Subdividing Three-Dimensional Riemannian Disks
SSUBDIVIDING THREE-DIMENSIONAL RIEMANNIAN DISKS
PARKER GLYNN-ADEY AND ZHIFEI ZHU
Abstract.
P. Papasoglu asked in [Pap13] whether for any Riemannian 3-disk M withdiameter d , boundary area A and volume V , there exists a homotopy S t contracting theboundary to a point so that the area of S t is bounded by f ( d, A, V ) for some function f . He further asks whether it is possible to subdivide M by a disk D into two regions ofvolume V / D is bounded by some function h ( d, A, V ).In this paper, we answer the questions above in the negative. We further prove thatgiven N > c ∈ (0 , g (cid:48) so that any 2-disk D subdividing( M, g (cid:48) ) into two regions of volume at least cV , the area of D is greater than N . We alsoprove that for any Riemannian 3-sphere M , there is a surface that subdivides the diskinto two regions of volume no less than V /
6, and the area of this surface is bounded by3 HF (2 d ), where HF is the homological filling function of M . Introduction
In this paper we prove the following results.
Theorem 1.1.
Let M be diffeomorophic to a 3-disk with boundary ∂M ∼ = S . For anynumber N > there exists a Riemannian metric g = g ( N ) on M such that: • The diameter of M satisfies diam( M, g ) ≤ . • The surface area of the boundary satisfies vol ( ∂M, g ) = 4 π. • The volume of M satisfies vol ( M, g ) ≤ . • For any homotopy F : S × [0 , → M between ∂M and a point p ∈ M there existssome t ∈ [0 , such that the area vol ( F ( S , t )) is greater than N . We will prove Theorem 1.1 by explicitly constructing metrics on the 3-disk. We will usea similar construction to prove the following:
Proposition 1.2.
For any
N > and c ∈ (0 , there exists a metric g (cid:48) = g (cid:48) ( c, N ) on the3-disk M such that the diameter d , volume V of M , and surface area A of ∂M are boundedabove by 10, but any smooth embedded disc D subdividing M into two regions of volumegreater than cV has area greater than N . Remark 1.3.
Panos Papasoglu and Eric Swenson independently constructed a Riemannian3-sphere which is hard to cut by a surface of any genus in [PS15]. Their expander basedconstruction implies Theorem 1.1 and Proposition 1.2; we give an alternative elementaryconstruction for genus zero case.
Remark 1.4.
Theorem 1.1 and Proposition 1.2 do not imply each other. In fact, inTheorem 1.1, the image F ( S , t ) can be an immersed sphere which does not necessarilysubdivide the 3-disk into two regions.We also show that given a Riemannian 3-sphere with diameter d and volume V , thereexists a surface that subdivides the 3-sphere into two regions of volume > V , and thearea of this surface is bounded in terms of d and the first homological filling function of themetric, which is defined below. a r X i v : . [ m a t h . DG ] A ug efinition 1.5. Given a Riemannian 3-sphere M with diameter d and volume V , letSA( M ) = inf H ⊂ M { vol ( H ) : M \ H = X (cid:116) X , vol ( X i ) > V for i = 1 , } be the subdivision area of M , where the infimum is taken over all embedded surfaces. Nowwe define HF ( (cid:96) ) = sup || z || ≤ (cid:96) (cid:18) inf ∂c = z vol ( c ) (cid:19) to be the first homological filling function. In the definition of HF ( (cid:96) ) the supremum istaken over all 1-cycles z satisfying vol ( z ) ≤ (cid:96) and the infimum computes the size of thesmallest 2-cycle c filling z = ∂c . Theorem 1.6.
For any Riemannian 3-sphere ( M, g ) with M diffeomorphic to S we have: SA( M ) ≤ (2 d ) where d is the diameter of ( M, g ) . Remark 1.7.
One can replace the constant in Definition 1.5 by − ε for any small ε >
0. The proof given below in Section 3 is straightforward to adapt to this more generalconstant. We use a fixed constant only for the sake of concreteness.The proof techinque we use to show Theorem 1.6 is an adaptation of technique firstdeveloped by Gromov in [Gro83, § M be a Riemannian manifold homeormophic to a 3-disk satisfying: (i)diam( M ) = d , (ii) area( ∂M ) = A , (iii) and vol ( M ) = V . Is it true that there is a homotopy S t : ∂M × [0 , → M such that: S = id ∂M and S is a point and vol ( S t ) ≤ f ( A, d, V ) forsome function f ?Question 2: Let M be as above. Is it true that there is a relative 2-disc D splitting M in totwo regions of volume at least V / D ) ≤ f ( A, d, V ) for some function f ?These questions were inspired by the work of Ye. Liokumovich, A. Nabutovsky, and R.Rotman in [LNR12]. They proved that any Riemannian 2-sphere ( M, g ) can be swept-out by curves of length at most 200 diam( M ) max (cid:26) , log √ Area( M )diam( M ) (cid:27) and showed that thisbound is optimal up to a constant factor. Liokumovich, Nabutovsky, and Rotman’s workwas related to the work of S. Frankel and M. Katz [FK93]. For further refinements of thatwork, see Liokumovich [Lio13] which constructs Riemannian 2-spheres which are hard tosweep out by 1-cycles.In this work, we give negative answers to Question 1 (Theorem 1.1) and Question 2(Proposition 1.2). We do, however, prove a positive result which majorizes the size of thedisk in Question 2 by the homological filling function and diameter of M (Theorem 1.6).Papasoglu’s Question 1 is a natural extension of the following question asked by Gromovin 1992: Consider all Riemannian metrics ( D , g ) on the 2-disk such that: (i) the length ofthe boundary is at most one and (ii) the diameter of the disk is at most one. Is there auniversal constant C such that: For every such metric there is a free homotopy of curveswhich contracts the boundary to a point through curves of length at most C ?Frankel and Katz answered Gromov’s question negatively in [FK93]. They construct ametric on the disc with a “wall” whose base is shaped like a regular binary tree with many odes. Combinatorial properties of the tree force any curve subdividing the nodes in to twoequal parts to meet the edges of tree many times. This curve will have to “climb over thewall” many times. This combinatorial obstruction forces any contraction of the boundaryof the disc to a point to contain a long curve. In our context we need to produce a largesurface in any contraction of the boundary of D to a point. We do so by constructig ametric which is concentrated around two solid tori embedded in D as a Hopf link. Thefact that the tori are linked forces any sweep-out of the 3-disk to meet one component oflink transversally and hence any sweep-out will contain a 2-cycle of large area.Lemma 2.4 tells us that any sweep-out of a 3-disk containing a pair of linked solid torimust contain an essential loop on the boundary of one of the tori. This essential loop willbound some disk in the solid torus. Our choice of the Riemannian metric on the solid toriforces any such disk to have a large area and this fact implies the assertion of Theorem 1.1and Proposition 1.2. Remark 1.8.
