the structure of the set of spheres and its applications · 2007-05-10 · the structure of the set...

The structure of the set of spheres andits applications

Jun O’Hara

[email protected]

Department of Mathematics, Tokyo Metropolitan University

Jun O’Hara 2007 – p.1/18

Table of ContentsJoint work with Rémi Langevin

The structure of the setS(q; n) of orientedq-spheres inSnfrom a conformal geometric viewpoint

Pseudo-Riemannian structureSymplectic structure ofS(0; n)

Applications (local object)Infinitesimal conformal arclengthThe real part of the infinitesimal cross ratio

Jun O’Hara 2007 – p.2/18

Motivation 1, Energy of knotsE

E(K)=�4+ZZK�Kn4� 1jx� yj2 � 1dK(x; y)2�dxdy

Functional on knots that blows up for singular knots withdouble points.

Motivation: To produce “optimal configuration" of knots asenergy minimizers.

It is invariant under Möbius transformations(Freedman-He-Wang, 94)

Jun O’Hara 2007 – p.3/18

Our strategyA conceptual illustration

Problem(Fukuhara, Sakuma)

Define an “energy ” e onthe space of knots.

Define a “canonical posi-tion ” for each knot type,which is an embeddingthat attains the mini-mum value of the “en-ergy” within its isotopyclass.

Jun O’Hara 2007 – p.4/18


The complement is theset of embeddings, i.e.the space of knots.

Jun O’Hara 2007 – p.4/18


Each “cell ” correspondsto a knot type, as twopoints in the space ofknots can be connectedby a path if and onlyif two correspondingknots are ambientisotopic.

Jun O’Hara 2007 – p.4/18

Our strategyOur strategy

Take an “energy”e : fknotsg ! R .

Suppose each “cell” issurrounded by an1-lyhigh energy wall.

Given a knot.

Jun O’Hara 2007 – p.4/18


Deform it along the gra-dient flow of the “en-ergy” e.

Jun O’Hara 2007 – p.4/18


If we are lucky theknot might reach ane-minimizer, which is a“canonical position” ofthat knot type.

Jun O’Hara 2007 – p.4/18

Our strategyRequired property of our functional

In order to keep theknot type unchangedduring the deformationprocess,crossing changesshould be avoided!

Jun O’Hara 2007 – p.4/18

Our strategyRequired property of our functional

Definition.e is self-repulsive

def.me(K) blows up as aknotK degenerates to asingular knot with dou-ble points.

We say thate isanenergy of knotsif it is self-repulsive.

Jun O’Hara 2007 – p.4/18

Motivation 2, Willmore Conjecture� : T 2 ! R 3 : a smooth embedding.�1; �2 : principal curvatures. Willmore functionalW is

W (�)=ZT 2��1 + �22�2dv=ZT 2��1 � �22�2dv

W is invariant under Möbius transformations.

Conjecture W (�) � 2�2. “="() �(T 2) is an image ofa torus of revolutionTp2;1 by a Möbius transformation.

Theorem (Bryant)W (�) = Area

�'�(T 2)�, where'� : T 2 ! S(2; 3) is themean sphere map :'� : T 2 3 p 7! � 2�1(p)+�2(p)(p) 2 S(2; 3),tangent to�(T 2) atp with curvature�1(p)+�2(p)2 .

Jun O’Hara 2007 – p.5/18

Minkowski spaceThe Minkowski spaceR 51 is R 5 with the indefinite pseudoinner producthx; yi=�x0y0 + x1y1 + � � �+ x4y4.PutL(v) = hv; vi andjvj = pjL(v)j.A non-zero vectorv is said to be

spacelike ifhv; vi > 0

lightlike if hv; vi = 0timelike if hv; vi < 0

Jun O’Hara 2007 – p.6/18

Minkowski spaceThe Minkowski spaceR 51 is R 5 with the indefinite pseudoinner producthx; yi=�x0y0 + x1y1 + � � �+ x4y4.PutL(v) = hv; vi andjvj = pjL(v)j.

