LINKING INTEGRALS ON S n

Transcription

1 LINKING INTEGRALS ON S n CLAY SHONKWILER Introduction In 4, Gauss gave an integral formula for the linking number of two closed curves in Euclidean 3-space Although Gauss paper does not include a proof of the formula, Moritz Epple 3 suggests that Gauss had in mind a degree-ofmap argument, though he undoubtedly also knew how to prove the formula using Ampère s Law Surprisingly, Gauss-type linking integrals have only recently been discovered in other spaces In a forthcoming paper, Dennis DeTurck and Herman Gluck use an electrodynamics approach to find linking integral formulas in the 3-sphere S 3 and in hyperbolic 3-space H 3 Their basic strategy is to prove that Maxwell s equations hold in those spaces and then prove their formulas using Ampère s Law This approach ends up being quite analytic, as they must solve for the Green and Biot-Savart operators In this paper, we pursue a more geometric approach to find Gauss-type linking integrals in S n for all n 2 Our integrals are equivalent to the integrals found via more analytic techniques in more recent work of DeTurck and Gluck 2 as well as those found via an alternative geometric technique by Greg Kuperberg 5 Notation Before stating our main result (Theorem ), we must first establish some notation Throughout, we will consider oriented, smooth, closed submanifolds K k, L l S n where k + l n, k l and S n is the round n-sphere which we will usually consider as embedded in the usual way in R n+ For reasons to be explained in Section 2, we will assume that K is disjoint both from L and from its antipodal image L Also, we think of K and L as being parametrized by local coordinates s (s,, s k ) and t (t,, t l ), respectively; that is, K {x(s)} and L {y(t)} for smooth functions x and y We will denote the distance in S n from x(s) to y(t) by α(s, t) Also, for any m we will let V,, V m signify the determinant of the m m matrix with columns given by the V i With that in mind, we will often need to make use of the expression () x(s), y(t), x(s) ds dt, s,, x(s) s k which we abbreviate as x, y, dx, dy, y(t),, y(t) t t l

2 2 CLAY SHONKWILER Finally, our linking integral will contain a convolution of sine functions, so we recall that, for any α (in practice, we will use α as defined above), sin k sin l (α) : 2π sin k (α β) sin l (β)dβ 2 Statement of main theorem With all of the above notation in mind, we can finally state the main theorem: Theorem With K, L, x, y and α as above, (2) Lk(K, L) + ( ) n Lk(K, L) ( )k sin k sin l (α) vol S n sin n x, y, dx, dy α where Lk(K, L) denotes the linking number of K and L and Lk(K, L) denotes the linking number of K with the antipodal image of L Remark 2 The fact that the lefthand side of (2) has two terms is the price we must pay for the fact that a straightforward application of the degree-of-map argument fails in S n 2 The degree-of-map argument First, recall the degree-of-map proof of the Gauss linking integral Let K and L be smooth, oriented, disjoint, simple closed curves in R 3 parametrized by s, t, 2π Then K {x(s)} and L {y(t)} for smooth functions x, y : S R 3 Define the map ϕ : S S S 2 given by ϕ(s, t) x(s) y(t) x(s) y(t) Let D be the projection of K and L to the xy-plane Then, assuming K and L are generic with respect to projection to the xy-plane, inverse images under ϕ of the north pole correspond exactly to points in the knot diagram D where K crosses over L In other words, since ϕ is a smooth map and (since K and L are generic with respect to projection to the xy-plane) the north pole is a regular value of ϕ, the degree of ϕ is equal to the linking number of K and L A bit of calculation shows that, if ω is a volume form which evaluates to on S 2, Lk(K, L) deg(ϕ) 4π ϕ ω dx 4π ds dy dt x y x y 3 dsdt 2 Why doesn t this work in S 3? Unfortunately, we can t use the same trick to find a linking integral in S 3 The key step in the above is that the configuration space of 2 distinct points in R 3, Conf 2 (R 3 ), deformation retracts to the 2-sphere (this is easy to see: translate so that the first point is at the origin, then scale until the second point has norm ) Therefore, the map ϕ contains all the homotopy information of the configuration map S S Conf 2 (R 3 ) given by (s, t) (x(s), y(t))

