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Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra, with a chosen normalized invariant bilinear form on a Lie algebra $B$. Its affine current algebra is the central extension of a Lie algebra $\widehat{L\mathfrak{g}}=(\mathfrak{g}\otimes \mathbb{C}[t,t^{-1}])\oplus \mathbb{C}K$ of the loop algebra, withInvariance of $B$ makes the extra term a Lie algebra two-cocycle, hence preserves the Jacobi identity. In a representation where $K=kI$, the scalar $k$ is the level. The central term vanishes for zero modes, which recover the original Lie algebra, but modifies the commutator of opposite nonzero modes.^[ref]

A restricted module has $J_{n}v=0$ for all sufficiently large $n$ for each vector $v$; this makes the Sugawara construction finite on each vector after normal ordering. Nontrivial unitary integrable positive-energy highest-weight representations have positive integer levels in the basic normalization. The zero-level integrable highest-weight representation is trivial. Adjoining the degree derivation gives the further extension with $[d,x\otimes t^n]=nx\otimes t^n$.

An algebra over a field $k$ is a vector space $A$ with a bilinear multiplication $A\times A\to A$. Bilinearity means $(ax+by)z=a(xz)+b(yz)$ and $z(ax+by)=a(zx)+b(zy)$ for $a,b\in k$. Associativity, commutativity, and a multiplicative identity are additional properties, not consequences of this definition.

The polynomial ring $k[X]$ and the matrix algebra $M_n(k)$ are associative unital examples; the latter is generally noncommutative. A Lie algebra is another kind of algebra over a field, with its bracket as multiplication, obeying antisymmetry and the Jacobi identity instead of associativity. Specifying which axioms are intended makes statements about all algebras precise.

The relation $[x,y]=-[y,x]$ for a bilinear map proposed as a Lie bracket. An Alternating Lie bracket always satisfies it. Conversely, setting $y=x$ gives $2[x,x]=0$, so antisymmetry implies alternation over a field of characteristic different from two.

In characteristic two, the sign disappears and this relation just says $[x,y]=[y,x]$. It does not force $[x,x]=0$, even when the Jacobi identity holds. For example, on a two-dimensional vector space with basis $u,v$, set $[u,u]=v$ and every other basis bracket to zero, then extend bilinearly. Every nested bracket vanishes, so the Jacobi identity holds, and the product is symmetric. Nevertheless it is not a Lie algebra, because $[u,u]\ne 0$.

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 first Bianchi identity follows by applying cyclic curvature commutators to the gradient of an arbitrary scalar field: its symmetric Hessian makes the third-derivative terms cancel, leaving $R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d=0$. For the second Bianchi identity, write the vector-bundle derivative as $D_a={\partial }_a+{\Gamma }_a$, so $[D_b,D_c]={\mathcal{R}}_{bc}$. The operator Jacobi identity gives the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ equal to zero. Restoring the connection terms on the two curvature-form indices gives precisely ${\nabla }_a{R}_{bcd}{}^e+{\nabla }_b{R}_{cad}{}^e+{\nabla }_c{R}_{abd}{}^e=0$: the additional terms cancel because the connection is torsion-free and curvature is antisymmetric in its first pair.

Specify the displayed rule on left-invariant vector fields and extend by the Leibniz rule to an affine connection on the Lie group. In a left-invariant vector field frame $L_i$, the extension is ${\nabla }_{L_i}(z^j{L}_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the Leibniz rule unless $\lambda =1$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesfor left-invariant arguments, where $B$ is the Killing form and $\mathrm{Ric}(X,Y)=\mathrm{tr}(Z\mapsto R(Z,X)Y)$. These identities determine the tensor fields everywhere. The endpoints $\lambda =0,1$ have zero curvature, with respectively left- and right-invariant parallel frames, and opposite torsion tensors. The midpoint $\lambda =1/2$ is torsion-free but need not have zero curvature.

The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant pseudo-Riemannian metric whose Levi-Civita connection is ${\nabla }^0$.

An adjoint-invariant symmetric bilinear form $B$ on a Lie algebra makes the displayed tensor alternating. If $B$ is invariant under the full group Adjoint representation, its left-translated three-form is also right-invariant. On left-invariant fields, the exterior derivative is $d\eta (W,X,Y,Z)=-2B(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])=0$ by the Jacobi identity. The Killing form supplies such a $B$ on every semisimple Lie group. An overall sign change in the lowered structure-constant convention negates the form but not its invariance or closedness.