� =fx 2 R 51jhx; xi= 1g :de Sitter space(“pseudo-sphere inR 51”)

Jun O’Hara 2007 – p.6/18


V =fx2 R 51jhx; xi=0g :light cone

Jun O’Hara 2007 – p.6/18


The 3-sphere is realized inR 51

asfl : line� V j 0 2 lgS3(1) = V \ fx0 = 1g

Jun O’Hara 2007 – p.6/18

Lorentz group and Möbius transf’nsThe Lorentz group (“pseudoorthogonal group” of R 51)O(4; 1) = fA 2M5(R ) j hAx;Ayi = hx; yi 8x; ygO(4; 1) acts onS3 �= fl : line� light conej 0 2 lg: Möbius transformation ofS3A Möb. transf. ofS3 stereo. proj.=) a Möb. transf. ofR 3[f1g

A Möbius transformation ofR 3 [ f1g can be expressed asa composition of inversions in spheres.

Jun O’Hara 2007 – p.7/18

de Sitter sp. as the set of spheres

S(2; 3) = f� j an oriented2-sphere inS3g

3�p

1 : 1~�

de Sitter space�

3p

Jun O’Hara 2007 – p.8/18


S(2; 3) = f� j an oriented2-sphere inS3g 3�p1 : 1~�

de Sitter space�

3p

Jun O’Hara 2007 – p.8/18


S(2; 3) = f� j an oriented2-sphere inS3g 3�p1 : 1~� compatible withO(4; 1)-action

de Sitter space� 3p

Jun O’Hara 2007 – p.8/18

Pseudo-Riemannian str. of�S3(1) 3 v1 = (v01; � � � ; v41); � � � ; v4 = (v04; � � � ; v44)“Lorentz exterior product”v1 ^ � � � ^ v4obtained from minors of

0�v01 � � � v41... . . . ...v04 � � � v441A

v1; � � � ; v4�(v1; � � � ; v4)v1 ^ v2 ^ v2 ^ v4jv1 ^ v2 ^ v2 ^ v4j �

(�) hu1 ^ u2 ^ u2 ^ u4; v1 ^ v2 ^ v2 ^ v4i= � det (hui; vji)ui; vj 2 R 51

Jun O’Hara 2007 – p.9/18


0�v01 � � � v41... . . . ...v04 � � � v441A

Supposev1; � � � ; v4 are not concircular.The sphere�(v1; � � � ; v4) corresponds tov1 ^ v2 ^ v2 ^ v4jv1 ^ v2 ^ v2 ^ v4j in � through the bijection

(�) hu1 ^ u2 ^ u2 ^ u4; v1 ^ v2 ^ v2 ^ v4i= � det (hui; vji)ui; vj 2 R 51

Jun O’Hara 2007 – p.9/18


0�v01 � � � v41... . . . ...v04 � � � v441A

Supposev1; � � � ; v4 are not concircular.The sphere�(v1; � � � ; v4) corresponds tov1 ^ v2 ^ v2 ^ v4jv1 ^ v2 ^ v2 ^ v4j in � through the bijection

(�) hu1 ^ u2 ^ u2 ^ u4; v1 ^ v2 ^ v2 ^ v4i= � det (hui; vji)

for ui; vj 2 R 51.Jun O’Hara 2007 – p.9/18

Pseudo-Riem. str. ofS(0; 3)PutS(0; 3) := fori. S0 � S3g �= S3 � S3 n 4.

Proposition 9 a pseudo-Riemannian structure (i.e.8T�S(0; 3) has an indefinite metric) of signature(3; 3).It is invariant under Möbius transformations.Proof: Two proofs:

Jun O’Hara 2007 – p.10/18


Proposition 9 a pseudo-Riemannian structure (i.e.8T�S(0; 3) has an indefinite metric) of signature(3; 3).It is invariant under Möbius transformations.Proof: The first proof:Plücker coordinates. S(0; 3)�=��0 2 � : ori.2-plane�R 51;� \ V transversely

�

: GrassmannianGr�(2; R 51).S(0; 3) �=M � 2^ R 51 �= R 106 .M = S96 \ fPlücker relationsg

