Transparent connections


 Theodora McDowell
 2 years ago
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2 The abelian case A definition (M, g) is a closed Riemannian manifold, d = dim M. E M is a rank n complex vector bundle with a Hermitian metric (i.e. a U(n)bundle). is a Hermitian (i.e. metric) connection on E. Definition: is said to be transparent if the parallel transport of along every closed geodesic of g is the identity. Problem: understand transparent connections. These connections are ghosts or invisible form the point of view of g. Obviously, transparent connections are invariant under gauge transformations.
3 The abelian case Motivation This is a typical inverse problem that relates to many others (DN maps, inverse spectral problems, derivatives of entropy...). When M is a bounded domain in R n with smooth boundary (or R n with suitable decay conditions), this problem has been studied by R. Novikov, Finch & Uhlmann, Sharafutdinov and G. Eskin among others.
4 The abelian case Here we will focus on the closed case. The role of the boundary will now be played by the dynamical properties of the geodesic flow of g. We can also replace the closed geodesics of g by another set of distinguished closed curves; for example the closed geodesics of an arbitrary affine connection on M. This finds applications in nonequilibrium statistical mechanics.
5 The abelian case We cannot expect to be able to understand transparent connections for an arbitrary g, but there are two important and opposite cases in which we can hope to be successful: g is a Zoll metric; g has negative curvature. In the closed case, there are results for M = S 2 and RP 2 with the round (Zoll) metric by R. Novikov and L. Mason (who introduced the term transparent connection ). appear naturally when trying to find finite dimensional families of solutions to the YangMillsHiggs equations D A Φ = F A in (2 + 1)space (Ward).
6 The abelian case Our setting g has negative curvature; E = M C n is a trivial vector bundle: this is not a serious restriction in the sense that our results will carry on to arbitrary bundles, but the assumption simplifies the presentation; d = 2, i.e. M is a surface. For E = M C n, the trivial connection is always transparent. One of the questions we wish to address is the following: Is the trivial connection the only transparent connection up to gauge equivalence?
7 The abelian case Connections A connection is given by a u(n)valued 1form A, which we view as a map A : TM u(n) which is linear in the velocities v T x M. Here = d + A. Recall that A and B are gauge equivalent if there exists a smooth function u : M U(n) such that B = u 1 Au + u 1 du. Thus A is gauge equivalent to the trivial connection iff there exists u : M U(n) such that du + Au = 0. Given γ : [a, b] M, the parallel transport along γ is obtained by solving the linear ODE: ṡ = A(γ, γ)s.
8 The abelian case The abelian case when n = 1, A = iθ, where θ Ω 1 (M). Transparent means θ 2πZ (T ) γ for every closed geodesic γ. A is gauge equivalent to the trivial connection iff there is u : M U(1) = S 1 such that 1 u du = iθ iff θ is closed and [θ/2π] H 1 (M, Z). Then, the question here is: does (T) = θ closed?
9 The abelian case The answer is YES. This follows (not immediately!) from the following result: if θ = 0 for all closed geodesics γ, then θ is exact. γ This was proved by Guillemin & Kazhdan (1980) for d = 2 and by Croke & Sharafutdinov (1998) for d 3.
10 Local uniqueness of the trivial connection A PDE classifying transparent connections Curvature Let F A be the curvature of = d + A. This is a u(n)valued 2form given by F A = da + A A. F A : M u(n), where is the Hodge star operator of (M, g). In other words, F A = ( F A )ω a, where ω a is the area form of (M, g). Let K be the Gaussian curvature of M. For each x M, ±i( F A ) K Id is a Hermitian matrix, which is positive definite for K < 0 and F A sufficiently small.
11 Local uniqueness of the trivial connection A PDE classifying transparent connections Local uniqueness Our first result is: Theorem A. Let (M, g) be a closed negatively curved orientable surface. Let A be a transparent connection on E = M C n, n 1. Then, if ±i( F A ) K Id is positive definite for all x M, A is gauge equivalent to the trivial connection. Question: Are there nontrivial transparent connections for n 2? Yes! LeviCivita on T M TM, so the theorem is optimal. In fact there are many other nontrivial ghosts.
12 Local uniqueness of the trivial connection A PDE classifying transparent connections A classification result T transparent connections modulo gauge. SM is the unit circle bundle of M with canonical frame {X, H, V } where X is the geodesic vector field, V is the vertical vector field and H = [V, X ]. Let f : SM u(n) be a smooth function. Consider the PDE: H(f ) + VX (f ) = [X (f ), f ]. Let H 0 be the set of all solutions to the PDE for which there is u : SM U(n) such that f = u 1 V (u). Note that U(n) acts on H 0 by conjugation. Theorem B. There is a 11 correspondence between T and H 0 /U(n).
13 Local uniqueness of the trivial connection A PDE classifying transparent connections Degree one SU(2)ghosts Suppose f depends only on the base point x M and takes values in su(2). Then the equation H(f ) + VX (f ) = [X (f ), f ] turns into 2 df = [df, f ]. For f nonconstant, this implies f 2 = I and df = df f. Its solutions correspond precisely with holomorphic maps f : M CP 1. For any of them one can show that there is u : SM SU(2) such that f = u 1 V (u) (locally u = cos θi + sin θf and V = / θ). The LeviCivita ghosts correspond to f being constant.
14 Obtaining new transparent connections Description of all the SU(2)ghosts for SU(2) Suppose f H 0 and f = b 1 V (b) where b : SM SU(2), so f defines a transparent connection A. Let g : M su(2) be a smooth map with det g = 1 (i.e. g 2 = I ). We can always find a : SM SU(2) such that g = a 1 V (a). Let u := ab : SM SU(2) and let F := (ab) 1 V (ab). A key calculation shows that F H 0 iff  d A g = (d A g)g.
