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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 four-dimensional diamond Lie algebra is the semidirect product of Lie algebras of the three-dimensional Heisenberg Lie algebra with a derivation having eigenvalues zero, one and minus one. With basis $c,d,p,q$, its nonzero brackets are $[d,p]=p$, $[d,q]=-q$, $[p,q]=c$. Its derived series of a Lie algebra has dimensions four, three, one and zero. The symmetric form with $B(c,d)=B(p,q)=1$ and all other basis pairings zero is a nondegenerate bilinear form and invariant. The alternating tensor $B([x,y],z)$ is the volume form on $⟨d,p,q⟩$, verifying invariance.

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.

The Chevalley-Eilenberg complex consists of alternating maps from ${\mathfrak{g}}^k$ to a Lie algebra representation $V$, with differential built from the bracket and module action. For trivial real coefficients, a two-cocycle is an alternating bilinear form $\kappa$ satisfying $\kappa ([X,Y],Z)+\kappa ([Y,Z],X)+\kappa ([Z,X],Y)=0$. Coboundaries have the form $b([X,Y])$, up to the differential's sign convention. Their quotient is $H^2(\mathfrak{g},\mathbb{R})$ and classifies central extensions of a Lie algebra.

A Lie algebra generator is an element of a chosen spanning or generating set for a Lie algebra. In a matrix representation, an infinitesimal group transformation has the form $1+i{\alpha }_a{t}^a+O({\alpha }^2)$ in terms of generator matrices $t^a$.

For trivial real coefficients, a Lie algebra two-cocycle is an alternating bilinear form satisfying $\kappa ([X,Y],Z)+\kappa ([Y,Z],X)+\kappa ([Z,X],Y)=0$. Adding a form $b([X,Y])$ does not change its Lie algebra cohomology class. The bracket $[(X,a),(Y,b)]=([X,Y],\kappa (X,Y))$ defines a central extension of a Lie algebra by $\mathbb{R}$ precisely when this cocycle identity holds. The constant discrepancy in a moment-map equivariance obstruction is such a cocycle.

Choose a basis $x,y$ of the two-dimensional abelian Lie algebra $\mathfrak{a}$, and lifts with the same names. Every alternating bilinear form is determined by $c=\alpha (x,y)$, and every such form is a Lie algebra two-cocycle because every Lie bracket in $\mathfrak{a}$ is zero. Thus every central extension of a Lie algebra has a basis $x,y,z$ withIf $c=0$, all Lie brackets vanish, so this is the direct sum of Lie algebras $\mathfrak{a}\oplus kz\cong k^3$. If $c\ne 0$, replace the central basis vector by $z^{\prime }=cz$. Then $[x,y]=z^{\prime }$, which is the Heisenberg Lie algebra ${\mathfrak{h}}_3(k)$. This change of basis shows that every nonzero $c$ gives the same Lie algebra isomorphism class.

These two Lie algebras cannot be isomorphic: their derived algebras have dimensions zero and one, respectively, and a Lie algebra isomorphism preserves the derived algebra. Moreover, the Heisenberg Lie algebra cannot be a direct sum of Lie algebras $kz\oplus \mathfrak{a}$, since that direct sum of Lie algebras would have zero Lie bracket. ThereforeHere “up to isomorphism” means the Lie algebra isomorphism class of the middle Lie algebra, allowing a change of the identifications at the ends of the short exact sequence. If instead equivalence of central extensions of a Lie algebra must induce the identity on both $kz$ and $\mathfrak{a}$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.

Let $A$ be a three-dimensional vector space with basis $p,q,z$. Define a bilinear map $\bullet :A\times A\to A$ bywith all other products of basis vectors zero. Equivalently,Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet (v\bullet w)$ vanish for every $u,v,w\in A$. This proves associativity, so $A$ is an associative algebra. The product is an alternating multilinear map, and each nested product in the Jacobi identity is zero, so the same product is a Lie bracket. It is nonzero because $p\bullet q=z\ne 0$.

There is no multiplicative identity element: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an identity element would require $1_A\bullet z=z$. ThusThis is an associative Lie product; its underlying Lie algebra is the Heisenberg Lie algebra.

Define the raw Killing form byFor a finite-dimensional real or complex Lie algebra in characteristic zero, it is nondegenerate exactly when the algebra is a semisimple Lie algebra. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ and $-K$ gives a Riemannian metric on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.

For a Lorentzian metric use $\mathfrak{sl}(2,\mathbb{R})$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces givewhose metric signature is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional Heisenberg Lie algebra has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.

For any adjoint-invariant nondegenerate metric on a Lie group, the Koszul formula on left-invariant fields simplifies toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus the metric $g=K$ is a Killing-form Einstein metric with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the Einstein manifold assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.

There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its torsion tensor on invariant fields isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These flat translation connections on a Lie group are not the torsion-free Levi-Civita connection; nonzero torsion permits flatness alongside the curved Killing metric.

