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A nondegenerate invariant symmetric bilinear form on a Lie algebra extends by translation to a metric preserved by both left and right translation. It may have indefinite signature. Its Levi-Civita connection is ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$ for left-invariant vector fields, and its Ricci tensor is $-K/4$. Choosing the nondegenerate Killing form of a semisimple Lie algebra yields a canonical indefinite Einstein manifold; compact groups instead admit the positive metric $-K$.

The product Riemannian metric of two round unit spheres has Ricci curvature equal to its metric, since the factor Ricci tensors equal their metrics and the product connection splits. Its directional Ricci curvatures are all one. A two-plane tangent to a single sphere has sectional curvature one, while a mixed plane has sectional curvature zero. This four-dimensional Einstein manifold therefore need not have constant sectional curvature.

An Einstein manifold has Ricci curvature equal to a constant multiple of its metric. In the Riemannian setting, the round unit sphere has $\Lambda =n-1$. This condition is weaker than constant sectional curvature in higher dimensions. The definition can also be applied to pseudo-Riemannian metrics, but positivity is assumed in the completeness theorems discussed here.

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.

An orthonormal coframe writes a metric as $g={\eta }_{ab}{\theta }^a\otimes {\theta }^b$ with constant diagonal signature matrix $\eta$. In Cartan's first structure equation, solveThese are the torsion-free and metric-compatibility conditions and determine the Levi-Civita connection. Then Cartan's second structure equation givesContract $R^a{}_{bad}$ to obtain the Ricci tensor and then contract again for the scalar curvature. This fixes the curvature convention; reversing the definition of the Riemann curvature tensor reverses the resulting curvature signs.

On $z>0$, choose ${\theta }^0=dt/z$, ${\theta }^1=dx/z$, ${\theta }^2=dy/z$, ${\theta }^3=dz/z$, with $\eta =\mathrm{diag}(-1,1,1,1)$. For $a=0,1,2$,A torsion-free metric connection therefore hasFor example, ${\Omega }^a{}_3=-d{\theta }^a=-{\theta }^a\wedge {\theta }^3$, while for $a,b<3$,Metric antisymmetry supplies the remaining components, giving uniformlyThus the metric has constant sectional curvature $-1$. In four dimensions,It is an Einstein manifold, with Einstein constant $-3$ in the stated curvature convention.

These are Poincare coordinates on anti-de Sitter spacetime of unit radius. Constant curvature gives the maximal ten-dimensional local isometry algebra $\mathfrak{so}(3,2)$. This can also be checked directly: set $u=(t,x,y)$, ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and $u_a={\eta }_{ab}{u}^b$. The ten independent Killing vector fields areTranslations and Lorentz transformations leave the numerator and $z$ unchanged. A dilation rescales numerator and denominator equally. For $K_a$, direct differentiation gives ${\mathcal{L}}_{K_a}{\eta }^{(4)}=4u_a{\eta }^{(4)}$ and $K_a(z)=2u_{a}z$, so the conformal factor cancels and ${\mathcal{L}}_{K_a}g=0$.

Hence the maximally extended Anti-de Sitter spacetime has connected isometry group locally $S{O}_0(3,2)$, with the appropriate covering group if one unwraps its time coordinate. The displayed coordinates cover only a patch. The ten local generators do not all give globally complete flows preserving that patch; for example special conformal flows can cross its horizon. The manifest complete patch symmetries include the boundary Poincare transformations and positive dilations. This distinguishes local maximal symmetry from the global isometries of a chosen coordinate domain.

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.

Fix the curvature conventionTo prove tensoriality, use $[hX,Y]=h[X,Y]-Y(h)X$ and the connection rules. The two $Y(h){\nabla }_{X}Z$ terms cancel, giving $R(hX,Y)Z=hR(X,Y)Z$. Antisymmetry in $X,Y$ gives linearity over smooth functions in the second input. Expanding the third input givesIt is therefore a smooth tensor of type $(1,3)$. Lowering the output with $g$ gives the type $(0,4)$ Riemann curvature tensor $R_4(X,Y,Z,W)=g(R(X,Y)Z,W)$.

For an independent pair $X,Y$, define sectional curvature byMetric compatibility gives $g(R(X,Y)Z,W)=-g(R(X,Y)W,Z)$ by applying $XY-YX-[X,Y]$ to $g(Z,W)$. Together with antisymmetry in $X,Y$, this shows that replacing the pair by $(aX+bY,cX+dY)$ multiplies both numerator and denominator by $(ad-bc{)}^2$. Thus the value depends only on the plane. Define Ricci curvature by $\mathrm{Ric}(Y,Z)={\sum }_{i}g(R(e_i,Y)Z,e_i)$ for any orthonormal basis; a trace is independent of the orthonormal basis.

For the curvature of the round unit sphere, the outward unit normal is the position vector $p$. The tangential projection of ambient differentiation is torsion-free and has metric compatibility, so uniqueness identifies it with $\nabla$. Ambient differentiation $D$ satisfies $D_{X}p=X$ andsince differentiating $g(Y,p)=0$ gives its normal component. The ambient curvature is zero. Take tangential components of $D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z=0$ to obtainHenceEvery two-plane has sectional curvature one. Tracing the first formula gives $\mathrm{Ric}(Y,Z)=ng(Y,Z)-g(Y,Z)$, and consequentlyThus the sphere is an Einstein manifold. When $n=1$ there are no tangent two-planes, the curvature tensor is zero and the same Ricci formula gives zero. The declared slot convention fixes all signs.