# Pu's inequality

In differential geometry, Pu's inequality is an inequality proved by Pao Ming Pu for the systole of an arbitrary Riemannian metric on the real projective plane RP2.

## Statement

A student of Charles Loewner's, P.M. Pu proved in a 1950 thesis (published in 1952) that every metric on the real projective plane ${\mathbb {RP} }^{2}$ satisfies the optimal inequality

$\operatorname {sys} ^{2}\leq {\frac {\pi }{2}}\operatorname {area} ({\mathbb {RP} }^{2}),$ where sys is the systole. The boundary case of equality is attained precisely when the metric is of constant Gaussian curvature.

## Reformulation

A more detailed explanation of this viewpoint may be found at the page Introduction to systolic geometry.

## Filling area conjecture

An alternative formulation of Pu's inequality is the following. Of all possible fillings of the Riemannian circle of length $2\pi$ by a $2$ -dimensional disk with the strongly isometric property, the round hemisphere has the least area.

To explain this formulation, we start with the observation that the equatorial circle of the unit $2$ -sphere $S^{2}\subset {\mathbb {R} }^{3}$ is a Riemannian circle $S^{1}$ of length $2\pi$ . More precisely, the Riemannian distance function of $S^{1}$ is induced from the ambient Riemannian distance on the sphere. Note that this property is not satisfied by the standard imbedding of the unit circle in the Euclidean plane. Indeed, the Euclidean distance between a pair of opposite points of the circle is only $2$ , whereas in the Riemannian circle it is $\pi$ .

We consider all fillings of $S^{1}$ by a $2$ -dimensional disk, such that the metric induced by the inclusion of the circle as the boundary of the disk is the Riemannian metric of a circle of length $2\pi$ . The inclusion of the circle as the boundary is then called a strongly isometric imbedding of the circle.

In 1983 Gromov conjectured that the round hemisphere gives the "best" way of filling the circle even when the filling surface is allowed to have positive genus.

## Isoperimetric inequality

Pu's inequality bears a curious resemblance to the classical isoperimetric inequality

$L^{2}\geq 4\pi A$ for Jordan curves in the plane, where $L$ is the length of the curve while $A$ is the area of the region it bounds. Namely, in both cases a 2-dimensional quantity (area) is bounded by (the square of) a 1-dimensional quantity (length). However, the inequality goes in the opposite direction. Thus, Pu's inequality can be thought of as an "opposite" isoperimetric inequality.