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A Lie bracket is alternating when $[x,x]=0$ for every $x$. This condition is imposed in every characteristic of a field. Bilinearity gives $0=[x+y,x+y]=[x,y]+[y,x]$, hence antisymmetry of a Lie bracket. Requiring alternation directly keeps the definition correct in characteristic two.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

The Jacobi identity is the compatibility condition on a Lie bracketTogether with an Alternating Lie bracket, it defines a Lie algebra. Unlike associativity, it relates three cyclic bracketings rather than equating two bracketings. Equivalently, using antisymmetry of a Lie bracket,Thus the linear map ${\mathrm{ad}}_x:y\mapsto [x,y]$ is a derivation of a Lie algebra: acting on a bracket is the sum of acting on each argument. This makes infinitesimal symmetries act consistently on each other's products. It also gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, the identity that makes the Adjoint representation of a Lie algebra a Lie algebra representation.

For the commutator in an associative algebra, the cancellation is directly visible:Each word occurs once with each sign. This proves the Jacobi identity without a restriction on the characteristic of a field.

For a computed check in the sl2 matrix commutator example,Here the last two terms are individually nonzero over characteristic zero; the identity records their cancellation. In the cross-product Lie algebra, the same cyclic cancellation is the vector triple product identity.

The historical connection is unusually concrete: Lie and Engel's volume I expands nested commutators of differential operators into twelve terms and cancels them cyclically, in Chapter 5, §26, equation (7). See Lie and Engel, §26, in Merker's translation, pp. 105–106.

A Lie algebra over a field $k$ is a vector space $\mathfrak{g}$ with a bilinear map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that is alternating and satisfies the Jacobi identity. Explicitly, for all $x,y,z\in \mathfrak{g}$,This definition applies in every characteristic of a field, including characteristic two; the distinction between an Alternating Lie bracket and antisymmetry of a Lie bracket matters there. See Pavel Etingof, Lie Groups and Lie Algebras, Definition 8.4 and Remark 8.5.

The motivating question is how to describe very small symmetries. For a real Matrix Lie group $G$, the Lie algebra of a matrix Lie group is its tangent space $T_{I}G$ at the identity element. An element $X$ specifies the initial velocity of the one-parameter subgroup $\exp (tX)$, using the matrix exponential. The Lie bracket records the first nonzero effect of changing the order of two small transformations. Even when first-order motions cancel, a second-order motion can remain: the small rotation commutator below leaves a rotation about a third axis.

This turns local symmetry questions into computations with linear maps. A basis and its structure constants of a Lie algebra determine every Lie bracket by bilinearity. For example, three matrices suffice for the sl2 matrix commutator example, while the familiar cross product describes the SO(3) Lie algebra. These descriptions help calculate rotations in three dimensions and build Lie algebra representations; the same bracket structure underlies angular momentum commutation relations. The striking feature is that a nonlinear family of motions has a local description by a vector space and one product. Global information still needs care: the Lie algebra alone does not distinguish the circle group from the additive real line.

Historically, the connection with differential equations is visible in Sophus Lie's Theorie der Transformationsgruppen, volume I (1888), prepared with Friedrich Engel. Chapter 6 interprets first-order differential operators as infinitesimal transformations, linking their action on functions to the transformations those functions admit. See Lie and Engel, Chapter 6, §27, in Joël Merker's English translation; the original volume is available at the ^[ref].

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.