The centralizer of a Lie subalgebra $\mathfrak{a}$ is the intersection of the centralizer of an element of a Lie algebra as that element ranges over $\mathfrak{a}$. It is a vector subspace and a Lie subalgebra: for $x,y$ commuting with $\mathfrak{a}$, the Jacobi identity gives $[[x,y],a]=[x,[y,a]]-[y,[x,a]]=0$. It is contained in the normalizer of a Lie subalgebra. If $\mathfrak{a}$ is an abelian Lie algebra, it is contained in its own centralizer.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

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.

The Lie derivative of a tensor field satisfies $[{\mathcal{L}}_X,{\mathcal{L}}_Y]={\mathcal{L}}_{[X,Y]}$. On smooth functions this is the definition of the Lie bracket of vector fields; on vector fields it follows from the Jacobi identity. The Leibniz rule and contraction compatibility extend the equality to tensor fields. The cyclic double-commutator identity also follows directly from associativity of operator composition.

The Jacobi identity for a Poisson bivector reduces to the displayed cyclic identity in every local chart. Necessity follows by applying the bracket to coordinate functions. Sufficiency follows because the second-derivative terms in an arbitrary Jacobiator cancel by antisymmetry, leaving this coefficient multiplying first derivatives of the three functions.

The vector space ${\mathbb{R}}^3$ with the cross product as Lie bracket, $[x,y]=x\times y$. Bilinearity and $x\times x=0$ give an Alternating Lie bracket; the vector triple product identity proves the Jacobi identity because the three expressions $x\times (y\times z)=y(x\cdot z)-z(x\cdot y)$ cancel cyclically, using the symmetric dot product.

Its relation to the SO(3) Lie algebra is explicit. For a vector $x=(x_1,x_2,x_3)$, define the skew-symmetric matrixThen $[\hat{x},\hat{y}]=\widehat{x\times y}$: acting on $v$, the left side is $x\times (y\times v)-y\times (x\times v)=(x\times y)\times v$. Every real $3\times 3$ skew-symmetric matrix occurs uniquely, so this hat map is a Lie algebra isomorphism. With the standard basis vectors $e_1,e_2,e_3$,and their matrix commutators are $[X,Y]=Z$, $[Y,Z]=X$, $[Z,X]=Y$. With active rotation matrices acting on column vectors, $\exp (t\hat{x})$ rotates through angle $t\Vert x\Vert$ about the oriented axis $x$ when $x\ne 0$. In particular $\exp (tZ)e_1=(\cos t,\sin t,0)$, fixing the positive-angle convention used in the small rotation commutator.

The map $x\mapsto -(i/2)x\cdot \sigma$ is also a Lie algebra isomorphism to the compact real SU(2) Lie algebra, using the Pauli matrix commutator identity. These two identifications concern Lie algebras; they do not identify the corresponding Lie groups globally.

The derived algebra is the linear span of all Lie brackets $[x,y]$ in a Lie algebra. The Jacobi identity makes it an ideal of a Lie algebra. Its quotient is the largest abelian Lie algebra quotient of $\mathfrak{g}$.

The Lie algebra over a commutative ring $R$ generated by a set $X$ subject only to bilinearity, alternation and the Jacobi identity. Every map from $X$ to the underlying set of an $R$-Lie algebra extends uniquely to a Lie algebra homomorphism. It can be constructed from bracketed words by quotienting by those identities; elements are represented by Lie polynomials.

The Jacobi identity for gauge covariant derivatives gives $D_{[\mu }{F}_{\nu \rho ]}=0$. In four dimensions this is $D_{\mu }{\tilde{F}}^{\mu \nu }=0$, with ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. It holds independently of the Yang-Mills equations.

The endomorphisms of a vector space form a Lie algebra under the commutator $[A,B]=AB-BA$. It is the Lie algebra of the general linear group; the Jacobi identity follows by expanding associative products.

A Gerstenhaber algebra has an associative product making it a graded commutative algebra and a degree-minus-one graded Lie bracket. In the left convention, $[u,vw]=[u,v]w+(-1{)}^{(|u|-1)|v|}v[u,w]$. The shifted degrees $|u|-1$ determine the signs of antisymmetry and the Jacobi identity. Hochschild cohomology and the exterior algebra of polynomial derivations are examples.

Over a field of characteristic different from two, the Jacobi identity of a Lie superalgebra includes parity signs. In particular, for odd $c$ and even $\varphi$ it gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$.