Orientation transport

Transcription

1 Orientation transport Liviu I. Nicolaescu Dept. of Mathematics University of Notre Dame Notre Dame, IN June S 1 -bundles over 3-manifolds: homological properties Let (Y, g) denote a compact, oriented Riemann 3-manifold without boundary. Denote by π : X Y a principal S 1 -bundle over Y, and by Z Y the associated 2-disk bundle. Set c := c 1 (Z) H 2 (Y, Z). Denote by t Z H 2 (Z, X; Z) the Thom class of Z Y, by j the inclusion X Z and by ζ : Y Z the natural inclusion. Using the Thom isomorphism H (Z) t Z H +2 (Z, X; Z), c = ζ t Z, and the long exact cohomological sequence of the pair (Z, X) we obtain the Gysin sequence π! H k 2 (Y, Z) c H k (Y, Z) π H k (X, Z) π! H k 1 (Y, Z) c If c is a torsion class we denote by ord(c) its order. Otherwise we set ord(c) = 0. The kernel of the map c : H 0 (Y, Z) H 2 (Y, Z) is ord(c) Z so for k = 1 we obtain an isomorphism H 1 (X, Z) = π H 1 (Y, Z) ord(c)z. For k = 2 we obtain a short exact sequence ( ) 0 H 2 (Y, Z)/ c H 2 (X, Z) ker H 1 (Y, Z) c H 3 (Y, Z) 0. The last group is free so the sequence is split. The image of the morphism H 1 (Y, Z) c H 3 (Y, Z) is a subgroup of H 3 (Y, Z) = Z so it has the form nz for some nonnegative integer n. We set deg c := n. Observe that deg c = 0 c is a torsion class ord(c) > 0. 1

2 For k = 3 we obtain a short exact sequence 0 Z/ deg c H 3 (X, Z) π! H 2 (Y, Z) 0. Homologically, the Thom isomorphism is described by ζ! : H (Z, X; Z) H 2 (Y, Z), H (Z, X; Z) σ σ [Y H 2 (Z, Z) = H 2 (Y, Z). We obtain the homological Gysin sequence H k (X, Z) j H k (Z, Z) ζ! H k 2 (Y, Z) π! H k 1 (X, Z) The morphism π!, also known as the tube map is described geometrically as follows. Represent σ H m (Y, Z) by an embedded oriented submanifold S. The total space of the restriction of the S 1 -bundle X Y to S is a (m + 1)-dimensional submanifold of X representing π! σ. If we use the isomorphism π : H (Z, Z) H (Y, Z) and we represent the Poincaré dual of c H 2 (Y, Z) by a link L Y then we can describe the Gysin sequence as H k (X, Z) π H k (Y, Z) L H k 2 (Y ) π! H k 1 (X, Z) 2 S 1 -bundles over 3-manifolds: geometric properties Denote by ˆd the exterior derivative on X. Denote by Θ Ω 2 (Y ) the g-harmonic 2-form on Y representing the first Chern class of the disk bundle Z Y. We denote by ϕ Vect(X) the infinitesimal generator of the S 1 -action on X ( ϕ f)(x) := d dt f(eit x), x X. We identify u(1)-the Lie algebra of U(1)-with ir. Now choose a u(1)-valued connection 1-form iϕ iω 1 (X) such that ϕ ϕ = 1, π Θ = i 2π ˆd(iϕ) π Θ = 1 2π ˆdϕ. For every r 0 we set ϕ r := rϕ and define a metric ĝ r on X by ĝ r = ϕ 2 r + π g. With respect to this metric the fibers of π : X Y have length 2πr. Choose an oriented orthonormal frame {e 1, e 2, e 3 } T Y defined on an open subset U Y and we denote by {e 1, e 2, e 3 } the dual coframe. We denote by Γ g = 0 A 3 A 2 A 3 0 A 1 A 2 A Ω 1 (U) so(3)

