}}(M)^2. \end{aligned}$$

Expressed in terms of the systolic area, this inequality takes the following form

$$\begin{aligned} \sigma (M) \ge \lambda + \frac{\mu }{\sigma (M)}, \end{aligned}$$

where $\lambda = \frac{\pi }{4}$ and $\mu = - \frac{\pi ^2 \chi (M)}{96} \ge 0$. That is, $\sigma (M)^2 - \lambda \, \sigma (M) - \mu \ge 0$. Hence,

$$\begin{aligned} \sigma (M) \ge \frac{\lambda + \sqrt{\lambda ^2 + 4 \mu }}{2}. \end{aligned}$$

It follows that

$$\begin{aligned} \sigma (M) \ge \frac{\pi }{8} \left( 1 + \sqrt{1-\tfrac{2}{3} \chi (M)} \right) \end{aligned}$$

(3.4)

for $\chi (M) \le -1$. Hence $\sigma (M)\ge \frac{\pi }{8}\left( 1+\sqrt{\tfrac{5}{3}}\right) \approx 0.899$ and the surface M is Loewner in this case. For the Klein bottle (where $\chi =0$), the minimal value of the systolic area over nonpositively curved metrics is attained by a square flat metric and is equal to 1. (Alternatively, the minimal value of the systolic area of a Riemannian Klein bottle is equal to $\frac{2 \sqrt{2}}{\pi } \approx 0.9$; see [1].) Hence, every closed nonpositively curved surface is Loewner. $\square$

4 Corollaries

In [8, 9], we computed the least value of the systolic area over all nonpositively curved metrics on the genus 2 surface $\Sigma _2$ and Dyck’s surface $3{\mathbb {R}\mathbb {P}}^2$. For the genus 3 surface, Calabi [3] presented a CAT(0) piecewise flat metric with the lowest systolic area we know of. See Remark 4.7 below. Combined with the systolic inequality Eq. (3.4), this leads to the following result.

Corollary 4.1

We obtain the following data, where the third column represents the minimal value of the systolic area over nonpositively curved metrics.

M	$\chi (M)$	$\sigma _{\le 0}(M)$	Source
${\mathbb {T}}^2$	0	$\frac{\sqrt{3}}{2} \approx 0.866$	[13]
${\mathbb {K}}^2$=$2{\mathbb {R}\mathbb {P}}^2$	0	1	Obvious
$3{\mathbb {R}\mathbb {P}}^2$	$-1$	1 + $\frac{(169-38 \sqrt{19})^{\frac{1}{2}}}{12} \approx 1.152$	[9]
$\Sigma _{2}$	$-2$	$3(\sqrt{2}-1) \approx 1.242$	[8]
$\Sigma _{3}$	$-4$	$< \frac{7 \sqrt{3}}{8} \approx 1.515$	[3]
$\Sigma _{g \ge 3}$	2–2g	$\ge \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{4g-1}{3}} \right) \ge 1.144$	(3.4)
$\Sigma _{g \ge 4}$	2–2g	$\ge \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{4g-1}{3}} \right) \ge 1.270$	(3.4)
$n {\mathbb {R}\mathbb {P}}^2$ (with $n \ge 4$)	2–n	$\ge \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{2n-1}{3}} \right) \ge 0.992$	(3.4)
$n {\mathbb {R}\mathbb {P}}^2$ (with $n \ge 7$)	2–n	$\ge \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{2n-1}{3}} \right) \ge 1.210$	(3.4)

We obtain the following consequences.

Corollary 4.2

Every closed nonpositively curved surface M other than a torus has a systolic area at least

$$\begin{aligned} \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{7}{3}} \right) \approx 0.992. \end{aligned}$$

Corollary 4.3

Every orientable nonpositively curved surface $\Sigma _g$ of genus $g \ge 2$ has a systolic area at least

$$\begin{aligned} \frac{\pi }{8} \left( 1 + \sqrt{\tfrac{11}{3}} \right) \approx 1.144. \end{aligned}$$

Definition 4.4

A Loewner disk of a closed nonsimply connected surface M with a Riemannian metric is a disk of radius $\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)$ and area at least $\frac{\sqrt{3}}{2} {{\,\mathrm{\textrm{sys}}\,}}(M)^2$.