Our construction of the metric above was inspired by D. Burago and S. Ivanov’sconstruction of a metric on the 3-torus ( T , g ) such that any homologically non-trivial 2-cycle in T has large area [BI98]. L. Guth remarked that such a construction should providean example of a sphere which is hard to bisect in [Gut07].1.1. Outline.
In Section 2, we will first construct a Riemannian metric g on the 3-disk andwe will then prove that g has the desired properties. Every contraction of the boundary ofthe disk will contain a surface of large area. In Section 3, we give a method of decomposingRiemannian 3-sphere ( M, g ) in to two regions with volume bounded below by a constantfraction of the total volume by a surface of area bounded by the homological filling functionand diameter of the M . That is, we will prove Theorem 1.1 and Proposition 1.2 in Section2 and Theorem 1.6 in Section 3.2. Proof of Theorem 1.1 and Proposition 1.2
The constructions of the metrics in Theorem 1.1 and Proposition 1.2 are similar. Let( D , g ◦ ) be a Euclidean 3-disk of radius one in R , where g ◦ is the standard Euclideanmetric on D . Consider a smooth embedding of a pair of solid tori f : T (cid:116) T → D into D which embed the solid tori as a thick Hopf link (See Figure 1). We modify the metricon the linked pair in the following way. DLT T T
Figure 1.
Embedding of linked solid tori.Let D be a hyperbolic 2-disk of unit radius and curvature − c , where c is positive. Thearea vol ( D ) of the disk is proportional to sinh( c ) /c . When c → ∞ we have vol ( D ) → ∞ .Let δ > N >
0, there exists an ε > ( D ) > N and ε · vol ( D ) < δ . Let C = D × [0 , ε ] be a cylinder with the product metric obtained rom the hyperbolic disc and the interval. The area of a horizontal cross section of C isvol ( D ) > N , but the volume vol ( C ) ≤ δ . Let T be a solid torus obtained by identifyingthe ends, D × { } and D × { ε } , of C . Let ˜ g be the induced metric on T . Note that thevolume vol ( T, ˜ g ) of the solid torus is bounded by δ .We denote by LT the image of the pair of linked solid tori in D . Let U be an openneighbourhood of LT . Let us define the metric ˜ g on LT to be ˜ g as above and then smoothlyextend it to the neighbourhood U . Let V be an open neighbourhood of D \ U such that V ∩ LT = ∅ . We take two smooth bump functions f and f on U and V respectively suchthat f + f = 1 and define the metric g on D to be f · ˜ g + f · g ◦ . Note that for any δ >
0, one can choose U and V so that the volume vol ( D , g ) is bounded by π + δ . Weclaim that, with a proper choice of the constants, the metric g has the desired properties inTheorem 1 . F : ∂D × [0 , → D a homotopy between theboundary ∂D and a point p ∈ D . By following lemma, it suffices to prove Theorem 1.1 inthe case where the intersection of F ( S , t ) and ∂LT is transverse for all but finitely many t ∈ [0 , Lemma 2.1.
Let X be an embedded submanifold of D . We write S = ∂D . For anyhomotopy F : S × [0 , → D of the boundary ∂D to a point, there exists a smoothhomotopy (cid:101) F : S × [0 , → D , such that max t ∈ [0 , | vol ( (cid:101) F ( S , t )) − vol ( F ( S , t )) | < ε and (cid:101) F ( S , t ) is transverse to X for all but finitely many t ∈ [0 , .Proof. We will use Thom Multijet Transversality Theorem ([ ? ], Theorem 4.13) to prove thisLemma. Let ι : X (cid:44) → D be the embedding of X . We denote the map F ( · , t ) : S ×{ t } → D by F t . Consider the map π : F − ( ι ( X )) → [0 , X × [0 , → [0 ,
1] to the domain F − ( ι ( X )). Note that for t ∈ [0 , F t (cid:116) ι is transverse if and only if t is a regular value of the projection π . The idea is that because[0 ,
1] is a one-dimensional manifold, we can perturb π to be a Morse function with distinctcritical values.Indeed, by Proposition 6.13 and Theorem 4.13 in [ ? ], the set of Morse functions all ofwhose critical values are distinct form a residual set in C ∞ ( F − ( ι ( X )) , [0 , F : S × [0 , → D , we first take a sequence of smooth homotopies F j approximating F . Then for each F j , we take a sequence of maps F jk approximating F j such that each F jk satisfies π : F − jk ( ι ( X )) → [0 ,
1] is Morse. And finally, we take ˜ F to be F kk for k sufficiently large. (cid:3) We will work with basic notions of geometric measure theory encompassed by the fol-lowing definitions from [Fed69]. The notion of flat norm was originally introduced by H.Whitney in [Whi12] and was used to describe the cycle spaces of manifolds in W. Fleming’spaper [Fle66]. We will use the following standard terminology:
Definition 2.2.
