MIN-MAX THEORY AND THE WILLMORE CONJECTURE

Transcription

1 MIN-MAX THEORY AND THE WILLMORE CONJECTURE FERNANDO CODÁ MARQUES - IMPA The content of these lecture notes are the applications of the min-max theory of Almgren and Pitts to the proof of the Willmore conjecture and the Freedman-He-Wang conjecture for links in euclidean three-space. These lecture notes were organized by Rafael Montezuma. Organization 1. Introduction 2. Canonical family and degree calculation 3. Min-max theory (Almgren and Pitts) 4. Ruling out great spheres 5. Proof of main theorems 6. Min-max theory and the energy of links 1. Introduction Fix topological type, ask what is the best immersion in R 3. For a genus g = 0 surface is the round sphere. What about the case of g = 1? Let R 3 closed, embedded, smooth surface of genus g, k 1 and k 2 the principal curvatures H = 1 2 (k 1 + k 2 ) the mean curvature K = k 1 k 2 the Gauss curvature The integral on of the Gauss curvature K is a topological invariant, (Gauss-Bonnet) Other quadratic integrand on is H 2 Kd = 2πχ(). (Willmore Energy) W() = and this is scaling invariant. H 2 d, Fact: W() is invariant under conformal transformations of R 3. (Blaschke, Thomsen 1920 s) Indeed, H 2 d = { (k1 k 2 ) 2 4 } + K d = 2πχ() Å 2 d,

2 2 FERNANDO CODÁ MARQUES - IMPA where Å is the trace free second fundamental form. Remark. Å 2 d is pointwise conformally invariant object. Therefore, F : R 3 R 3 conformal map W(F ()) = W(). Willmore (1960 s): W() 4π = W(Round sphere). Willmore Conjecture (1965): If R 3 is a torus, then W() 2π 2. Example 1.1. Let 2 be the torus obtained by the revolution of a circle with center at distance 2 of the axis of revolution and radius 1. Then W( 2 ) = 2π2. Let π : S 3 \ {p} R 3 the stereographic projection, S 3 \ {p} and = π() R 3. By some calculations, we have H 2 d = (1 + H 2 )d. Definition 1.2. For S 3 the Willmore energy is W() = (1+H2 )d. Remark. If S 3 is minimal, then W() = area(). Example 1.3. The equator S1 2 (0) is minimal and has area 4π. The Clifford Torus ˆ = S 1 (1/ 2) S 1 (1/ 2) S 3 is minimal and has area 2π 2. Moreover, π(ˆ) = 2. Li and Yau (1982): If is not embedded, then W() 8π. Fix v R 4 and consider the vector field x v (x), given by tangential projection on T x S 3. It is a conformal killing vector field in S 3 and generates a flow φ t of centered dilations. By conformal invariance of Willmore energy, W() = W(φ t ()) = (1 + Hφ 2 )dφ t() t() area(φ t ()). φ t() If we choose v and let t, we have W() 4π. In case F : S 3 is immersion and p satisfies #F 1 (p) = k, then W() 4πk. Theorem A. Let S 3 closed, embedded, genus g 1. Then W() 2π 2, and W() = 2π 2 if and only if is the Clifford torus up to Conf(S 3 ). the Willmore Conjecture! Theorem B. Let S 3 closed, embedded, genus g 1. If is minimal, then area() 2π 2, and area() = 2π 2 if and only if is the Clifford torus up to Iso(S 3 ).

3 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 3 Example 1.4 (Min-max example). Let S R n a submanifold and h its height function. Consider γ 0 : [0, 1] S with γ 0 (0) = p and γ 0 (1) = q and Π the homotopy class of γ 0 relative to {0, 1}. We define the min-max invariant c = inf sup h(γ(t)). γ Π t [0,1] The aim of min-max theory is to detect a critical point x 0 for h of index 1 as c = h(x 0 ), for some path γ 0. Remark. In order to detect index k critical points we need to do min-max over k-parameters families. Can think of the equator (great sphere) as a min-max minimal surface of 1-dimensional sweep out of S 3. Question: Can we produce Clifford torus as a min-max minimal surface? (one need 5-parameters) The main ingredients on the proof are (I) For each S 3, there exists canonical family (v,t) S 3, (v, t) B 4 [ π, π], with (0,0) = and area( (v,t) ) W(). (II) [Urbano, 1990] If S 3 minimal, index() 5, g 0, then is a Clifford torus (index 5) or a great sphere (index 1). (III) Min-max theory for the area functional (Almgren-Pitts). 2. Canonical Family and Degree Calculation We begin this section by introducing some notation: S 3 surface B 4 R 4 open unit ball and B 4 = S 3 B 4 R (Q) = {x R4 : x Q < R} B r (q) = {x S 3 : d(x, q) < r}, d is spherical distance For each v B 4 we consider F v : S 3 S 3 the conformal map defined by (1) F v (x) = 1 v 2 (x v) v. x v 2 It is a centered dilation fixing ±v/ v. Note that if v p S 3, then F v (x) p, for every x S 3, x p. Let S 3 \ = A A, the connected components, and N the unit normal vector to pointing out A. Consider the images by F v A v = F v (A), A v = F v (A ) and v = F v () = A v. The unit normal vector to v is given by N v = DF v (N)/ DF v (N). We also consider in S 3 the signed distance function to v { d(x, v ) if x / A d v (x) = v d(x, v ) if x A v. This is a Lipschitz function, smooth near v.

