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A Riemannian or pseudo-Riemannian metric is Einstein when its Ricci tensor is a constant scalar multiple of the metric. Constant-sectional-curvature metrics satisfy this condition with $\Lambda =(d-1)k$. The Killing-form metric on a real semisimple Lie group is another example, with $\Lambda =-1/4$ for $g=B$ in the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$.

Let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. A real semisimple Lie algebra has nondegenerate $B$, and its invariance gives $B([X,Y],Z)=B(X,[Y,Z])$. Translating $B$ defines a bi-invariant pseudo-Riemannian metric $g=B$ on the Lie group.

For left-invariant vector fields, all derivatives of their pairwise inner products vanish. The Koszul formula reduces toNondegeneracy gives ${\nabla }_{X}Y=[X,Y]/2$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity now givesTaking the trace of $X\mapsto R(X,Y)Z=-\frac{1}{4}\mathrm{ad}Z\mathrm{ad}Y(X)$ yields the Ricci tensorThus the Killing-form Einstein metric makes the group an Einstein manifold, allowing an indefinite metric. Its scalar curvature in this convention is $-\dim G/4$.

For a connected compact semisimple Lie group, $B$ is negative definite: the adjoint operators are skew-adjoint for an invariant positive inner product, so $B(X,X)=\mathrm{tr}((\mathrm{ad}X{)}^2)<0$ for $X\ne 0$. The conventional positive Riemannian metric is instead $g_{+}=-B$. Its Ricci tensor is $g_{+}/4$, and its sectional curvature isIt can vanish on commuting two-planes; positivity is not automatic in every direction. With the literal negative metric $B$, the corresponding sectional-curvature signs are reversed.

For a noncompact real semisimple group, a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ has $B$ negative on $\mathfrak{k}$ and positive on $\mathfrak{p}$. The Killing metric is indefinite and cannot be made positive by an overall sign. Its Einstein constant remains $-1/4$ for $g=B$, but that does not assert a uniform sign for sectional curvature of all nondegenerate planes. For connected semisimple groups, negative-definite $B$ characterizes the compact case. Disconnected groups require a separate condition on the number of components before compactness follows.

For the Lie algebra $\mathfrak{g}$ of a Lie group $G$, let ${\mathrm{ad}}_X(Y)=[X,Y]$. The Killing form is the symmetric bilinear formFor a finite-dimensional real Lie algebra, the Cartan criterion for semisimplicity says that $B$ is nondegenerate precisely when $\mathfrak{g}$ is a semisimple Lie algebra. Thus a semisimple Lie group supplies the required condition. Compactness alone is insufficient: an abelian Lie algebra, including that of a torus, has zero Killing form.

The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Cyclic invariance of the trace therefore givesThere is also invariance under the full Adjoint representation of a Lie group: ${\mathrm{ad}}_{{\mathrm{Ad}}_{a}X}={\mathrm{Ad}}_a{\mathrm{ad}}_X{\mathrm{Ad}}_{a^{-1}}$, so $B({\mathrm{Ad}}_{a}X,{\mathrm{Ad}}_{a}Y)=B(X,Y)$.

Define a pseudo-Riemannian metric by translating $B$ from the identity:It is smooth and nondegenerate. Left translations preserve it by construction. Right translation, expressed in left-translated tangent coordinates, acts by ${\mathrm{Ad}}_{b^{-1}}$, so it also preserves the metric. This constructs a bi-invariant pseudo-Riemannian metric.

For left-invariant vector fields, the derivative terms in the Koszul formula vanish. Invariance of $B$ then givesUse the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. Substituting the Levi-Civita connection and using the Jacobi identity yieldsThe Ricci tensor is the trace of $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]={\mathrm{ad}}_Z{\mathrm{ad}}_{Y}X$, it follows thatThis is the Killing-form Einstein metric. Rescaling to $g=cB$, with nonzero constant $c$, gives Einstein metric constant $-1/(4c)$.

For a compact semisimple Lie group, $B$ is negative definite and $-B$ is a positive-definite bi-invariant Riemannian metric. For example, $SU(2)$ has topology $S^3$ and metric signature $(0,3)$ for $B$, while $SO(3)$ has the same Lie algebra and metric signature, but topology ${\mathbb{RP}}^3$. The special linear group $SL(2,\mathbb{R})$ has metric signature $(2,1)$ and topology $S^1\times {\mathbb{R}}^2$; its universal covering Lie group has topology ${\mathbb{R}}^3$ and the same Killing form. The real Lie algebra of $SL(2,\mathbb{C})$ has metric signature $(3,3)$, and the group has topology $S^3\times {\mathbb{R}}^3$. Thus the Killing form distinguishes compact from noncompact directions in these semisimple examples, but does not determine global topology or distinguish different covering groups.

Use ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. For an actual infinitesimal left Lie group action, define ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$. Its fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ with this convention.

Existence and its global obstruction. Preservation of the symplectic form gives ${\mathcal{L}}_{{\xi }_P}\omega =0$. By Cartan's magic formula, ${\iota }_{{\xi }_P}\omega$ is closed. Locally it equals $d{\mu }_{\xi }$; globally this requires its class in $H_{\mathrm{dR}}^1(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and defineThis gives the component Hamiltonians of a weakly Hamiltonian action. The assumption in the question does not alone guarantee a global moment map: on $T^2$ with $\omega =dx\wedge dy$, translation in $x$ preserves $\omega$, but its contraction is $dy$, whose integral around the $y$ circle is nonzero. It is not an exact one-form. Thus no global real-valued component Hamiltonian exists for that circle action. The intended construction is valid locally, or globally with this exactness condition, for example if $H_{\mathrm{dR}}^1(P)=0$.

