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Pu's inequality

From Wikipedia, the free encyclopedia
Inequality in differential geometry
An animation of theRoman surface representingRP2 inR3

Indifferential geometry,Pu's inequality, proved byPao Ming Pu, relates thearea of an arbitraryRiemannian surface homeomorphic to thereal projective plane with thelengths of the closed curves contained in it.

Statement

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A student ofCharles Loewner, Pu proved in his 1950 thesis (Pu 1952) that every Riemannian surfaceM{\displaystyle M} homeomorphic to thereal projective plane satisfies the inequality

Area(M)2πSystole(M)2,{\displaystyle \operatorname {Area} (M)\geq {\frac {2}{\pi }}\operatorname {Systole} (M)^{2},}

whereSystole(M){\displaystyle \operatorname {Systole} (M)} is thesystole ofM{\displaystyle M}.The equality is attained precisely when the metric has constantGaussian curvature.

In other words, if allnoncontractible loops inM{\displaystyle M} have length at leastL{\displaystyle L}, thenArea(M)2πL2,{\displaystyle \operatorname {Area} (M)\geq {\frac {2}{\pi }}L^{2},} and the equality holds if and only ifM{\displaystyle M} is obtained from a Euclidean sphere of radiusr=L/π{\displaystyle r=L/\pi } by identifying each point with its antipodal.

Pu's paper also stated for the first timeLoewner's inequality, a similar result for Riemannian metrics on thetorus.

Proof

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Pu's original proof relies on theuniformization theorem and employs an averaging argument, as follows.

By uniformization, the Riemannian surface(M,g){\displaystyle (M,g)} isconformally diffeomorphic to a round projective plane. This means that we may assume that the surfaceM{\displaystyle M} is obtained from the Euclidean unit sphereS2{\displaystyle S^{2}} by identifying antipodal points, and the Riemannian length element at each pointx{\displaystyle x} is

dLength=f(x)dLengthEuclidean,{\displaystyle \mathrm {dLength} =f(x)\mathrm {dLength} _{\text{Euclidean}},}

wheredLengthEuclidean{\displaystyle \mathrm {dLength} _{\text{Euclidean}}} is the Euclidean length element and the functionf:S2(0,+){\displaystyle f:S^{2}\to (0,+\infty )}, called theconformal factor, satisfiesf(x)=f(x){\displaystyle f(-x)=f(x)}.

More precisely, the universal cover ofM{\displaystyle M} isS2{\displaystyle S^{2}}, a loopγM{\displaystyle \gamma \subseteq M} is noncontractible if and only if its liftγ~S2{\displaystyle {\widetilde {\gamma }}\subseteq S^{2}} goes from one point to its opposite, and the length of each curveγ{\displaystyle \gamma } is

Length(γ)=γ~fdLengthEuclidean.{\displaystyle \operatorname {Length} (\gamma )=\int _{\widetilde {\gamma }}f\,\mathrm {dLength} _{\text{Euclidean}}.}

Subject to the restriction that each of these lengths is at leastL{\displaystyle L}, we want to find anf{\displaystyle f} that minimizes the

Area(M,g)=S+2f(x)2dAreaEuclidean(x),{\displaystyle \operatorname {Area} (M,g)=\int _{S_{+}^{2}}f(x)^{2}\,\mathrm {dArea} _{\text{Euclidean}}(x),}

whereS+2{\displaystyle S_{+}^{2}} is the upper half of the sphere.

A key observation is that if we average several differentfi{\displaystyle f_{i}} that satisfy the length restriction and have the same areaA{\displaystyle A}, then we obtain a better conformal factorfnew=1n0i<nfi{\displaystyle f_{\text{new}}={\frac {1}{n}}\sum _{0\leq i<n}f_{i}}, that also satisfies the length restriction and has

Area(M,gnew)=S+2(1nifi(x))2dAreaEuclidean(x){\displaystyle \operatorname {Area} (M,g_{\text{new}})=\int _{S_{+}^{2}}\left({\frac {1}{n}}\sum _{i}f_{i}(x)\right)^{2}\mathrm {dArea} _{\text{Euclidean}}(x)}
1ni(S+2fi(x)2dAreaEuclidean(x))=A,{\displaystyle \qquad \qquad \leq {\frac {1}{n}}\sum _{i}\left(\int _{S_{+}^{2}}f_{i}(x)^{2}\mathrm {dArea} _{\text{Euclidean}}(x)\right)=A,}

and the inequality is strict unless the functionsfi{\displaystyle f_{i}} are equal.

A way to improve any non-constantf{\displaystyle f} is to obtain the different functionsfi{\displaystyle f_{i}} fromf{\displaystyle f} usingrotations of the sphereRiSO3{\displaystyle R_{i}\in SO^{3}}, definingfi(x)=f(Ri(x)){\displaystyle f_{i}(x)=f(R_{i}(x))}. If weaverage over all possible rotations, then we get anfnew{\displaystyle f_{\text{new}}} that is constant over all the sphere. We can further reduce this constant to minimum valuer=Lπ{\displaystyle r={\frac {L}{\pi }}} allowed by the length restriction. Then we obtain the obtain the unique metric that attains the minimum area2πr2=2πL2{\displaystyle 2\pi r^{2}={\frac {2}{\pi }}L^{2}}.

Reformulation

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Alternatively, every metric on the sphereS2{\displaystyle S^{2}} invariant under the antipodal map admits a pair of opposite pointsp,qS2{\displaystyle p,q\in S^{2}} at Riemannian distanced=d(p,q){\displaystyle d=d(p,q)} satisfyingd2π4area(S2).{\displaystyle d^{2}\leq {\frac {\pi }{4}}\operatorname {area} (S^{2}).}

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

Filling area conjecture

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An alternative formulation of Pu's inequality is the following. Of all possible fillings of theRiemannian circle of length2π{\displaystyle 2\pi } by a2{\displaystyle 2}-dimensional disk with the strongly isometric property, the roundhemisphere has the least area.

To explain this formulation, we start with the observation that the equatorial circle of the unit2{\displaystyle 2}-sphereS2R3{\displaystyle S^{2}\subset \mathbb {R} ^{3}} is aRiemannian circleS1{\displaystyle S^{1}} of length2π{\displaystyle 2\pi }. More precisely, the Riemannian distance functionofS1{\displaystyle 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 isonly2{\displaystyle 2}, whereas in the Riemannian circle it isπ{\displaystyle \pi }.

We consider all fillings ofS1{\displaystyle S^{1}} by a2{\displaystyle 2}-dimensional disk, such that the metric induced by the inclusion of the circle as the boundary of the disk is the Riemannianmetric of a circle of length2π{\displaystyle 2\pi }. The inclusion of the circle as the boundary is then called a strongly isometric imbedding of the circle.

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

Isoperimetric inequality

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Pu's inequality bears a curious resemblance to the classicalisoperimetric inequality

L24πA{\displaystyle L^{2}\geq 4\pi A}

forJordan curves in the plane, whereL{\displaystyle L} is the length of the curve whileA{\displaystyle 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.

See also

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References

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1-systoles of surfaces
1-systoles of manifolds
Higher systoles
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