Filling Area I

M. Heuer ([email protected]) June 14, 2011

In this lecture, we will be learning about the filling volume of a Rieman- nian manifold. For a compact surface embedded in R3, this can easily be imagined as filling up the inside of it with water and hence calculating the volume of the resulting 3-dimensional manifold (but with a different metric than the usual one). However, we will use a more general approach by considering n-dimensional manifolds and a so-called filling by a manifold of higher dimension having the surface as its boundary. In particular, we will learn how to calculate the filling volume for spheres. Moreover, we will get to know an estimate of the displacement of a point in relation to the area of the hyperelliptic surface.

1 Preliminaries

– As we have seen in the preceding lectures, for a Riemann surface Σ and a hyperelliptic involution J, we have a conformal branched 2-fold covering Q :Σ → S2 of the sphere S2

2 π – Pu’s inequality: systπ1(G) ≤ 2 area(G) – A closed Riemannian surface Σ is called ovalless real if it possesses a ﬁxed point free, antiholomorphic involution τ.

2 Orbifolds

Deﬁnition 2.1. – An action of a group G on a topological space X is called faithful if there exists an x ∈ X for all g ∈ G with g 6= e, e being the neutral element, such that g · x 6= x. – A group action of a group G on a topological space X is cocompact if X/G is compact.

Seminar “Systolic Geometry“, SS 2011, University of G¨ottingen

1 2 2 Orbifolds

We will now get to know a more general definition of a manifold. Definition 2.2. A smooth n-dimensional orbifold is a pair (X,A) corresponding to a Hausdorff paracompact topological space X called the underlying space and an atlas A which consists of

– a covering {Ui ⊂ X|i ∈ I} of X such that the open sets Ui are closed under ﬁnite intersection;

n – a collection of open sets {Vi ⊂ R |i ∈ I};

– ﬁnite groups of diﬀeomorphisms {Γi|i ∈ I} acting on Vi whereas the Vi are invariant under a faithful action of Γi;

– a collection of homeomorphisms ϕi : Ui → Vi/Γi. Furthermore, A has the following properties:

– For every inclusion Ui ⊂ Uj there is a smooth embedding ψij : Vi → Vj called a gluing map and a monomorphism fij :Γi → Γj such that ψij is Γi-equivariant, i.e. it commutes with the action of Γi. −1 – The gluing maps are compatible with the atlas, i.e. ψij = ϕj ◦ ϕi. An orbifold is called good if the underlying space X can be obtained by the quotient of a manifold M and a group Γ where the action of Γ is discrete and cocompact. An orbifold is bad if this is not the case. Remark 2.3. If the group Γ is a symmetry operation which does not possess a ﬁxed points, then X is a manifold. If there are one or more ﬁxed point, the result is an orbifold with singular points. Example 2.4.

– a manifold with boundary, where the Γi are the trivial groups; 2 – Consider the upper hemisphere of the sphere S . By choosing a point w0 and its corresponding antipodal point, we can fold the boundary together by keeping these two points fixed and gluing the opposite semicircles together. The result is an orbifold which looks like an American football (cf. fig. 1). We will consider the last example in more detail later on. Definition 2.5. A Riemannian metric G on an orbifold (X,A) is the usual metric on X\Σ0 where Σ0 denotes the set of singular points of the orbifold. When G is lifted to the neighborhoods Vi, we obtain a Γi-invariant metric for all Vi, i ∈ I and thus for the orbifold (X,A). Definition 2.6. A pullback metric on an orbifold X corresponding to a differentiable map Q : X → Σ where (Σ, G) is a Riemannian manifold, is defined as

∗ (Q G)(v, w) = G(dQ(v), dQ(w)) = G(Q∗(v),Q∗(w)) ∀v, w ∈ TpX. When we have a metric H = Q∗G, then Q is called a local isometry. 3 To ﬁll a circle 3

Figure 1: The football as an orbifold

3 To ﬁll a circle

If we have a Riemannian manifold N n, we can fill it in by using a manifold Xn+1 which has N as its boundary. The filling volume of N is then defined as the volume of X.

Deﬁnition 3.1. Let X be a Riemannian manifold with metric G.

1. G is called a complete Riemannian metric if (X, G) is complete.

2.A distance function D is deﬁned by

D(p, q) = inf length{piecewise smooth paths between p and q}.

If G is complete, then there always exists a geodesic between p and q for all p, q ∈ X.