Matching the Lie algebra. On each connected component, the moment-map equivariance obstruction is the constantIndeed, the identities $[X_f,X_g]=-X_{\{f,g\}}$ and $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ show that its differential vanishes. The Jacobi identity for the Poisson bracket saysChanging ${\mu }_{\xi }$ by a linear constant $b(\xi )$ changes $c$ to $c-b([\xi ,\eta ])$. Consequently the necessary and sufficient infinitesimal condition is the cocycle must be a coboundary, so that the constants can be chosen withEquivalently its class in $H^2(\mathfrak{g},\mathbb{R})$ vanishes. For a connected group this infinitesimal equivariance integrates to equivariance under its coadjoint action. A disconnected group can impose additional conditions. Here “coincides” means representing the specified Lie brackets; it does not guarantee injectivity when the group action has ineffective generators.

Proof using the Killing form. A nondegenerate Killing form $B$ makes $\mathfrak{g}$ semisimple. First, it is perfect: the $B$-orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ is the center by invariance, and the center is zero by nondegeneracy. For symplectic fields,by Cartan's magic formula and the antihomomorphism sign. Every commutator therefore has an exact contraction. Perfectness supplies component Hamiltonians for every generator, even when $P$ is not simply connected: this is Hamiltonian existence for a semisimple symplectic action.

It remains to eliminate the cocycle, not merely to name an existence theorem. Define a linear operator $D$ on $\mathfrak{g}$ by $B(D\xi ,\eta )=c(\xi ,\eta )$. Antisymmetry of $c$ makes $D$ skew for $B$. Combining its cocycle identity with invariance of $B$ givesChoose $Z\in \mathfrak{g}$ by $B(Z,\xi )=\mathrm{tr}(D{\mathrm{ad}}_{\xi })$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,{\mathrm{ad}}_{\xi }]={\mathrm{ad}}_{D\xi }$. Cyclicity of the matrix trace then givesThus $D={\mathrm{ad}}_Z$, and $c(\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes $c$, proving the required condition. This applies to semisimple Lie groups, including $SU(n)$ for $n\ge 2$, $SO(3)$ and $SL(n,\mathbb{R})$ for $n\ge 2$. It is nondegeneracy, not positive definiteness, that matters. Groups with a nonzero central Lie algebra, such as $U(3)$, do not satisfy this Killing-form criterion, although particular actions can still have equivariant moment maps.

The isotropic oscillator. On $T^{\ast }{\mathbb{R}}^3$, putThe action $z\mapsto Uz$, $U\in U(3)$, preserves the Hermitian norm and the symplectic form, and therefore preserves $H$. A Hermitian matrix $A$ parametrizes the Lie algebra generator $\xi =-iA$. Its unitary moment map of an isotropic oscillator isSince $\{z_j,{\bar{z}}_k\}=-i{\delta }_{jk}$, direct differentiation yieldsThe matrix $-i[A,B]$ is Hermitian and corresponds precisely to $[{\xi }_A,{\xi }_B]$; no central defect remains. The full moment map can be identified, using the pairing with Hermitian matrices, with the rank-one matrix $z{z}^{†}$. The central generator $A=I$ gives ${\mu }_I=H$, whose flow is $z(t)=e^{-it}z(0)$. Globally $U(3)=(SU(3)\times U(1))/{\mathbb{Z}}_3$, so the “$U(1)$ factor” refers to this central circle, rather than a literal direct-product decomposition.

The real orthogonal subgroup acts geometrically by $(x,p)\mapsto (Rx,Rp)$, the cotangent lift of a diffeomorphism of ${\mathbb{R}}^3$. If $K$ is real antisymmetric and $A=iK$, then ${\mu }_A=-x^{T}Kp$. For $Kv=\Omega \times v$, this is $\Omega \cdot (x\times p)$, so its three components are the usual angular momentum $L_i={ϵ}_{ijk}{x}_j{p}_k$.

The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions spanThe diagonal ones are the individual oscillator energies; their sum is $H$. The off-diagonal ones correlate two oscillator modes. Their flows mix positions and momenta and are not cotangent lifts of the obvious spatial rotation action: for a real symmetric $A$, $\dot{x}=Ap$ and $\dot{p}=-Ax$, rather than $\dot{x}=Kx$, $\dot{p}=Kp$. All nine quadratic functions commute with $H$, since the entire unitary group action preserves the oscillator Hamiltonian.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

Declare a global frame of left-invariant vector fields parallel on a semisimple Lie group. This affine connection is flat and preserves a left-invariant metric tensor because its frame components are constant. If $[e_a,e_c]=C_a{}^b{}_c{e}_b$, its torsion tensor is $T_a{}^b{}_c=-C_a{}^b{}_c$. For fixed $c$, the matrix with entries $T_a{}^b{}_c$ is $\mathrm{ad}{e}_c$. Its trace vanishes because a semisimple Lie algebra is perfect and adjoints of brackets are commutators. The trace of the product for $c,d$ is the Killing form $K_{cd}$, proving both contractions. This flat metric connection differs from the curved torsion-free Levi-Civita connection of the Killing metric.