Lie Groups and Lie Algebras, Part I

Prerequisites: Linear algebra (matrix exponentials, commutators), some exposure to group theory (what a group is, what the identity element is), and basic quantum mechanics (angular momentum).

If you’ve ever tried learning quantum field theory, you’ve probably hit a wall that sounds something like: “Consider the Lie algebra of the Lorentz group…” — and suddenly every textbook assumes you already know what that means.

This post is the first in a series that builds the language of Lie groups and their representations from the ground up. The goal is not to be rigorous in the way a mathematician would demand, but to be honest — to tell you what each object is, why we care about it, and where the subtleties hide.

Definition: Lie Algebra#

At its core, a Lie group is a group whose elements depend continuously on some set of parameters. Think of rotations: you can rotate by any angle $\theta$ , and as $\theta$ varies smoothly, so does the rotation. This is in contrast to discrete groups (like the group $\{+1, -1\}$ under multiplication), where you can list all the elements.

Because the group elements depend on continuous parameters, we can do calculus on them. In particular, we can expand a group element near the identity and extract what are called generators — the infinitesimal building blocks of the group. The commutation relations between these generators define the Lie algebra:

$[T^a, T^b] = i f^{ab}{}_c \, T^c$

where $T^a, T^b$ are generators, and $f^{ab}{}_c$ are called the structure constants.

A few important points:

The structure constants are independent of representation. This is crucial. No matter how you choose to represent the generators (as $2\times2$ matrices, $3\times3$ matrices, differential operators, etc.), the structure constants $f^{ab}{}_c$ are always the same. They are intrinsic to the algebra itself.
This relation is exact, not an approximation. When we derive the commutator from the group multiplication law (which we will do shortly), it emerges at second order in the expansion parameters. One might worry: don’t higher-order terms modify the algebra? They don’t. The reason is that the Lie algebra captures the full local structure of the group near the identity. Higher-order terms in the Baker-Campbell-Hausdorff expansion are entirely determined by repeated commutators — they add no new information beyond what $[T^a, T^b] = if^{ab}{}_c T^c$ already encodes.
The structure constants define the Lie algebra. Two Lie groups can look very different globally but have the same Lie algebra. For instance, $SU(2)$ and $SO(3)$ share the same algebra, even though $SU(2)$ is a double cover of $SO(3)$ . The note to carry forward: the algebra captures the local structure, not the global topology.

Putting it differently: the problem of finding all matrix representations of a Lie algebra is the algebraic problem of finding all possible matrix solutions $T^a_R$ satisfying $[T^a_R, T^b_R] = if^{ab}{}_c T^c_R$ .

Abelian Lie Groups#

A group is called abelian if all its elements commute:

$g_1 \cdot g_2 = g_2 \cdot g_1 \quad \text{for all } g_1, g_2$

Since the elements commute, the generators must also commute amongst themselves. This means:

$[T^a, T^b] = 0 \quad \text{for all } a, b$

and therefore, for an abelian Lie group, all structure constants vanish: $f^{ab}{}_c = 0$ .

Representations of Abelian Lie Groups#

The representation theory of abelian Lie groups has a beautifully simple structure:

Any $d$ -dimensional abelian Lie algebra is isomorphic to the direct sum of $d$ one-dimensional abelian Lie algebras.

This is equivalent to saying:

All irreducible representations of abelian groups are one-dimensional.

Why? If two generators commute, Schur’s lemma tells us they can be simultaneously diagonalized. In an irreducible representation, each generator must act as a scalar (a $1\times1$ matrix). So the irreducible representations are all one-dimensional, each labeled by the eigenvalue of the single generator.

The classic example is $U(1)$ : the group of phase rotations $e^{i\alpha}$ . It has one generator, and every irreducible representation is labeled by a single number — the charge $q$ . In quantum electrodynamics, this is the electric charge.

Non-Abelian Lie Groups#

When generators do not all commute — that is, when at least some $f^{ab}{}_c \neq 0$ — the group is called non-abelian. This is where Lie theory becomes rich and, frankly, where most of the physics lives. The Standard Model is built on the non-abelian groups $SU(3) \times SU(2) \times U(1)$ .

Definition: Casimir Operators#

In a non-abelian algebra, individual generators don’t commute with each other. But we can ask: are there combinations of generators that commute with every generator?

Operators constructed from the generators $T^a$ that commute with all the generators are called Casimir operators.

In other words, a Casimir operator $C$ satisfies:

$[C, T^a] = 0 \quad \text{for all } a$

These operators have two key properties:

In each irreducible representation, the Casimir is proportional to the identity matrix. This is a direct consequence of Schur’s lemma: any operator that commutes with all generators in an irreducible representation must be a multiple of the identity.
The proportionality constants label the representation. Since the Casimir takes a definite value on each irreducible representation, these values serve as quantum numbers that identify the representation.