3 LINKING INTEGRALS ON S n 3 On the other hand, the configuration space of two distinct points in S 3, Conf 2 (S 3 ), is three-dimensional To see this, suppose (p, q) Conf 2 (S 3 ) Let γ be a geodesic from p to q Then simply translate q along γ until it is antipodal to p In this way, we see that Conf 2 (S 3 ) deformation retracts to the anti-diagonal S 3 The S 3 equivalent of the map ϕ defined above should, therefore, be a map φ : S S S 3 However, any such map is clearly homotopically trivial and, therefore, its degree tells us nothing about the linking number 22 How to fix this argument Another way to think of the map ϕ : S S S 2 defined above is as a map ϕ : S S UR 3, the unit tangent bundle of R 3, given by (3) (s, t) ( x(s), ) x(s) y(t) x(s) y(t) If we let η dx dy dz be the standard volume form on R 3 and let ω be a volume form on S 2, then, since UR 3 is trivial (ie homeomorphic to R 3 S 2 ), eta ω is a volume form on UR 3 which restricts to the volume form on the fiber Then, if ϕ is as defined in (3), Lk(K, L) deg(ϕ) If we want to define an S 3 version of ϕ, one reasonable thing to try is to define f(p, q) (p, v), where v T p S 3 is the unit vector at p pointing in the direction of the geodesic from p to q Of course, if q is antipodal to p, there are infinitely many geodesics from p to q, so this map is not well-defined However, if we add the restriction that our closed curves K and L in S 3 must not only be smooth and disjoint, but also disjoint from the other s antipodal image, then there is no problem (locally) a map f in this way More precisely, for K {x(s)} and L {y(t)}, we define f : S S US 3 by f(s, t) (x(s), v(s, t)), where v(s, t) T p S 3 is the unit vector tangent to the unique geodesic from x(s) to y(t) In fact, it is just as straightforward to define such an f : K L US n for any n Therefore, in analogy with the argument outlined above in the case of R 3, our goal is to find an SO(n + )-invariant (n )-form ω n on US n which restricts to the volume form on the S n fiber Then we will show that (4) Lk(K, L) + ( ) n Lk(K, L) vol S n f ω n Unfortunately, as we will see, such a form only exists when n is odd Therefore, the proof of Theorem will split into two cases The case where n is odd is addressed in Section 3: we will find the form ω n in Section 3, prove (4) in Section 32 and derive (2) in Section 33 A completely different

4 4 CLAY SHONKWILER argument is used in Section 4 to show that, if Theorem is true for odd spheres, it is also true for even spheres 3 Theorem for odd spheres 3 Invariant forms on the unit tangent bundle of S n For positive n, let US n denote the unit tangent bundle of the sphere S n The goal in this section is to prove the following proposition: Proposition 3 For odd n, there exists a closed SO(n + )-invariant (n )-form ω n on US n which restricts to the volume form on the fibers of US n Almost as important as the statement of the proposition is the formula for ω n, which is given below in (5) Note that the fact that n is odd is essential: a simple application of the Gysin sequence demonstrates that the derham cohomology of US n is given by { HdR k H (USn dr k ) (S2n ) for n odd HdR k (Sn S n ) for n even so there s no point to looking for such a closed form in US n for n even Proof of Proposition 3 First, we establish some terminology Let E,, E n+ denote the standard basis of R n+ and let M i,j (θ) SO(n + ) be the oneparameter group of isometries of R n+ which spin the x i x j -plane by an angle θ and leave the complement of this plane fixed Let E i,j be the tangent vector to this family of rotations at the identity Then {E i,j i j n + } forms a basis of the Lie algebra so(n + ) Let V i,j denote the left-invariant vector field on SO(n + ) determined by E i,j and let Φ i,j denote the dual left-invariant -form Now, consider the point (E, E 2 ) US n The tangent vectors to the vertical fiber at this point have {E 2,j 2 < j n + } as a basis, while {E,j < j n + } forms a basis for the horizontal tangent vectors Let ϕ i,j be the -forms on US n SO(n + )/SO(n ) corresponding to the Φ i,j defined above (of course, ϕ i,j only makes sense for i, 2) We are interested in finding a closed SO(n+)-invariant (n )-form on US n which restricts to the volume form ϕ 2,3 ϕ 2,n+ on the fiber Unfortunately the fiber form ϕ 2,3 ϕ 2,n+ is not closed on US n However, it is invariant under the SO(n ) action which fixes the x x 2 - plane in R n+ Moreover, we can make the fiber form closed by averaging it over the SO(2)-action which spins the x x 2 -plane as follows: (5) 2π ω n 2π (cos βϕ sin n,3 sin βϕ 2,3 ) (cos βϕ,n+ sin βϕ 2,n+ )dβ, βdβ Thus defined, ω n certainly restricts to the fiber form Also, the extra SO(2) averaging makes it the pullback of an SO(n + )-invariant form in