Def. and prop. 3.2. Let N be a compact, connected, orientable manifold of dimension n ≥ 1 with a metric d which is not necessarily Riemannian. For n ≥ 2 we have n n+1 FillVol(N , d) = infgVol(X , G), where X is an arbitrary ﬁxed manifold satisfying ∂X = N and G is a complete Rie- mannian metric such that its induced distance function D restricted to the boundary

∂X is bounded below by d, i.e. D|∂X (p, q) ≤ d(p, q) ∀p, q ∈ N. Remark 3.3. In particular, for n ≥ 2 the ﬁlling volume does not depend on the topology of X with ∂X = N. Therefore, we can consider the cylinder X = N ×[0, ∞), since its boundary is N.

Idea for a proof. To show the independence of the filling manifold, we have to take a n long journey. First, we consider a singular n−simplex σn : ∆ → X. From this we construct a singular complex and consider a singular cycle which we “fill in” with a singular chain for which the boundary is the cycle. We then calculate the volume of the chain. Then this has to be generalized for arbitrary manifolds. For the proof in detail, cf. [Gro1983]. 4 3 To fill a circle

The ﬁlling volume cannot easily be calculated, hence it is not known for any Rie- mannian metric. However, we have an interesting Conjecture 3.4. 1 FillVol(Sn, can) = Vol(Sn+1, can), 2 where can is the canonical metric on the spheres of curvature +1. We will prove later in corollary 3.8 that the conjecture holds for orientable ﬁllings of genus 0 and 1. Conjecture 3.5. Let Σ be an orientable surface of even genus and G a Riemannian metric on Σ which allows an isometric involution τ that reverses orientation. Then there exists a point p ∈ Σ with dist (p, τ(p))2 π G ≤ . area(G) 4

Example 3.6. For the sphere Σ = S2 with the canonical round metric G we have area(G) = 4π (proven in the lecture about the Gauss-Bonnet theorem) and τ = −id, we have distG(x, −x) = π. Theorem 3.7. (Optimal displacement bound). Let (Σ, τ, J) an ovalless real hyperelliptic surface of even genus g and G a Riemannian metric conformal to the original metric. 1. Then there is a point p ∈ Σ which satisﬁes dist (p, τ(p))2 π G ≤ area(G) 4

2. In particular, there exists a curve joining p and τ(p) whose length is at most π 1/2 ( 4 area(G)) . This curve consists of at most g + 1 smooth curves γi.

3. Every γi is a geodesic with respect to the singular constant curvature metric τ◦J τ◦J AF(Σ , wi), where wi is a Weierstrass point and the set of fixed points Σ of the involution τ ◦ J is a circle. The proof will be discussed in section 5. Corollary 3.8. The filling area conjecture, i.e. conjecture 3.4 in the case n = 1 is satisfied by all orientable fillings of the circle with genus 0 or 1.

Proof. In the lecture about the Gauss-Bonnet theorem, we already calculated the area of the 2-sphere, which is 4π. Hence, it has to be shown that FillVol(S1,X)=2π.

Firstly, we study the case of a genus 0 surface. We consider the 2-dimensional disc such that points at the boundary, i.e. on the sphere, cannot be connected by a curve of smaller length than in the metric on S1. 3 To ﬁll a circle 5

Figure 2: Fillings of the circle

Let’s assume the conjecture is false. Then there exists a metric G on D2 with 2 areaG(D ) < 2π. Furthermore, we want that G factorizes to a metric H on RP2 in order to be able to 2 ∼ 2 use Pu’s inequality. We have RP = D /∼ where “∼“ identifies antipodal points on the boundary S1, i.e. (x, y) ∼ (−x, −y) ∀ (x, y) ∈ ∂D2. Since S1 is one-dimensional, 2 2 2 it does not have any influence on the area of D , hence areaG(D ) = areaH(RP ), where H denotes the quotient metric on RP2. We assume here that the metric G is compatible with the quotient metric (that this is well-defined). 2 2 We regard a loop in RP satisfying length(b) = sysπ1(RP , H). Let a be a lift of the loop to D2 with length(a) = length(b). Let’s analyse at first the special case that a is a lift to the boundary S1 and let G0 denote its induced metric. By Pu’s inequality, we get

length(a)2 length(b)2 π 2 = 2 ≤ . areaG(D ) areaH(RP ) 4

1 2 On the other hand, areaG0 (S ) ≤ areaG(D ), thus

length(a)2 length(a)2 length(a)2 < 2 ≤ 1 2π areaG(D ) areaG0 (S )

2 which leads to a contradiction since length(a)2 < π ⇒ length(a) < √π , but a connects 2 2 two antipodal points, so we also have length(a) ≥ π. If now a is a loop connecting two points p, τ(p) ∈/ S1, we have to verify that length(a) ≥ π in order for the proof to remain true. This can be seen in ﬁgure (3a): By projecting the curve a to the boundary S1, we can see that the length of the projection of a is 2π, hence the actual length is ≥ 2π.