4 4 FERNANDO CODÁ MARQUES - IMPA Definition 2.1 (Canonical Family). For v B 4 and t [ π, π], define (v,t) = A (v,t), where A (v,t) = {x S 3 : d v (x) < t} are open subsets of S 3. Remark. Consider ψ (v,t) : v S 3 given by ψ (v,t) (y) = exp y (tn v (y)), i.e., ψ (v,t) (y) = cos(t)y + sin(t)n v (y). Then (v,t) ψ (v,t) ({Jac ψ (v,t) 0}). In particular, we conclude that (v,t) are 2-rectifiable subsets of S 3. Area Estimate of (v,t). Theorem 2.2. We have, for every (v, t) B 4 [ π, π], (2) area( (v,t) ) W(). Moreover, if is not a geodesic sphere and area( (v,t) ) = W(), then t = 0 and v is a minimal surface. Proof. Consider {e 1, e 2 } T y v orthonormal basis such that DN v y = k i (v)e i, i = 1, 2, where k i (v) are the principal curvatures of v. Since Dψ (v,t) y (e i ) = (cos(t) k i (v) sin(t))e i, we have Jac ψ (v,t) = (cos(t) k 1 (v) sin(t))(cos(t) k 2 (v) sin(t)) By a simple calculation, this gives = cos 2 (t) (k 1 + k 2 ) cos(t) sin(t) + k 1 k 2 sin 2 (t). Lemma 2.3. If H(v) is the mean curvature of v, we have Jac ψ (v,t) = (1 + H(v) 2 ) (sin(t) + H(v) cos(t)) 2 (k 1(v) k 2 (v)) 2 sin 2 (t). 4 Using this lemma, the area formula, and the conformal invariance of the Willmore energy we obtain area( (v,t) ) area ( ψ (v,t) ({Jac ψ (v,t) (p) 0}) ) (Jac ψ (v,t) ) d v {Jac ψ (v,t) 0} (1 + H(v) 2 ) d v sin 2 (k 1 (v) k 2 (v)) t v v 2 d v 4 = W() sin2 t 2 Å 2 d. Suppose that equality holds for some (v, t) and that is not a geodesic sphere. Then from conformal invariance we have (k 1 k 2 ) 2 (k 1 (v) k 2 (v)) 0 d = 4 v 2 d v for all v B 4 4 and so t = 0, t = π, or t = π. The last two cases are impossible because (v,π) = and (v,t) = and thus t = 0. This means area( v ) = W( v ) and so v is a minimal surface.

5 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 5 Blow-up Argument. Next we analyse what happens to (v,t) as v S 3. The notation for symmetric difference here is, as usual, X Y = (X \ Y ) (Y \ X). Assume v n B 4 and t n [ π, π], (v n, t n ) (v, t) B 4 [ π, π]. The case v / is divided in three: Claim 1. If v B 4, vn v smoothly and vol (A (vn,t n) A (v,t) ) 0. Claim 2. If v A, vn v, A vn B π (v) and vol (A (vn,t n) B π+t (v)) 0. Claim 3. If v A, vn v, A vn B 0 ( v) and vol (A (vn,t n) B t ( v)) 0. Problem: If v n v = p. For example, if v n = (1 1/n)v, we have A vn B π/2 ( N). But this convergence only happens because the angle that v n makes with N(p) also converge. In fact, the limit of F vn (A) depends only on the angle of convergence. Let D+(r) 2 = {s = (s 1, s 2 ) R 2 : s < r, s 1 0}. Consider ε > 0 small so that Λ : D 2 +(3ε) B 4 is a diffeomorphism to a tubular neighbourhood of in B 4, given by Λ(p, s) = (1 s 1 )(cos(s 2 )p + sin(s 2 )N(p)). We also use Ω r for Λ( D 2 +(r)). Conformal images of balls and sectors. Let p, N S 3 and r > 0, so that p, N = 0. In this section we study the conformal images of sectors of S 3, (3) (p, N, r) = S 3 \ {B r (cos(r)p + sin(r)n) B r (cos(r)p sin(r)n)}, and of balls (4) B π/2 ( N) = B 4 2 ( N) S 3. Consider the following notation. For sufficient small ε 0, v = (1 s)(cos(t)p+ sin(t)n) with 0 < s ε 0 and t ε 0, t/s Q = p 1 N 1 + (t/s) (t/s) 2 S3 and ( ) t/s R = (t/s) 2 Proposition 2.4. Then, there exists C > 0 so that: (i) B 4 (Q) R C (s,t) S3 F v (B 4 2 ( N) S 3 ) B 4 (Q) S3 R+C (s,t) (ii) F v ( (p, N, r)) B 4 R+C (s,t) (Q) \ B4 R C (s,t) (Q).

