Lie Groups and Representation Theory

Group Representation

Let $G$ be a group. A group representation $(\rho, V)$ is given by a vector space $V$ and a homomorphism $\rho: G \rightarrow GL(V)$.

• A representation is said to be irreducible if for any subspace $W \subset V$ which is invariant by $\rho$, or $\rho(G)(W) \subset W$, then $W = \{0\}$ or $W = V$.
• The space $V$ is called the carrier space of the representation
• If $G$ is compact, let $\pi: G \rightarrow U(\mathcal{H})$ be a continuous unitary representation on a Hilbert space $\mathcal{H}$. Then, $\mathcal{H}$ is unitarily isomorphic to a Hilbert direct sum of finite-dimensional irreducible unitary representations of $G$. (Peter-Weyl Theorem) \[ \mathcal{H} \cong \widehat{\bigoplus}_{\alpha \in A} V_{\alpha}, \]

  where each $V_\alpha$ is a finite-dimensional irreducible unitary representation of $G$.

  • Unitary operator is a bounded linear operator $U: H \rightarrow H$ on a Hilbert space $H$ that satisfies $U^*U = UU^* = I$, where $U^*$ is the adjoint of $U$, and $I$ is identity operator.

Lie Group and Lie Algebra

A Lie group $G$ is:

• A group $(G, \cdot)$
• A smooth (i.e. $C^\infty$) manifold
• Maps \begin{gather*} m: G \times G \rightarrow G, \quad m(g,h) = gh \\ \iota: G \rightarrow G, \quad \iota(g) = g^{-1} \end{gather*}

  are smooth

A homomorphism of Lie groups $\Phi: G \to H$ is a smooth map that is also a group homomorphism. Specifically, if $(G, \cdot)$ and $(H, *)$ are the groups, then:

• $\Phi(x \cdot y) = \Phi(x) * \Phi(y)$ for all $x, y \in G$.
  

Let $G$ be a lie group and $e$ be its identity element. The Lie algebra of $G$ is

the tangent space of $G$ at $e$ together with it's canonical Lie bracket.

Denote by $\mathfrak{X}(G)$ the space of smooth vector fields on $G$.

• A vector field $X \in \mathfrak{X}(G)$ assigns to each point $p \in G$ a tangent vector $X_p \in T_pG$.
• Given a smooth function $f \in C^\infty(G)$, the vector field $X$ acts as a derivation: \[X(f)(p) = \text{directional derivative of $f$ at $p$ along $X_p$}\]

  In local coordinates ($x_1, \dots, x_d)$,

  \[X = \sum_{i=1}^d X^i \frac{\partial}{\partial x_i}, \quad X(f) = \sum_{i=1}^d X^i \frac{\partial f}{\partial x_i}\]

  This gives $X(f) \in C^\infty(G)$.

• Given two vector fields $X, Y \in \mathfrak{X}(G)$, their Lie bracket $[X,Y]$ is the new vector field defined by
  \[[X,Y](f) = X(Y(f)) - Y(X(f)) \quad \text{for all }f \in C^\infty(G)\]
  • $[X,Y]$ is still a smooth vector field
  • The bracket satisfies:
    • Bilinearity: $[aX+bY, Z] = a[X,Z] + b[Y,Z]$
    • Antisymmetry: $[X,Y] = -[Y,X]$
    • Jacobi identity: $[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0$
      

For any tangent vector $X_0 \in \mathfrak{g} = T_eG$, there exists (typically left-invariant) vector field $X \in \mathfrak{X}(G)$ such that $X(e) = X_0$. Similarly, for $Y_0 \in \mathfrak{g}$, choose a vector field $Y$ with $Y(e) = Y_0$.

We can then define the Lie bracket on $\mathfrak{g}$ by

This is well-defined [TODO], hence, $(\mathfrak{g}, [\cdot, \cdot])$ is a Lie algebra. \newline

If $G \subset GL_n(\mathbb{C})$ is a matrix Lie group (a closed subgroup), then:

• The Lie algebra $\mathfrak{g}$ can be identified with a subspace $\mathfrak{gl}_n(\mathbb{C})$, the space of all $n \times n$ complex matrices.
• For $X, Y \in \mathfrak{g}$, the Lie bracket is given by the commutator \[[X,Y] = XY-YX\]

  where $XY$ is the usual matrix product.