For a Riemannian manifold M , the space of Lipschitz k -chains in M withcoefficients in G , an abelian group, is C k ( M ; G ) = { (cid:80) a i f i : f i : ∆ k → M a Lipschitz map , a i ∈ G } . We write Z k = ker ∂ k for the cycles in C k , where ∂ k : C k → C k − is the boundary map.When G = Z or Z and c = (cid:80) a i f i ∈ C k ( M ; G ) we define the k -mass of c to be:mass k ( c ) = (cid:88) i | a i | (cid:90) ∆ k f ∗ i ( d vol g )We endow the space of integral cycles with the flat norm given by: || c || = inf d ∈ C k +1 mass k ( c + ∂d ) + mass k +1 ( d ) e define the flat distance between two cycles c , c to be the flat norm || c − c || .We give a definition of sweep-out which we first encountered in [Gut07]. Definition 2.3.
Let M be an n -dimensional manifold and G an abelian group. Denotethe space of integral flat cycle with coefficient group G on M by Z ∗ ( M ; G ). A ( n − k )-dimensional family of k -cycles z is a continuous map from an ( n − k )-simplicial complex K to Z k ( M ; G ). We say z is a sweep-out of M , if z induces a nontrivial gluing homomorphism.The gluing homomorphism is constructed in the following way.We pick a fine triangulation of K so that if v j and v j +1 are two neighbouring vertices inthe triangulation of K then the k -cycle ( z ( v j ) − z ( v j +1 )) bounds a ( k + 1)-chain in M .We map the edge [ v j , v j +1 ] to this ( k + 1)-chain. Inductively, one can map each i -simplex∆ i of K to a ( k + i )-chain in M as long as the triangulation is sufficiently fine. This inducesa chain map between the simplicial complex of K and singular complex of M . This chainmap induces, on homology, the gluing homomorphism z ∗ : H n − k ( K ; Z ) → H n ( M ; Z ). Onecan check that z ∗ is independent of choice of the chains if the triangulation is fine enoughand the fillings are chosen appropriately.In this work we will only use sweep-outs of T by families of 1-cycles. For more detaileddiscussions of sweep-outs, one may refer to [Gut07, Section 1], [BS10, Section 2] and [Alm62].Note that the homotopy F induces a degree one map ˜ F : ( S × [0 , / ( S × { , } ) → D /∂D . By restricting the map F to F − ( ∂LT ), the transversality of F ( S , t ) (cid:116) ∂LT im-plies that the one parameter family of 2-cycles F ( S , t ) induces a sweep-out of the embeddedpair of tori ∂LT by 1-cycles. Indeed, for t ∈ [0 , F ( S , t ) (cid:116) ∂LT defines a continuousfamily of 1-cycles. When we consider the image of this family of 1-cycles under the gluinghomomorphism, we will obtain precisely the degree one map ˜ F . We will prove that these1-cycles contain a meridian of one of the tori in ∂LT , which bounds a disk with large area.To begin with, we first show in Lemma 2.4 that the family of 1-cycles must contain at leastone non-contractible loop on the torus.Let T be Riemannian 2-torus and z : S → Z ( T ; Z ) a one-parameter family of 1-cyclesbased at a constant loop which is a sweep-out of T . Let us parameterize S by [0 ,
1] anddefine z t = z ( t ) so that z = z is the constant loop. Note that for every t , the 1-cycle z t is obtained from the the intersection between a 2-cycle and a torus, thus we may represent z t as a finite sum z t = (cid:80) n ( t ) i =1 a t,i f t,i , where each a t,i ∈ Z and f t,i : S → T . We show thefollowing: Lemma 2.4.
Let T be Riemannian 2-torus and z t = (cid:80) n ( t ) i =1 a t,i f t,i be as above. If the familyof cycles z t is a sweep-out of T then there exists some t ∈ [0 , and ≤ i ≤ n ( t ) , suchthat the image f t ,i ( S ) is a non-contractible loop in T .Proof. We proceed by contradiction. Suppose for all t , every component f t,i of z t iscontractible. We show that z induces a trivial gluing homomorphism z ∗ : H ( S ; Z ) → H ( T ; Z ).More precisely, the family z t taken as a map z : S → Z ( T ; Z ), represents a class[ z ] in the fundamental group π ( Z ( T ; Z )). We would like to represent the class [ z ] bya sum (cid:80) Ni =1 c i , where each c i : S → Z ( T ; Z ) is a continuous family of circles in T .If every loop is contractible, we show that every c i corresponds to a spherical class in H ( T ; Z ) ∼ = π ( Z ( T ; Z )) which is a contradiction. See the Figure 4 for the visual intuitionof the proof.The maps c i are obtained in the following way. We first construct a one-dimensionalsimplicial complex P from z . Recall that for every t ∈ [0 ,
1] the cycle z t is a union of t := (cid:80) n ( t ) i =1 | a t,j | circles. We say that the topology of z t changes if: (i) a circle f t,j vanishesor (ii) several circles merge; we define these terms below. Definition 2.5.