6 6 FERNANDO CODÁ MARQUES - IMPA If v n v = p we can write v n = Λ(p n, (s n1, s n2 )), with p n p and s n 0. Let B pn = B 4 2 ( N) S 3. Suppose also the angle convergence s n2 (5) lim = k [, ]. n s n1 Remark. Choose r 0 > 0 sufficient small for which B r0 (cos(r 0 )p sin(r 0 )N(p)) A, and A S 3 \B r0 (cos(r 0 )p + sin(r 0 )N(p)). Since then, we have A B pn (p n, N(p n ), r 0 ), and thus F vn (A B pn ) F vn ( (p n, N(p n ), r 0 )). Using the above result,we have where F vn (A) F vn (B pn ) B 4 R n+c s n (Q n) \ B 4 R n C s n (Q n), k Q n p 1 N(p) S3 1 + k k 2 and ( ) k R n k 2 Then vol(f vn (A) F vn (B pn )) 0. But F vn (B pn ) converges if we have angle convergence s n2 /s n1 k [, ]. Summarizing, Proposition 2.5. Consider a sequence (v n, t n ) B 4 [ π, π] converging to (v, t) B 4 [ π, π]. (i) If v B 4 then (ii) if v A then lim n vol ( A (vn,t n) A (v,t) ) = 0. lim vol ( A (vn,t n n) B π+t (v) ) = 0 and, given any δ > 0, (vn,t n) B π+t+δ (v) \ B π+t δ (v) for all n sufficiently large. (iii) if v A then lim vol ( A (vn,t n n) B t ( v) ) = 0 and, given any δ > 0, (vn,t n) B t+δ ( v) \ B t δ ( v) for all n sufficiently large. (iv) If v = p and s n2 v n = Λ(p n, (s n1, s n2 )) with lim = k [, ], n s n1 then lim vol ( A (vn,t n n) B rk +t(q p,k ) ) = 0 and, given any δ > 0, (vn,t n) B rk +t+δ(q p,k ) \ B rk +t δ(q p,k ) for all n sufficiently large.

7 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 7 Reparametrizing Map T : B 4 B 4. Take a smooth function φ : [0, 3ε] [0, 1] such that φ([0, ε]) = 0, strictly increasing in [ε, 2ε] and φ([2ε, 3ε]) = 1. Define T : B 4 B 4 by { 4 T (v) = v if v B \ Ω3ε Λ(p, φ( s )s) if v = Λ(p, s) Ω 3ε. T is continuous and collapses Ω ε onto preserving s n2 /s n1. Moreover, T : B 4 \ Ω ɛ B 4 is a homeomorphism. This map allows us to make a blow up argument and reparametrize the canonical family in such a way it can be continuously extended to the whole B 4 [ π, π]. For each v B 4 \ Ω ε, put C(v, t) = (T (v),t). Using the results in the previous section we know that as v (B 4 \ Ω ε ), C(v, t) converges to some geodesic sphere (round sphere). Extend C to Ω ε being constant radially, along s 2 = constant. By construction, we have (6) C(v, t) is continuous in volume in B 4 [ π, π] and (7) C(v, t) = B r(v)+t (Q(v)), for v B 4, where the center map Q : S 3 S 3 is given by T (v) if v A \ Ω ɛ Q(v) = T (v) if v A \ Ω ɛ s ε p ε 2 s 2 ε N(p) if v = cos(s)p + sin(s)n(p), s [ ε, ε]. Theorem 2.6. deg(q) = genus(). Proof. Let dv and dvol be the volume forms of S 3 and R 4, respectively. First, regard that Q is piecewise smooth. The idea is to calculate the degree by integrating Q (dv ) on S 3, and using the formula Q (dv ) = deg(q) S 3 dv. S 3 In order to proceed the calculation, divide S 3 on A \Ω ε, A\Ω ε and Ω ɛ S 3. in the first two we have Q (dv ) = A \Ω ε ( T ) (dv ) = A \Ω ε dv = vol(a ) A and, analogously, A\Ω ε Q (dv ) = vol(a). Let G : [ ε, ε] S 3 Ω ε be the diffeomorphism G(p, t) = cos(t)p + sin(t)n(p) and define Q = Q G : [ ε, ε] S 3, so that Q(p, t) = t ε ε p 2 t 2 N(p). ε