5 LINKING INTEGRALS ON S n 5 the Grassmann manifold G 2 R n+ of oriented 2-planes through the origin of R n+ Since the Grassmann manifold is a symmetric space, all invariant forms are closed, so we see that ω n is the desired closed, SO(n + )- invariant form on US n 32 Linking numbers via pullbacks of invariant forms Now that we have our closed, SO(n + )-invariant (n )-form ω n on US n for odd n, we want to prove the following proposition: Proposition 32 Suppose that ω 2n is a closed, SO(2n + 2)-invariant (2n)- form on US 2n+ such that ω 2n restricts to the volume form on the fiber Moreover, suppose K k and L l are smooth, oriented, closed submanifolds of S 2n+ such that k + l 2n and K and L are disjoint from each other and from the other s antipodal image Let K {x(s)} where s (s,, s k ) and L {y(t)} where t (t,, t l ) Then (6) Lk(K, L) Lk(K, L) vol S 2n f ω 2n where Lk(, ) denotes the linking number of two manifolds, L denotes the antipodal image of L and f : K L US n is given by f(s, t) (x(s), v(s, t)) where v(s, t) is the unit vector in T x(s) S 2n+ tangent to the unique geodesic from x(s) to y(t) Remark: (6) is, spiritually, the exact analogue of Gauss linking integral in R 3, in which the linking number of two curves is given by /4π times the integral of the pullback of the volume form on S 2 (ie the fiber of the unit tangent bundle to R 3 ) The assumption that K and L are disjoint not only from each other but from the other s antipodal image is required so that the vector v (and thus the map f) is well-defined To prove Proposition 32, we will construct a singular annulus A bounded by L and L such that Lk(K, L) Lk(K, L) is equal to the signed intersection number of K with A Then, we will demonstrate that the signed intersection number of K with A is equal to the right hand side of (6) Lemma 33 Let K and L be submanifolds of S 2n+ satisfying the hypotheses of Proposition 32 Then Lk(K, L) Lk(K, L) K A where A is a singular manifold of dimension l + such that A L ( L) and K A denotes the signed intersection number of K with A Proof Let h : S 2n+ CP n be the Hopf map and let H denote the corresponding collection of oriented great circles in S 2n+ Each point y(t) of L lies on some oriented Hopf circle C(t) H, as does its antipodal image

