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For a Lie group $G$, Left translation on a Lie group and right translation on a Lie group are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the Lie algebra gives global left-invariant vector fields $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, declaring the $F_a$ parallel gives another flat affine connection ${\nabla }^{+}$, with $T^{+}(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^{+}(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the flat translation connections on a Lie group. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.

The average ${\nabla }^0=({\nabla }^{-}+{\nabla }^{+})/2$ is again an affine connection and satisfies ${\nabla }_X^{0}Y=\frac{1}{2}[X,Y]$ for left-invariant fields. Its torsion tensor is zero. With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesDefine the Ricci tensor by $\mathrm{Ric}(X,Y)=\mathrm{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, this giveswhere $K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$ is the Killing form. Thus the canonical torsion-free connection on a Lie group is generally curved. Its exact flatness criterion is $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$: Abelian and two-step nilpotent groups are exceptions. A vanishing Ricci tensor alone does not prove flatness.

For a semisimple Lie algebra, the Killing form is nondegenerate. Its adjoint invariance gives $K([X,Y],Z)+K(Y,[X,Z])=0$, so translating it defines a bi-invariant pseudo-Riemannian metric $g_K$. For left-invariant fields,A metric-compatible torsion-free connection is uniquely the Levi-Civita connection; therefore ${\nabla }^0$ is its Levi-Civita connection. Moreover,The special feature is a canonical nondegenerate metric and an Einstein manifold structure. The metric need not be positive definite: noncompact semisimple groups generally require indefinite signature. For compact semisimple groups, using $-K$ gives a bi-invariant Riemannian metric with $\mathrm{Ric}=\frac{1}{4}{g}_{-K}$ instead.

Identify the ambient real four-dimensional vector space with $2\times 2$ real matrices byFor $A,B\in SL(2,\mathbb{R})$, the linear action $X\mapsto AX{B}^{-1}$ preserves this quadratic form. It gives a group homomorphism to the special orthogonal group of signature $(2,2)$; its determinant on matrix space is $(\det A{)}^2(\det B{)}^{-2}=1$. The group $SL(2,\mathbb{R})$ is connected: an oriented Gram–Schmidt factorization writes every element as a rotation times an upper triangular matrix with positive diagonal and determinant one, and each factor varies in a connected space. Thus the image lies in the identity component $S{O}_0(2,2)$.

If $AX{B}^{-1}=X$ for every $X$, taking $X=I$ gives $A=B$. A matrix commuting with every real $2\times 2$ matrix is scalar, and its determinant-one condition then gives $A=B=\pm I$. The kernel is therefore exactly $\{(I,I),(-I,-I)\}$. Infinitesimally the action is $X\mapsto aX-Xb$, with $a,b$ traceless. If this vanishes for every $X$, then $a=b$ and it is scalar and traceless, hence zero. The differential is injective; source and target Lie algebras each have dimension six, so it is an isomorphism. The image is therefore an open subgroup of the connected target, which must be the entire target. ConsequentlyThe finite quotient is by the diagonal central subgroup, not by a sign in just one factor. The connected-component subscript is a genuine qualification of the source's displayed formula: the full $SO(2,2)$ is disconnected and cannot be isomorphic to a quotient of the connected product.

Let the anti-de Sitter radius be $\ell >0$. Its quadric is $-u^2-v^2+x^2+y^2=-{\ell }^2$, or $\det X={\ell }^2$. Thus the map $X\mapsto g=X/\ell$ identifies this quadric smoothly and globally with $SL(2,\mathbb{R})$. This is the group model of three-dimensional anti-de Sitter spacetime. The induced Lorentzian metric is bi-invariant, since the two matrix multiplication actions preserve the ambient quadratic form. At the identity, tangent matrices $a,b$ are traceless andIndeed, for a traceless matrix $a$, $\det a=-\frac{1}{2}\mathrm{Tr}(a^2)$, and polarizing $-\det (\ell a)$ gives this metric. For ${\mathfrak{sl}}_2(\mathbb{R})$ the Killing form is $K(a,b)=4\mathrm{Tr}(ab)$: in the basis $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$, giving $K(H,H)=8$, $K(E,F)=4$ and the remaining independent pairings zero. Therefore $g_{\ell }={\ell }^{2}K/8$.

The canonical torsion-free connection on a Lie group is the Levi-Civita connection of this bi-invariant pseudo-Riemannian metric. Its Ricci tensor, already calculated from the double bracket, isSubstitution into the vacuum Einstein field equations givesThus the group metric solves the three-dimensional vacuum equations with negative cosmological constant.

The antipodal map $X\mapsto -X$ is $g\mapsto -g$, multiplication by the central element $-I$. Hence the antipodal quotient is $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$, with the same local curvature and cosmological constant. To see the global time issue, useThe induced metric isOn the quadric $t$ has period $2\pi$; curves with fixed $\rho ,\phi$ are closed timelike curves. The antipodal identification is $(t,\phi )\sim (t+\pi ,\phi +\pi )$, and does not eliminate this problem. At $\rho =0$, even the half-period time shift closes after the antipodal quotient.

If global time is not identified, take the universal cover, with $t\in \mathbb{R}$. The topology of $SL(2,\mathbb{R})$ is $S^1\times {\mathbb{R}}^2$, as follows from the same rotation–triangular decomposition, so the universal cover is diffeomorphic to ${\mathbb{R}}^3$ and carries the lifted group structure $\widetilde{SL(2,\mathbb{R})}$. The metric and Einstein equation are unchanged locally. Here $t$ is a global time function with timelike gradient, excluding closed causal curves. Unwrapped anti-de Sitter spacetime is the universal covering group, rather than the original embedded quadric or its antipodal quotient; retaining the antipodal time-shift identification would wrap time again.