The set { f t,j } j ∈ J of circles merge at t = s , if there is a map z (cid:48) s : (cid:87) i ∈ I S i → T such that the flat distance lim t → s (cid:13)(cid:13) z (cid:48) s − (cid:80) i ∈ I f t,i (cid:13)(cid:13) = 0 for some finite index set I ⊂{ , , . . . , k s } , where (cid:107) · (cid:107) is the flat norm defined in Definition 2.2. And we say that aparticular circle f t,i vanishes if lim t → s (cid:107) z (cid:48) s − f t,i (cid:107) = 0 where z (cid:48) s is a constant map.Note that by Lemma 2.1, we may perturb the family { z t } so that: for every t , the topologyof z t changes at most once at t . We define the 0-skeleton of P as follows. For t = 0 and 1,we define one vertex v (resp. v ) that corresponds to the constant loops z (resp. z ). Wecreate a vertex v t , if for any sufficiently small ε >
0, we have | k t − k t + ε | (cid:54) = 0.And the 1-skeleton of P is defined as follows. Let v t and v t be two consecutive vertices,where t < t . For t ∈ ( t , t ), we have that k t is a constant and there are k t circles. Foreach f t,i which changes topology at t and t , we define an edge e i connecting v t and v t .Let us parameterize the edge by e i : [ t , t ] → P . (See Figure 2.) Note that if z t is inducedby a Morse function then P is just the corresponding Reeb graph. ttt v t v t e t v t e e f t,1 f t,2 z’ t z’ t z’ t ,1 z’ t ,2 Pt v t z ... ... Figure 2.
Constructionof P . Here there are twomerges at t and t . Anda circle vanishes at t . G G Figure 3.
Contractionof the circles at a vertex.A vertex v s of our graph at t = s either corresponds to a constant loop or, for someindex set I ⊂ { , , . . . , k s } , there is a map z (cid:48) s : (cid:87) i ∈ I S i → T such that the flat distancelim t → s (cid:107) z (cid:48) s − (cid:80) i ∈ I f t,i (cid:107) = 0. In the latter case, let ι j : S j (cid:44) → (cid:87) i ∈ I S i be the naturalinclusion map for j ∈ I . By our assumption, every z (cid:48) s ( ι j ( S j )) is contractible in T . Weparameterize this contraction of by G j : S × [0 , δ ] → T , where δ is a small positive number, G j ( S × { } ) = z (cid:48) s ( ι j ( S j )) and G j ( S × { δ } ) is a point in T . (See Figure 3.)For each edge e i : [ t , t ] → P , we will define the corresponding map c i : [0 , → Z ( T ; Z )in the following way.If t = 0 or t = 1 then, by our assumption, the vertices v and v correspond to a singleconstant loop and we may define c i ( t ) or c i ( t ) to be this loop. For every t ∈ ( t , t ), wewill define c i ( t ) to be the cycle f t,i : S → T .Suppose t (cid:54) = 0. When t = t , if e i ( t ) is a vertex corresponding to a constant loop ,define c i ( t ) to be this loop. Otherwise, there is a merged circle z (cid:48) t : (cid:87) i ∈ I S i → T at t .Let G : (cid:0)(cid:87) i ∈ I S (cid:1) × [0 , δ ] → T be a contraction such that (cid:107) G ( ι j ( S ) × [0 , δ ]) (cid:107) = (cid:107) G j (cid:107) , forevery j ∈ I , where G j is the map defined above. We note that this map G keeps track of See the vertex v t in Figure 2 ll the individual contractions of circles given to us by hypothesis. It is straightforward tocheck that such a G exists since we can reparameterize each G j without changing the flatnorm. Let f : S → (cid:87) i ∈ I S be a degree one surjection, and now define c i | [ t − δ,t ] ( t ) to bethe cycle G ( f ( · ) , t − t ) : S → T . = + + + G + ...... Figure 4.
Representing the class [ z ] by (cid:80) c i . We insert two copies of thedisks obtain from G at each vertex.Similarly, when t (cid:54) = 1, one can define c i | [ t ,t + δ ] ( t ) to be the cycle G ( f ( · ) , t − t ) : S → T for the corresponding contraction G . This completes the construction.Let c = (cid:80) i c i : [0 , → Z ( T ; Z ), where the sum is over the set of all edges in P .(See Figure 4.) Observe that c represents the same element as z in π ( Z ( T ; Z )). Tosee this note that, up to a reparameterization, the image of z − c corresponds to a unionof spheres in T obtained by gluing two copies of the contracting disks G at each vertex.Because every spherical class is trivial in H ( T ; Z ) the spheres represent the trivial elementin π ( Z ( T ; Z )) under Almgren’s isomorphism.Finally, we show that each c i induces a trivial gluing homomorphism. Indeed, let η > S so that if u j and u j +1 are two consecutive vertices in thesubdivision then the flat distance between c i ( u j ) and c i ( u j +1 ) is less than η . We take η to besufficiently small so that c i induces a well-defined gluing homomorphism. See Definition 2.3.Because each c i ( u j ) is a contractible loop in T one may homotope c i ( u j ) to c i ( u j +1 )for consecutive vertices in the fine subdivision of S . Because c i (0) = c i (1) is a constantloop one may contract c i ( u j ) to the constant loop through [0 , u j ] and [ u j , S ] under the gluing homomorphism c i ∗ : H ( S ) → H ( T ) can be represented by a map from the sphere S → T , where the S is obtained by gluing the two contractions of c i ( u j ). This shows that c i ∗ is trivial andfurther implies z ∗ is trivial, which is a contradiction. (cid:3) We now use the linking property of the solid tori LT to show that these 1-cycles indeedcontain a meridian circle of ∂LT . Recall that T i is the connected component of LT and C i is the cylinder D × [0 , ε ], where D is a hyperbolic 2-disk. Definition 2.6.