6 6 CLAY SHONKWILER y(t) Let S(t) denote the oriented semi-circle from y(t) to y(t) which follows the orientation of C(t) Now, define A : S(t), t oriented so that A L ( L) Note that A can be quite singular; for example, of L is itself one of the Hopf circles, then A coincides with L However, note that the left hand side of (6) is invariant under homotopies which keep K disjoint from both L and L and, therefore, under short homotopies Moreover, the right hand side of (6) is invariant under short homotopies since ω 2n is a closed form Therefore, we are free to perturb L slightly so that it intersects all the Hopf circles transversally As a result, h L : L CP n is an immersion We can perturb L a bit more so that this immersion has only finitely many double points, at all of which two branches meet transversally, and no triple points Since the images of L, L and A under h all coincide, the result of these perturbations is that A is an immersed (l + )-manifold with one arc of double points for each isolated double point in h(l) Now, we claim that Lk(K, L) Lk(K, L) K A Given the construction of A, this is straightforward Let D be an (l + )-cycle in S 2n+ bounded by L Then A D is an (l + )-cycle bounded by L Hence, by definition of the linking number, Lk(K, L) K D and Lk(K, L) K (A D) Therefore, Lk(K, L) Lk(K, L) K A, as desired Now, we want to show that K A is equal to the right hand side of (6): Lemma 34 Under the hypotheses of Proposition 32, vol S 2n f ω 2n K A, with A defined as in Lemma 33 Proof Let H be the Hopf fibration of S 2n+ considered above Let v H be the Hopf vector field on S 2n+ ; that is, for p S 2n+, v H (p) is the unit vector at p tangent to the oriented Hopf circle through p Then S 2n+ H : {(p, v H (p))} US 2n+ is a section of the unit tangent bundle S 2n US 2n+ π S 2n+

7 LINKING INTEGRALS ON S n 7 and meets each fiber π (p) transversally at the single point (p, v H (p)) Now, each fiber π (p) is a cycle representing the generator of H 2n (US 2n+ ; Z) Z and each section of US 2n+ (eg S 2n+ H ) is a cycle representing the generator of H 2n+ (US 2n+ ; Z) Z Choose orientations so that the intersection number of π (p) with S 2n+ H is + Since ω 2n represents a 2n-dimensional cohomology class of US 2n+ which restricts to the volume form on each fiber and since each fiber generates the 2n-dimensional homology of US 2n+, we see that, for any 2n-cycle c 2n, c 2n vol S 2n ω 2n c 2n S 2n+ H Applied to the 2n-cycle c 2n f(k L), this formula tells us that (7) vol S 2n f ω 2n vol S 2n ω 2n f(k L) S 2n+ H f() Since f(k L) {(x(s), v(s, t))}, we see that f(k L) meets S 2n+ H whenever v(s, t) v H (x(s)), which occurs precisely when K intersects A Therefore, by our choices of orientations, vol S 2n f ω 2n K A, To complete the proof of Proposition 32, we simply note that combining Lemma 33 and Lemma 34 yields: Lk(K, L) Lk(K, L) K A f(k L) S 2n+ H vol S 2n f ω 2n 33 Deriving the convolution formula With Proposition 32, it now simply remains to show that the right hand side of (6) is equal to the right hand side of (2) when n is odd in order to complete the proof of Theorem for odd n In other words, we want to prove the following proposition: Proposition 35 Suppose n is odd Let K k, L l S n satisfy the conditions of Theorem Let ω n be given by (5) and let f be as defined in Proposition 32 Then (8) vol S n f ω ( )k vol S n sin k sin l (α) sin n x, y, dx, dy α Proof Recall that K {x(s)}, that L {y(t)} where s (s,, s k ) and t (t,, t l ) and that v(s, t) is the unit vector at x(s) pointing to y(t) Since ω n is SO(n + )-invariant, we are free work at the point (s, t ) (, ) and to assume that x() E and v(, ) E 2 (where the E i are the standard basis for R n+ )