8 8 FERNANDO CODÁ MARQUES - IMPA In this reparametrization, the third integral can be calculated as Q (dv ) = Q (dv ). Ω ε S 3 [ ε,ε] If {e 1, e 2 } T p orthonormal basis with DN p (e i ) = k i e i, we have Q (dv )(e 1, e 2, t ) = dv (DQ(e 1 ), DQ(e 2 ), DQ( t )) = dvol(dq(e 1 ), DQ(e 2 ), DQ( t ), Q) ( = t ε ε + 2 t 2 )( k 1 t ε ε ε + 2 t 2 ) ( 1) k 2 ε ε 2 t. 2 The Gauss curvature of satisfies K = 1 + k 1 k 2, the Gauss equation, hence ( 1 Q (dv ) = k 1 k 2 ε 2 t 2 t 2 (k 1 + k 2 )t + )dtd ε 2 t 2 [ ε,ε] = π 2 ε 2 (K 1)d π 2 d = π 2 (2g 2). Adding the results, Q (dv ) = vol(s 3 ) + π 2 (2g 2) = 2π 2 g = g dv. S 3 S 3 3. Geometric Measure Theory The intent of this section is to introduce a few geometric measure theory notions that will appear latter. Let (M 3, g) be an orientable Riemannian 3-manifold isometrically embedded in R L. Consider the notation I k (M) for the space of k-dimensional integral currents in R L with support in M, i.e., k-dimensional rectifiable sets with integer multiplicities and orientations chosen H k -almost everywhere. A k-dimensional integral current T can also be interpreted as a continuous functional that operates in Ω k c (R L ), the space of compactly supported differential k-forms provided with the comass norm, by integration T (φ) = φ, ξ θdh k, φ Ω k c (R L ), M where θ and ξ are the multiplicity and orientation maps of T, respectively. The boundary operator : I k (M) I k 1 (M) is defined as T (φ) = T (dφ). The set of currents without boundary is denoted by Z k (M). Example 3.1. Oriented closed k-dimensional surfaces are the basic examples of k-dimensional integral currents.

9 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 9 The notion of volume extends to currents. For each T I k (M) we define its mass to be M(T ) = sup{t (φ) : φ Ω k c (R L ) and φ 1}, where φ is the comass norm. We can introduce then the mass norm M(S 1, S 2 ) = M(S 1 S 2 ). The flat metric defined below in the space of currents has the important property that its induced topology is the weak topology F(S 1, S 2 ) = inf{m(s 1 S 2 Q) + M(Q) : Q I k+1 (M)}. We use also the notation F(T ) = F(T, 0). Note that F(T ) M(T ). Theorem 3.2. The map C : B 4 [ π, π] Z 2 (S 3 ) is well defined and continuous in the flat topology. In order to prove the map is well defined recall that And for the continuity claim use C(v, t) = U(v, t) C(v, t) = 0. F(C(v 1, t 1 ), C(v 2, t 2 )) = F( (U(v 1, t 1 ) U(v 2, t 2 ))) = vol(u(v 1, t 1 ) U(v 2, t 2 )). 4. Min-max Theory Theorem 4.1 (Almgren-Pitts). Every compact Riemannian manifold (M n, g), n 7, contains a smooth embedded minimal hypersurface M. Those are obtained by min-max methods with only 1-parameter and the proof can be founded in Existence and regularity of minimal surfaces in Riemannian manifolds, by Pitts. The min-max theory we would like to have. We would like to run min-max with maps Φ : I n Z 2 (M 3 ) continuous in the flat topology, defined on the n-dimensional cube I n = [0, 1] n. If Π is the homotopy class of Φ relative the boundary, define the width of Π by (8) L(Π) = inf{l(φ ) : Φ Π}, where L(Φ ) = sup{m(φ (x)) : x I n }. In this context, the result we would like to have is Min-max Theorem: If L(Π) > sup{m(φ(x)) : x I n }, then there exists a smooth embedded minimal surface (possibly disconnected, with integer multiplicities), in M so that L(Π) = area(). Moreover, if {φ i } i Π is such that L(φ i ) L(Π), then there exists x i I n with F(, φ i (x i )) 0.