Let us identify the solid tori T i with C i / (( x, ∼ ( x, ε )), x ∈ D . Let F : S × I → D be a homotopy contracting the boundary ∂D to a point. We say theimmersed sphere F ( S , t ) intersects ∂T i along the essential circle ∂D , if the projection map p D : F ( S , t ) ∩ C i → D × { } to the base of the cylinder C i is surjective. (See Figure 5.) Inparticular, the restriction p D : ∂ ( F ( S , t ) ∩ C i ) → ∂D × { } is also surjective. Lemma 2.7.
Let f : T (cid:116) T → D be a topological embedding of a pair of solid tori into D and such that the image LT is a Hopf link. Let F : S × [0 , → D be a homotopy D C i Figure 5.
Intersect along the essential circle ∂D . contracting ∂D ∼ = S to a point. Suppose that for almost all t ∈ [0 , , the intersection F ( S , t ) (cid:116) ∂LT is transverse. Let us identify each solid torus with a cylinder C i = D × [0 , ε ] .Then there exists some t ∈ [0 , such that F ( S , t ) intersects one of the ∂T and ∂T alongthe essential circle ∂D .Proof. Let F t = F ( · , t ) : S → D . For almost all t ∈ [0 ,
1] we write z t = F t ( S ) ∩ ∂LT as aunion of circles ∪ n t k =1 S t,k for some integer n t > A i = ∂D × { } in ∂T i be an essential circle along the meridian of the torus and B i an essential circle along the equator. We identify the torus ∂T i in the usual way with asquare whose boundary is A i and B i and define p A i : ∂T i → A i and p B i : ∂T i → B i be theprojection maps of this square to its boundary edges.When we analyze the behaviour of the circles S t,k one of the following three cases mustoccur:(1) The circles S t,k are contractible in ∂LT for every t .(2) For some t ∈ [0 , { S t ,k } n t k =1 are not all contractible in ∂LT but themap p A i : S t ,k → A i has degree 0 for every k = 1 , , . . . , n t and i = 1 , t and k the circle S t ,k is non-contractible in ∂LT and the map p A i : S t ,k → A i has non-zero degree for i = 1 or 2. (1)(2)(3) Figure 6.
Examples of the three cases.Note that (1) is impossible because of Lemma 2.4. If t is the time such that case(3) happens, then we are done. Indeed, without loss of generality, let us assume that p A : S t ,k → A has non-zero degree. Note that the circle S t ,k bounds a disk ∆ in F t ( S ) and the projection p A : S t ,k → A is the restriction of p : ∆ ∩ T → D to its oundary. We will check that p is surjective. Consider the following commutative diagram. . . . (cid:47) (cid:47) H (∆) (cid:15) (cid:15) (cid:47) (cid:47) H (∆ /S t ,k ) ˜ p ∗ (cid:15) (cid:15) ∼ = (cid:47) (cid:47) H ( S t ,k ) ( p a, ) ∗ (cid:15) (cid:15) (cid:47) (cid:47) H (∆) (cid:15) (cid:15) (cid:47) (cid:47) . . .. . . (cid:47) (cid:47) H ( D ) (cid:47) (cid:47) H ( D/A ) ∼ = (cid:47) (cid:47) H ( A ) (cid:47) (cid:47) H ( D ) (cid:47) (cid:47) . . . The map p is surjective since, by the long exact sequence above, the middle map ˜ p :∆ /S t ,k → D/A induced by p has non-zero degree.Finally we show that case (2) is impossible. In case (2), there exists some non-contractiblecircle S t ,k of z t . Since the degrees of the maps p A and p A are both zero, the non-contractibility of S t ,k in ∂LT implies one of the maps p B and p B has non-zero degree.Let us assume p B has non-zero degree. This implies that the linking number between S t ,k and B is non-zero. Let ∆ be a disk bounded by S t ,k in F t ( S ).Then the intersection ∆ ∩ T is non-empty. Because, otherwise, we may unlink S t ,k with B by contracting S t ,k in ∆. In fact, this further implies B intersects some non-contractible circle S t ,k in ∆ ∩ ∂T . However, by our assumption, the map p A : S t ,k → A has degree 0. This is a contradiction, because in this case, by contracting its image under p A , we may homotope S t ,k such that the intersection between S t ,k and B is empty. (cid:3) We may now complete the proof of Theorem 1.1.
Proof of Theorem 1.1.
Fix any
N >
0. Our choice of N fixes a specific hyperbolic disc D .We choose δ = δ ( N ) > D , g ) be theRiemannian 3-disk we’ve constructed. Recall the linked tori LT consisting of T and T ina Hopf link. See Figure 1. Let F : S × [0 , → D be a homotopy from S ∼ = ∂D to apoint p ∈ D By our construction, the volume vol ( D , g ) is bounded by δ + π ≤ d of ( D , g ) is also bounded. Indeed, let U be the tubu-lar neighbourhood of LT as above. By construction, for any point x ∈ LT , the distancedist( x, ∂LT ) ≤ y ∈ D \ U we have dist( y, ∂D ) ≤
1. Therefore, with an appropri-ate choice of U and δ , we have dist( x, ∂D ) ≤ η , for some small η >
0. And then thediameter d ≤ · η + π <
10, since every pair of points in ∂D can be connected by acurve of length at most π .Finally, we show that there exists a t ∈ [0 ,
1] such that the area of the intersectionof the image of F with LT satisfies vol ( F t ( S ) ∩ LT ) ≥ N . By Lemma 2.7, there is a t ∈ [0 ,
1] such that F t ( S ) intersects one of the tori ∂T or ∂T along the essential circle ∂D . Without loss of generality let us assume the projection map p : F t ( S ) ∩ T → D issurjective. Because the metric g on the cylinder D × [0 , ε ] is a product metric, the projectionmap p is area decreasing. That is, vol ( F t ( S ) ∩ T ) ≥ vol ( D ) ≥ N , which completes theproof. (cid:3) Using a similar argument, we are able to prove Proposition 1.2. Recall that in Proposition1.2, we would like to show that given any large number N and a fixed c ∈ (0 , g (cid:48) on the 3-disk M with bounded diameter, volume, and boundary area. And forany smooth boundary relative embedded disk f : ( D , S ) → ( M, ∂M ) that subdivides M into two regions of volume at least c · vol ( M ) we have that the area of the disk f ( D ) isgreater than N . Proof of Proposition 1.2.