Secondly, we prove the conjecture for the case g = 1. We obtain an oriented 2-fold cover by gluing together the boundary with the antipodal map τ as in the case above. (One can see that this is actually oriented by moving a trihedron from one sheet to another with aid of ﬁgure (3b). The orientation remains the same when crossing the sphere.) The resulting manifold X is a surface of genus 2 and therefore, we can apply the optimal displacement bound theorem. τ is isometric on S1 because there it is the antipodal map. On the cover, it stays isometric. 6 4 Near optimal surfaces

We now follow the same procedure as above. We take a curve satisfying length(a) = sysπ1(X). We then assume area(X) < 2π and for a curve connecting two points on the sphere, we have the same contradiction. If p, τ(p) ∈/ S1, ﬁgure (3b) shows that length(a) ≥ π by applying the same reasoning as above.

Figure 3: The ﬁlling area conjecture

4 Near optimal surfaces

2 2 2 2 Let G¯0 be the standard round metric on the 2-sphere, namely ds =dθ + sin θdϕ . The 2-sphere can be viewed as the union of the complex plane and the point at inﬁnity S2 = C ∪ {∞}.

Theorem 4.1. The pullback metric G0 on C of the metric on the sphere obtained by stereographic projection can be written as

4 G = (dx2 + dy2) 0 (1 + x2 + y2)2 where x and y are the real and imaginary parts of z ∈ C respectively. 4 Near optimal surfaces 7

We now take a closer look at the orbifold discussed above having the shape of an American football. This is reasonable in order to prove our inequality since its systolic ratio is nearly optimal. Later on, we will consider an optimally short loop on the football which will be part 2 of the proof of the optimal displacement bound theorem. The metric on AF (π, π) will be denoted by GAF . Proposition 4.2. The football can be described in the two equivalent ways: 1. The ﬁrst description of AF (π, π) is the orbifold obtained from the sphere as described in example 2.4.

2. Consider the rotation of the sphere around the z-axis in R3 by π. This results in 2 a double cover D :(S , G¯0) → AF (π, π) which is locally an isometry away from the poles. Proof. 1. The western hemisphere of the sphere can be considered as the upper half of the complex plane, H2. The positive and negative rays of the x-axis can be identified with the rays of the semicircles going from the south to the north pole (cf. image 5). Gluing together the positive and negative rays corresponds to gluing together the boundary of the sphere. The points at infinity will be identified with {+∞} which corresponds to the north pole. Let w be the complex coordinate in H2. There exists a conformal map c : (H2, w) → (S2, z) which identifies the orbifold with the 2-sphere with complex coordinate z. Since the sterographic projection is also conformal (cf. figure 4 which illustrates that perpendicular angles are mapped to perpendicular angles), c¯ ϕ we can write (H2, w) → (C ∪ {+∞}, z) → (S2, z) and just consider the map c¯ : w 7→ z = w2 which we have already seen in the preceding lecture which can 2 also be viewed as a mapc ¯ : H /∼ = AF (π, π) → C ∪ {+∞}. 2 2 This is well-defined, since H /∼ = {[w] | w = (u, v) ∈ H , (u, 0) ∼ (−u, 0)} and w2 = (u + iv)2 = v2 − v2 + i(2uv). It can bee seen thatc ¯ : [(u, v)] 7→ (u2 − v2, 2uv) = (x, y) does not depend on the representative. 2 Thus, we have a conformal map from H /∼ to the sphere, where ∼ is the equiv- alence relation described above. 2. We obtain the double cover as follows. The western hemisphere is mapped onto the orbifold AF (π, π) as before. The eastern hemisphere is then mapped onto the western hemisphere. Subsequently, the (new) western hemisphere is also mapped onto the football by the quotient map. Hence, we have a double cover of the orbifold. We have an isometry away from the poles because the metric is obtained by the pullback of the metric on the sphere.

