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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.

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 ambient quadratic form $Q=-(X^0{)}^2-(X^4{)}^2+(X^1{)}^2+(X^2{)}^2$. The printed sine coordinate has a duplicated superscript: it must be $X^2$, not another $X^1$. With that correction the embedding lies on $Q=-1$. Put $s=\sqrt{1+r^2}$. Differentiating the two timelike coordinates giveswhereas the two spatial coordinates give $(d{X}^1{)}^2+(d{X}^2{)}^2=d{r}^2+r^{2}d{\theta }^2$. Their difference is the induced Lorentzian metricThe polar coordinate $\theta$ collapses at $r=0$; this is a coordinate degeneracy, not a degeneracy of the metric.

For the matrix model of anti-de Sitter three-space, defineThen $\det M=-Q$. This linear identification of ambient space with real two-by-two matrices identifies $Q=-1$ with $SL(2,\mathbb{R})$. Its inverse recovers each $X$ by sums and differences of matrix entries, so the identification is global. The identity corresponds to $X^4=1$ and the other coordinates zero. For a traceless tangent matrix $U$, the induced squared norm isPolarization gives $g_I(U,V)=\frac{1}{2}\mathrm{tr}(UV)$.

The Killing form of ${\mathfrak{sl}}_2(\mathbb{R})$ is $B(U,V)=4\mathrm{tr}(UV)$. This normalization can be checked without assuming it: for $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. Taking traces of their three-dimensional adjoint matrices gives $B(H,H)=8$, $B(E,F)=B(F,E)=4$, and all other basis pairings zero. These agree with $4\mathrm{tr}(UV)$.

Both left and right multiplication by matrices in $SL(2,\mathbb{R})$ preserve $-\det M$, hence its polarized ambient bilinear form and the induced metric. Equality at the identity therefore propagates throughout the group:In particular the Killing-form Einstein metric calculation gives $\mathrm{Ric}=-2g$, consistent with unit-radius three-dimensional Anti-de Sitter spacetime.

Every element of $SO(2,2)$ preserves the ambient quadratic form, hence the hyperboloid and its induced metric, and so acts by isometries. Define the left-right double cover of SO0(2,2) byThis is a group homomorphism, since successive actions multiply the two factors in the same order. It preserves $\det M$; as a transformation of the four-dimensional matrix space it has determinant $(\det A{)}^2(\det B^{-1}{)}^2=1$. Thus it takes values in $SO(2,2)$.

If its action is the identity, applying it to $M=I$ gives $A=B$. Commuting with every matrix then forces $A$ to be scalar. The condition $\det A=1$ leaves only $A=I$ or $A=-I$. HenceThe differential is $(a,b):M\mapsto aM-Mb$. A zero differential implies $a=b$ and that $a$ is scalar; tracelessness makes $a=b=0$. It is therefore an isomorphism onto the target Lie algebra, since both dimensions are six. The image is an open subgroup. Since $SL(2,\mathbb{R})$ is connected, the image is connected and is exactly the identity component $S{O}_0(2,2)$: an open subgroup of a connected group is also closed and must be the whole group. Thus the precise covering statement isIt is not onto the disconnected full $SO(2,2)$. For example, conjugation $M\mapsto DM{D}^{-1}$, $D=\mathrm{diag}(1,-1)$, sends $(X^0,X^4,X^1,X^2)$ to $(-X^0,X^4,X^1,-X^2)$ and belongs to the other component of $SO(2,2)$. It reverses the orientation of the negative two-plane and is not in the image of the connected source.

The Adjoint representation of a Lie group of $SL(2,\mathbb{R})$ preserves its Killing form of metric signature $(2,1)$ and gives a second double cover,Its kernel is the scalar centre; its differential is injective and both dimensions are three, so the same connected-image argument proves surjectivity onto the identity component. The full $SO(2,1)$ again has an additional component. Thus $S{O}_0(2,1)$ is locally the same Lie group geometry as $SL(2,\mathbb{R})$, with its Killing metric, and globally is its central quotient.

With the printed periodic time, the group model has topology $S^1\times {\mathbb{R}}^2$ and closed timelike curves, since varying $t$ at fixed spatial position gives a timelike circle. The causally unwrapped version of Anti-de Sitter spacetime instead has $t\in \mathbb{R}$ and is the universal covering Lie group. It must not be substituted silently for the periodic spacetime specified here.

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.