Jun O’Hara 2007 – p.10/18


Proposition 9 a pseudo-Riemannian structure (i.e.8T�S(0; 3) has an indefinite metric) of signature(3; 3).It is invariant under Möbius transformations.Proof: The second proof:

Homogeneous space:S(0; 3) �= GrassmannianGr�(2; R 51) = f�jori. timelike2-plane; transv.g.SO(4; 1) acts onR 51,SO(1; 1) acts on� andSO(3) acts on�?.Gr�(2; R 51) �= SO(4; 1)=SO(3)� SO(1; 1).Jun O’Hara 2007 – p.10/18

Plücker coordinatesS0 = fv1; v2g,v1 = (v01; � � � ; v41); v2 = (v02; � � � ; v42) 2 S3.S0 = S3(1) \� = Spanhv1; v2i � R 51.

v1 ^ v2 = (� � � ; pij; � � � ) 2 R 10 (10 = 5C2)pij = ��vi1 vj1vi2 vj2�� v01 � � � v41v02 � � � v42�

(�)hu1 ^ u2; v1 ^ v2i = � det(hui; vji)hp; pi = �p234 � � � � � p212 + p204 + � � �+ p201v1; v2 2 S3(1) (v1 6= v2)L(v1 ^ v2) = hv1 ^ v2; v1 ^ v2i > 0S0 = fv1; v2g v1 ^ v2jv1 ^ v2j

Jun O’Hara 2007 – p.11/18

Plücker coordinatesS0 = fv1; v2g,v1 = (v01; � � � ; v41); v2 = (v02; � � � ; v42) 2 S3.S0 = S3(1) \� = Spanhv1; v2i � R 51.v1 ^ v2 = (� � � ; pij; � � � ) 2 R 10 (10 = 5C2),wherepij = ��vi1 vj1vi2 vj2

�� is a minor of

�v01 � � � v41v02 � � � v42�

(�)hu1 ^ u2; v1 ^ v2i = � det(hui; vji)hp; pi = �p234 � � � � � p212 + p204 + � � �+ p201v1; v2 2 S3(1) (v1 6= v2)L(v1 ^ v2) = hv1 ^ v2; v1 ^ v2i > 0S0 = fv1; v2g v1 ^ v2jv1 ^ v2j

Jun O’Hara 2007 – p.11/18

Plücker coordinatesS0 = fv1; v2g,v1 = (v01; � � � ; v41); v2 = (v02; � � � ; v42) 2 S3.S0 = S3(1) \� = Spanhv1; v2i � R 51.v1 ^ v2 = (� � � ; pij; � � � ) 2 R 10 (10 = 5C2),wherepij = ��vi1 vj1vi2 vj2

�� is a minor of

�v01 � � � v41v02 � � � v42�

The pseudo-Riemannian str. is given by(�), i.e.hu1 ^ u2; v1 ^ v2i = � det(hui; vji). Thenhp; pi = �p234 � � � � � p212 + p204 + � � �+ p201.If v1; v2 2 S3(1) (v1 6= v2) thenL(v1 ^ v2) = hv1 ^ v2; v1 ^ v2i > 0.S0 = fv1; v2g is given by

v1 ^ v2jv1 ^ v2j . Jun O’Hara 2007 – p.11/18

Symplectic str. ofS(0; 3)S(0; 3) �= S3 � S3 n� �= T �S3 (we fix a metric ofS3)

(x; y) 7! �TxS3 3 v 7! px(y) � v�px : S3 n fxg ! R 3 �= TxS3stereographic projection

!M(q1; � � � ; qm) M(q1; � � � ; qm; p1; � � � ; pm) T �MT �(q1;�� ;qm)M 3 v =X pidqi!M =X dqi ^ dpi� = P pidqi 1 !M = �d�

Jun O’Hara 2007 – p.12/18

Symplectic str. ofS(0; 3)S(0; 3) �= S3 � S3 n� �= T �S3 (we fix a metric ofS3)