Given a positive number N and c ∈ (0 , ε > M, g ) as they are in Theorem 1.1 such that the diameter d , volume V and the oundary area of the disk is bounded by six, but the area of the cross sections of the solidtori LT is greater than N/c . Let us modify the metric g in the following way.We first “concentrate” the volume of M in the pair of linked solid tori LT so that if f : D → M is a subdividing disk, it also subdivides LT into two regions of volume atleast c · vol ( LT ). Given any η >
0, we may choose δ > h : M → R a smoothfunction with h | LT = 1 and 0 < h | M \ U ≤ δ , where U is a neighbourhood of LT . Forany p ∈ M , u, v ∈ T p M , define g (cid:48) ( u, v ) = h ( p ) g ( u, v ). Denote by d (cid:48) , V (cid:48) the diameter andvolume of ( M, g (cid:48) ) respectively. Then d (cid:48) ≤ d , and V (cid:48) ≤ V , because 0 ≤ h ≤
1. Let δ be sufficiently small so that | vol ( M, g (cid:48) ) − vol ( LT, g (cid:48) ) | < η . Suppose that f : D → M subdivides ( M, g (cid:48) ) into two regions R and R such that vol ( R i , g (cid:48) ) > cV (cid:48) , for i = 1 , ( LT ∩ R i , g (cid:48) ) ≥ cV (cid:48) − η .We will then prove that in this case f ( D ) must have large area. Let T and T be theconnected component of LT . Each of them is a solid torus. Without loss of generality, wemay assume that the intersection between f ( D ) and the boundary ∂LT = ∂T (cid:116) ∂T istransverse. Let (cid:70) Kj =1 S j be the intersection, where each S j is an embedded circle in ∂LT .As in the proof of Theorem 1.1, if a circle S j is not contractible in ∂T i , for i = 1 or 2, thenthe intersection between f ( D ) and LT has area greater than N/c .Let us now assume that all S j are contractible in ∂T i and consider the cylinders C and C ( ∼ = D × [0 , ε ]) which are identified with the solid tori T and T respectively. The disk f ( D ) subdivides each C i into two regions. We use the notations R ∩ C i and R ∩ C i todenote these regions. Let ρ : C ∪ C → [0 , ε ] be the projection onto the interval [0 , ε ], thenthe volume of R i ∩ LT satisfies:vol ( R i ∩ LT ) = vol ( R i ∩ ( C ∪ C ))= (cid:90) ε vol ( ρ − ( t ) ∩ R i ) dt ≤ ε · max t ∈ [0 ,ε ] vol ( ρ − ( t ) ∩ R i ) . We will now use the fact that T and T are linked to show the following projectionequality. Note that T and T are identified with the cylinders C and C . Let p : C ∪ C → D ∪ D be the projection to the base disk D . We show that: Lemma 2.8 (The Projection Equality for Links) . If f : ( D , ∂D ) → ( D , ∂D ) is a smoothembedding that separates D into two closed connected regions R and R as above and theintersection f ( D ) ∩ ∂LT is contractible in ∂LT then p ( R i ∩ LT ) = p ( ∂ ( R i ∩ LT )) for atleast one of i ∈ { , } , where LT = T (cid:116) T .Proof of Lemma 2.8. We suppose that both equalities fail. In the argument that followswe will show that either T and T are unlinked or one of the circles in f ( D ) ∩ LT isnon-contractible in LT . We may only work with R and T because our hypotheses aresymmetric. First note that ∂ ( R ∩ T ) ⊆ R ∩ T . If p ( R ∩ T ) (cid:54) = p ( ∂ ( R ∩ T )) then wemay pick y ∈ D such that p − ( y ) ∩ ∂ ( R ∩ T ) = ∅ . Since p − ( y ) ⊂ T is connected anddoes not intersect ∂ ( R ∩ T ), it must be contained within a single connected component ofInt( R ∩ T ).Now, consider R . If p ( R ∩ LT ) (cid:54) = p ( ∂ ( R ∩ LT )) then, as before, we may pick y ∈ D such that p − ( y ) ∩ ∂ ( R ∩ LT ) = ∅ . By the connectedness of p − ( y ) there are two cases,either p − ( y ) ⊂ Int( R ∩ T ) or p − ( y ) ⊂ Int( R ∩ T ).In the first case, suppose that p − ( y ) ⊂ Int( R ∩ T ). Note that p − ( y ) and p − ( y ) arecobordant inside T . Thus there is an annulus A with ∂A = S y (cid:116) S y and an embedding : A → T such that α ( S y ) = p − ( y ) and α ( S y ) = p − ( y ). We have that p − ( y ) ⊂ Int( R )and p − ( y ) ⊂ Int( R ). It follows that the annulus α ( A ) meets the disc f ( D ) along a circle.The contraction of this circle in f ( D ) meets an essential circle S in ∂T . Because, otherwise,one can contract the circle p − ( y ) in the solid torus T by homotoping p − ( y ) to S and thencontracting it in ∂T . This contradicts to our hypothesis that each circle in f ( D ) ∩ ∂LT iscontractible in ∂LT .In the second case, suppose that p − ( y ) ⊂ Int( R ∩ T ). If this holds, then p − ( y ) and p − ( y ) can be separated by a disk. That is, we can contract p − ( y ) in R and p − ( y ) in R , and thus unlink LT . (cid:3) We now complete the proof using the above lemma to obtain a lower bound for the areaof f ( D ). Without lost of generality we assume that Lemma 2.8 holds for the region R .Then ε · vol ( f ( D ) ∩ LT ) ≥ ε · vol { p ( f ( D ) ∩ ( C ∪ C )) } = ε · vol { p ( ∂ ( R ∩ ( C ∪ C ))) } = ε · vol { p ( R ∩ ( C ∪ C )) }≥ ε · max t ∈ [0 ,ε ] vol ( ρ − ( t ) ∩ R ) ≥ vol ( R ∩ LT ) > cv (cid:48) − η, Therefore, we obtain that vol ( f ( D )) ≥ cv (cid:48) /ε − η/ε > N − η/ε, for arbitrarily small η > (cid:3) Proof of Theorem 1.6
