Linking integrals in three-dimensional geometries D. DeTurck - PowerPoint PPT Presentation

Linking integrals in three-dimensional geometries D. DeTurck University of Pennsylvania March 7, 2014 D. DeTurck Tetrahedral Geometry/Topology Seminar 1 / 32

Gauss linking integral Carl Friedrich Gauss, in a half-page paper dated January 22, 1833, gave an integral formula for the linking number in Euclidean 3-space, x − y ˆ d x ds × d y Link ( K 1 , K 2 ) = dt · 4 π | x − y | 3 ds dt . K 1 × K 2 D. DeTurck Tetrahedral Geometry/Topology Seminar 2 / 32

General goal Our goal is to define geometrically natural linking integrals for each of the eight homogeneous three-dimensional geometries S 2 × R , H 2 × R , R 3 , S 3 , H 3 , Nil , Sol , SL (2 , R ) and at least some of their higher-dimensional generalizations. “Geometrically natural” in this context means that the integrands should be invariant under orientation-preserving isometries of the ambient spaces. D. DeTurck Tetrahedral Geometry/Topology Seminar 3 / 32

Another expression for the Gauss integral Another way to write Gauss’s formula is to define the double-form ω 1 , 1 Φ 1 , 1 ( x , y ) = 4 π | x − y | 3 on R 3 × R 3 , where ω 1 , 1 = ( y 1 − x 1 )( dx 2 ⊗ dy 3 − dx 3 ⊗ dy 2 ) + ( y 2 − x 2 )( dx 3 ⊗ dy 1 − dx 1 ⊗ dy 3 ) + ( y 3 − x 3 )( dx 1 ⊗ dy 2 − dx 2 ⊗ dy 1 ) . Thinking of the curves K 1 and K 2 as maps from S 1 into R 3 , we define ˆˆ S 1 × S 1 X ∗ Φ 1 , 1 Link ( K 1 , K 2 ) = where X = ( x , y ) is the product mapping. D. DeTurck Tetrahedral Geometry/Topology Seminar 4 / 32

Double forms • A double-form is a differential form on M × M which can be viewed either as a differential form on the first factor with coefficients being differential forms on the second factor, or vice versa. • A ( p , q )-form has of degree p over the first M factor and of degree q over the second. For example, a (2 , 1)-form on R 3 × R 3 can be expressed as: f ( x , y ) dx 2 ∧ dx 3 ⊗ dy 1 + · · · and so on for nine terms. • For such forms, we have exterior derivatives d x and d y which commute with each other and have other standard properties such as d 2 x = d 2 y = 0, etc. D. DeTurck Tetrahedral Geometry/Topology Seminar 5 / 32

Linking integrals We wish to calculate the linking number of K and L via an integral of the form ˆ Link ( K , L ) = Φ 1 , 1 ( x , y ) , K × L where x ∈ K and y ∈ L and Φ 1 , 1 is an appropriately-chosen isometry-invariant (1,1)-form on ( M ) × ( M ). The form Φ 1 , 1 will be singular along the diagonal ∆ of ( M ) × ( M ), but will be smooth otherwise. D. DeTurck Tetrahedral Geometry/Topology Seminar 6 / 32

Link-homotopy invariance If there is a (2,0)-form Ψ 2 , 0 ( x , y ) having the property that d y Ψ 2 , 0 = d x Φ 1 , 1 as (2,1)-forms on M × M ) \ ∆, define the ordinary 1-form ˆ Ω 1 ( x ) = Φ 1 , 1 ( x , y ) L on M \ L by integrating Φ 1 , 1 over the curve L for each x �∈ L . Then we will have ˆ ˆ ˆ d x Ω 1 = d x Φ 1 , 1 ( x , y ) = d x Φ 1 , 1 ( x , y ) = d y Ψ 2 , 0 ( x , y ) = 0 , L L L by Stokes’s Theorem. So Ω 1 is a closed 1-form (in x ) on M \ L , and the value of ˆ ˆ Ω 1 ( x ) = Φ 1 , 1 ( x , y ) = Link ( K , L ) K K × L depends only on the homology class of K within M \ L , so K can be deformed without affecting the value of the linking integral as long as K never meets L . D. DeTurck Tetrahedral Geometry/Topology Seminar 7 / 32

