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The left Schouten-Nijenhuis bracket on polynomial polyvector fields is the degree-minus-one Gerstenhaber bracket extending the commutator of derivations and $[D,f]=D(f)$ by graded antisymmetry and the left graded Leibniz rule. For $\Pi ={\partial }_X\wedge {\partial }_Y$, this convention gives $[h\Pi ,f]=h(f_Y{\partial }_X-f_X{\partial }_Y)$ and $[F{\partial }_X+G{\partial }_Y,h\Pi ]=(D(h)-h(F_X+G_Y))\Pi$. A right insertion convention reverses the degree-two-with-function formula.

The symmetric second-order formula, also called Strang splitting, isfor bounded operators, with the constant controlled by nested commutators.

In the sl2 Lie algebra, if $Xv=0$ and $Hv=mv$, then the commutator relations give $H{Y}^{j}v=(m-2j)Y^{j}v$ and $X{Y}^{k+1}v=(k+1)(m-k)Y^{k}v$. Induct using $XY=YX+H$. This controls the raising operator on a string formed by the lowering operator and yields the classification of finite-dimensional sl2 representations.

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.

The special linear Lie algebra ${\mathfrak{sl}}_n$ consists of the trace-zero $n\times n$ matrices with the commutator bracket.

The real Lie algebra of the special unitary group consists of traceless skew-Hermitian matrices. Differentiating unitarity and determinant one gives the two constraints; exponentiating any such matrix proves their sufficiency. The commutator is again skew-Hermitian and traceless. Its real dimension is $n^2-1$, and its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$.

For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

A square matrix is strictly upper triangular if every entry on or below its diagonal is zero. A product of $n$ such $n$-by-$n$ matrices vanishes, so they form a nilpotent Lie algebra under the commutator. Multiplication by an upper triangular matrix preserves strict upper triangularity and zero trace.

For bounded matrices, differentiating the symmetrized product gives $F^{\prime }-(A+B)F=\frac{1}{2}([e^{tB},A]e^{tA}+[e^{tA},B]e^{tB})$. The variation-of-constants formula turns this into an integral expression for $F-e^{t(A+B)}$. This retains the noncommutativity of the factors explicitly; cancellation of their leading opposite commutators explains why the averaged product is more accurate than either ordering alone.

For a nonsingular skew-symmetric unitary $J$, if each skew-Hermitian control generator obeys $X^{T}J+JX=0$, then so do their real linear combinations and commutators. Differentiating $U^{T}JU$ shows that it stays equal to $J$. The reachable group is therefore contained in a unitary conjugate of the compact symplectic group. This is preservation of an alternating bilinear form, not commutation with $J$.

For traceless generators, the tensor product of Lie algebra representations has generators $t_a\otimes I+I\otimes s_a$. The trace of their summed squares is the sum of the two factor Casimir traces; the cross term vanishes because $\mathrm{tr}{t}_a=\mathrm{tr}{s}_a=0$. Decomposition into irreducible representations, with multiplicities counted, gives the displayed identity. Without tracelessness an extra term $2{\sum }_a\mathrm{tr}{t}_a\mathrm{tr}{s}_a$ is present. A one-dimensional abelian representation of charge one tensored with itself has Casimir four, not the sum two; semisimple Lie algebras avoid this problem because every generator is a sum of commutators.

For traceless matrices $T,S$ transforming by conjugation under SU(3), the trace $\mathrm{tr}(TS)$ supplies a singlet, the commutator $[T,S]$ an antisymmetric octet, and the traceless anticommutator a symmetric octet. Symmetrizing the two upper and two lower indices and removing traces supplies the 27. Contracting an antisymmetric lower pair with the invariant alternating tensor and symmetrizing the remaining three upper indices supplies the 10, and the conjugate construction supplies the conjugate 10. These nonzero equivariant maps exhaust the dimensions: the symmetric part has $36=1+8+27$, the antisymmetric part $28=8+10+10$.

A timelike basis polarization in covariant photon quantization has positive Minkowski metric squared length, conventionally ${ϵ}^{0\mu }=(1,0)$ with signature $(+,-,-,-)$. It is distinct from the spatial longitudinal polarization ${ϵ}^{3\mu }=(0,\hat{\mathbf{p}})$. The extra minus sign in the covariant oscillator commutator makes its one-photon state a negative-norm photon state. Neither this temporal polarization nor an isolated longitudinal photon survives the Gupta-Bleuler null-state quotient.

Every commutator of square matrices is traceless by cyclicity of the trace. Conversely, off-diagonal matrix units and differences of diagonal units are themselves commutators and span all traceless matrices. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.

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.

Let $H_0=\mathrm{diag}(-{\omega }_1,0,-{\omega }_2)$ and $H_1=d_1(E_{12}+E_{21})+d_2(E_{23}+E_{32})$, with nonzero $d_j$, distinct positive ${\omega }_j$. The squared adjoint action of $-i{H}_0$ has different eigenvalues $-{\omega }_1^2,-{\omega }_2^2$ on the two adjacent skew-Hermitian couplings. Linear combinations of $-i{H}_1$ and its double commutator with $-i{H}_0$ isolate both couplings. Further commutators give their real antisymmetric partners, two traceless diagonal generators, and both $1$–$3$ generators, hence the special unitary Lie algebra $\mathfrak{su}(3)$. The drift has nonzero trace; subtracting its traceless part supplies $iI$, giving the Lie algebra of the unitary group $\mathfrak{u}(3)$ and unitary operator controllability under the usual freely timed real-control assumptions.

For a free real scalar field, split $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$ into annihilation and creation parts. Moving the creation operators left gives $\varphi (x)\varphi (y)=:\varphi (x)\varphi (y):+[{\varphi }^{(+)}(x),{\varphi }^{(-)}(y)]$. The commutator is the Wightman function. Choosing either time order gives the displayed identity with the Feynman propagator. It is an identity of operator-valued distributions; coincident-point products need a suitable prescription.