All brackets belong to the line $\mathbb{C}c$, and $[p,q]=c$ shows this line is exactly the first derived algebra. Since $c$ is central, the derived series of a Lie algebra isThe lower central series of a Lie algebra is likewiseFor completeness, the center of a Lie algebra is $\mathbb{C}c$: an element $ap+bq+dc$ commutes with $p$ only if $b=0$, and with $q$ only if $a=0$. The upper central series of a Lie algebra is therefore $0\subset \mathbb{C}c\subset \mathfrak{g}$, because the quotient by $\mathbb{C}c$ is abelian. Thus the Heisenberg Lie algebra is nilpotent of class two and solvable of derived length two. These properties are not mutually exclusive.

In the ordered basis $(K_1,K_2,J)$, the three Adjoint representation matrices areTheir cross-product traces vanish, while their squared traces are $-2\lambda$, $-2\lambda$ and $-2{\lambda }^2$. ConsequentlyThis is the Killing form for cyclic three-generator brackets with constants $(1,\lambda ,\lambda )$.

For $\lambda >0$, the form is nondegenerate and negative-definite. The algebra is semisimple and compact, isomorphic to the SU(2) Lie algebra. For $\lambda <0$, it is nondegenerate with signature $(2,1)$, so the algebra is semisimple and noncompact, isomorphic to ${\mathfrak{sl}}_2(\mathbb{R})$.

For $\lambda =0$, the only nonzero bracket is $[K_1,K_2]=J$, with $J$ central. This is the Heisenberg Lie algebra, which is nilpotent and neither semisimple nor compact. Compactness is ruled out even though the Killing form is merely degenerate: $A_{K_1}$ is nonzero and nilpotent, whereas in a compact Lie algebra every adjoint map is skew-adjoint for a positive inner product. Skew-adjoint nilpotent operators vanish.

For a Lie algebra $\mathfrak{g}$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions areThese are respectively an Abelian Lie algebra, a solvable Lie algebra and a nilpotent Lie algebra. The second and third sequences are the derived series of a Lie algebra and the lower central series of a Lie algebra.

An Abelian Lie algebra has ${\gamma }_2=0$, and a nilpotent Lie algebra is a solvable Lie algebra: induction gives ${\mathfrak{g}}^{(j)}\subseteq {\gamma }_{j+1}$. Thus all the implications are generated byNone of the reverse implications holds. The Heisenberg Lie algebra with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has ${\gamma }_2=\mathbb{C}z$, ${\gamma }_3=0$. The two-dimensional affine Lie algebra of the line with $[h,e]=e$ has ${\mathfrak{g}}^{(1)}=\mathbb{C}e$ and ${\mathfrak{g}}^{(2)}=0$, but ${\gamma }_j=\mathbb{C}e$ for every $j\ge 2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.

The Lie theorem says that a finite-dimensional Lie algebra representation of a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic zero has a common eigenvector whenever its representation space is nonzero. Equivalently it admits an invariant complete flag, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb{C}$; these hypotheses are essential.

To prove the common-eigenvector assertion, induct on $\dim \mathfrak{g}$. The zero algebra is immediate. Since $\mathfrak{g}$ is nonzero and solvable, its derived algebra is proper. Choose a codimension-one Lie algebra ideal $\mathfrak{h}$ containing it, and write $\mathfrak{g}=\mathfrak{h}\oplus \mathbb{C}x$. The Lie algebra $\mathfrak{h}$ is solvable, so induction supplies $v\ne 0$ and a linear functional $\lambda :\mathfrak{h}\to \mathbb{C}$ with $hv=\lambda (h)v$.

Consider the finite-dimensional cyclic subspace $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Until the first linear dependence, these powers form a basis. The identityand $[h,x]\in \mathfrak{h}$ show by induction, simultaneously for all $h\in \mathfrak{h}$, that $W$ is $\mathfrak{h}$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda (h)$. It is also $x$-invariant by construction. HenceThe trace of a commutator is zero, and characteristic zero gives $\lambda ([x,h])=0$.

The simultaneous eigenspace $E_{\lambda }=\{w:hw=\lambda (h)w\text{ for every }h\in \mathfrak{h}\}$ is nonzero and $x$-invariant, sinceOver an algebraically closed field, $x{|}_{E_{\lambda }}$ has an eigenvector, which is therefore a common eigenvector for all of $\mathfrak{g}$. This finishes induction. Apply the same assertion to the quotient representation by its invariant line, and then to successive quotients. A basis adapted to the resulting complete flag gives the stated simultaneous triangularization of a Lie algebra representation, completing the proof of the Lie theorem.

A nilpotent Lie algebra is a solvable Lie algebra, so the inclusion $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ satisfies the Lie theorem and is upper triangular in a suitable basis, for finite-dimensional complex $V$.

This is insufficient to prove the Engel theorem. Its matrix version starts with a Lie subalgebra of nilpotent endomorphisms and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every Adjoint representation endomorphism. Merely upper triangular matrices can have nonzero diagonal entries: the one-dimensional algebra $\mathbb{C}{I}_V$ is an Abelian Lie algebra and a nilpotent Lie algebra, but acts by a nonnilpotent identity matrix. This is nilpotent Lie algebras need not act nilpotently. Moreover, in the abstract Engel theorem nilpotence of the algebra is a conclusion, so assuming it first to invoke the Lie theorem would be circular. Abstract nilpotence and nilpotence of each representing matrix are different conditions.