In this last section, we prove Theorem 1.6. Given a Riemannian 3-sphere (
M, g ) withdiameter d and volume V , consider the hypersurfaces Σ in M that subdivide M into twoconnected components, M \ Σ = R (cid:116) R , with both parts satisfying vol ( R i ) > V . Weclaim that for any small δ >
0, there exists such a subdividing surface Σ with areavol (Σ) ≤ (2 d + δ ) + o ( δ )By taking δ →
0, we obtain the result in Theorem 1.6. Our argument is derived fromsimilar filling arguments found in Nabutovsky and Rotman’s work on minimal hypersur-faces [NR06]. We prove the claim above by contradiction: we show that if there is no suchsurface then the fundamental class [ M ] ∈ H ( M ; Z ) is zero. During the proof we will tryto construct a 4-chain which has the fundamental class of M as its boundary; the proof issimilar to a standard coning argument. Proof of Theorem 1.6.
Suppose there is no subdividing hypersurface Σ satisfying the volumebound above. Let us choose a triangulation of M such that the length of each edge of thetriangulation is at most δ , where δ > M (cid:52) = (cid:80) σ ijkl , where each σ ijkl is a continuous map σ ijkl : ∆ → M . Note that M (cid:52) represents the fundamental classof M ; that is, [ M (cid:52) ] = [ M ] ∈ H ( M ; Z ). We label the zero skeleton by e i : ∆ → M , and v i = e i (∆ ).We cone off inductively, skeleton by skeleton. Let M ( i ) (cid:52) denote the i -skeleton of thetriangulation of M . In what follows we will abuse notation and add simplices to M (cid:52) inorder to complete it to a filling of [ M ]. We begin with the zero skeleton. Pick some genericpoint v (cid:63) in M and add it to M (0) (cid:52) . We continue to the one skeleton. We add the following 1-chains to M (cid:52) : For every v α with α (cid:54) = (cid:63) we let e α(cid:63) : ∆ → M be a minimal geodesic from v α o v (cid:63) . Let e αβ denote an edge in the original triangulation M (cid:52) . Note that mass ( e αβ ) ≤ δ and mass ( e α(cid:63) ) ≤ d . We continue to the two skeleton. We add the following 2-chains to M (cid:52) : First note that for every v α and v β adjacent in the triangulation with α, β (cid:54) = (cid:63) wehave that z αβ(cid:63) = e αβ + e β(cid:63) + e α(cid:63) is a 1-cycle of mass at most 2 d + δ . Thus, by the definitionof HF there is a 2-chain φ αβ(cid:63) such that ∂φ αβ(cid:63) = z αβ(cid:63) and mass ( φ αβ(cid:63) ) ≤ HF (2 d + δ ). Welet e ijk be the restriction of the simplex σ ijkl to the face containing the vertices { v i , v j , v k } .We let z αβγ(cid:63) = e αβγ + φ αγ(cid:63) + φ α(cid:63)β + φ γβ(cid:63) be the cycle we’ve constructed. Note thatmass ( z αβγ(cid:63) ) ≤ (2 d + δ ) + o ( δ ).We now continue to the three skeleton. In the rest of the proof, we will work in Z -homology to avoid orientation issues. In Lemma 3.1 we will show that, under our assumptionabout subdividing surfaces, each of the 2-cycles z αβγ(cid:63) can be replaced by a 2-cycle ˆ z αβγ(cid:63) in Z -coefficient with the same support which, in addition, bounds a small volume 3-chainˆ φ αβγ(cid:63) . In Lemma 3.1 we will show that mass( ˆ φ αβγ(cid:63) ) ≤ V . Let us assume, for now,that Lemma 3.1 holds. We will use the small volume fillings of the chains z αβγ(cid:63) to fill thefundamental class of M .We now continue to the four skeleton. Consider the 3-chain formed by taking all thesmall volume fillings and the original 3-simplex from the triangulation of M :ˆ z ijkl(cid:63) = ˆ φ ijk(cid:63) + ˆ φ jkl(cid:63) + ˆ φ kli(cid:63) + ˆ φ ijl(cid:63) + σ ijkl One can check that ∂ ˆ z ikjl(cid:63) = 0 ∈ C ( M ; Z ). By construction we have mass (ˆ z ijkl(cid:63) ) ≤ V + o ( δ ). Thus, there is a point not in the support of ˆ z ijkl(cid:63) and we obtain that ˆ z ijkl(cid:63) = ∂ ˆ φ ijkl(cid:63) for some ˆ φ ijkl(cid:63) ∈ C ( M ; Z ). We then have that[ M ] = (cid:104)(cid:88) σ ijkl (cid:105) = (cid:104)(cid:88) ˆ z ijkl(cid:63) (cid:105) = (cid:104) ∂ (cid:16)(cid:88) ˆ φ ijkl(cid:63) (cid:17)(cid:105) since (cid:80) ˆ φ ijk(cid:63) = 0. This last equality holds since the codimension-1 faces cancel in pairs.We now have that [ M ] = 0 which is a contradiction. (cid:3) Lemma 3.1.