15 Obtaining new transparent connections Description of all the SU(2)ghosts Description of all the SU(2)ghosts The connection A induces an operator A and defines a holomorphic structure on the trivial bundle M C 2.  d A g = (d A g)g says precisely that the ieigenspace of g is a A holomorphic line bundle. Theorem C. Any transparent SU(2)connection is obtained by applying a finite number of as described above.
16 A nonabelian cocycle The nonabelian Livsic theorem Cocycles Let SM be the unit sphere bundle of (M, g) and φ t : SM SM the geodesic flow. Let C : SM R U(n) be the unique solution to C t = A(φ t(x, v))c, C(x, v, 0) = Id. C is a U(n)cocycle over φ t : C(x, v, t + s) = C(φ s (x, v), t)c(x, v, s). A transparent means that C(x, v, T ) = Id whenever φ T (x, v) = (x, v) (closed orbits).
17 A nonabelian cocycle The nonabelian Livsic theorem The nonabelian Livsic theorem A cocycle C : SM R U(n) is said to be cohomologically trivial if there exists a continuous u : SM U(n) such that C(x, v, t) = u(φ t (x, v))u 1 (x, v). The Livsic theorem (1972). A cocycle C is cohomologically trivial iff C(x, v, T ) = Id whenever φ T (x, v) = (x, v). Regularity (Nitica & Török, 1998). If C is C k, then u is C k ε for any small ε > 0 (for k = 1,, ω, we may take ε = 0).
18 A nonabelian cocycle The nonabelian Livsic theorem Summarizing: if A is transparent, there exists a smooth function u : SM U(n) such that C(x, v, t) = u(φ t (x, v))u 1 (x, v). Differentiating with respect to t and setting t = 0 we obtain: X (u) + Au = 0. Let V be the vertical vector field, i.e. the infinitesimal generator of the circle action of the bundle π : SM M. Goal: to show that V (u) = 0 for Theorem A, or that u must be a polynomial in the velocities for Theorem C (finite degree).
19 A nonabelian cocycle The nonabelian Livsic theorem Why V (u) = 0 implies Theorem A If V (u) = 0, then u is independent of v, so in fact u(x, v) = ũ(π(x, v)), for some smooth ũ : M U(n). But X (ũ π)(x, v) = dũ x (v) and X (u) + Au = 0 implies that A is gauge equivalent to the trivial connection! The heart of the argument is then to show that if u : SM U(n) satisfies the linear PDE X (u) + Au = 0, then V (u) = 0 under the hypothesis of Theorem A.
20 The operators of Guillemin and Kazhdan Modified operators Sketch of proof that V (u) = 0 Let M n (C) be the set of n n complex matrices. Given functions u, v : SM M n (C) we consider the inner product u, v = trace (u v ) dµ. SM The space L 2 (SM, M n (C)) decomposes orthogonally as a direct sum L 2 (SM, M n (C)) = n Z H n where iv acts as n Id on H n. We can write A = A 1 + A 1, where A 1 = (A iv (A))/2 H 1, A 1 = (A + iv (A))/2 H 1.
21 The operators of Guillemin and Kazhdan Modified operators Sketch of proof that V (u) = 0 The operators of Guillemin and Kazhdan Introduce the first order elliptic operators η + := (X ih)/2, η := (X + ih)/2. Clearly X = η + + η. We have η + : H n H n+1, η : H n H n 1, (η + ) = η.
22 The operators of Guillemin and Kazhdan Modified operators Sketch of proof that V (u) = 0 Modified operators To deal with X (u) + Au = 0, introduce the operators We also have µ + := η + + A 1,, µ := η + A 1. µ + : H n H n+1, µ : H n H n 1, (µ + ) = µ. X (u) + Au = 0 is now µ + (u) + µ (u) = 0. Calculating [µ +, µ ] one gets the key property: µ + (f ) 2 = µ (f ) 2 + (i F A Kn Id)f, f /2 µ (f ) 2 + c f 2, for f H n, n 1, c > 0 constant.
23 The operators of Guillemin and Kazhdan Modified operators Sketch of proof that V (u) = 0 V (u) = 0 Expand u = u n to get µ + (u n 1 ) + µ (u n+1 ) = 0. We have to show that u n = 0 for all n 0. The key property above shows that µ + (u n+1 ) 2 µ + (u n 1 ) 2 for n 2. But µ + (u n+1 ) 2 0 as n, thus µ + (u n 1 ) = 0 for all n 2. Since µ + is injective in H n for n 1, we have u n = 0 for all n 1 and we are done!
24 Higgs fields Suppose Φ : M u(n) is given. The pair (A, Φ) determines a cocycle C obtained by solving C t = {A(φ t(x, v)) + Φ(π φ t (x, v))}c, C(x, v, 0) = Id. (A, Φ) transparent means that C(x, v, T ) = Id whenever φ T (x, v) = (x, v) (closed orbits).
25 Transparent SU(2)pairs can also be classified using a suitable Bäcklund transformation. There is a also a finiteness result for the Fourier series of the relevant u given by the Livsic theorem. However the Bäcklund transformation requires to work with SO(3) as structure group. To obtain transparent SU(2)pairs of degree one needs to perform two Bäcklund SO(3)transformations starting from the trivial pair.
26 Most of what was said above works for arbitrary bundles but we have the following topological constraint coming from the Livsic theorem suitably used: Theorem. Let E be a complex vector bundle over a closed negatively curved orientable surface of genus g. Then E admits a transparent connection if and only if 2 2g divides the degree of E (= the first Chern class of E).
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