8 8 CLAY SHONKWILER Although x(s) does not depend on t, assume, for the moment, that x is a function of s and t Then, to first order, x(s, t) E + (a 2, s + + a 2,k s k + b 2, t + + b 2,l t l )E (a n+, s + + a n+,k s k + b n+, t + + b n+,l t l )E n+ v(s, t) (c, s + + c,k s k + d, t + + d,l t l )E + E 2 +(c 3, s + + c 3,k s k + d 3, t + + d 3,l t l )E 3 + +(c n+, s + + c n+,k s k + d n+, t + + d n+,l t l )E n+ Since v(s, t) T x(s,t) S n, x(s, t), v(s, t) (where this is the inner product in R n+ ) Differentiating this relation with respect to the s i and the t j yields to following system of equations: a 2,i + c,i b 2,j + d,j for i,, k and j,, l From their first-order expansions, we see that differentiating x and v gives: x s i a 2,i E a n+,i E n+ v s i c,i E + c 3,i E c n+,i E n+ x t j b 2,j E b n+,j E n+ v t j d,j E + d 3,j E d n+,j E n+ Therefore, ( partial f ω n,, partial ),,, ω n (A,, A k, B,, B l ) s s k t t l where ( x A i, v ) s i s i (a 2,i E a n+,i E n+, c,i E + c 3,i E c n+,i E n+ ) (a 2,i E a n+,i E n+, a 2,i E + c 3,i E c n+,i E n+ ) a 2,i (E 2, E ) + a 3,i (E 3, ) + + a n+,i (E n+, ) + c 3,i (, E 3 ) + + c n+,i (, E n+ ) a 2,i e,2 + a 3,i e,3 + + a n+,i e,n+ + c 3,i e 2,3 + + c n+,i e 2,n+ and, similarly, ( x B j, v ) t j t j b 2,j e,2 + b 3,j e,3 + + b n+,j e,n+ + d 3,j e 2,3 + + d n+,j e 2,n+ for i,, k and j,, l

9 LINKING INTEGRALS ON S n 9 Of course, x(s, t) x(s) so x t j, meaning that b ν,η for all ν 2,, n + and η,, k Therefore, ω n (A,, A k, B,, B l ) ω n (a 2, e,2 + a 3, e,3 + + a n+, e,n+ + c 3, e 2,3 + + c n+, e 2,n+, (9) a 2,k e,2 + a 3,k e,3 + + a n+,k e,n+ + c 3,k e 2,k + + c n+,k e 2,n+, d 3, e 2,3 + + d n+, e 2,n+, d 3,l e 2,3 + + d n+,l e 2,n+ ) Having analyzed f ω n as much as we can without actually writing down ω n, we now recall from (5) that ω n : 2π sin k+l βdβ 2π sin n βdβ 2π + 2π 2π 2π 2π + sin n βdβ () sin n βdt 2π + 2π 2π 2π n (cos βϕ,3 sin βϕ 2,3 ) (cos βϕ,n+ sin βϕ 2,n+ )dβ cos n βϕ,3 ϕ,n+ dβ cos n 2 β sin βξ n 2 dβ cos n 3 β sin 2 βξ n 3 dβ cos 2 β sin n 3 βξ 2 dβ cos β sin n 2 βξ dβ sin n βϕ 23 ϕ 2,n+ dβ 2π ( ) i cos n i β sin i βξ n i dβ i n 2π ( ) n i cos i β sin n i βξ i dβ, i where ξ a is the sum of all possible simple (n )-forms in the ϕ i,j with a terms of the form ϕ,i and n a terms of the form ϕ 2,j Note that, in (9), there are only k entries with terms of the form e,i, so ξ η (A,, A k, B,, B l ) for all η > k