10 10 FERNANDO CODÁ MARQUES - IMPA Almgren-Pitts min-max theory. We briefly describe Almgren-Pitts min-max theory comparing this with the continuous theory. We need the following notation. I(1, k) denotes the cell complex on I 1 whose 1-cells and 0-cells are, respectively, [0, 3 k ], [3 k, 2 3 k ],..., [1 3 k, 1], and [0], [3 k ],..., [1 3 k ], [1]. I(n, k) is the n-dimensional cell complex on I n given by I(n, k) = I(1, k)... I(1, k) (n times). I(n, k) 0 is the subset of all 0-cells, or vertices, of I(n, k). Continuous Φ : I n Z 2 (M) Discrete S = {φ i } i, φ i : I(n, k i ) 0 Z 2 (M) Φ continuous in flat topology M(φ i (x) φ i (y)) < δ i, [x, y] I(n.k i ) 1 L(Φ) = sup{m(φ(x)) : x I n } L(S) = lim i sup{m(φ i (x)) : x I(n, k i ) 0 } Φ Φ homotopy φ i δi φ i and δ i 0 L(Π) = {L(Φ ) : Φ Π} L(Π) = {L(S ) : S Π} 5. Ruling out great spheres (Topological Argument) The min-max family C : B 4 [ π, π] Z 2 (S 3 ) constructed above gives, up to a reparametrization, a map Φ defined on the 5-cube. Theorem 5.1. Let S 3 embedded, closed, genus g surface. Then, the map Φ : I 5 Z 2 (S 3 ) satisfies the following (i) Φ is continuous in the flat topology. (ii) Φ(I 4 0) = Φ(I 4 1) = 0. (iii) for every x I 4, {Φ(x, t)} t [0,1] is a standard sweep out of S 3 by oriented geodesic spheres. (iv) there exists a map Q : I 4 S 3, called the center map, so that Φ(x, 1/2) = B π/2 (Q(x)), for all x I 4. This map has degree g. Also, sup{m(φ(x)) : x I 5 } W(). Consider Π the homotopy class of Φ. This homotopy class is considered in the discrete min-max context, but here we explain the arguments using the continuous language, supposing we have the continuous version of Min-max Theorem stated in the previous section. Remark. L(Π) 4π = least area a minimal surface can have in S 3. Theorem 5.2. If g 1, then L(Π) > 4π.

11 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 11 Before proving that, we introduce some notation. We denote by τ V 2 (S 3 ) the space of unoriented great spheres. For each element in τ we can associate its center, this gives a diffeomorphism between τ and RP 3. This equivalence has the property that if we push the F -metric of τ V 2 (S 3 ) to RP 3, the obtained metric is equivalent to the standard metric of RP 3. Associated with Φ by forgetting the orientations of currents in Z 2 (S 3 ) we have a map Φ : I 4 {1/2} RP 3 τ, given by Φ (x, 1/2) = Φ(x, 1/2). The degree calculation gives Φ ( I 4 {1/2}) = 2g H 3 (RP 3 ). Proof. Suppose, by contradiction, that L(Π) = 4π. sequence {φ i } i Π so that Then, there exists a sup area(φ i (x)) 4π + 1 x I 5 i. Recall φ i = Φ on I 5. Let ε 0 > 0 small (to be chosen later) and take δ > 0, such that y = (x, t) J δ = I 4 [1/2 δ, 1/2 + δ] F ( Φ(y), Φ(x, 1/2) ) ε 0. Consider 0 < ε 1 ε 0 with the property y I 4 I, F( Φ(y), τ) ε 1 y J δ. If we could find a 4-manifold R so that R = I 4 {1/2} and a continuous function f : R RP 3 with f R = Φ, then we would have Φ ( I 4 {1/2}) = f ( R) = f(r) = 0 in H 3 (RP 3, Z). Instead of that, we first construct a sequence of 4-dimensional R(i) I 5 with support( R(i)) I 4 I. It is constructed so that φ i (x) is sufficient close to τ for any x R(i), when i is large. We begin with A(i) = {x I 5 : F( φ i (x), τ) ε 1 }, A(i) A(i) the connected component of I 4 0. Claim 4. A(i) (I 4 {1}) =, if i is large. Proof. If not, up to a subsequence, can find γ i : [0, 1] I 5 with γ i (0) I 4 0, γ i (1) I 4 1 and γ i ([0, 1]) A(i). Since φ i Π, for every i, we have [φ i γ i ] = Π 1 does not depend on i, and 4π L(Π 1 ) L(φ i γ i ) 4π + 1/i. Then L(Π 1 ) = 4π and the sequence {φ i γ i } i is optimal in Π 1. By Minmax theorem, we find a smooth embedded minimal surface S 3 and a sequence x i I so that area() = 4π and F( (φ i γ i )(x i ), ) 0. The only minimal surface in S 3 with area 4π are great spheres, so φ i (γ i (x i )) approximates to τ, and this is a contradiction, since γ i (x i ) A(i).