The number of independent Casimir operators equals the rank of the Lie algebra (the dimension of its maximal abelian subalgebra, also known as the Cartan subalgebra). For $SU(2)$ , the rank is 1, so there is one Casimir. For $SU(3)$ , the rank is 2, so there are two.

Example: Angular Momentum#

The angular momentum algebra is:

$[J^i, J^j] = i\epsilon^{ijk} J^k$

This is the Lie algebra of $SU(2)$ (equivalently, $\mathfrak{so}(3)$ ). The single Casimir operator is:

$\mathbf{J}^2 = (J^x)^2 + (J^y)^2 + (J^z)^2$

You can verify that $[\mathbf{J}^2, J^i] = 0$ for any $i$ — it commutes with all three generators.

On an irreducible representation, $\mathbf{J}^2$ takes the value:

$\mathbf{J}^2 = j(j+1)\,\mathbf{I}$

where $j = 0, \tfrac{1}{2}, 1, \tfrac{3}{2}, 2, \ldots$ labels the representation. Each value of $j$ gives a $(2j+1)$ -dimensional irreducible representation. This is exactly the story of angular momentum quantization in quantum mechanics — and now you see it’s not a peculiarity of angular momentum, but a structural consequence of the $SU(2)$ Lie algebra.

Definition: Linear Representation#

We’ve been using the word “representation” loosely. Let’s be precise.

A (linear) representation $R$ of a group is an operation that assigns a linear operator $D_R(g)$ to each abstract group element $g$ .

Properties#

The map $g \mapsto D_R(g)$ must respect the group structure:

Identity: $D_R(e) = \mathbf{1}$ , where $e$ is the identity element of the group. (For a Lie group with continuous parameter $\theta$ , we have $g(0) = e$ , so $D_R(g(0)) = \mathbf{1}$ .)
Homomorphism: $D_R(g_1) D_R(g_2) = D_R(g_1 g_2)$ , so the mapping preserves the group multiplication law.

In other words, a representation is a homomorphism from the group to the group of linear operators on some vector space. The representation might not be injective (one-to-one) — multiple group elements could map to the same operator. When the map is injective, the representation is called faithful.

Basis of the Representation#

The vector space on which the operators $D_R(g)$ act is called the basis (or carrier space) of the representation $R$ .

In a matrix representation, the operators $D_R(g)$ are $n \times n$ matrices $(D_R(g))^i{}_j$ , and they act on an $n$ -dimensional vector space. The indices $i, j = 1, \ldots, n$ label the components of vectors in this space.

Reducible vs. Irreducible Representations#

A representation is reducible if there exists a basis in which every $D_R(g)$ is simultaneously block-diagonal:

$D_R(g) = \begin{pmatrix} D_{R_1}(g) & 0 \\ 0 & D_{R_2}(g) \end{pmatrix}$

for all group elements $g$ . In this case, the representation decomposes as a direct sum: $R = R_1 \oplus R_2$ . The vector space splits into invariant subspaces that don’t talk to each other under the group action.

A representation is irreducible if no such decomposition exists — there is no invariant subspace other than $\{0\}$ and the whole space. Irreducible representations (often called irreps) are the building blocks of representation theory. Any representation can, under fairly general conditions, be decomposed into a direct sum of irreducible ones. This is the analog of decomposing a vector into components along basis vectors.

Equivalent Representations#

Two representations $D_R(g)$ and $D_{R'}(g)$ are equivalent if they are related by a similarity transformation:

$D_{R'}(g) = S \, D_R(g) \, S^{-1}$

for some fixed invertible matrix $S$ and for all $g$ . Equivalent representations are, physically speaking, the same representation written in a different basis — they encode the same physics, just in different coordinates on the vector space.

Deriving the Commutator from the Group: Where $[T^a, T^b]$ Comes From#

We stated the Lie algebra relation $[T^a, T^b] = if^{ab}{}_c T^c$ at the beginning. But where does it actually come from? Here’s the quick derivation.

Consider a Lie group element near the identity, parameterized by a small parameter $\alpha^a$ :

$g(\alpha) \approx \mathbf{1} + i\alpha^a T^a + \frac{1}{2}(i\alpha^a T^a)^2 + \ldots$

Now consider the combination $g(\alpha)\,g(\beta)\,g(\alpha)^{-1}\,g(\beta)^{-1}$ . For an abelian group this would just be the identity. For a non-abelian group, it isn’t — and the deviation from the identity tells us about the commutator.

