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Chapter 6 the Derivative of Exp and Dynkin's Formula

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Chapter 6 the Derivative of Exp and Dynkin's Formula Chapter 6 The Derivative of exp and Dynkin’s Formula 6.1 The Derivative of the Exponential Map We know that if [X, Y ]=0,thenexp(X + Y )=exp(X)exp(Y ), but this generally false if X and Y do not commute. For X and Y in a small enough open subset, U,containing0, we know that exp is a diffeomorphism from U to its image, so the function, µ: U U U, given by × → µ(X, Y )=log(exp(X)exp(Y )) is well-defined and it turns out that, for U small enough, it is analytic. Thus, it is natural to seek a formula for the Taylor expansion of µ near the origin. This problem was investigated by Campbell (1897/98), Baker (1905) and in a more rigorous fashion by Hausdorff(1906). These authors gave recursive identities expressing the Taylor expansion of µ at the origin and the corresponding result is often referred to as the Campbell-Baker-HausdorffFormula. F. Schur (1891) and Poincar´e(1899) also investigated the exponential map, in particular formulae for its derivative and the problem of expressing the function µ. However, it was Dynkin who finally gave an explicit formula (see Section 6.3) in 1947. The proof that µ is analytic in a suitable domain can be proved using a formula for the derivative of the exponential map, a formula that was obtained by F. Schur and Poincar´e. Thus, we begin by presenting such a formula. First, we introduce a convenient notation. If A is any real (or complex) n n matrix, the following formula is clear: × 1 ∞ Ak etAdt = . 0 (k +1)! k=0 If A is invertible, then the right-hand side can be written explicitly as ∞ k A 1 A = A− (e I), (k +1)! − k=0 217 218 CHAPTER 6. THE DERIVATIVE OF EXP AND DYNKIN’S FORMULA and we also write the latter as eA I ∞ Ak − = . ( ) A (k +1)! ∗ k=0 eA I Even if A is not invertible, we use ( )asthedefinitionof − . ∗ A We can use the following trick to figure out what (dX exp)(Y )is: d d (d exp)(Y )= exp(X + Y )= dR (1), X d d exp(X+Y ) =0 =0 since by Proposition 5.2, the map, s R is the flow of the left-invariant vector exp s(X+Y ) field (X + Y )L on G. Now, (X + Y→)L is an -dependent vector field which depends on 1 in a C fashion. From the theory of ODE’s, if p v(p)isasmoothvectorfielddepending 1 → in a C fashion on a real parameter and if Φt denotes its flow (after time), then the map Φ is differentiable and we have → t t ∂Φt ∂v (x)= dΦ(x)(Φt s) (Φs(x))ds. ∂ t − ∂ 0 See Duistermaat and Kolk [53], Appendix B, Formula (B.10). Using this, the following is proved in Duistermaat and Kolk [53] (Chapter 1, Section 1.5): Proposition 6.1 Given any Lie group, G, for any X g, the linear map, ∈ d exp : g T G, is given by X → exp(X) 1 ad X s ad X e I d expX =(dRexp(X))1 e ds =(dRexp(X))1 − ◦ 0 ◦ ad X 1 ad X s ad X I e− =(dL ) e− ds =(dL ) − . exp(X) 1 ◦ exp(X) 1 ◦ ad X 0 Remark: If G is a matrix group of n n matrices, we see immediately that the derivative of left multiplication (X L X = AX×)isgivenby → A (dLA)X Y = AY, for all n n matrices, X, Y .Consequently,foramatrixgroup,weget × ad X I e− d exp = eX − . X ad X 1 sA Now, if A is a real matrix, it is clear that the (complex) eigenvalues of 0 e ds are of the form eλ 1 − (= 1 if λ =0), λ where λ ranges over the (complex) eigenvalues of A.Consequently,weget 6.2. THE PRODUCT IN LOGARITHMIC COORDINATES 219 Proposition 6.2 The singular points of the exponential map, exp: g G, that is, the set → of X g such that d exp is singular (not invertible) are the X g such that the linear ∈ X ∈ map, ad X : g g, has an eigenvalue of the form k2πi, with k Z and k =0. → ∈ Another way to describe the singular locus,Σ,oftheexponentialmapistosaythatit is the disjoint union Σ= kΣ1, k Z 0 ∈−{ } where Σ1 is the algebraic variety in g given by Σ = X g det(ad X 2πi I)=0 . 1 { ∈ | − } For example, for SL(2, R), ab Σ = sl(2) a2 + bc = π2 , 1 c a ∈ | − − a two-sheeted hyperboloid mapped to I by exp. − ead X I Let g = g Σbe the set of X g such that − is invertible. This is an open subset e − ∈ ad X of g containing 0. 