12 12 FERNANDO CODÁ MARQUES - IMPA In particular, A(i) I 5 ( I 4 I) (I 4 0). Define R(i) = ( A(i)) int(i 5 ). Therefore F( φ i (x), τ) = ε 1 on R(i) and, if i is large enough, R(i) J δ. Take C(i) = A(i) ( I 4 I). Because ( A(i)) = 0 and A(i) = R(i) C(i) (I 4 0), we have C(i) = R(i) (I 4 0) [ R(i)] = [ I 4 {1/2}] in H 3 (J δ, Z). Dealing with the discrete case we have some extra problems, for instance: (1) surfaces along R(i) are close to τ, but not exactly in τ. (2) the cycles R(i) and I 4 {1/2} are homologous and close, but not necessarily equal. Need a projection argument to conclude. Definition 5.3. Let Φ : I 4 I RP 3 τ be the continuous map given by Φ(x, t) = Φ(x, 1/2). In particular, Φ( I 4 {1/2}) = 2g in H 3 (RP 3, Z). Claim 5. There exists a continuous f i : R(i) τ with f i R(i) = Φ R(i). Proof. We can choose a triangulation of R(i) so that if [x, y] 1 and x, y 0, then F( φ i (x), φ i (y) ) ε 0. First define f i : 0 τ by f i (x) = Φ(x), if x R(i), and f i (x) = y, if x / R(i), so that F( φ i (x), y) = F( φ i (x), τ). If x, y 0 are adjacent, the triangle inequality gives F(f i (x), f i (y)) 3ε 0, If ε 0 is sufficient small (depending only on size of convex balls in RP 3 with standard geometry), then the map f i can be extended to = R(i). In the end, we have Φ [ R(i)] = (f i ) [ R(i)] = [ (f i ) (R(i))] = 0 in H 3 (RP 3, Z). On the other hand, we already had used the degree calculation to obtain Φ [ R(i)] = Φ [ I 4 {1/2}] = 2g in H 3 (RP 3, Z). Contradiction if g 1. Then L(Π) > 4π. 6. Proof of Theorems(A and B) Theorem B. S 3 embedded, closed, g 1, minimal. Then area() 2π 2, and area() = 2π 2 if and only if is isometric to the Clifford Torus. Proof. Consider F 1 = {S S 3 : S is embedded, closed, g 1, minimal}. There exists F 1 (Appendix), satisfying (9) area() = inf S F 1 area(s).

13 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 13 Urbano (1990): If S 3, H = 0, index() 5, then is the Clifford torus or the great sphere. Claim 6. Take as (9), then index() 5. The general index bound is missing in current min-max theory: minmax with k parameters, then index() k. If t is a variation of the minimal surface = 0, the second variation is given by d 2 (10) dt 2 area( t) t=0 = ψlψd, where L is the Jacobi operator, ψn is the variational vector field and N is the unit normal vector to. Recall we constructed a Canonical family (v,t), (v, t) B 4 [ π, π], with the area estimate area( (v,t) ) W() = area(). Consider the variational vector fields of (v,t) at (0, 0): ψ i = N, e i, i = 1, 2, 3, 4 and ψ 5 = 1, where e i are the canonical directions in R 4 and let E = span{ψ 1,..., ψ 5 }. Lemma 6.1. (11) unless is totally geodesic. ψlψd < 0, ψ E \ {0}, Suppose index() 6. So, there exists ϕ C () with ϕlϕd < 0 and ϕlψ i d = 0, for all i = 1,..., 5. Choose vector field X on S 3 with X = ϕn on, and let Γ s be its flow. Define f(v, t, s) = area(γ s ( (v,t) )), for v < δ, t < δ and s < δ, then f(0, 0, 0) = area(), Df(0, 0, 0) = 0, D 2 f(0, 0, 0) < 0, and then f has a strict local maximum at the origin. Replace (v,t) with (v,t) = Γ β(v,t)( (v,t) ), where β(v, t) is chosen satisfying area( (v,t) ) < area(), if (v, t) < δ and (v,t) = (v,t), if (v, t) δ. Therefore, (12) area( (v,t) ) < area(), (v, t) B4 [ π, π]. Recall the equality in area( (v,t) ) W() implies t = 0 and v is minimal. Remark. minimal area( v ) = area() 4 v,n(x) 2 d. x v 4