10 CLAY SHONKWILER Now, ξ (A,, A k, B,, B l ) ϕ 2,3 ϕ 2,n+ (A,, A k, B,, B l ) is simply c 3, c n+, c det 3,k c n+,k, d 3, c n+, d 3,l d n+,l which is equal to In turn, since v x, v, v,, v, v,, v s s k t t l y cos αx sin α () ( ) k cosk α sin n α, this is just x, y, x,, x, y,, y s s k t t l Similarly, ξ 2 (A,, A k, B,, B l ) is given by (2) a a 3, a 3, a n+, n+, c c a 3,2 a n+,2 3,2 c n+,2 3, c n+, a c 3,3 c n+,3 3,3 a n+,3 c 3,4 c n+,4 c 3,k 2 c n+,k 2 a det +det + +det 3,k a n+,k c 3,k c n+,k a c d 3, d 3,k c 3,k a n+,k n+,k n+, d d 3, d 3, d n+, n+, d 3,k d n+,l d d 3,k d 3,k d n+,l n+,l (ie all determinants where there are two rows of a s in the first k rows and the rest of the first k rows consist of c s) In turn, the first term of (2) is just x, v, x, x, v,, v, v,, v, s s 2 s 3 s k t t l which, using the definition of v, reduces to ( ) k 2 cosk 2 α sin n 2 α x, y, x s,, x s k, y t,, y t l The other terms in (2) reduce to the same thing, so (3) ξ 2 (A,, A k, B,, B l ) ( ) k 2 ( ) k 2 cosk 2 α sin n 2 α x, y, x,, x, y,, y s s k t t l

11 LINKING INTEGRALS ON S n In fact, in general, (4) ξ i (A,, A k, B,, B k ) ( k i ) ( ) k i cosk i α sin n i α for i k and ξ i (A,, A k, B,, B l ) for i > k Therefore, combining () and (4), (5) ω n (A,, A k, B, B l ) x, y, x,, x, y,, y s s k t t l 2π sin n βdβ ϕ k,l(α) x, y, x,, x, y,, y s s k t t l where k ( ) k 2π ϕ k,l (α) ( ) k i ( ) n i cos i β sin n i βdβ cosk i α i i sin n i α 2π k (( ) ) k ( ) k sin i α cos k i α sin n i β cos i β dβ i sin n α ( ) k 2π i k (( ) ) k sin i α cos k i α sin 2k i β cos i β sin l k βdβ i sin n α i where we ve used the facts that n k + l and n is even In turn, note that the sum is simply the binomial expansion of (sin α sin β cos β + cos α sin 2 β) k Since and 2π sin α sin β cos β + cos α sin 2 β sin(α + β) sin β, sin k (α + β) sin l βdβ 2π sin k (α β) sin l β 2 sin k sin l, we see that (5) is equal to: ( ) k 2π ( sin α sin β cos β cos α sin 2 β ) k sin l k βdβ 2π sin n βdβ ( )k 2 sin k sin l (α) 2π sin n βdβ sin n α ( )k sin k sin l (α) π sinn βdβ sin n α Thus we arrive at the desired (8): vol S n f ω n ( )k vol S n completing the proof of Proposition 35 x, y, x sin n x, y, x α,, x, y,, y s s k t t l x, y, x,, x, y,, y s s k t t l sin k sin l (α) sin n x, y, dx, dy, α,, x, y,, y s s k t t l

12 2 CLAY SHONKWILER Now, combining Proposition 32 with Proposition 35 immediately gives Theorem for odd n 4 Moving from odd to even spheres Now that we ve proved Theorem for odd spheres, it remains only to show that it is also true for even spheres To that end, it is sufficient to prove the following proposition: Proposition 4 The truth of Theorem for a given odd value of n implies that (2) holds for oriented smooth closed submanifolds K k, L l of S n where k + l n 2 and k l We will prove this proposition by taking K and L in S n, embedding them in S n by suspending L and then showing that the linking formula which holds in S n, when applied to K and the suspension of L, gives the desired formula for linking in S n 4 The setup In general, let Σ denote suspension That is, for a manifold X, ΣX is the suspension of X We know that, for any n, ΣS n is homeomorphic to S n Given an orientation of S n, we will orient ΣS n S n by using the orientation of S n followed by the orientation of the quarter circle from S n to the north pole of S n For a smooth submanifold L S n, a copy of ΣL sits inside ΣS n S n by adding the quarter circles from each point of L to the north and south poles of S n ΣL is smooth in S n except at the north and south poles and we orient ΣL by using the orientation of L followed by that of the quarter circle from a point of L to the north pole of S n In such a way, ΣL is a cycle so linking numbers make sense Moreover, since ΣL is smooth except at two points, our linking integrals will make sense as well Moreover, for K k, L l S n, (6) Lk(K, L) Lk(K, ΣL) where on the lefthand side we re taking the linking number in S n and on the righthand side we re taking linking in S n Using a minus sign to denote antipodal images, Σ( L) and ΣL coincide as subsets of S n but have opposite orientations Therefore, from (6), Thus, it follows that Lk(K, L) Lk(K, Σ( L)) Lk(K, ΣL) (7) Lk(K, L) + ( ) n Lk(K, L) Lk(K, ΣL) + ( ) n Lk(K, ΣL), where, again, the lefthand side deals with linking in S n and the righthand side with linking in S n The strategy is to use (7) to prove Proposition 4