Proposition 4.3. The orbifold metric GAF can be expressed in terms of the standard metric G0 on the complex plane with a singular conformal factor: 1 + O(r) G = G AF 4r 0 8 4 Near optimal surfaces

Figure 4: Conformality of the stereographic projection

Figure 5: The western hemisphere of S2 identiﬁed with H2

with lim O(r) = 0 and r = |z| = px2 + y2. r→0

Proof. We consider the map c : H2 → C ∪ {+∞}, w 7→ z = w2 which can also be 2 viewed as a mapc ¯ : H /∼ = AF (π, π) → C ∪ {+∞} as shown in the proof above. 4 Near optimal surfaces 9

∗ We calculate the pullback metric GAF =c ¯ G0:

4 c¯∗G =c ¯∗ (dx2 + dy2) 0 (1 + x2 + y2)2 4((d(u2 − v2))2 + (d(2uv))2) = , wherec ¯∗(x) = x(¯c(w)) = pr (¯c(w)) = u2 − v2, (1 + (u2 − v2)2 + 4u2v2)2 1 andc ¯∗(y) = 2uv 16[(udu − vdv)2 + (vdu + udv)2] = (1 + (u2 − v2)2 + 4u2v2)2 16[u2du2 + v2dv2 − 2uvdudv + v2du2 + u2dv2 + 2uvdudv] = (1 + (u2 − v2)2 + 4u2v2)2 16(u2 + v2)(du + dv)2 = (1 + (u2 − v2)2 + 4u2v2)2 4(u2 + v2) · (1 + u2 + v2) = µ ·G , where µ(u2 + v2) = . 0 (1 + (u2 − v2)2 + 4u2v2)2

To obtain the assertion, we calculate:

4r(1 + r2) µ(r2) = (1 + (u2 − v2)2 + 4u2v2)2 4r2(1 + r2) = (1 + u4 + v4 − 2u2v2 + 4u2v2)2 4r2(1 + r2) = (1 + u4 + v4 + 2u2v2)2 4r2(1 + r2) = (1 + (u2 + v2)2)2 4r2(1 + r2) = . (1 + r4)2

To conclude the proof, we use the property that the American football is home- omorphic to the sphere. We had constructed it by identifying the boundary of the hemisphere. Since the 2-sphere consists of two hemispheres, we can describe the home- omorphism by doubling up the angles. We can imagine the angle on the football going (horizontally) only from 0 to π, instead of 2π, since we only used the western hemisphere. Therefore, we can use the map z 7→ z2. Luckily, this is exactly the same mapc ¯ which we had before. Hence, we can pull back the metric GAF . Recall that GAF lives on the tangent space of the upper hemisphere 2 ∗ 2 ∗ H . Sincec ¯ : Tz2 C ∪ {∞} → Tz H /∼, we get G0|H2 =c ¯ (GAF ). The restriction on H2 is necessary here because the mapc ¯∗ only goes to T H2. 10 5 The optimal displacement bound theorem: Step 1 of the proof

∗ 2 With the relationc ¯ (G0)(w) = µ(|w| )G0 which we had found above, we obtain: 1 c¯∗(G ) = c¯∗(G ) AF µ(|w|2) 0 1 = c¯∗(G ) µ(|c¯(z)| 0 1 =c ¯∗ G µ(|z|) 0 1 + O(|z|)) =c ¯∗ (G ) . 4|z| 0 The factor µ is now in the required form since: 1 (1 + r2)2 1 + O(r) = =! µ(r) 4r(1 + r)2 4r (1 + r2)2 ⇒ = 1 + O(r) (1 + r)2 (1 + r2)2 ⇒ O(r) = − 1 (1 + r)2 ⇒ lim O(r) = 0. r→0 We obtain the assertion by locally inverting c∗.

5 The optimal displacement bound theorem: Step 1 of the proof 5.1 Idea The idea is the following. We have an ovalless real hyperelliptic surface (Σ, τ, J) of even genus g. Since the hyperelliptic involution J induces a 2-fold covering of S2, we will prove the estimate by considering a quotient metric on the sphere. Furthermore, we will work with a metric G conformal to the original metric such that its area is A.

In the ﬁrst step of the proof, we will show that one can assume that τ and J are 2 isometries of G. As a result we will obtain an induced metric G1 = f G0 on the sphere with area A/2 whose pullback (not applied to the Weierstrass points) is our metric G on Σ.

