Commutator ($[A,B]$)

The commutator of two elements of an associative algebra over a field isIt measures how much the order of multiplication matters: $[A,B]=0$ exactly when $AB=BA$. It is a bilinear map, it is alternating since $AA-AA=0$, and associativity makes it satisfy the Jacobi identity. Hence every associative algebra becomes a Lie algebra on the same vector space when equipped with this product. This holds in every characteristic of a field.

For square matrices, this is the matrix commutator, and it is defined even for singular matrices. The group commutator of invertible matrices is instead $AB{A}^{-1}{B}^{-1}$: it equals the identity element when the matrices commute, whereas $AB-BA$ equals the zero matrix. The Lie bracket from local group commutators explains how differentiation connects the two constructions.

Over a field of characteristic zero, on the degree-$n$ homogeneous polynomials use $e_k=x^{n-k}{y}^k$. Then $P{e}_k=k{e}_{k-1}$ and $Q{e}_k=(n-k)e_{k+1}$; both maps are nilpotent of index $n+1$. Their commutator is diagonal in this basis with distinct eigenvalues $n,n-2,\dots ,-n$. Thus the raising and lowering operators are not diagonalizable for $n\ge 1$, while their commutator is.

If two linear maps commute and $Av=\lambda v$, then $A(Bv)=B(Av)=\lambda Bv$. Hence $B$ preserves each eigenspace of $A$. On a one-dimensional eigenspace, $Bv$ is a scalar multiple of $v$, so each eigenvector for a simple eigenvalue of $A$ is also an eigenvector of $B$. For repeated eigenvalues the invariant eigenspace can have higher dimension; arbitrary vectors in it need not individually be eigenvectors of $B$.

Suppose linear operators $A,B$ obey $[A,B]=cB$, where $c$ is a scalar, on a common invariant operator domain. The commutator product rule givesFor $n=0$ both sides vanish. This identity explains how powers of a creation operator shift the number operator eigenvalue.

For linear operators on a common invariant domain, $[A,BC]=[A,B]C+B[A,C]$. The identity follows by adding and subtracting $BAC$ in the commutator.

If $D=[A,B]$ commutes with $A$, then $[A^n,B]=nD{A}^{n-1}$. A convergent power series therefore gives $[F(A),B]=D{F}^{\prime }(A)$ on a common domain where termwise operations are justified. For unbounded linear operators, the domain condition is essential.

Over a field $k$, the trace-zero $2\times 2$ matrices form the special linear Lie algebra ${\mathfrak{sl}}_2(k)$. The matrix trace of a commutator is zero, so this is a Lie subalgebra of the general linear Lie algebra. A basis isIn fact $\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=aH+bE+cF$, uniquely. The individual products areSubtracting in the stated order givesFor characteristic different from two, this is the standard sl2 triple; see Etingof, Section 11.4. The products remain valid in characteristic two, but there $H=I$ is central and $2E=2F=0$, so the familiar characteristic zero structure changes.

The following table gives all nine basis brackets. Each entry is $[\text{row},\text{column}]$; the diagonal vanishes by alternation, and exchanging row and column reverses the sign.

Bilinearity now computes any commutator, for exampleThe three structure constants of a Lie algebra relations encode all of this multiplication. They also show why the sl2 triple is useful in Lie algebra representations: the actions of $E$ and $F$ move between eigenvalues of the action of $H$ in steps of two.

For real-linear operators $[A,B]=I$ and a nonzero vector $y$ with $Ay=0$, induction gives $[A,B^i]=i{B}^{i-1}$, hence the displayed formula. Applying $A^n$ to a nontrivial finite dependence with highest term $c_n{B}^{n}y$ leaves $n!c_{n}y$, a contradiction. Thus $y,By,B^{2}y,\dots$ are linearly independent, and their span is invariant under both operators. Characteristic zero is essential to the factorial argument.

For bounded linear operators, or formally in an associative algebra,Differentiate the two exponentials using the product rule: the result is $e^{\lambda A}[A,B]e^{-\lambda A}$. Induction gives the $n$th derivative $e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$, and the Taylor series at zero proves the formula. For bounded operators convergence follows from $\Vert {\mathrm{ad}}_A^n(B)\Vert \le (2\Vert A\Vert {)}^n\Vert B\Vert$. For unbounded quantum operators, use a common invariant domain and justify the series there, or use the finite-dimensional system of commutators when their span closes.

For finite-dimensional complex linear operators with $C=[A,B]$ commuting with both $A$ and $B$, differentiating $e^{-vB}A{e}^{vB}$ gives the constant derivative $C$. Hence $e^{-B}A{e}^B=A+C$, and conjugating the matrix exponential gives $e^{-B}{e}^{-A}{e}^B=e^{-A-C}$. Since $[A,C]=0$, multiplication by $e^A$ proves the identity. The commutator order determines the minus sign.

The matrix trace of $[A,B]=AB-BA$ is zero, since $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$. This obstructs a matrix with nonzero trace from being a single commutator.

Expand both sides using the definition of a commutator: the middle terms $BAC$ cancel, leaving $ABC-BCA$. Thus commutation with a fixed operator is a derivation of an algebra. In particular, $[A,[B,C]]=[[A,B],C]+[B,[A,C]]$, which gives a short way to derive a Lie algebra representation from generator actions on an underlying algebra.