3 the 1-form describing the Levi-Civita connection with respect to the frame {e 1, e 2, e 3 }. From Cartan s structural equations we deduce e 1 e 1 d e 2 e 3 = Γ g e 2 e 3. (2.1) Set f 0 = f 0 (r) = ϕ r, f i = π e i, i = 1, 2, 3, so that {f 0, f 1, f 2, f 3 } is a ĝ r -orthonormal co-frame. We denote by {f 0 = f 0 (r), f 1, f 2, f 3 } the dual frame and by ˆΓ r the connection 1-form describing the Levi-Civita connection ˆ r of the metric ĝ r. ˆΓr is also characterized by Cartan s structural equations ˆd f 0 f 1 f 2 f 3 = ˆΓ r f 0 f 1 f 2 f 3. Using (2.1) and the equality ˆdf 0 = ˆdϕ r = 2πrΘ we deduce f 0 2πrΘ ˆd f 1 f 2 = A 3 f 2 + A 2 f 3 A 3 f 1 A 1 f 3 = ˆΓ r f 3 A 2 f 1 + A 1 f 2 We set and we write f 0 f 1 f 2 f 3 Θ = Θ 23 e 2 e 3 + Θ 31 e 3 e 1 + Θ 12 e 1 e 2, Θ ij = Θ ji, [ 0 0 ˆΓ r = 0 π + Γ }{{} :=ˆΓ 0 0 rξ 0 1 rξ 0 2 rξ 0 3 rξ rξ 1 2 rξ 1 3 rξ 2 0 rξ rξ 2 3 rξ 3 0 rξ 2 1 rξ } {{ } := rξ, rξ α β = rξ β α.. (2.2) The bundle T X admits a ĝ r -orthogonal decomposition T X = f 0 π T Y and as such it is equipped with a metric connection ˆ 0 = f 0 f0 π g. The 1-form describing this connection with respect to the frame {f α } is ˆΓ 0. Then Using (2.2) we deduce f 0 2πrΘ rξ f 1 f 2 = 0 0 f 3 0 ˆ r = ˆ 0 + r Ξ. rξ 0 1 f 1 + r Ξ 0 2 f 2 + r Ξ 0 3 f 3 = 2πrΘ =: Ψ 0 rξ 1 0 f 0 + r Ξ 1 2 f 2 + r Ξ 1 2 f 3 = 0 =: Ψ 1 rξ 2 0 f 0 + r Ξ 2 1 f 1 + r Ξ 2 3 f 3 = 0 =: Ψ 2 rξ 3 0 f 0 + r Ξ 3 1 f 1 + r Ξ 3 2 f 2 = 0 =: Ψ 3 (2.3) 3

4 Set rξ α β = rξ α βγ f γ, Ψ α = 1 Ψ α βγ 2 f β f γ, Ψ α βγ = Ψα γβ. Arguing as in [1, we deduce rξ α βγ = 1 2 β,γ ( ) Ψ α βγ + Ψβ γα Ψ γ αβ We deduce so that rξ 0 ij = πrθ ij, 1 i, j 3, rξ 0 i = πr j Θ ij f j = πrf i Θ. Next, observe that for 1 i, j, k 3 we have r Ξ i jk = 0 so that Hence rξ = πr Consider the isometry Now set rξ i j = r Ξ i j0f 0 = 1 2 Ψ0 ijf 0 = πrθ ij f 0 0 f 1 Θ f 2 Θ f 3 Θ f 1 Θ 0 Θ 12 f 0 Θ 13 f 0 f 2 Θ Θ 21 f 0 0 Θ 23 f 0 f 3 Θ Θ 31 f 0 Θ 32 f 0 0, f 0 = rϕ. L r : (T X, ĝ r ) (T X, ĝ 1 ), ϕ r ϕ, f i f i, i = 1, 2, 3. r := L r ˆ r L 1 r, r [0, 1. This is a connection on T X, compatible with the metric ĝ 1. Its torsion is nontrivial. Lemma 2.1. With respect to the ĝ 1 -orthonormal frame ϕ, f 1, f 2, f 3 we have decomposition r = ˆ 0 + r Ξ, that is, if V = 3 α=0 V α f α Vect(X), f 0 = ϕ we have r V = ˆ 0 V + α,β=0 rξ β αv α f β. In particular, lim r = ˆ 0. r 0 4

5 Proof. For α > 0 and V Vect(X) we have L r ˆ r V L 1 r f α = L r ˆ r V f α = L r ˆ 0 V f α + L r = L r ˆ 0 V f α + L r ( 1 r r Ξ 0 α(v ) ϕ ) + rξ β α(v )f β β=0 rξ β α(v )f β β=1 = ˆ 0 V f α πr(v f α Θ) ϕ ˆ 0 V f α as r 0. L r ˆ r V L 1 r ϕ = L r ˆ r V f 0 = L ˆ 0 r f 0 + πr (V f i Θ)f i ˆ 0 V ϕ as r 0. Recall (see [1, 4.1.5) that the exterior derivative ˆd : Ω (X) Ω +1 (X) can be described as the composition i=1 C (Λ T X) ˆ 1 C (T X Λ T ε X) C (Λ +1 T X), (2.4) where ε : T X Λ T X Λ +1 T X denotes the exterior multiplication. Denote by d r the operator obtained by replacing in (2.4) the connection ˆ 1 with the connection r. 3 The ASD operator on S 1 -bundles over 3-manifolds Denote ˆ the Hodge -operator on (X, ĝ 1 ) and by the Hodge operator on Y. The ASD operator on (X, ĝ r ) is the first order elliptic operator Set ASD = 2 ˆd + ˆd : Ω 1 (X) Ω 2 +(X) Ω 0 (X). E := R π T Y = R f 0 π T Y, We identify as above Λ 1 T X and (Λ 0 Λ 2 +)T X with E as follows. As in [2, Ex we have an ĝ 1 -isometry T X E = R f 0 π T Y, a a 0 a H, a 0 := f 0 a, a H = a a 0 f 0. To produce an identification of (Λ 0 Λ 2 +)T X with E we use the ĝ 1 -isometry 2f0 : Λ 2 +T X π T Y. If ω is a 2-form on X, so that ω = f 0 η + θ, rθ = 0 then ˆ ω = f 0 θ + η, ω + = 1 2 ( ) f 0 (η + θ) + (θ + η) 5

