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# Download Lecture 1 - Lie Groups and the Maurer-Cartan equation

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Lecture 1 - Lie Groups and the Maurer-Cartan equation January 11, 2013 1 Lie groups A Lie group is a differentiable manifold along with a group structure so that the group operations of products and inverses are differentiable. Let G be a Lie group. An element A ∈ G induces three standard diffeomorphisms LA , RA , AdA :→ G (1) namely left translation, right translation, and the adjoint transformation AdA = LA RA . Note that LA RB = RB LA , so left and right translation commute. A Lie group has a standard element, the identity e ∈ G. Given any vector v ∈ TE G, we can use left translation (right translation, if we wanted) to produce a global vector field by vA = L∗A v vA ∈ TA G, (2) where L∗A Te G → TA G is the push-forward along the diffeomorphism LA . Any vector field v so that vBA = L∗B vA , vBA ∈ TBA G, vA ∈ TV G (3) is called left-invariant (similarly for right-invariant vector fields). Because L∗A L∗B = L∗AB , such fields are well defined, and determined by a vector at any point of the manifold. Given v, w ∈ Te G, after extending them to left-invariant fields, we define the Lie bracket of the two vectors to be the topological bracket of the left-invariant fields, restricted to the vector at e. Since LA is a diffeomorphism, we have L∗A [v, w] = [L∗A v, L∗A w] (4) for any vector fields, so that the bracket of two left-invariant fields is a left-invariant field. It is thus convenient to identity the algebra on the tangent space at the identity with the 1 algebra of left-invariant vector fields on the manifold G. Since this is a Lie subalgebra of the Lie algebra of all differentiable vector fields under the bracket, the Jacobi identity and antisymmetry hold, so we have a lie algebra g canonically associated with the group G, with dim g = dim G. We have two canonical representations on g. the first is the standard adjoint representation adv : g → g given by adv w = [v, w]. Note that given any A ∈ G we have Ad∗A : Te G → Te G, which is a linear map. To see this is a representation, first extend ∗ v, w ∈ Te G to left-invariant fields. Note that AdA = L∗A RA , that L∗A leaves any left∗ invariant field invariant, and RA takes left-invariant fields to left-invariant fields (by the ∗ commutativity of L∗B (any B) and RA ). Finally since AdA : G → G is a diffeomorphism Ad∗A [v, w] = [Ad∗A v, Ad∗A w] (5) It is customary to drop the ∗ and just write AdA , with context distinguishing whether it is a map Te G → Te G or G → G. 2 2.1 Left-fields, Right-actions, and exponentials Basics Any vector v ∈ g ≈ Te G extends uniquely to a left-invariant vector field, which integrates out to a 1-parameter family of diffeomorphisms ϕt : G → G (6) ϕt (A) = LA ϕt (e) = Rϕt (e)−1 A. (7) which has the property that which is simply the integrated version of vA = L∗A v. Therefore left-invariant fields integrate out to right-translations, or, put more accurately, the flow of a left-invariant vector field is a flow by right translations. We also have the group law Rϕs+t (e)−1 A = ϕt+s (A) = ϕt (ϕs (A)) = Rϕt (e)−1 Rϕs (e)−1 A (8) so that ϕt (e)ϕs (e) = ϕs+t (e) (note this also implies ϕ−t (e) = ϕt (e)−1 ). Similarly, and flow by a 1-parameter subgroup of right translations determines a leftinvariant vector field. The path ϕt (e) (9) through the origin is called the exponential map, denoted exp(tv). It obeys the group law exp(tv)exp(sv) = exp((t + s)v), as shown above. 