Expanding each factor to the relevant order and multiplying out, the first-order terms cancel (because $g \cdot g^{-1} = \mathbf{1}$ ), and at second order we find:

$g(\alpha)\,g(\beta)\,g(\alpha)^{-1}\,g(\beta)^{-1} \approx \mathbf{1} - \alpha^a \beta^b [T^a, T^b] + \ldots$

Since the left side is a group element (the group is closed under multiplication), and it’s near the identity, it must be expressible in terms of generators:

$\approx \mathbf{1} + i\gamma^c T^c + \ldots$

for some parameters $\gamma^c$ . Comparing the two expressions:

$[T^a, T^b] = if^{ab}{}_c T^c$

where the structure constants $f^{ab}{}_c$ are defined by the relationship $\gamma^c = -f^{ab}{}_c \,\alpha^a \beta^b$ . The commutator of generators is itself a linear combination of generators — the algebra closes.

This is the key insight: the Lie algebra is the infinitesimal version of the group multiplication law.

Example: Deriving a Generator — Rotations about the $z$ -axis#

All of this is quite abstract, so let’s see it work in a concrete case. Consider a rotation of a wavefunction $\psi(\theta^z)$ about the $z$ -axis by an infinitesimal angle $\delta\theta_0$ :

$\psi(\theta^z + \delta\theta_0) = \psi(\theta^z) + \frac{d\psi(\theta^z)}{d\theta^z}\,\delta\theta_0 + \ldots$

We recall that the angular momentum operator along $z$ is:

$J^z = -i\frac{d}{d\theta^z}$

Substituting:

$\psi(\theta^z + \delta\theta_0) = \psi(\theta^z) + iJ^z\,\psi(\theta^z)\,\delta\theta_0 = (1 + iJ^z\,\delta\theta_0)\,\psi(\theta^z)$

This is the infinitesimal rotation: $(1 + iJ^z \delta\theta_0)$ acting on the wavefunction.

Building a Finite Rotation#

To rotate by a finite angle $\theta_0$ , we can compose $N$ infinitesimal rotations, each of size $\delta\theta_0 = \theta_0/N$ :

$\psi(\theta^z + \theta_0) = \lim_{N\to\infty}\left(1 + iJ^z\frac{\theta_0}{N}\right)^N \psi(\theta^z) = e^{iJ^z\theta_0}\,\psi(\theta^z)$

This is the exponential map — the bridge from the Lie algebra (the generator $J^z$ ) to the Lie group (the finite rotation $e^{iJ^z\theta_0}$ ).

Sign Convention#

For a general rotation parameterized by the angle vector $\boldsymbol{\theta} = (\theta^x, \theta^y, \theta^z)$ , the rotation operator is conventionally written as:

$R(\boldsymbol{\theta}) = \exp(-i\,\mathbf{J}\cdot\boldsymbol{\theta})$

where $\mathbf{J} = (J^x, J^y, J^z)$ .

Wait — we just derived $e^{+iJ^z\theta_0}$ , and now there’s a minus sign? This is not a mistake; it’s a choice of convention. The sign depends on whether you think of the rotation as acting on the coordinates (passive) or on the physical system (active):

Active transformation (rotating the physical state): The operator acting on a state $|\psi\rangle$ in the Hilbert space is $e^{-i\mathbf{J}\cdot\boldsymbol{\theta}}$ .
Passive transformation (rotating the coordinate axes): The wavefunction transforms as $\psi \to e^{+i\mathbf{J}\cdot\boldsymbol{\theta}}\psi$ .

The two are inverses of each other, related by $\boldsymbol{\theta} \to -\boldsymbol{\theta}$ . Most QFT textbooks (Schwartz, Peskin & Schroeder, Weinberg) use the active convention with the minus sign. Lancaster & Blundell also use the minus sign. Pick one, state it clearly, and be consistent throughout.

Connection to What Comes Next#

Everything we’ve built here — generators, commutation relations, representations, the exponential map — forms the language we need for the Lorentz group.

In the next post, we’ll define the Lorentz group $O(3,1)$ as the group of transformations preserving the spacetime interval, extract its generators (three rotations $J^i$ and three boosts $K^i$ ), and discover that these six generators satisfy a very specific algebra:

$\Lambda = \exp\!\left(-\tfrac{i}{2}\,\omega_{\mu\nu}\,J^{\mu\nu}\right)$

The fact that this algebra is non-abelian — and non-compact — will have profound consequences for representation theory, ultimately leading us to spinors, the Dirac equation, and the classification of all particles by mass and spin.

Next post: Part II: The Lorentz Group — Definition, Structure, and Representations

This post is based on my own self-study notes that I created in 2022 in order to get a deeper understanding of all of this.