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

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)