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For a left-invariant metric on a Lie group, the one-parameter subgroup ${\gamma }_{\xi }(t)=\exp (t\xi )$ is a geodesic exactly whenfor every $\eta$ in the Lie algebra. The Koszul formula gives $⟨{\nabla }_{\xi }\xi ,\eta ⟩=-⟨\xi ,[\xi ,\eta ]⟩$.

A Riemannian manifold is homogeneous when its isometry group acts transitively on points. For example, a Lie group with a left-invariant metric is homogeneous under left translations. Homogeneity gives the same local metric-ball geometry at every point, which yields the completeness of homogeneous Riemannian manifolds. It does not imply two-point homogeneity: that stronger property also controls directions and pairs at equal distance.

The Maurer-Cartan form of the translation-dilation group of the plane has these coefficients in its dilation–translation basis, with $a=1,2$. Left multiplication rescales each coordinate differential and $\rho$ equally, proving they are left-invariant differential forms. Their dual left-invariant vector fields are $\rho {\partial }_{\rho }$ and $\rho {\partial }_{x^a}$. The Maurer-Cartan equation is $d{\lambda }^0=0$, $d{\lambda }^a=-{\lambda }^0\wedge {\lambda }^a$. The tensor ${\sum }_j{\lambda }^j\otimes {\lambda }^j$ is a positive definite left-invariant metric.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

Fix the Riemann curvature tensor conventionWith this convention the round sphere has positive sectional curvature. Write $R(X,Y,Z,W)=g(R(X,Y)Z,W)$. We use skew-symmetry in the first and last pairs, pair interchange $R(X,Y,Z,W)=R(Z,W,X,Y)$, and the first Bianchi identity. These are the standard curvature symmetries of the Levi-Civita connection.

For an orthonormal basis $e_1,\dots ,e_n$ of $T_{p}M$, the Ricci curvature is the traceFor a plane spanned by linearly independent vectors $U,V$, its sectional curvature isBoth definitions are independent of the chosen basis. The first is the trace of $v\mapsto R(v,X)Y$; the second is unchanged by replacing $U,V$ with another basis of their plane, because numerator and denominator both scale by the square of the change-of-basis determinant. For $U=e_i,V=e_j$, $i\ne j$, the denominator is one. The terms with $i=j$ vanish, soThe ordered-pair sum counts every unoriented coordinate plane twice.

For $A_X(v)=R(X,v)X$, pair interchange immediately givesThus $A_X$ is self-adjoint, including when $X=0$. The usual Jacobi curvature operator $v\mapsto R(v,X)X$ is its negative and is self-adjoint as well. Keeping these two signs distinct will also fix the Jacobi field equation below.

The Bonnet-Myers theorem states that a connected, complete $n$-dimensional Riemannian manifold, $n\ge 2$, with $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$ has diameter at most $\pi /\sqrt{\kappa }$ and is compact. Completeness is required; a pointwise positive Ricci tensor without a uniform positive lower bound would not suffice for this stated conclusion.

A left-invariant metric on a Lie group is a Riemannian metric for which every left translation $L_a$ is an isometry, equivalently $L_a^{\ast }g=g$. It is determined by an inner product on the Lie algebra $\mathfrak{g}=T_{e}G$. If the metric is also right-invariant, this inner product is invariant under the adjoint action; differentiating yieldsUse left-invariant vector fields and the permitted formula ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$. The Jacobi identity givesAdjoint skew-symmetry then impliesThis is the Ricci curvature of a bi-invariant Riemannian metric. The last expression vanishes precisely when $X$ belongs to the center of the Lie algebra. If that center is zero, it is positive for every nonzero $X$. Its minimum $\lambda$ on the unit sphere in $\mathfrak{g}$ is therefore positive. Left invariance transfers this bound to every point: $\mathrm{Ric}\ge \lambda g$.

Work in the identity component $G^0$, since ${\pi }_1(G,e)={\pi }_1(G^0,e)$. It is compact and hence complete. Give its universal cover ${\tilde{G}}^0$ the lifted metric. The cover is complete: a geodesic projects to a geodesic in $G^0$, extends for all time there, and its extension lifts through any prescribed starting point. The local isometry also preserves Ricci curvature, so the same uniform bound holds on the cover. Applying Bonnet-Myers with $\kappa =\lambda /(n-1)$ makes ${\tilde{G}}^0$ compact. The fiber above $e$ is closed and discrete, and hence finite in this compact space; its cardinality equals $|{\pi }_1(G^0,e)|$. Consequently ${\pi }_1(G,e)$ is finite. This is the finite fundamental group from a uniform positive Ricci bound. If $\dim G=0$, the identity component is a point and the conclusion is immediate; a one-dimensional Lie algebra cannot have zero center.

The Maurer-Cartan form is $g^{-1}dg={\lambda }^j{X}_j$. Direct matrix multiplication givesThese are left-invariant differential forms: for a constant group element $a$, $(ag{)}^{-1}d(ag)=g^{-1}dg$. Equivalently, left translation sends $(\rho ,x)$ to $(a\rho ,b+ax)$, so all three coordinate differentials and the denominator acquire the same factor $a>0$.

The left-invariant coframe of the translation-dilation group is pointwise linearly independent because $\rho >0$. Consequentlyis positive definite and unchanged under every left translation. It therefore defines the required left-invariant metric, with precisely the coordinate expression obtained from this sum. The notation $d{\rho }^2$ in that metric means $d\rho \otimes d\rho$, not the differential of ${\rho }^2$.

Take $u=v=l_{\xi }$ and $w=l_{\eta }$ in the identity from part (e). For a left-invariant metric, all three scalar products are constant and $[l_{\xi },l_{\xi }]=0$, soThe curve ${\gamma }_{\xi }$ has velocity $l_{\xi }$, hence is a geodesic exactly when ${\nabla }_{l_{\xi }}{l}_{\xi }=0$. Nondegeneracy of the inner product now givesThis is the geodesic-vector criterion for a left-invariant metric.

The metric is the left-invariant metricIts right-invariant vector fields areTheir flows act by left translations, which preserve a left-invariant metric. Directly,so both are Killing vector fields and generate one-parameter isometry groups.

There is an additional Killing field. Put $y=e^a>0$ and $x=b$; thenthe hyperbolic plane of constant curvature $-1$. Its isometry algebra is three-dimensional, whereas the space of right-invariant fields here is two-dimensional. For example, the third independent Killing field can be writtenwhich is not right invariant. Hence the answer is yes.

The multiplication $(x^{\prime },y^{\prime },z^{\prime })(x,y,z)=(x^{\prime }+e^{-z^{\prime }}x,y^{\prime }+e^{z^{\prime }}y,z^{\prime }+z)$ defines the three-dimensional Sol Lie group. Its left-invariant vector fields $e^{-z}{\partial }_x,e^z{\partial }_y,{\partial }_z$ have dual coframe $e^{z}dx,e^{-z}dy,dz$. Their sum of squares is a left-invariant metric, $d{s}^2=e^{2z}d{x}^2+e^{-2z}d{y}^2+d{z}^2$.