(x; y) 7! �TxS3 3 v 7! px(y) � v�px : S3 n fxg ! R 3 �= TxS3stereographic projection

Local expression of the canonical symplectic form!M .(q1; � � � ; qm) local coordinates ofnM(q1; � � � ; qm; p1; � � � ; pm) local coordinates ofT �M

such thatT �(q1;�� ;qm)M 3 v =X pidqi.!M =X dqi ^ dpi. It is exact:� = P pidqi : tautological1-form. !M = �d�.Jun O’Hara 2007 – p.12/18

Infinitesimal cross ratioGeometric definition. Letx; x+ dx; y; y + dy 2 K.

�(x; x+dx; y; y+dy) �=!C [f1gx; x+dx; y; y+dy~x ~x+ fdx ~y; ~y + fdy 2 C

(x; y)(~x+ fdx)� ~x(~x+ fdx)� (~y + fdy) : ~y � ~x~y � (~y + fdy) � fdxfdy(~x� ~y)2 :�

Jun O’Hara 2007 – p.13/18


� = �(x; x+ dx; y; y + dy) :2-sphere throughx; x+ dx; y; y + dy.



Jun O’Hara 2007 – p.13/18

Infinitesimal cross ratioGeometric definition. Letx; x+ dx; y; y + dy 2 K.� = �(x; x+ dx; y; y + dy)??ya stereographic

projection

�=C [ f1g



Jun O’Hara 2007 – p.13/18


Through a stereographic projection�(x; x+dx; y; y+dy) �=!C [f1g,x; x+dx; y; y+dy can be identifiedwith ~x, ~x+ fdx, ~y; ~y + fdy 2 C .


Jun O’Hara 2007 – p.13/18


Through a stereographic projection�(x; x+dx; y; y+dy) �=!C [f1g,x; x+dx; y; y+dy can be identifiedwith ~x, ~x+ fdx, ~y; ~y + fdy 2 C .

Definition. Let theinfinitesimal cross ratio of a knot,(x; y), be the cross ratio(~x+ fdx)� ~x(~x+ fdx)� (~y + fdy) : ~y � ~x~y � (~y + fdy) � fdxfdy(~x� ~y)2 :

Remark The four complex numbers are not uniquely determined.But the cross ratio is well-defined. We need the orientation of �.

Jun O’Hara 2007 – p.13/18

Conformal angle and

Definition. (Doyle and Schramm)

C(x; x; y)K x y�K(x; y)C(x; x; y) C(y; y; x)

x ydxdyjx� yj2 �K(x; y)

(x; y) =ei�K(x; y) dxdyjx� yj2

Jun O’Hara 2007 – p.14/18

Conformal angle and


LetC(x; x; y) be a circletangent toK atx thoughy.

�K(x; y)C(x; x; y) C(y; y; x)

x ydxdyjx� yj2 �K(x; y)


Jun O’Hara 2007 – p.14/18

Conformal angle and


LetC(x; x; y) be a circletangent toK atx thoughy.LetC(y; y; x) be a circletangent toK aty throughx.

�K(x; y)C(x; x; y) C(y; y; x)

x y dxdyjx� yj2 �K(x; y)


Jun O’Hara 2007 – p.14/18

Conformal angle and



Let �K(x; y) be the angle betweenC(x; x; y) andC(y; y; x).Call it theconformal anglebetweenx andy.

dxdyjx� yj2 �K(x; y)


Jun O’Hara 2007 – p.14/18

Conformal angle and



Let �K(x; y) be the angle betweenC(x; x; y) andC(y; y; x).Call it theconformal anglebetweenx andy.

The absolute value of the infinitesimal cross ration is equal todxdyjx� yj2 . The argument of is equal to�K(x; y).Proposition. (x; y) =ei�K(x; y) dxdyjx� yj2 .