A closed hyperbolic surface of sufficiently small systole does not contain any Loewner disk. All disks of radius half the systole have area close to that of a Euclidean disk of the same radius. However, the existence of Loewner disks is guaranteed in the following cases.

Corollary 4.5

Let $\Sigma _g$ be a nonpositively curved surface of genus $g\ge 2$.

• (1) Let $C=\frac{\pi ^2}{12(2 \sqrt{3}-\pi )}\approx 2.5502$. If $\sigma (\Sigma _g) \le C(g-1)$ then $\Sigma _g$ contains a Loewner disk.

• (2) If $\Sigma _g$ is systolically extremal then it contains a Loewner disk.

Proof

The first part of the corollary is immediate from Proposition 3.3 with $r=\frac{1}{2} {{\,\mathrm{\textrm{sys}}\,}}(\Sigma _g)$.

For the second part, consider a rectangle $R=[0,2g-1] \times [0,1]$ in the plane. Subdivide each edge of the rectangle into unit length segments. The rectangle R can be seen as a 4g-gon with unit sides labeled circularly $a_1,\cdots ,a_{2g},\bar{a}_1,\cdots ,\bar{a}_{2g}$. Identifying every side $a_i$ of this 4g-gon with its corresponding side $\bar{a}_i$ by a plane translation, we obtain a translation surface which is a piecewise flat genus g surface $\Sigma _g$ with a single conical singularity (which can be approximated by a nonpositively curved metric); see [5, p. 21]. It has area $2g-1$ and systole 1. Therefore, every nonpositively curved systolically extremal surface of genus $g\ge 3$ has systolic area at most $2g-1$ and contains a Loewner disk by the first part of the corollary. For $g=2$, we use the minimal value of the systolic area given in Corollary 4.1. $\square$

Remark 4.6

Similarly, one can show that every nonpositively curved systolically extremal surface of negative Euler characteristic contains a Loewner disk.

Remark 4.7

Observe that nonpositively curved surfaces of genus $g \ge 4$ have a systolic area greater than the minimal systolic area of a nonpositively curved surface of genus 2, which is equal to $3(\sqrt{2}-1) \approx 1.242$; see [8]. It would be interesting to know if this still holds true in genus 3, showing a monotonicity of the minimal systolic area in terms of the genus for low genera. Observe that the best value of the systolic area we know of in genus 3 is given by a CAT(0) piecewise flat surface in the conformal class of the Klein quartic described by Calabi [3] and is equal to

$$\begin{aligned} \frac{7 \sqrt{3}}{8} \approx 1.515. \end{aligned}$$

(4.1)

This metric is a critical with respect to some metric variations (see [14]) and might be extremal among all metrics without any curvature assumption. It is not surprising that such a metric is piecewise flat since optimal CAT(0) metrics are flat with finitely many singularities in every genus by [10]. Note that Calabi’s surface has only a slightly better systolic area than the one given by the triangle hyperbolic surface (2, 3, 12) of the same genus described by Schmutz [15] (and conjectured extremal among hyperbolic metrics), which is equal to

$$\begin{aligned} \frac{2\pi }{{{\,\textrm{arcsinh}\,}}(2+\sqrt{3})^2} \approx 1.528. \end{aligned}$$

It would be interesting to find the optimal CAT(0) metric in the conformal class of the surface described by Schmutz (or at least find a good approximation by piecewise flat metrics) to see if it has a lower systolic area than the one of Calabi’s surface.

Similarly, nonpositively curved surfaces $n{\mathbb {R}\mathbb {P}}^2$ with $n \ge 7$ have a systolic area greater than the minimal systolic area of a nonpositively curved Dyck’s surface $3{\mathbb {R}\mathbb {P}}^2$, which is equal to

$$\begin{aligned} 1 + \frac{(169-38 \sqrt{19})^{\frac{1}{2}}}{12} \approx 1.152 \end{aligned}$$

(see [9]) and the question is open for $n=4,5,6$.