A Systolic Inequality with Remainder in the Real Projective Plane

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Lemma 2.1. The maps p and q on (M,gM ) are Riemannian submersions, over the unit sphere S2 R3. ⊆ Proof. For the projection p, given (v,w) M, the vector (0, n) as defined above is tangent to the fiber p−1(v).∈ The other two vectors, (n, 0) and (w, v), are thus an orthonormal basis for the subspace of T(v,w)M normal to− the fiber, and are mapped by dp to the orthonormal basis n, w 2 of TvS . The projection p maps the fiber q−1(w) onto a great circle of S2. This map preserves length since the unit vector (n, 0), tangent to the −1 2 fiber q (w) at (v,w), is mapped by dp to the unit vector n TvS . The same comments apply when the roles of p and q are reversed.∈ In the following proposition, integration takes place respectively over great circles C S2, over the fibers in M, over S2, and over M. The integration⊆ is always with respect to the volume element of the given Riemannian metric. Since p and q are Riemannian submersions by Lemma 2.1, we can use Fubini’s Theorem to integrate over M by integrating first over the fibers of either p or q, and then over S2; cf. [18, Lemma 4]. By the remarks above, if C = p(q−1(w)) and f : S2 R → then −1 f p = f. q (w) ◦ C R R SYSTOLIC INEQUALITY WITH REMAINDER IN PROJECTIVE PLANE 3

Proposition 2.2. Given a continuous function f : S2 R+, we de- ﬁne m R by setting m = min f : C S2 a great circle→ . Then ∈ { C ⊆ } m2 1 R 2 f f 2, π ≤ 4π ZS2 ≤ ZS2 where equality in the second inequality occurs if and only if f is constant. Proof. Using the fact that M is the total space of a pair of Riemannian submersions, we obtain 1 f = f p ZS2 ZS2 2π Zp−1(v) ◦ 1 = f p 2π ZM ◦ 1 = f p 2π ZS2 Zq−1(w) ◦ 1 m =2m, ≥ 2π ZS2 proving the ﬁrst inequality. By the Cauchy–Schwarz inequality, we have 2 1 f 4π f 2, ZS2 · ≤ ZS2 proving the second inequality. Here equality occurs if and only if f and 1 are linearly dependent, i.e., if and only if f is constant.

2 1 2 We deﬁne the quantity V by setting V = 2 f 2 f . Then f f S − 4π S Proposition 2.2 can be restated as follows. R R Corollary 2.3. Let f : S2 R+ be continuous. Then → 2 2 m f Vf 0, ZS2 − π ≥ ≥ and Vf =0 if and only if f is constant. Proof. The proof is obtained from Proposition 2.2 by noting that a b c if and only if c a c b 0. ≤ ≤ − ≥ − ≥ We can assign a probabilistic meaning to Vf as follows. Divide the area measure on S2 by 4π, thus turning it into a probability measure µ. A function f : S2 R+ is then thought of as a random variable with → 1 expectation Eµ(f)= 4π S2 f. Its variance is thus given by R 2 2 2 1 2 1 1 Varµ(f)= Eµ(f ) Eµ(f) = f f = Vf . − 4π Z 2 − 4π Z 2 4π S S 4 MIKHAILG.KATZANDTAHLNOWIK

The variance of a random variable f is non-negative, and it vanishes if and only if f is constant. This reproves the corresponding properties of Vf established above via the Cauchy–Schwarz inequality. Now let g0 be the metric of constant Gaussian curvature K = 1 2 2 2 on RP . The double covering ρ: S (RP ,g0) is a local isometry. Each projective line C RP2 is the→ image under ρ of a great circle of S2. ⊆ Proposition 2.4. Given a function f : RP2 R+, we deﬁne m¯ R by setting m¯ = min f : C RP2 a projective→ line . Then ∈ { C ⊆ } R 2¯m2 1 2 f f 2, π ≤ 2π ZRP2 ≤ ZRP2 where equality in the second inequality occurs if and only if f is constant. Proof. We apply Proposition 2.2 to the composition f ρ. Note that ◦ we have −1 f ρ = 2 f and 2 f ρ = 2 2 f. The condition ρ (C) ◦ C S ◦ RP for f toR be constant holdsR since fRis constant ifR and only if f ρ is constant. ◦

2 2 1 2 1 For RP we deﬁne V¯ = 2 f 2 f = V ◦ . We obtain f RP − 2π RP 2 f ρ the following restatement ofR Proposition 2.4.R Corollary 2.5. Let f : RP2 R+ be a continuous function. Then → 2 2 2¯m f V¯f 0, ZRP2 − π ≥ ≥ where V¯f =0 if and only if f is constant. 1 RP2 Relative to the probability measure induced by 2π g0 on , we 1 1 2 ¯ have E(f)= 2π RP f, and therefore Var(f)= 2π Vf , providing a probabilistic meaningR for the quantity V¯f , as before. By the uniformization theorem, every metric g on RP2 is of the 2 form g = f g0 where g0 is of constant Gaussian curvature +1, and 2 + 2 the function f : RP R is continuous. The area of g is 2 f , → RP and the g-length of a projective line C is C f. Let L be the shortestR length of a noncontractible loop. Then LR m¯ wherem ¯ is deﬁned in Proposition 2.4, since a projective line in≤RP2 is a noncontractible RP2 2L2 ¯ loop. Then Corollary 2.5 implies area( ,g) π Vf 0. RP2 2L2 ¯ − ≥ ≥ If area( ,g)= π then Vf = 0, which implies that f is constant, by Corollary 2.5. Conversely, if f is a constant c, then the only geodesics 2 2L 2 are the projective lines, and therefore L = cπ. Hence π = 2πc = SYSTOLIC INEQUALITY WITH REMAINDER IN PROJECTIVE PLANE 5 area(RP2). We have thus completed the proof of the following result strengthening Pu’s inequality. Theorem 2.6. Let g be a Riemannian metric on RP2. Let L be the shortest length of a noncontractible loop in (RP2,g). Let f : RP2 R+ 2 → be such that g = f g0 where g0 is of constant Gaussian curvature +1. Then 2L2 area(g) 2π Var(f), − π ≥ where the variance is with respect to the probability measure induced 1 2L2 by 2π g0. Furthermore, equality area(g)= π holds if and only if f is constant.

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M. Katz, Department of Mathematics, Bar Ilan University, Ramat Gan 5290002 Israel E-mail address: [email protected] T. Nowik, Department of Mathematics, Bar Ilan University, Ramat Gan 5290002 Israel E-mail address: [email protected]