13 LINKING INTEGRALS ON S n 3 S n ȳ ᾱ θ S n K x α y L ΣL Figure The suspension of L in S n A sketch of the situation is given in Figure In the sketch, α is the S n distance from x(s) K to y(t) L, ᾱ is the S n distance from x(s) K to a point ȳ ΣL and θ ranges from π/2 at the south pole to π/2 at the north pole Note that, from the spherical law of cosines, we have (8) cos ᾱ cos θ cos α 42 Proof of Proposition 4 Let K {x(s)} and L {y(t)} be parametrizations of K and L, where s (s,, s k ) and t (t,, t l ) Then x(s) (x(s), ) and ȳ(t, θ) (cos θy(t), sin θ) are parametrizations of K and ΣL, respectively, in S n, where θ π/2, π/2 Let ᾱ be the distance in S n from x to ȳ The fact that Theorem holds for n combined with (7) tells us that: (9) Lk(K, L) + Lk(K, L) ( )k vol S n K ΣL sin k sin l+ (ᾱ) sin n x, ȳ, d x, dȳ ᾱ First, let s analyze the term x, ȳ, d x, dȳ By definition, x, ȳ, d x, dȳ x, ȳ, x,, x, ȳ,, ȳ, ȳ dsdtdθ s s k t t l θ

14 4 CLAY SHONKWILER Now, the determinant on the right is equal to the determinant of the matrix x x n cos θy cos θy n sin θ ( ( x s x ) s )n ( ( x s k x ) s k ( ( )n, cos θ y t cos θ y ) t )n ( ( cos θ y t l cos θ y ) t l )n sin θy sin θy n cos θ where, eg, ( ) x s 3 x denotes the 3rd coordinate of the vector s R n In turn, expanding along the last column, we see that this determinant is equal to (2) sin θ sin 2 θ cos l θ x, x,, x s + cos θ, sin θy t l, cos θ y,, cos θ y s k t x, cos θy, x,, x, cos θ y,, cos θ y s s k t t l x, y, x,, x, y,, y s s k t t l Therefore, (9) simplifies as (2) Lk(K, L)+Lk(K, L) ( )k vol S n cos l θ +cos 2 θ cos l θ x, y, x x, y, x s,, x s k, y t,, y t l ( θπ/2 θ π/2 sin k sin l+ (ᾱ) sin n ᾱ,, x, y,, y s s k t t l cos l θdθ Now, let s shift our focus to the convolution in the above formula By definition, sin k sin l+ (ᾱ) (22) π π i sin k (ᾱ β) sin l+ βdβ (sin ᾱ cos β cos ᾱ sin β) k sin l+ βdβ ) x, y, dx, dy k ( k π )( ) i sin k i ᾱ cos i ᾱ sin l+i+ β cos k i βdβ i k ( k π )( ) i sin k i ᾱ cos i θ cos i α sin l+i+ β cos k i βdβ, i i