6 2f0 ω + = 1 2 (η + θ) Via the above identifications we can regard the ASD operator with a differential operator C (E) C (E). We will locally represent the sections of E as linear combinations a 0 f 0 + a 1 f 1 + a 2 f 2 + a 3 f 3, f 0 = ϕ. }{{} :=a H d 0 [a 0, a 1, a 2, a 3 = ˆda β f β + β=0 a j π Γ j k f k where Γ 1 2 = A 3, Γ 3 1 = A 2, Γ 2 3 = A 1 and Γ i j = Γj i. Set for simplicity d H = j=1 f j 0 fj : Ω (X) Ω +1 (X), ϕ a H = j=1 ( ϕ a j )f j. j=1 Observe that Then d H (π ω) = π dω, ω Ω (Y ). d 0 (a 0 f 0 + a 1 f 1 + a 2 f 2 + a 3 f 3 ) = f 0 ( d H a 0 + ϕ a H ) + d H a H 2f0 ( 2 d + 0 ) = ( d H a 0 + ϕ a H ) + d H a H. Next we look at the differential operator d 0 : Ω 0 (X) Ω 1 (X) = ϕ ϕ + d H. Since ϕ generates a 1-parameter group of ĝ 1 -isometries we deduce ĝ1 ϕ ϕ = ϕ and d 0(a 0 ϕ + a H ) = ϕ a 0 + d Ha H. = 0 so that If we define ASD 0 := d 0 2 d + 0 : C (E) C (E) [ [ a0 ϕ a 0 + d H a H a H d H a 0 + ϕ a H + d H a H [ [ [ 1 0 a0 0 d = 0 1 ϕ + H a H d H d H = [ [ 0 ϕ + d H d H d H }{{} :=S [ a0 [ a0 a H a H. 6

7 Similarly, if W is metric vector bundle on Y and A is a metric connection on W then we get a differential operator d A : Ω (W ) Ω +1. We can pull back the bundle W and the connection A on X. Denote by r,a the connection on T X W obtained by twisting 0 with π A and then similarly We obtain twisted ASD-operators and as above we deduce We set Then d H,A = f j A,0. j=1 ASD A,r : Ω 1 (π W ) Ω 0 (W ) Ω 2 +(W ). ASD A,0 = [ [ 0 ϕ + d H,A d H,A d H,A }{{} :=S A P A := ϕ + S A. ker ASD A,0 = ker P A = ker A AA A, ind A W = ind ASD A,0. The operators ϕ and S W commute so that We deduce that if a = a 0 + a H ker P A then P AP A = 2 ϕ + S 2 A. ϕ a = 0, S A a = 0. This shows that the pullback by π induces an isomorphism π : ker S A ker P A. A similar argument shows that if A t is a path of metric connections on W then the orientation transport along the path P At is equal to Now observe that the difference ( 1) SF (S A t ) D A,r := ASD A,r ASD A,0 is a zeroth order operator which converges to zero in in any C k -norm. We denote by OT A,r the orientation transport along the path t ASD A,(1 t)r 7

8 which connects ASD A,r to ASD A,0. Since ind ASD A,r = ind ASD A,0 we deduce that if ker S A = 0 then ker ASD A,r = 0 for all 0 r 1. In particular OT A,r = 0, 0 < r 1. Suppose {A t ; t [0, 1} is a path of connections on W such that ker S Aj = 0 for j = 0, 1. Then for every r > 0 we have For r sufficiently small we deduce OT (ASD At,r) = OT A0,r( 1) SF (S A t ) OT A1,r. OT (ASD At,r) = ( 1) SF (S A t ) Now it remains to see that the operator ASD At,r is conjugate (via L r with the usual ASDoperator defined using the metric ĝ r of radius r and the twisting connection A t ). References [1 L.I. Nicolaescu: Lectures on the Geometry of Manifolds, World Sci. Pub. Co [2 L.I. Nicolaescu: Notes on Seiberg-Witten theory, Graduate Studies in Math, vol. 28, Amer. Math. Soc.,