Chapter 2: Lie Brackets and Commutators

The matrix commutator and its connection to physics

The Matrix Commutator

The most basic example of a Lie bracket is the matrix commutator.

Definition: Matrix Commutator

For $n \times n$ matrices $A$ and $B$, the commutator is defined by

$$[A, B] \triangleq AB - BA.$$

Theorem: The Commutator Satisfies the Lie Bracket Axioms

The matrix commutator satisfies the Lie bracket axioms (alternation and the Jacobi identity).

Proof

Alternation: $[A, A] = AA - AA = 0$.　✓

Jacobi identity: we verify by direct computation. First, expand $[A, [B, C]]$.

\begin{align*} [A, [B, C]] &= [A, BC - CB] \\ &= A(BC - CB) - (BC - CB)A \\ &= ABC - ACB - BCA + CBA \end{align*}

Similarly, expand $[B, [C, A]]$.

\begin{align*} [B, [C, A]] &= [B, CA - AC] \\ &= B(CA - AC) - (CA - AC)B \\ &= BCA - BAC - CAB + ACB \end{align*}

And expand $[C, [A, B]]$.

\begin{align*} [C, [A, B]] &= [C, AB - BA] \\ &= C(AB - BA) - (AB - BA)C \\ &= CAB - CBA - ABC + BAC \end{align*}

Adding the three.

\begin{align*} &[A, [B, C]] + [B, [C, A]] + [C, [A, B]] \\ &= (ABC - ACB - BCA + CBA) \\ &\quad + (BCA - BAC - CAB + ACB) \\ &\quad + (CAB - CBA - ABC + BAC) \\ &= ABC - ABC + BCA - BCA + CAB - CAB \\ &\quad - ACB + ACB - BAC + BAC - CBA + CBA \\ &= 0 \end{align*}

Every term cancels in pairs, giving $0$.　□

Meaning of the Commutator

The commutator $[A, B] = AB - BA$ measures the "degree of non-commutativity" between $A$ and $B$.

  • $[A, B] = 0$ ⟺ $AB = BA$ ($A$ and $B$ commute)
  • $[A, B] \neq 0$ ⟺ $A$ and $B$ do not commute

Geometric meaning: the residual gap of a round trip

Regarding $A$ and $B$ as generators of infinitesimal transformations and writing them as $e^{tA} \approx I + tA$ for a small parameter $t$, one has

$$e^{tA} \, e^{tB} \, e^{-tA} \, e^{-tB} = I + t^2 [A, B] + O(t^3).$$

That is, performing the round trip "advance by $A$ → advance by $B$ → retreat by $A$ → retreat by $B$" leaves a residual gap of order $t^2$ equal to the commutator $[A,B]$. The Lie bracket $[X, Y]$ of vector fields $X, Y$ on a manifold is defined through the same geometric picture, and the matrix commutator is its special case.

The commutator is not merely an algebraic expression but a geometric quantity that measures the non-commutativity of infinitesimal transformations. This is the underlying motivation for using Lie algebras in physics and geometry.

Example: 2×2 matrices

Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Then

$$AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ $$[A, B] = AB - BA = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$$

Since $[A, B] \neq 0$, $A$ and $B$ do not commute.

Example: diagonal matrices commute

Products of diagonal matrices commute, so the commutator of any two diagonal matrices is always $0$.

Connection to Physics

In quantum mechanics, the commutator plays a fundamental role.

Commutation relations in quantum mechanics

The position operator $\hat{x}$ and the momentum operator $\hat{p}$ satisfy the canonical commutation relation:

$$[\hat{x}, \hat{p}] = i\hbar.$$

Here $\hbar$ is Planck's constant divided by $2\pi$ (the reduced Planck constant).

This commutation relation is at the root of the uncertainty principle, and it expresses the essential difference between quantum and classical mechanics.

Commutation relations of angular momentum

The components $L_x$, $L_y$, $L_z$ of angular momentum satisfy

$$[L_x, L_y] = i\hbar L_z, \quad [L_y, L_z] = i\hbar L_x, \quad [L_z, L_x] = i\hbar L_y.$$

These correspond to the commutation relations of $\mathfrak{so}(3)$, the Lie algebra of 3-dimensional rotations.

The Poisson Bracket

Classical mechanics has a structure analogous to the Lie bracket.

What is phase space?

To describe the motion of a classical particle completely, one needs both the position $q$ and the momentum $p$. The space with these as coordinates is called phase space.

  • One-dimensional particle: phase space is two-dimensional, $(q, p)$.
  • Three-dimensional particle: phase space is six-dimensional, $(q_1, q_2, q_3, p_1, p_2, p_3)$.

A function on phase space $f(q, p)$ takes position and momentum as arguments. For example, the energy $H(q, p) = \dfrac{p^2}{2m} + V(q)$ is a function on phase space.