2 Furthermore, τ induces an isometry τ0 on S which reverses orientation, for instance, the equator remains fixed and points on the southern hemisphere are mapped to points on the northern hemisphere and the other way round. Hence, ramification points are mapped onto ramification points and therefore, each hemisphere contains g + 1 of them. 5 The optimal displacement bound theorem: Step 1 of the proof 11

5.2 Proof ¯ 1 ∗ Lemma 5.1. Let F be an angle-preserving involution of Σ and G = 2 (G + F G) the averaged metric. Then for p, q ∈ Σ we have 2 1 2 dist ¯(p, q) ( [dist (p, q) + dist (F (p),F (q))]) G ≥ 2 G G . area(G¯) area(G) Proof. Since F is angle-preserving, the metric F ∗G is conformal to G, i.e. F ∗G = f 2G ¯ 1 2 where f is a smooth positive function. Then G can be written as 2 (f + 1)G and we have Z 1 2 area(G¯) = (f + 1) dvolG Σ 2 Z Z 1 2 = f dvolG + 1 dvolG 2 Σ Σ 1 = (area(F ∗(G)) + area(G)) 2 = area(G). As F :Σ → Σ is an angle-preserving involution, we have F 2 = 1 and f 4 = 1. The volume element remains unchainged since detF ∗G = detG·(detF )2 = detG. Therefore, R 2 R if we had without loss of generality f dvolG < dvolG, as a consequence, we would have Z Z Z Z Z 4 2 2 dvolG = f dvolG = f dvolF ∗G = f dvolG < dvolG which leads to a contradiction. For the numerator, consider a curve γ : [0, 1] → Σ with γ(0) = p and γ(1) = q, hence 0 F (γ) is a curve from F (p) to F (q). Fur a curve γ we have the relation F∗(γ (t)) = (F ◦ γ)(t). Hence,

1 Z p 0 0 LG(γ) = G(γ (t), γ (t)) dt, 0 1 s Z f 2 + 1 L ¯(γ) = G(γ0(t), γ0(t)) dt, G 2 0  1 1  1 1 Z Z (L (γ) + L (F ◦ γ)) = pG(γ0(t), γ0(t)) dt + pG((F ◦ γ)0(t), (F ◦ γ)0(t)) dt 2 G G 2   0 0  1 1  1 Z Z = pG(γ0(t), γ0(t)) dt + pG(F (γ)0(t),F (γ)0(t)) dt 2  ∗ ∗  0 0 1 Z 1 + f = pG(γ0(t), γ0(t)) dt. 2 0 12 References

Furthermore, we have

f 2 − 2f + 1 = (f − 1)2 ≥ 0 ⇒ 2f 2 + 2 ≥ f 2 + 2f + 1 | : 4 f 2 + 1 f 2 + 2f + 1 ⇒ ≥ |p() 2 4 r f 2 + 1 f + 1 ⇒ ≥ . 2 2 By choosing γ as the minimal path between p and q in G¯, we obtain

distG¯ = LG¯(γ) 1 ≥ (L (γ) + L (F ◦ γ) 2 G G 1 ≥ [dist (p, q) + dist (F (p),F (q))] 2 G G which proves the lemma.

Remark 5.2. By taking F = τ and q = τ(p), we get a τ-invariant metric G− conformal to G such that 2 2 dist − (p, q) dist (p, q) G ≥ G area(G−) area(G) By applying the preceding lemma to G−, F = J and q = τ(p), we get a J-invariant metric G¯, conformal to G.

We can now assume that J and τ are isometries to show the inequality, since if the inequality is true for G¯, it is also true for G. We will see the rest of the proof in the next lecture.

References

[1] M. Gromov. Filling Riemannian Manifolds, volume 18 of J. Diﬀ. Geom. 1983.

[2] Michael Kapovich. Hyperbolic Manifolds and Discrete Groups, volume 183 of Progress in Mathematics. Birkh¨auser,Boston, Boston, MA, 2001.

[3] Mikhail G. Katz. Systolic geometry and topology, volume 137 of Mathematical Surveys and Monographs. American Mathematical Society, Providence, RI, 2007. With an appendix by Jake P. Solomon.

[4] Wikipedia. http: //en.wikipedia.org/wiki/covering space (access: May 4, 2011).

[5] Wikipedia. http: //en.wikipedia.org/wiki/group action (access: May 4, 2011).

[6] Wikipedia. http: //en.wikipedia.org/wiki/local isometry (access: June 13, 2011). References 13

[7] Wikipedia. http: //en.wikipedia.org/wiki/orbifold (access: June 3, 2011). [8] Wikipedia. http: //en.wikipedia.org/wiki/riemannian manifold (access: May 10, 2011).

Images

Figure 4: http : //en.wikipedia.org/wiki/F ile : CartesianStereoP roj.png (access: June 6, 2011)