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A Lie algebra is abelian when every Lie bracket vanishes.

A scalar field transforming in the Adjoint representation of a Lie algebra transforms by a Lie bracket with the infinitesimal gauge parameter.

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.

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 $[u,u]\ne 0$.

Lower the output index of the structure constants of a Lie algebra with its Killing form. The Lie bracket makes the result antisymmetric in the first two indices, while invariance gives $\kappa ([t_a,t_b],t_c)=\kappa (t_a,[t_b,t_c])$, making it antisymmetric in the last two indices. These two adjacent transpositions imply total antisymmetry. Nondegeneracy of the Killing form is unnecessary.

// Target: lie-theory.bigb

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.

For an almost complex structure, closure of $T^{1,0}$ under the Lie bracket is equivalent to the absence of a $(0,2)$ component in $d\alpha$ for every $(1,0)$-form $\alpha$. Conjugation exchanges the bracket condition with closure of $T^{0,1}$. For sections $U,V$ of $T^{0,1}$, the exterior derivative identity gives $d\alpha (U,V)=-\alpha ([U,V])$. Its vanishing for all $(1,0)$-forms is exactly $[U,V]\in T^{0,1}$.

A central ideal is a linear subspace of the center of a Lie algebra. It is a Lie algebra ideal because its Lie bracket with every element is zero. Passing to the quotient Lie algebra removes this central subspace. For ${\mathfrak{sl}}_p$ in characteristic $p$, the identity matrix has zero trace and spans a central ideal.

A Chevalley basis of a complex semisimple Lie algebra consists of suitable coroot elements and one vector $e_{\alpha }$ in each root space, chosen so all Lie bracket structure constants are integers. In particular $[e_{\alpha },e_{\beta }]=N_{\alpha ,\beta }{e}_{\alpha +\beta }$ when the sum is a root, with $N_{\alpha ,\beta }=\pm (p+1)$ where $\beta -p\alpha$ is the last negative-direction root in the root string. For independent roots with no root sum the bracket is zero. The integral lattice makes reduction to other fields possible.

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.

Choose five Hermitian four-by-four matrices obeying the Euclidean Clifford algebra relations ${\gamma }_a{\gamma }_b+{\gamma }_b{\gamma }_a=2{\delta }_{ab}I$. For $M_{ab}=E_{ab}-E_{ba}$, the displayed anti-Hermitian matrices satisfy the Special orthogonal Lie algebra Lie bracket relations. A direct proof first establishes $[\rho (M_{ab}),{\gamma }_c]={\delta }_{bc}{\gamma }_a-{\delta }_{ac}{\gamma }_b$, then applies the commutator product rule to $[{\gamma }_c,{\gamma }_d]$. One explicit choice is ${\gamma }_1={\sigma }_1\otimes I$, ${\gamma }_2={\sigma }_2\otimes I$, ${\gamma }_3={\sigma }_3\otimes {\sigma }_1$, ${\gamma }_4={\sigma }_3\otimes {\sigma }_2$, ${\gamma }_5={\sigma }_3\otimes {\sigma }_3$, with the Pauli matrices. The Cartan subalgebra operators $-i\rho (M_{12})$ and $-i\rho (M_{34})$ have simultaneous eigenvalues $(\pm 1/2,\pm 1/2)$, identifying the four-dimensional spinor representation. A full $2\pi$ rotation acts as $-I_4$, proving it does not descend to $SO(5)$.

On a coordinate trivialization of a principal bundle, the right-invariant vector fields $R_a(\gamma )=T_a\gamma$ satisfy $[R_a,R_b]=-c^d{}_{ab}{R}_d$. With $A=A_i^a{T}_{a}d{x}^i$, the displayed lifts project to the commuting coordinate fields and contract to zero with the principal connection. Taking their Lie brackets gives the local curvature of a principal connection coefficients $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i+[A_i,A_j]$. Thus these lifts commute exactly when the connection is flat. Arbitrary base fields need not commute, and flatness alone does not supply a global coordinate frame or remove holonomy of a connection.