6.2 The Product in Logarithmic Coordinates Since the map, ead X I X − → ad X is invertible for all X g = g Σ, in view of the chain rule, the inverse of the above map, ∈ e − ad X X , → ead X I − 2 is an analytic function from ge to gl(g, g). Let ge be the subset of g ge consisting of all (X, Y ) such that the solution, t Z(t), of the differential equation × → dZ(t) ad Z(t) = (X) dt ead Z(t) I − with initial condition Z(0) = Y ( g ), is defined for all t [0, 1]. Set ∈ e ∈ µ(X, Y )=Z(1), (X, Y ) g2. ∈ e The following theorem is proved in Duistermaat and Kolk [53] (Chapter 1, Section 1.6, Theorem 1.6.1): 220 CHAPTER 6. THE DERIVATIVE OF EXP AND DYNKIN’S FORMULA 2 Theorem 6.3 Given any Lie group G with Lie algebra, g, the set ge is an open subset of g g containing (0, 0) and the map, µ: g2 g, is real-analytic. Furthermore, we have × e → exp(X)exp(Y )=exp(µ(X, Y )), (X, Y ) g2, ∈ e where exp: g G.Ifg is a complex Lie algebra, then µ is complex-analytic. → We may think of µ as the product in logarithmic coordinates. It is explained in Duister- maat and Kolk [53] (Chapter 1, Section 1.6) how Theorem 6.3 implies that a Lie group can be provided with the structure of a real-analytic Lie group. Rather than going into this, we will state a remarkable formula due to Dynkin expressing the Taylor expansion of µ at the origin. 6.3 Dynkin’s Formula As we said in Section 6.3, the problem of finding the Taylor expansion of µ near the origin was investigated by Campbell (1897/98), Baker (1905) and Hausdorff(1906). However, it was Dynkin who finally gave an explicit formula in 1947. There are actually slightly different versions of Dynkin’s formula. One version is given (and proved convergent) in Duistermaat and Kolk [53] (Chapter 1, Section 1.7). Another slightly more explicit version (because it gives a formula for the homogeneous components of µ(X, Y )) is given (and proved convergent) in Bourbaki [22] (Chapter II, 6, Section 4) and Serre [136] (Part I, Chapter IV, Section 8). We present the version in Bourbaki§ and Serre without proof. The proof uses formal power series and free Lie algebras. Given X, Y g2,wecanwrite ∈ e ∞ µ(X, Y )= zn(X, Y ), n=1 where zn(X, Y )isahomogeneouspolynomialofdegreen in the non-commuting variables X, Y . Theorem 6.4 (Dynkin’s Formula) If we write µ(X, Y )= n∞=1 zn(X, Y ), then we have 1 z (X, Y )= (z (X, Y )+z (X, Y )), n n p,q p,q p+q=n with m 1 ( 1)m+1 − (ad X)pi (ad Y )qi (ad X)pm zp,q (X, Y )= − (Y ) m pi! qi! pm! p1+ +pm=p i=1 q1+ +···qm 1=q 1 ··· − − p +q 1,pm 1,m 1 i i≥ ≥ ≥ 6.3. DYNKIN’S FORMULA 221 and m 1 ( 1)m+1 − (ad X)pi (ad Y )qi zp,q (X, Y )= − (X). m pi! qi! p1+ +pm 1=p 1 i=1 − q1+··· +qm 1=−q p +···q 1,m− 1 i i≥ ≥ As a concrete illustration of Dynkin’s formula, after some labor, the following Taylor expansion up to order 4 is obtained: 1 1 1 1 µ(X, Y )=X + Y + [X, Y ]+ [X, [X, Y ]] + [Y,[Y,X]] [X, [Y,[X, Y ]]] 2 12 12 − 24 +higherorderterms. Observe that due the lack of associativity of the Lie bracket quite different looking ex- pressions can be obtained using the Jacobi identity. For example, [X, [Y,[X, Y ]]] = [Y,[X, [Y,X]]]. − There is also an integral version of the Campbell-Baker-Hausdorffformula, see Hall [70] (Chapter 3). 222 CHAPTER 6. THE DERIVATIVE OF EXP AND DYNKIN’S FORMULA Chapter 7 Bundles, Riemannian Manifolds and Homogeneous Spaces, II 7.1 Fibre Bundles We saw in Section 2.2 that a transitive action, : G X X,ofagroup,G,onaset, · × → X,yieldsadescriptionofX as a quotient G/Gx,whereGx is the stabilizer of any element, x X.InTheorem2.26,wesawthatifX is a “well-behaved” topological space, G is a ∈ “well-behaved” topological group and the action is continuous, then G/Gx is homeomorphic to X.InparticulartheconditionsofTheorem2.26aresatisfiedifG is a Lie group and X is a manifold. Intuitively, the above theorem says that G can be viewed as a family of “fibres”, Gx,allisomorphictoG, these fibres being parametrized by the “base space”, X, and varying smoothly when x moves in X. We have an example of what is called a fibre bundle, in fact, a principal fibre bundle.
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