Link-homotopy invariance • Likewise, if there is a (0,2)-form Ψ 0 , 2 ( x , y ) having the property that d x Ψ 0 , 2 = d y Φ 1 , 1 as (1,2)-forms on ( M × M ) \ ∆ we can deform L as long as it never meets K and not affect the value of the linking integral. D. DeTurck Tetrahedral Geometry/Topology Seminar 8 / 32

Link-homotopy invariance • Likewise, if there is a (0,2)-form Ψ 0 , 2 ( x , y ) having the property that d x Ψ 0 , 2 = d y Φ 1 , 1 as (1,2)-forms on ( M × M ) \ ∆ we can deform L as long as it never meets K and not affect the value of the linking integral. • Our goal will be to produce isometry-invariant forms Φ 1 , 1 , Ψ 2 , 0 and Ψ 0 , 2 and to show that they satisfy the differential equations d y Ψ 2 , 0 = d x Φ 1 , 1 and d x Ψ 0 , 2 = d y Φ 1 , 1 . D. DeTurck Tetrahedral Geometry/Topology Seminar 8 / 32

R n , S n , H n Write x = ( x 0 , x 1 , . . . , x n ) and y = ( y 0 , y 1 , . . . , y n ) for points in R n +1 , and � x , y � ± for the inner product on R n +1 given by � x , y � ± = x 0 y 0 ± ( x 1 y 1 + x 2 y 2 + · · · + x n y n ) . We will then view S n ⊂ R n +1 as S n = { x ∈ R n +1 | � x , x � + = 1 } , H n ⊂ R n +1 as H n = { x ∈ R n +1 | � x , x � − = 1 , x 0 > 0 } , and R n ⊂ R n +1 as R n = { x ∈ R n +1 | x 0 = 1 } . D. DeTurck Tetrahedral Geometry/Topology Seminar 9 / 32

Group actions There are natural group actions on R n +1 that restrict to groups of isometries on S n , H n and R n , namely the standard actions of SO ( n + 1), SO + (1 , n ) and E ( n ) respectively, where E ( n ) is the group of Euclidean motions of R n . An element of E ( n ) is given by the matrix  1  0    ,   v R  where R ∈ SO ( n ) and v ∈ R n , so this matrix rotates R n according to the matrix R and then translates by v . D. DeTurck Tetrahedral Geometry/Topology Seminar 10 / 32

Basic invariant forms Let x and y be two points in R n +1 . For each k and ℓ satisfying k + ℓ = n − 1, define a differential form ω k ,ℓ as follows: For v 1 , v 2 , . . . , v k ∈ T x R n +1 and w 1 , w 2 , . . . , w ℓ ∈ T y R n +1 , ω k ,ℓ ( x , y ; v 1 , . . . , v k ; w 1 , . . . , w ℓ ) = � x , y , v 1 , . . . , v k , w 1 , . . . , w ℓ � . Formally think of ω k ,ℓ as the determinant 1 k ! ℓ ! � x , y , d x , . . . , d x , d y , . . . , d y � D. DeTurck Tetrahedral Geometry/Topology Seminar 11 / 32

More forms Two other families of forms for k + ℓ = n : α k ,ℓ ( x , y ; v 1 , . . . , v k ; w 1 , . . . , w ℓ ) = � x , v 1 , . . . , v k , w 1 , . . . , w ℓ � , β k ,ℓ ( x , y ; v 1 , . . . , v k ; w 1 , . . . , w ℓ ) = � y , v 1 , . . . , v k , w 1 , . . . , w ℓ � . Forms restrict in a natural way to M , are invariant under the action of G . Also write: 1 α k ,ℓ = k ! ℓ ! � x , d x , . . . , d x , d y , . . . , d y � 1 β k ,ℓ = k ! ℓ ! � y , d x , . . . , d x , d y , . . . , d y � . Define exterior differential operators, d x and d y . Check that 1 d x ω k ,ℓ = k ! ℓ ! � d x , y , d x , . . . , d x , d y , . . . , d y � = − ( k + 1) β k +1 ,ℓ and 1 k ! ℓ ! � x , d y , d x , . . . , d x , d y , . . . , d y � = ( − 1) k ( ℓ +1) α k ,ℓ +1 . d y ω k ,ℓ = D. DeTurck Tetrahedral Geometry/Topology Seminar 12 / 32