Suppose that ( M, g ) is a Riemmanian 3-sphere such that for all embeddedsurfaces Σ ⊂ M we have the following: If mass (Σ) ≤ (2 d ) and M \ Σ = R (cid:116) R then mass ( R i ) ≤ V for i = 1 or i = 2 . Given such an ( M, g ) we have that for any z = z αβγ ◦ as above there is ˆ z ∈ Z ( M, Z ) such that: (i) supp( z ) = supp(ˆ z ) and (ii)there is ˆ φ ∈ C ( M ; Z ) satisfying ∂ ˆ φ = ˆ z and mass ( ˆ φ ) ≤ V .Proof. Let z be as above,. First note that z is piecewise smooth by the Fleming RegularityLemma [Fle62], since it is realized as a finite union of mass-minimizing surfaces in dimensionthree. We choose a sufficiently fine triangulation of the image of z so that each simplexof the triangulation is a smoothly embedded surface f i . We note that z = (cid:80) ε i f i where f i : ∆ → M is smooth and ε i ∈ Z . By the mass -minimality of z we have that ε i = ± Z -coefficients.Consider ˆ z = (cid:80) f i ∈ C ( M ; Z ). We may perturb the image of ˆ z so that it is in generalposition in M . That is, the image of ˆ z consists of: regular points, double arcs, triple points,and branch points [Car95, Chapter 4].) We have mass (ˆ z ) = mass ( z ) because | ε i | = 1and the cycles have the same support. By construction, ∂ ˆ z = 0. Thus there is ˆ φ suchthat ˆ z = ∂ ˆ φ since H ( M ; Z ) = 0.We show that there must be a small volume filling of ˆ z . Suppose that mass ( ˆ φ ) ≥ V .Let ˆ ψ be a chain supported in M \ supp ˆ φ satisfying ∂ ˆ ψ = ˆ z . We will prove that ass ( ˆ ψ ) < V by contradiction. We will use ˆ z , ˆ φ and ˆ ψ to construct a subdividingsurface whose area is at most 3 HF (2 d + δ ) + o ( δ ). That is, we will show there is a surfaceΣ such that: M \ Σ = R (cid:116) R with vol ( R i ) ≥ V and vol (Σ) ≤ (2 d + δ ) + o ( δ ).We now construct the surface Σ. We will first describe how to replace ˆ z with a unionof closed embedded surfaces. Since ˆ z is a piecewise smooth 2-cycle in S we may pick anopen metric ball B ( p, η ) ⊂ S \ supp(ˆ z ) such that S \ B ( p, η ) is homeomorphic to theclosed unit ball in R . Let ρ be this homeomorphism. Then ρ is C = C ( p, η, g )-bilipschitzsince its domain and target are both compact.Consider image ρ (ˆ z ) in R . We want to replace this cycle with a union of closed surfaces.Let U ε = ∂ { x ∈ R : d ( x, ρ (ˆ z ) ≤ ε } be the boundary of the ε − neighbourhood of ρ (ˆ z ).By Ferry [Fer76] we know that this is an embedded 2-manifold for an open dense set of ε ∈ R + . We choose ε to be sufficiently small and we pick V ε to be a connected componentof U ε which deformation retracts onto the image of the cycle ρ (ˆ z ). Then ˆ z ε = ρ − ( V ε ) is aunion of closed embedded surfaces. Since ρ is C -bilipschitz we can pick ε small enough sothat | mass (ˆ z ) − mass (ˆ z ε ) | ≤ ε . tubez Σ Figure 7.
Construction of Σ. First take surfaces in a neighbourhood of ˆ z .Then connect the components by small tubes.We now consider the surface Σ which is obtained in the following way: Add thin tubes toˆ z ε so as to form one connected component. See Figure 7. We may do this while adding atmost ε to both the surface area of ˆ z ε , and volume of supp( ˆ φ ). We let Σ be the boundaryof supp( ˆ φ ε ) union the thin tubes.We have vol (Σ) ≤ (2 d + δ ) + o ( δ ) + 2 ε , and M \ Σ = R (cid:116) R . Note thatvol ( R i ) > V − o ( ε ). Taking ε → (cid:3) Further Work: Theorem 1.6 provides evidence for a positive answer to the following question:Question. Is there a universal constant C such that any Riemannian manifold M diffeomor-phic to S admits a sweep-out by 2-cycles such that the area of each cycle is bounded by C · HF (2 d )?A positive constructive answer to the question above would provide an effective con-struction of the minimal surfaces in [NR06]. It is also desirable to have a continuous orparameterized version of Theorem 1.6. Acknowledgements
The authors are grateful to their thesis advisor Regina Rotman for suggesting this prob-lem and numerous discussions. Parker Glynn-Adey thanks Robert Young, his co-advisor, or support and patience while working on this project. We also thank Alexander Nab-utovsky for discussing this work with us. This work was partially supported by NSERCgrants held by Regina Rotman and Robert Young. We thank the anonymous referee forproviding many useful comments. References [Alm62] Frederick Justin Almgren. The homotopy groups of the integral cycle groups.
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