2 2.2 Adjoint actions Consider the differential of the adjoint action along exp(tv) on the vector w d Adexp(tv) w dt t=0 = d R∗ w L∗ dt t=0 exp(tv) exp(tv) (10) A vector is determined by its action on functions, and this action is given by d d f Adexp(tv) exp(sw) ds s=0 dt t=τ d d = Lexp((t+τ )v) Lexp(sw) Lexp(−(t+τ )v) f (e) ds s=0 dt t=0 d d = Lexp(τ v) Rexp(τ v) Lexp(tv) Lexp(sw) Lexp(−tv) f (e) ds s=0 dt t=0 d v(Lexp(sw) f )(e) − Lexp(sw) (vf )(e) = Lexp(τ v) Rexp(τ v) ds s=0 = Lexp(τ v) Rexp(τ v) (v(wf )(e) − w(vf )(e)) . (11) d Adexp(tv) = Adexp(tv) ◦ adv dt (12) d Adexp(tv) = adv dt t=0 (13) so that Note in particular that On the other hand given any operator D, we can (aside from convergence issues) create the operator Exp(D) by Exp(D) = and note that d dt Exp(tD) ∞ X 1 n D n! n=0 (14) = Exp(tD)D = D Exp(tD). Because the solution to d M = M ◦ adv dt (15) with initial conditions M (0) = Id is unique, we have Adexp(tv) = Exp(t adv ). 3 (16) 3 3.1 The Canonical 1-form g-valued p-forms A Lie group, as a differentiable manifold, has a cotangent bundle and other associated bundles. A g-valued p-form α is simply a section of the bundle ^p ⊗g (17) Vp Vp ∗ (where is an abbreviation for T M ). Letting {gi }ni=1 be a basis for g, a g-valued p-form α can be written α = αi ⊗ gi (18) = αji 1 ...jp dxj1 ∧ · · · ∧ dxjp ⊗ gi where the αi are p-forms and the αji 1 ...jp are functions. Due to the bracket product on g, we can define a product on Vp Vq α∈ ⊗g and β ∈ ⊗g, we define ^p+q [α, β] ∈ ⊗g L n p=0 Vp ⊗ g. Letting to be the g-valued (p + q)-form given by 1 X (−1)sqn π [α(vπ1 , . . . , vπp ), β(vπp+1 , . . . , vπp+q )] [α, β](v1 , . . . , vp+q ) = p!q! π∈σ (19) (20) p+q where the sum is taken over all permutations π in the symmetric group σp+q on p + q many symbols. Or, writing α = αi ⊗ gi and β = β j ⊗ gj , we have [α, β] = αi ∧ β j ⊗ [gi , gj ] (21) so the “bracket” on g-valued p-forms is a combination of exterior differentiation and the bracket operation. Sometime Vp+q Vp+qyou will see symbols like α ∧ β, which is not technically a section of ⊗g but of ⊗g ⊗ g. For instance, if α, β are g-valued 1-forms, we have [α, β](v, w) = [α(v), β(w)] − [α(w), β(v)] (22) Note in particular that [α, α](v, w) = 2[α(v), α(w)]. Finally we have an “exterior differential” ^ ^p p+1 d:Γ ⊗g →Γ ⊗g 4 (23) Vp given on α = αi ⊗ gi ∈ ( ⊗ g) by d(α) = dαi ⊗ gi . (24) Differential geometers will recognize this as covariant exterior differentiation on the flat bundle g → G × g → G. It is easily checked that the exterior derivative works well with the bracket operation: d[α, β] = [dα, β] + (−1)|α| [α, dβ] (25) where α, β are g-valued forms of order |α| and |β|. 3.2 The trivial, but not-so-trivial, canonical 1-form Let θ be a the g-valued 1-form that assigns to a vector v ∈ TA G is associated element of g. In some sense, θ is an identity automorphism. It is common to write θA = A−1 dA. (26) This really is an abuse of the notation, but correctly indicates that tangent vectors at A are brought back to vectors at e via left-translation along A−1 . If G is a group of matrices, then A−1 dA has a precise meaning as written. Because the Lie algebra bracket and the topological bracket are equivalent, note that θ[v, w] = [v, w] = [θ(v), θ(w)]. (27) Write θ = θi ⊗ gi . Letting {gj }nj=1 be basis vectors extended to left-invariant fields, note that θi (gj ) = δji . Then dθ(gj , gk ) = (dθi )(gj , gk ) · gi = gj θi (gk ) − gk θi (gj ) − θi ([gj , gk ]) gi = gj (δki ) − gk (δji ) − θi ([gj , gk ]) gi = −θi ([gj , gk ])gi (28) = −θ[gj , gk ] = −[θ(gj ), θ(gk )] 1 = − [θ, θ] (gj , gk ) 2 Having checked it on basis vectors, we arrive at the Maurer-Cartan structure equation: dθ + 1 [θ, θ] = 0. 2 5 (29)