Jun O’Hara 2007 – p.14/18

Infinitesimal cross ratio and the energyEProposition (Doyle and Schramm’s cosine formula)

E(K) = ZZK�Kn4 1� os �K(x; y)jx� yj2 dxdy:

This is another proof of the conformal invariance ofE

(x; y) = ei�K(x;y) dxdyjx� yj2 :

E(K) = ZZK�Kn4(jj � <e):

Jun O’Hara 2007 – p.15/18

Infinitesimal cross ratio and the energyEProposition (Doyle and Schramm’s cosine formula)

E(K) = ZZK�Kn4 1� os �K(x; y)jx� yj2 dxdy:

This is another proof of the conformal invariance ofE

Recall(x; y) = ei�K(x;y) dxdyjx� yj2 :Proposition E(K) = ZZ

K�Kn4(jj � <e):

Jun O’Hara 2007 – p.15/18

<e as signed area element

Put� : K �K n 4 ,! S3 � S3 n 4 �=! S(0; 3).Then�(K �K n 4) is a surface inS(0; 3).Its area element is given bys��h�x; �xi h�x; �yih�y; �xi h�y; �yi

��dx^dy=p�h�x; �yi2 dx^dy

becauseh�x; �xi = h�y; �yi = 0.Define the “(imaginary) signed area element” by� = h�x; �yi dx^dy.

� = 2<e: h ; ijT�(K�Kn�)(1; 1) (0; 0)= 0 h�x; �yi dx^dy = 2 os � dx^dyjx� yj2

Jun O’Hara 2007 – p.16/18

<e as signed area element

Put� : K �K n 4 ,! S3 � S3 n 4 �=! S(0; 3).Then�(K �K n 4) is a surface inS(0; 3).Its area element is given bys��h�x; �xi h�x; �yih�y; �xi h�y; �yi

��dx^dy=p�h�x; �yi2 dx^dy

becauseh�x; �xi = h�y; �yi = 0.Define the “(imaginary) signed area element” by� = h�x; �yi dx^dy.

Theorem. � = 2<e:Corollary. The signature ofh ; ijT�(K�Kn�) is either(1; 1) or (0; 0).

Proof: Trace= 0, h�x; �yi dx^dy = 2 os � dx^dyjx� yj2 Jun O’Hara 2007 – p.16/18

<e and a can. symplectic formS(0; 3) �= S3 � S3 n� �= T �S3T �S3 has the canonical symplectic form!S3Proposition Let � : K �K n 4 ,! S3 � S3 n 4be the inclusion. Then<e(x; y) = �12��!S3 .

Especially,<e is exact.

Corollary. LetC1 [ C2 be a2-component link. Then thesigned area of�(C1 � C2) � S(0; 3) is equal to0.

Jun O’Hara 2007 – p.17/18

Osc. circles& conformal arc-lengthLetC = f(S1) be a curve inR 3, � (t) be an osculatingcircle ofC atf(t). The setf� (t)g forms a curvef (t)gin S(1; 3) � R 104 via a stereographic projection. is a null curve, i.e.h 0(t); 0(t)i � 0, hencethe lenghtL( ) = limP j (ti+1)� (ti)j = 0.

L12 limPpj (ti+1)� (ti)j = �= 4p12 :� d� = 4p�02 + �2� 2 dss C � �C

h ; i

Jun O’Hara 2007 – p.18/18

Osc. circles& conformal arc-lengthLetC = f(S1) be a curve inR 3, � (t) be an osculatingcircle ofC atf(t). The setf� (t)g forms a curvef (t)gin S(1; 3) � R 104 via a stereographic projection. is a null curve, i.e.h 0(t); 0(t)i � 0, hencethe lenghtL( ) = limP j (ti+1)� (ti)j = 0.

Theorem The “L12 -length” of satisfieslimPpj (ti+1)� (ti)j = �= 4p12 :

Here,� is the “conformal arc-length" given byd� = 4p�02 + �2� 2 ds,wheres is the arc-length ofC, and� and� are thecurvature and the torsion ofC.Proof: By Taylor expansions of both sides and theconformal angle using the pseudo inner producth ; i.Jun O’Hara 2007 – p.18/18

the structure of the set of spheres and its applications · 2007-05-10 · the structure of the set...

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