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Topic 6 of 8

Lie Algebras

The Lie algebra $\mathfrak{g}$ is the tangent space to the Lie group $G$ at the identity. It is a vector space equipped with a bracket $[\cdot,\cdot]$ that captures the non-commutativity of the group. For physicists, the generators are the observables — charge, angular momentum, color — and their commutation relations determine the entire representation theory.

Key Concepts

Lie Algebra

\mathfrak{g}

The tangent space

T_e G

at the identity of a Lie group, equipped with the Lie bracket

[X,Y]

. As a vector space,

\dim\mathfrak{g} = \dim G

Generator

T^a = -i\,\partial g / \partial\alpha_a|_{\alpha=0}

. For unitary groups, generators are Hermitian. The

\{T^a\}

are a basis for

\mathfrak{g}

Structure Constants

f^{abc}

defined by

[T^a, T^b] = if^{abc}T^c

. Totally antisymmetric:

f^{abc}=-f^{bac}

. Completely characterize the Lie algebra.

Rank

The dimension of the maximal abelian subalgebra (Cartan subalgebra). For

\mathfrak{su}(n)

: rank

= n-1

. Equals the number of simultaneously diagonalizable generators.

Killing Form

B(X,Y) = \text{Tr}(\text{ad}_X \circ \text{ad}_Y)

. Negative definite for compact semisimple Lie algebras. Used to raise/lower structure constant indices.

Key Equations

Lie Algebra Commutation Relations

[T^a,\, T^b] = i\,f^{abc}\,T^c

The structure constants

f^{abc}

are antisymmetric and satisfy the Jacobi identity.

Jacobi Identity

[A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0

Automatically satisfied in any Lie algebra. Corresponds to the associativity constraint.

Normalization (Fundamental Rep)

\mathrm{Tr}(T^a T^b) = \tfrac{1}{2}\,\delta^{ab}

Standard normalization for generators in the fundamental representation of

SU(n)

Worked Example

Structure Constants of $\mathfrak{su}(2)$

Problem

The generators of $SU(2)$ are $T^i = \sigma^i/2$ . Using the known Pauli matrix commutator $[\sigma^i, \sigma^j] = 2i\varepsilon^{ijk}\sigma^k$ , find $f^{123}$ .

Solution

$[T^1, T^2] = [\sigma^1/2,\, \sigma^2/2] = \tfrac{1}{4}[\sigma^1,\sigma^2] = \tfrac{1}{4}(2i\sigma^3) = \tfrac{i\sigma^3}{2} = iT^3$ .

From the definition $[T^a,T^b] = if^{abc}T^c$ : $[T^1,T^2] = if^{123}T^3$ .

Comparing: $iT^3 = if^{123}T^3$ , so $f^{123} = 1$ .

More generally, $f^{ijk} = \varepsilon^{ijk}$ for $\mathfrak{su}(2)$ (the Levi-Civita symbol).

Answer

f^{123} = 1

Practice

Exercises

7 problems

1 of 7

For $SU(2)$ generators with $f^{ijk}=\varepsilon^{ijk}$ , what is $f^{231}$ ?

Your answer

2 of 7

The rank of $\mathfrak{su}(n)$ is $n-1$ . What is the rank of $\mathfrak{su}(5)$ ?

Your answer

3 of 7

For $T^i = \sigma^i/2$ , the normalization gives $\mathrm{Tr}(T^a T^b) = \frac{1}{2}\delta^{ab}$ . What is $2\,\mathrm{Tr}((T^1)^2)$ ?

Unlock Exercise 3