Lecture 19 - Lie Group

#math

A Lie group is a smooth manifold together with a group structure i.e whose elements are points of a $C^{\infty}$ manifold of finite dim (though not necessarily), with the condition that the group operation holds:

Composition

\begin{array}{r} G \times G ⟶ G \\ (g, h) ⟶ g h \end{array}

Inverse map

\begin{array}{r} G ⟶ G \\ g ⟶ g^{- 1} \end{array}

The inverse map has two characteristic features, viewing it on a manifold it is a diffeomorphism from $G$ to $G$ , and viewing it as a group it is an automorphism.

Let g be a fixed element of a Lie group, the the mapping:

\begin{array}{r} L_{g} : G \to G \\ h \to g h \end{array}

is called the left translation, and by definition of the Lie group $L_{g}$ is a $C^{\infty}$ mapping.. Equivalently we can define the right translation:

\begin{array}{r} R_{g} : G \to G \\ h \to h g \end{array}

The adjoint map is defined to be:

\begin{array}{r} A d (g) : G \to G \\ h \to g h g^{- 1} \end{array}

Let $X$ be a vector field on $G$ . The vector field is left invariant if it is $L_{g}$ invariant for all $a \in G$ .

X_{a g} = {L_{a}}_{*} (X_{g}) = {X |}_{L_{a} (g)} = d L_{a} (X_{g})

Theorem

Let $τ$ be a tangent vector to a Lie group $G$ at the identity e, then $\exists$ a unique left invariant $C^{\infty}$ vector field $X$ on G such that:

X_{g} = {L_{g}}_{*} (τ)

Let $G$ be a Lie group, the the left invariant vector field on $G$ form a Lie algebra with the product defined by a bracket operation:

[X_{1}, X_{2}] = X_{1} X_{2} - X_{2} X_{1}

Let $G_{1}$ and $G_{2}$ be lie groups and $ϕ : G_{1} \to G_{2}$ be a homomorphism of the the Lie groups, then $ϕ_{*} : G_{1} \to G_{2}$ is a homomorphism of the Lie algebra. That is:

ϕ_{*} [X_{1}, X_{2}] = [ϕ_{*} X_{1}, ϕ_{*} X_{2}] = [Y_{1}, Y_{2}]

Where $X_{1}, X_{2} \in G_{1}$ and $Y_{1}, Y_{2} \in G_{2}$ .

Representation Theory

Adjoint Representation is defined to be:

\begin{array}{r} A d : G \to A u t o m o r p h i s m (G) \\ A d (g) \to g * g^{- 1} = D C_{g} |_{e} \end{array}

It has the property that:

A d (g) [X, Y] = [A d (g) X, A d (g) Y]

We can also define the derivation map:

\begin{array}{r} a d = D (A d) |_{e} \\ X \to [X, \cdot] \end{array}

Example

If $A d (e^{t Y}) (\cdot) = e^{t Y} \cdot e^{- t Y}$
Then $\frac{d}{d t} | e^{t Y} X e^{- t Y} | = [Y, X] = a d (Y)$

The derivation map has the property that:

a d ([X, Y]) = [a d (X), a d (Y)]

Now we study dual of the Lie algebra, in particular the function:

\begin{aligned} φ : G \to G^{*} \\ a \to φ (a) & =< a, \cdot > \end{aligned}

The functional derivative of $f$ at $μ \in G^{*}$ defined as the unique element $\frac{δ f}{δ μ}$ of $G$ :

lim_{ϵ \to 0} \frac{1}{ϵ} [f (μ + ϵ δ μ) - f (μ)] =< δ μ, \frac{δ f}{δ μ} >

The Lie poisson bracket is defined to be:

{f_{1}, f_{2}} =< [a_{1}, a_{2}], \cdot >

Where $f_{i} = ϕ (a_{i})$ . In other words:

{[a_{1}, \cdot], [a_{2}, \cdot]} =< [a_{1}, a_{2}], \cdot >

Now let $G, H$ be two Lie groups, we are interested in studying $G \times H$ :

G \times H \to_{Lie Algebra}^{} G \times H

We can naively define the bracket product to be:

[(X_{1}, Y_{1}), (X_{2}, Y_{2})] = ([X_{1}, X_{2}], [Y_{1}, Y_{2}])

Though this is not of particular interest, in the language of differential equations this is like studying two systems of diff eqs that are totally disconnected from one another. Thus we introduce the semi direct product.

H = G ⋉ V = {(g, v) : g \in G, v \in V}

with the product defined to be $(g_{1}, v_{1}) \cdot (g_{2}, v_{2}) = (g_{1} g_{2}, v_{1} + g_{1} v_{2})$ with the inverse existing to be:

(g, v)^{- 1} = (g^{- 1}, - g^{- 1} v)

Example

Let us consider $S O (n) ⋉ R^{n}$ this is known to be the special euclidean group $S E (n)$ .
$S E (n)$ is a subgroup of $G L (n + 1)$ . The elements have the representation to be:

(R, γ) \to (\begin{matrix} R & p^{T} \\ 0 & 1 \end{matrix})

Of particular interest $S O (2, 1) ⋉ R^{2, 1}$ is related to group operations that preserves rotation of $R^{3}$ that preserves the metric:

J = (\begin{matrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix})

The Lie algebra of a semi direct product has a bracket product to be:

[(A_{1}, u_{1}), (A_{2}, u_{2})] = ([A_{1}, A_{2}], A_{2} u_{1} - A_{1} u_{2})

Let $f, g : G^{*} \to R$ (real valued functions of the dual space). Then we can define:

\dot{f} = {f, g}_{\pm} (μ) = \pm < μ, [\frac{δ f}{δ μ}, \frac{δ g}{δ μ}] >= \mp < μ, a d_{\frac{δ f}{δ μ}} \frac{δ g}{δ μ} >= \mp < a d_{\frac{δ g}{δ μ}}^{*}, \frac{δ f}{δ μ} >

Thus:

\dot{u} = u, h = \mp a d_{\frac{δ h}{δ μ}}^{*} μ

Now let us consider semidirect product space's Lie algebra then:

{f, g} (μ, p) = \pm < μ, [\frac{δ f}{δ μ}, \frac{δ g}{δ μ}] > \pm < p, \frac{δ f}{δ μ} \frac{δ g}{δ p} - \frac{δ g}{δ μ} \frac{δ f}{δ p} >