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.

For a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$, its defining representation is its given action on the vector space $V$. The Lie bracket identity $[X,Y]v=X(Yv)-Y(Xv)$ verifies the representation property. For the special linear Lie algebra, diagonal trace-zero matrices have coordinate weights of a representation $L_i$ on the standard basis; for a Special orthogonal Lie algebra, the same action preserves the defining symmetric bilinear form.

An algebraic derivation of an algebra over a field is a linear map satisfying the Leibniz rule for its multiplication: $D(xy)=D(x)y+xD(y)$. This definition makes sense even when the multiplication is not associative. A derivation of an algebra uses an associative product; a derivation of a Lie algebra uses the Lie bracket as its product.

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 $\mathfrak{g}$.

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over an algebraically closed field. If every operator $\mathrm{ad}(x)$ of the Adjoint representation of a Lie algebra is diagonalizable, then $\mathfrak{g}$ is abelian. For if $[x,y]=\lambda y$ with $\lambda \ne 0$, then $\mathrm{ad}(y)x=-\lambda y\ne 0$ but $(\mathrm{ad}y{)}^{2}x=0$. This contradicts $\ker T^2=\ker T$ for a diagonalizable endomorphism $T$. Therefore every adjoint operator has only the eigenvalue zero and is diagonalizable, so every adjoint operator is zero. Hence every Lie bracket vanishes.

In the $(A^1,A^2,A^3,B)$ basis with gauge transformations $U=e^{i\alpha Q}$ and $Q=T_3+Y$, the non-Abelian connection transforms by conjugation. The Lie brackets $[T_3,T_1]=i{T}_2$ and $[T_3,T_2]=-i{T}_1$ give $\delta A^1=\alpha A^2$, $\delta A^2=-\alpha A^1$, with neutral $A^3,B$. Thus $\delta A=i\alpha Q_{\mathrm{gauge}}A$ gives the displayed Hermitian matrix. The fields $W^{\pm }=(A^1\mp i{A}^2)/\sqrt{2}$ transform as $e^{\pm i\alpha }{W}^{\pm }$, while both neutral combinations are uncharged. This separates the Adjoint representation on the gauge fields from the doublet charge matrix $\mathrm{diag}(1,0)$ on the Higgs field.

Two central extensions of a Lie algebra $0\to \mathfrak{z}\to E\to \mathfrak{g}\to 0$ and $0\to \mathfrak{z}\to E^{\prime }\to \mathfrak{g}\to 0$ are equivalent if a Lie algebra isomorphism $E\to E^{\prime }$ commutes with both maps in the short exact sequences, inducing the identity on the specified $\mathfrak{z}$ and $\mathfrak{g}$. This is finer than asking only that $E$ and $E^{\prime }$ be isomorphic as Lie algebras.

For $\mathfrak{z}=kz$, choose linear sections and identify the two vector spaces with $\mathfrak{g}\oplus kz$. A linear map compatible with both ends necessarily has the form $(x,uz)\mapsto (x,(u+t(x))z)$ for a linear functional $t$. If the two Lie algebra two-cocycles are $\alpha ,{\alpha }^{\prime }$, preservation of the Lie bracket is exactly ${\alpha }^{\prime }(x,y)=\alpha (x,y)+t([x,y])$. Hence equivalence means that their difference is a Lie algebra two-coboundary. If changes of the identifications at the ends are allowed, an automorphism of a Lie algebra at either end can identify further classes. For a two-dimensional abelian Lie algebra and a one-dimensional central line, the finer classes are parametrized by $k$, whereas the middle Lie algebras have only the zero and nonzero classes, represented by the abelian Lie algebra $k^3$ and the Heisenberg Lie algebra.

Inside the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the span of iterated Lie brackets of elements of $V$ of lengths at most $N$. Its exponential is the connected simply connected free nilpotent Lie group of step $N$, with tensor multiplication. The bracket is the commutator in the tensor algebra. Its grading defines dilations that multiply degree $k$ by ${\lambda }^k$.