14 14 FERNANDO CODÁ MARQUES - IMPA Since 0 = is minimal, if v is minimal, v B 4, then area( v ) = W( v ) = W() = area(), and we conclude v, N = 0 along. Because of that, is invariant under flow of v. From the canonical family we constructed C(v, t) via a blow-up argument and Φ defined on the 5-cube. Analogous considerations from (v,t) gives C (v, t) and Φ : I 5 Z 2 (S 3 ), with sup M(Φ (x)) sup{area( (v,t) ) : (v, t) B4 [ π, π]} < area(). x I 5 Apply min-max theory to Π, the homotopy class of Φ, to obtain a minimal surface so that 4π < area( ) = L(Π ) < area() 2π 2 < 8π, where the 2π 2 estimate is the area of Clifford torus, as a surface in F 1. The multiplicity of is 1, otherwise would be area( ) 8π. Also, has genus( ) 1, because there is a theorem by Almgren that guarantees the only minimal surfaces of genus g = 0 in S 3 are great spheres. Contradiction, since F 1 and area( ) < area(). This implies index() 5. By Urbano, is a Clifford torus and Theorem B is proved. Theorem A. S 3 closed, embedded, genus g 1. Then W() 2π 2 and W() = 2π 2 if and only if is Clifford torus up to Conf(S 3 ). Proof. Let Φ : I 5 Z 2 (S 3 ) the min-max family constructed from and Π its homotopy class. Applying min-max theory to Π, we obtain a minimal surface with 4π < area( ) = L(Π) W (). If has multiplicity greater than 1, then area( ) 8π > 2π 2. If not, the lower bound on the area implies genus( ) 1. In any case, area( ) 2π 2, and we have W () 2π 2. Suppose W () = 2π 2. Recall area( (v,t) ) W () = 2π 2. Claim 7. For some (v, t) B 4 [ π, π], we have area( (v,t) ) = W (). Otherwise, the min-max family Φ constructed from has L(Φ) < W () = 2π 2. We can run min-max again and contradic Theorem B. Because of equality area( (v,t) ) = W (), we conclude t = 0 and v is a minimal surface. Then area( v ) = W ( v ) = W () = 2π 2. By rigidity part of Theorem B, v = F v () must be isometric to the Clifford torus, with F v Conf(S 3 ). 7. Min-max theory and the energy of links A 2-component link in R 3 is a pair of rectifiable closed curves γ i : S 1 R 3, i = 1, 2, so that γ 1 (S 1 ) γ 2 (S 1 ) =.

15 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 15 Möbius Cross Energy. E(γ 1, γ 2 ) = Freedman-He-Wang(1994). S 1 S 1 γ 1 (s) γ 2 (t) γ 1 (s) γ 2 (t) 2 dsdt. F Conf(R 3 ) E(F γ 1, F γ 2 ) = E(γ 1, γ 2 ). In the same paper they also conjecture Conjecture: The energy of any non-trivial link should be at least 2π 2. Example 7.1 (Standard Hopf link). The stereographic projection of γ(s) = (cos s, sin s, 0, 0) S 3 and γ(t) = (0, 0, cos t, sin t) S 3 is a 2-component link in R 3, called standard Hopf link. Its Möbius energy is given by E( γ, γ) = 2π 2. He: The minimizer is isotopic to standard Hopf link. We introduce now the Gauss map and linking number of a link in R 3. Let (γ 1, γ 2 ) be a 2-component link in R 3. The Gauss map g = G(γ 1, γ 2 ) : S 1 S 1 S 2 of (γ 1, γ 2 ) is defined as g(s, t) = γ 1(s) γ 2 (t) γ 1 (s) γ 2 (t) S2. The linking number of (γ 1, γ 2 ) is defined to be the degree of the Gauss map, and the notation is lk(γ 1, γ 2 ). This can also be given via the Gauss formula lk(γ 1, γ 2 ) = 1 det(γ 1 (s), γ 2 (t), γ 1(s) γ 2 (t)) 4π S 1 S 1 γ 1 (s) γ 2 (t) 3 dsdt. We conclude then the first lower bound to the Möbius energy: E(γ 1, γ 2 ) 4π lk(γ 1, γ 2 ). In particular, if lk(γ 1, γ 2 ) = 1 or 1, we have E(γ 1, γ 2 ) 4π. Theorem 7.2. If (γ 1, γ 2 ) R 3 has lk(γ 1, γ 2 ) = 1 or 1, then E(γ 1, γ 2 ) 2π 2. Moreover, if E(γ 1, γ 2 ) = 2π 2 then (γ 1, γ 2 ) is a conformal image of standard Hopf link. Remark. Analogy with the Willmore problem: R 3 closed surface, genus g, W() = H2 d is conformal invariant. In this case we know Marques-Neves: g 1 W() 2π 2.