  • Differential operator that measures the "failure" of two operations to commute
  • IMPORTANT: For matrix Lie groups, intrinsic Lie bracket coincides with commutator of matrices
    .

For a general Lie group $G$:

• A smooth group homomorphism $\gamma: \mathbb{R} \rightarrow G$, which satisfies $\gamma(t+s)=\gamma(t)\gamma(s)$, is called a one-parameter subgroup.
• Its derivative at $0$ is a tangent vector:

  \[\gamma'(0)\in T_eG = \mathfrak{g}\]

The exponential map is

defined as follows:

• Given $X \in \mathfrak{g}$, there exists a unique one-parameter subgroup $\gamma_X : \mathbb{R} \rightarrow G$ such that $\gamma_X'(0) = X$
• Then \[\exp(X) := \gamma_X(1)\]

  In other words: $\exp(X)$ is when you start at the identity, walk along that special path defined by $X$, and stop exactly when time $t = 1$.

  • Can be further extended for any $t$, as curve $\gamma_X(t) = \exp(tX)$ is the unique one-parameter subgroup with $\gamma_X(0)$ and $\gamma_X'(0) = X$.
    

For matrices $X \in \mathfrak{gl}_n(\mathbb{C})$, we also have matrix exponential:

• When $G$ is a matrix Lie group and the Riemannian metric is compatible, the abstract $\exp: \mathfrak{g} \rightarrow G$ coincides with the matrix exponential restricted to $\mathfrak{g}$.
  • When viewing $G$ as a Riemannian manifold with a suitable metric, you can define \[\exp_{\text{Riem}}: T_eG \cong \mathfrak{g} \rightarrow G\]

    by taking the unique geodesic (locally length-minimizing path) $\gamma_X(t)$ with $\gamma_X(0) = e$ and $\gamma_X'(0) = X$.
    
    In this setup, $\exp(X) = \exp_{\text{Riem}}(X)$.
    

Assume $G$ is a Lie group with a Riemannian metric $\langle \cdot, \cdot\rangle_G$.

• Left-invariant matrix: for all $g \in G$, left multiplication $L_g: G \rightarrow G$, $L_g(h)=gh$ is an isometry.
• Right-invariant matrix: for all $g \in G$, right multiplication $R_g: G \rightarrow G$, $L_g(h)=hg$ is an isometry.
• Bi-invariant metric: invariant under both left and right translations.

If $G$ is compact and you start with any left-invariant metric $\langle \cdot, \cdot \rangle_G$, you can build a bi-invariant metric by averaging over the group using the (left-invariant) Haar measure $\mu$:

For each $p \in G$ and $X_p, Y_p \in T_pG$,

For a bi-invariant metric on a matrix Lie group, the exponential map from Riemannian geometry coincides with group exponential, which matches the matrix exponential.

Baker-Campbell-Hausdorff (BCH) Formula:

has solution $Z = X + Y + \frac{1}{2}[X, Y] + \frac{1}{12}([X, [X, Y]] - [Y, [X, Y]]) + \dots$

If $G$ is compact and connected, then:

• The exponential map \[\exp: \mathfrak{g} \rightarrow G\]

  is surjective: every group element $g \in G$ can be written as $g = \exp(X)$ for some $X \in \mathfrak{g}$.

So the Lie algebra's exponential map reaches all of $G$.

Let

• $G,H$ be Lie groups,
• $\Phi: G \rightarrow H$ be a Lie group homomorphism,
• $\mathfrak{g} = T_eG$, $\mathfrak{h}=T_{e_H}H$ their Lie algebras
• $\varphi = d\Phi(e): \mathfrak{g} \rightarrow \mathfrak{h}$ the derivative at the identity.

Then:

This shows that the exponential map is "intrinsic." If you translate a problem from one group to another (e.g., representation theory), the exponential map translates perfectly with it.

SE(3) Group

It acts on a point $p \in \mathbb{R}^3$ via the transformation:

As a manifold, $\mathrm{SE}(3)$ is $\mathrm{SO}(3) \times \mathbb{R}^3$, so the degree-of-freedom is $6$.