Definition: Poisson Bracket

For functions $f$ and $g$ on phase space, the Poisson bracket is defined by

$$\{f, g\} = \displaystyle\sum_i \left( \dfrac{\partial f}{\partial q_i}\dfrac{\partial g}{\partial p_i} - \dfrac{\partial f}{\partial p_i}\dfrac{\partial g}{\partial q_i} \right).$$

The Poisson bracket also satisfies the Lie bracket axioms. Indeed, under quantization $\{f, g\} \to \dfrac{1}{i\hbar}[\hat{f}, \hat{g}]$, the classical Poisson bracket corresponds to the quantum mechanical commutator.

Basic Properties of the Commutator

Proposition: Properties of the Commutator

  • Bilinearity: $[aA + bB, C] = a[A, C] + b[B, C]$.
  • Anticommutativity: $[A, B] = -[B, A]$.
  • Jacobi identity: $[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0$.
  • Product rule (Leibniz rule): $[A, BC] = [A, B]C + B[A, C]$.

Proof

Bilinearity (in the first argument):

\begin{align*} [aA + bB, C] &= (aA + bB)C - C(aA + bB) \\ &= aAC + bBC - aCA - bCB \\ &= a(AC - CA) + b(BC - CB) \\ &= a[A, C] + b[B, C]. \end{align*}

The argument for the second argument is analogous.

Anticommutativity:

\begin{align*} [A, B] &= AB - BA \\ &= -(BA - AB) \\ &= -[B, A]. \end{align*}

Jacobi identity: see the proof above.

Product rule:

\begin{align*} [A, BC] &= A(BC) - (BC)A \\ &= ABC - BCA. \end{align*}

On the other hand,

\begin{align*} [A, B]C + B[A, C] &= (AB - BA)C + B(AC - CA) \\ &= ABC - BAC + BAC - BCA \\ &= ABC - BCA. \end{align*}

The two are equal.　□

The last item, the product rule, shows that $[A, \cdot]$ behaves like a derivation.

Summary

  • The matrix commutator $[A, B] = AB - BA$ is the canonical example of a Lie bracket.
  • The commutator measures the "degree of non-commutativity".
  • The commutation relations of quantum mechanics carry a Lie algebra structure.
  • The Poisson bracket shares the same algebraic structure and, under quantization, corresponds to the commutator.

Further Reading

  • Chapter 3: Matrix Lie Algebras — See the commutator at work in concrete examples like $\mathfrak{gl}(n)$, $\mathfrak{sl}(n)$, $\mathfrak{so}(n)$, $\mathfrak{u}(n)$.
  • Chapter 4: Subalgebras and Ideals — Subspaces closed under the bracket (subalgebras) and the stronger notion of ideals.
  • Systematic treatment of structure constants and the Jacobi identity: Humphreys, Introduction to Lie Algebras and Representation Theory, Chapters 1–2 (a standard introduction).
  • Physics applications (angular momentum, canonical commutation): Sakurai, Modern Quantum Mechanics, Chapters 1–3.

Frequently Asked Questions

What is the Lie bracket (commutator)?

The Lie bracket $[X,Y]$ is the fundamental operation of a Lie algebra. For matrix Lie algebras, the commutator $[A,B]=AB-BA$ realizes the Lie bracket. It measures the "non-commutativity" of the matrix product; $[A,B]=0$ means that $A$ and $B$ commute (and are simultaneously diagonalizable). In quantum mechanics, $[\hat{x},\hat{p}]=i\hbar$ is the mathematical statement of the position-momentum uncertainty principle.

What does bilinearity of the Lie bracket mean?

It means that the Lie bracket $[X,Y]$ is linear in each argument: $[aX+bY,Z]=a[X,Z]+b[Y,Z]$ and $[X,aY+bZ]=a[X,Y]+b[X,Z]$ for $a,b\in\mathbb{F}$. Consequently, the entire bracket of the algebra is determined by specifying the brackets of a basis $\{e_1,\ldots,e_n\}$: $[e_i,e_j]=\sum_k f^k_{ij}e_k$, where $f^k_{ij}$ are the structure constants.

What are structure constants?

For a basis $\{e_i\}$ of a Lie algebra, the constants $f^k_{ij}$ defined by $[e_i,e_j]=\sum_k f^k_{ij}e_k$ are called structure constants. Antisymmetry gives $f^k_{ij}=-f^k_{ji}$, and the Jacobi identity yields $\sum_l(f^l_{ij}f^m_{lk}+f^l_{jk}f^m_{li}+f^l_{ki}f^m_{lj})=0$. For $\mathfrak{su}(2)$, the structure constants are the Levi-Civita symbol $\varepsilon_{ijk}$.