15 using (8) Plugging this into (2) gives (23) Lk(K, L) + Lk(K, L) ( )k vol S n where ψ(ᾱ) (24) since θπ/2 k θ π/2 i k i LINKING INTEGRALS ON S n 5 ψ(ᾱ)x, y, dx, dy ( ) k ( ) i cosl+i θ cos i α βπ i sin l+i+2 sin l+i+ β cos k i βdβ dθ ᾱ β ( k θπ/2 )( ) i i θ π/2 cos l+i θ cos i α ( cos 2 θ cos 2 α) l+i+2 2 (25) sin ᾱ cos 2 ᾱ cos 2 θ cos 2 α dθ βπ To simplify a bit more, we make use of the following lemma: Lemma 42 For m odd, θπ/2 θ π/2 cos m 2 θ ( cos 2 θ cos 2 α) m/2 dθ sin m α π β sin m 2 ϕdϕ The proof of this lemma consists entirely of freshman calculus substitutions, so we omit it Since l + i and k i have opposite signs and the β integral in (24) is zero when k i is odd, Lemma 42 implies that (23) is equal to (26) ( ) k vol S n (27) k i sin l+i+ β cos k i βdβ dθ ( k )( ) i cosi α π sinl+i ϕdϕ βπ i sin l+i+ sin l+i+ β cos k i βdβx, y, dx, dy β On the other hand, we want to show that (26) is equal to ( ) k vol S n ( )k vol S n ( )k vol S n k i sin k sin l (α) sin n x, y, dx, dy α π (sin α cos β cos α sin β)k sin l βdβ sin n x, y, dx, dy α ( ) k ( ) i cos i ( α βπ i sin l+i+ α β ) sin l+i β cos k i βdβ x, y, dx, dy Since we want to show that (26) is equal to (27), the nicest possible situation would be if the terms in the sums were equal term-by-term In fact, as the following lemma demonstrates, this is true

16 6 CLAY SHONKWILER Lemma 43 For any choices of k, l and i such that k and l have opposite signs, (28) π sinl+i ϕdϕ βπ vol S n sin l+i+ β cos k i βdβ β Proof Note that proving (28) is equivalent to proving that (29) ( π ) ( π ) sin l+i ϕdϕ sin l+i+ β cos k i βdβ Recall that, for any positive integers ν, κ, λ with λ even, and Therefore, (29) becomes π vol S ν vol S ν sin ν θdθ vol S κ+λ+ vol Sκ vol s λ 2 π βπ vol S n sin l+i β cos k i βdβ β vol π Sn vol S n sin l+i β cos k i βdβ sin κ θ sin λ θdθ vol S l+i+ vol S k+l+2 vol S l+i 2vol S l+i+ vol S k i vol Sn vol S k+l+ vol S n 2vol S l+i vol S k i, which is clearly true, since k + l + n and k + l + 2 n With Lemma 43 in hand, we see that (26) and (27) are equal term-byterm, which demonstrates that Lk(K, L) + ( ) n sin k sin l (α) Lk(K, L) vol S n sin n x, y, dx, dy, α completing the proof of Proposition 4 Thus, we have showed that (2) holds in both odd and even spheres, completing the proof of Theorem References Dennis DeTurck and Herman Gluck: Electrodynamics and the Gauss linking integral on the 3-sphere and in hyperbolic 3-space Preprint, 25 arxiv:math/5388v 2 Dennis DeTurck and Herman Gluck: Linking integrals via double forms Preprint, 27 3 Moritz Epple: Orbits of asteroids, a braid, and the first link invariant Math Intelligencer, 2():45 52, Carl Friedrich Gauss: Integral formula for linking number In Zur mathematischen theorie der electrodynamische wirkungen (Collected Works, Vol 5), p 65 Koniglichen Gesellschaft des Wissenschaften, Göttingen, 2 edition, Greg Kuperberg: From the Mahler conjecture to Gauss linking forms Preprint, 26 arxiv:math/694v2 DRL 3E3A, University of Pennsylvania address: shonkwil@mathupennedu