Multiplying the matrices gives the Heisenberg group lawDifferentiating at the identity gives all strictly upper-triangular matrices. With $X=E_{12}$, $Y=E_{23}$ and $Z=E_{13}$, the commutator givesThis is the Heisenberg Lie algebra.

The right Maurer-Cartan form isThus the right-invariant coframe of the real Heisenberg group is ${\sigma }^1=dx$, ${\sigma }^2=dy$, ${\sigma }^3=dz-ydx$. Under a right translation,we have $d{x}^{\prime }=dx$, $d{y}^{\prime }=dy$, and $d{z}^{\prime }-y^{\prime }d{x}^{\prime }=dz-ydx$. All three right-invariant differential forms are preserved.

Every right-invariant Riemannian metric is determined by an arbitrary inner product at the identity. In this coframe its most general expression isEquivalently,Products here are symmetric products. If a pseudo-Riemannian metric tensor is intended, replace positive definiteness by nondegeneracy. Since every right translation preserves the coframe, it is an isometry. The action is faithful, and $a\mapsto R_{a^{-1}}$ is a group homomorphism, embedding a copy of $G$ into the isometry group.

The one-parameter right translations produce the Killing frame for a right-invariant Heisenberg metric:Their flows are respectively $(x+t,y,z)$, $(x,y+t,z+xt)$ and $(x,y,z+t)$. Each preserves $h$, and their Lie brackets of vector fields are $[K_X,K_Y]=K_Z$, with the other two zero. This explicitly realizes the Heisenberg Lie algebra. These Killing vector fields are left-invariant vector fields; the vector fields dual to the right-invariant coframe are instead ${\partial }_x+y{\partial }_z,{\partial }_y,{\partial }_z$, and should not be substituted for these generators.

For the diagonal case, putThe Kaluza-Klein decomposition along a Killing field uses $V=h(K_X,K_X)=\alpha +\chi y^2$. Completing the square in $dx$ gives the Heisenberg metric Kaluza-Klein reduction:These depend only on the quotient coordinates $(y,z)$ and satisfy $h=V(dx+A{)}^2+\gamma$. For a diagonal indefinite metric tensor, the same expressions hold wherever $V\ne 0$; a null Killing vector field cannot be treated with this completed-square decomposition.

The derived series gives a direct test, with no structural theorem needed. By bilinearity and antisymmetry of the Lie bracket, every bracket in $\mathfrak{h}$ is a multiple of $z$, and $z=[x,y]$ actually occurs. ThereforeConsequently this Heisenberg Lie algebra is a nonzero solvable Lie algebra. It is an ideal of a Lie algebra in itself, since $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{h}$. Its solvable radical is therefore all of $\mathfrak{h}$, contradicting the first criterion in part (a). Hence

The line $\mathbb{R}S$ is a nonzero central ideal, so the Heisenberg Lie algebra is not simple. It is two-step nilpotent and hence solvable; a nonzero solvable Lie algebra is not semisimple. It is therefore neither simple nor semisimple.

For this Heisenberg Lie algebra,because $c$ belongs to the center. Thus its lower central series of a Lie algebra is $\mathfrak{g},\mathbb{C}c,0$, so $\mathfrak{g}$ is a two-step nilpotent Lie algebra.

On $\mathbb{C}[x]$, the operators $d/dx$, multiplication by $x$, and the identity satisfy $[d/dx,x]=1$. Sending $a,b,c$ to these operators gives a faithful irreducible infinite-dimensional representation of the Heisenberg Lie algebra.

Let $G$ be the real upper unitriangular three-by-three group, let $Z$ be its centre, and let $Z_0$ be the integer subgroup of $Z$. Every finite-dimensional continuous complex representation of $G/Z_0$ kills the central circle $Z/Z_0$, and hence is not faithful.

If $D:\mathfrak{b}\to \mathrm{Der}(\mathfrak{a})$ is a Lie algebra homomorphism into the derivations of a Lie algebra, the semidirect product has underlying vector space $\mathfrak{a}\oplus \mathfrak{b}$ and bracket $[(a,b),(a^{\prime },b^{\prime })]=([a,a^{\prime }]+D_b{a}^{\prime }-D_{b^{\prime }}a,[b,b^{\prime }])$. The homomorphism and derivation identities ensure the Jacobi identity. The first factor is an ideal; the second is a subalgebra whose action records the twisting. The diamond Lie algebra arises by letting a one-dimensional algebra act on a Heisenberg Lie algebra.

// Target: linear-algebra.bigb

Each term is the preimage of the center of a Lie algebra after quotienting by the preceding term. Equivalently, $Z_{i+1}=\{x:[\mathfrak{g},x]\subseteq Z_i\}$. A Lie algebra is nilpotent exactly when this series reaches the whole algebra after finitely many steps. For the Heisenberg Lie algebra, the series is $0\subset \mathbb{C}c\subset \mathfrak{g}$.