16 16 FERNANDO CODÁ MARQUES - IMPA 2π 2 Theorem. Let Φ : I 5 Z 2 (S 3 ) a map with the following properties: (1) Φ is continuous in the flat norm (2) Φ(I 4 {0}) = Φ(I 4 {1}) = 0 (3) x I 4, {Φ(x, t)} t [0,1] is a standard foliation of S 3 by round spheres centered at Q(x) S 3 (4) x I 4, Φ(x, 1/2) = B π/2 (Q(x)) (5) Q : I 4 S 3 has deg(q) 0. Then, there exists y I 5 so that M(Φ(y)) 2π 2. Fix (γ 1, γ 2 ) with lk(γ 1, γ 2 ) = 1. We produce Φ : I 5 Z 2 (S 3 ) satisfying (1) (5) and M(Φ(x)) E(γ 1, γ 2 ), for every x I 5. Recall the notation g = G(γ 1, γ 2 ) for the Gauss map of the link. The key calculation here is the following estimate of the jacobian of g: Jac(g) (s, t) γ 1 (s) γ 2 (t) γ 1 (s) γ 2 (t) 2. Consider g (S 1 S 1 ) Z 2 (S 3 ) given by g (S 1 S 1 )φ = S 1 S g φ. The 1 key calculation is useful to estimate the mass of this current M(g (S 1 S 1 )) Jac(g) dsdt E(γ 1, γ 2 ). S 1 S 1 We introduce some notation and observation before constructing Φ. Let p R 4, λ > 0 and define D λ,p (x) = λ(x p) + p, the dilation of λ centered at p. If (γ 1, γ 2 ) Sr 3 (p) R 4, λ > 0 and g = G(γ 1, D λ,p γ 2 ), then M( g (S 1 S 1 )) E(γ 1, γ 2 ). For every v B 4, we consider the conformal map F v : R 4 \ {v} R 4 given as Note that F v (x) = x v x v 2. F v (S 3 1(0)) = S v 2 (c(v)), where c(v) = Finally, for every v B 4 and z (0, 1) take and b(v, z) = 2z 1 (1 v 2 (0, ) + z)(1 z) a(v, z) = 1 + (1 v 2 )b(v, z) v 1 v 2. ( ) 1 1 v 2,. Definition of the family: For every (v, z) B 4 (0, 1) define (13) C(v, z) = g (v,z) (S 1 S 1 ) Z 2 (S 3 ), where (14) g (v,z) = G(F v γ 1, D c(v),a(v,z) F v γ 2 ).

17 MIN-MAX THEORY AND THE WILLMORE CONJECTURE 17 By construction, we know the masses of this family has an upper bound M(C(v, z)) E(γ 1, γ 2 ), for every (v, z) B 4 (0, 1). Remark. C(v, 0) = 0 = C(v, 1). Need to extend C to B 4 [0, 1] and determine the boundary behaviour. Fix v S 3 \ (γ 1 (S 1 ) γ 2 (S 1 )) and observe that (15) F v (S 3 1(0) \ {v}) = P v = {x R 4 : x, v = 1/2}. If x R 4 \ {v} and w B 4 v S 3, then D a(w,z),c(w) F w (x) = a(w, z)(f w (x) c(w)) + c(w) F v (x) b(z)v, where b(z) = (2z 1)/z(1 z) (, ). The Gauss map G(F v γ 1, F v γ 2 ) parametrizes B π/2 ( v) with degree 1, so we can write C(v, 1/2) = G(F v γ 1, F v γ 2 ) (S 1 S 1 ) = lk(γ 1, γ 2 ) B π/2 ( v) = B π/2 (v). Then the center map has degree 1. The problem we have to deal with is that C(v, z) does not parametrize a round sphere if z 1/2. But it is contained in a hemisphere, then it is possible to take a length decreasing retraction R(v, λ, t) onto B r(z) (v). By doing this it is possible to apply the 2π 2 -Theorem. References [1] F. Codá Marques and A. Neves, Min-max theory and the Willmore conjecture. To appear in Annals of Mathematics. [2] I. Agol, F. Codá Marques and A. Neves, Min-max theory and the energy of links. preprint arxiv: [math.gt]