Distance functions • Let σ = σ ( x , y ) = � x , y � + . The geodesic distance α between two points x and y on S n is α = arccos σ . • Likewise, if σ = � x , y � − , the geodesic distance α between two points x and y on H n is α = arccosh σ . • And of course the distance between two points x and y on R n is α = (( y − x ) · ( y − x )) 1 / 2 and we let σ = 1 2 ( y − x ) · ( y − x ) in this case. D. DeTurck Tetrahedral Geometry/Topology Seminar 13 / 32

Linking integrands To construct linking integrands, need an identity of the form: d x [ ϕ ( σ ) ω 1 , 1 ] = d y [ ψ ( σ ) ω 2 , 0 ] , where σ is a function of the distance α ( x , y ) between x and y . On S 3 and H 3 , use σ = � x , y � ± , so that σ = cos α and σ = cosh α respectively. On S 3 (and H 3 ), we have d x [ ϕ ( σ ) ω 1 , 1 ] = ϕ ′ ( σ ) [ α 2 , 1 − σβ 2 , 1 ] − 2 ϕ ( σ ) β 2 , 1 and d y [ ψ ( σ ) ω 2 , 0 ] = − ψ ′ ( σ ) [ β 2 , 1 − σα 2 , 1 ] + ψ ( σ ) α 2 , 1 . D. DeTurck Tetrahedral Geometry/Topology Seminar 14 / 32

Differential equations for ϕ and ψ Because α 2 , 1 and β 2 , 1 are independent we need the coefficients of both to be equal for these quantities to be equal. So ϕ and ψ must satisfy: ϕ ′ ( σ ) − σψ ′ ( σ ) − ψ ( σ ) = 0 σϕ ′ ( σ ) + 2 ϕ ( σ ) − ψ ′ ( σ ) = 0 . Derive that ϕ satisfies the second-order equation (1 − σ 2 ) ϕ ′′ − 5 σϕ ′ − 4 ϕ = 0 . ( ∗ ) Likewise (1 − σ 2 ) ψ ′′ − 5 σψ ′ − 3 ψ = 0 . Thus we can find functions ψ and χ so that d x [ ϕω 1 , 1 ] = d y [ ψω 2 , 0 ] and d y [ ϕω 1 , 1 ] = d x [ χω 0 , 2 ] . We refer to this fact as the Key Lemma . D. DeTurck Tetrahedral Geometry/Topology Seminar 15 / 32

Solution on R 3 On R 3 the differential equation for ϕ ( σ ) analogous to ( ∗ ) is: ϕ ( σ ) = C 1 2 σϕ ′′ + 5 ϕ ′ = 0 which has general solution σ 3 / 2 + C 2 . We want ϕ to decay to zero when σ → ∞ , so C 2 = 0. By considering a single nontrivial example, we get that C 1 = 1 / vol ( S 2 ) so ω 1 , 1 Link ( K , L ) = 1 ˆˆ 4 π σ 3 / 2 K × L just as Gauss did. D. DeTurck Tetrahedral Geometry/Topology Seminar 16 / 32

Solution on S 3 On S 3 , √ 1 − σ 2 − σ arccos σ σ ϕ = c 1 (1 − σ 2 ) 3 / 2 + c 2 . (1 − σ 2 ) 3 / 2 We need σ ↓− 1 ϕ ( σ ) lim to be finite, and considering a single non-trivial example (two orthogonal linked great circles), we conclude that 1 c 1 = 1 c 2 = ϕ (0) = and 4 π. 4 π 2 This determines the linking integral on S 3 . D. DeTurck Tetrahedral Geometry/Topology Seminar 17 / 32