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If a Lie group has an Adjoint representation of a Lie group element with a real eigenvector of eigenvalue whose modulus differs from one, it has no bi-invariant Riemannian metric. Conjugation is a composition of left and right translations, hence would preserve the positive identity inner product. Applying it to that eigenvector gives $\Vert v\Vert =|\lambda |\Vert v\Vert$, a contradiction. The orientation-preserving affine group of the real line supplies a connected example: conjugation by dilation of factor two multiplies the infinitesimal translation by two.

Give the Lie algebra of a compact Lie group an inner product invariant under the Adjoint representation. For a maximal torus $T$ with Lie algebra $\mathfrak{t}$, its Lie-algebra centralizer is $\mathfrak{t}$: any additional commuting vector would generate, together with $T$, a larger compact connected abelian subgroup. The torus weight decomposition therefore supplies $H\in \mathfrak{t}$ whose centralizer is exactly $\mathfrak{t}$, by avoiding finitely many nonzero-weight hyperplanes. For any $X$, maximize $⟨{\mathrm{Ad}}_{g}X,H⟩$ on the group. Its directional derivative gives $⟨Z,[{\mathrm{Ad}}_{g}X,H]⟩=0$ for every $Z$, so ${\mathrm{Ad}}_{g}X\in \mathfrak{t}$. For a connected compact group, the bi-invariant Riemannian metric and Hopf-Rinow theorem make the exponential onto; exponentiating this conjugacy puts every element in a conjugate of $T$.

A group distance is bi-invariant when both left and right translations are isometries. Conjugation then fixes the identity element and is an isometry, so each metric ball there is invariant under every inner automorphism. A compatible bi-invariant group metric therefore requires arbitrarily small conjugation-invariant neighbourhoods. This concerns a distance function, not merely a bi-invariant Riemannian metric tensor.

Integral curves through the identity of left-invariant vector fields are one-parameter subgroups, by flow uniqueness and left translation. They extend for all time because a fixed local existence interval translates to every point. Under a bi-invariant Riemannian metric they are geodesics by the Levi-Civita connection of a bi-invariant metric. Geodesic uniqueness proves that these are all geodesics through the identity.

For a Lie group with a bi-invariant Riemannian metric, vary $\gamma (t)=e^{tA}$ by conjugation: $F(s,t)=e^{sX}{e}^{tA}{e}^{-sX}$. Its Jacobi field is $J_X(t)=X{e}^{tA}-e^{tA}X$. If $e^A$ is central, all these fields vanish at both endpoints. Their space has dimension $\dim \mathfrak{g}-\dim \mathfrak{z}(A)$, since the kernel is the centralizer of an element of a Lie algebra.

For left-invariant vector fields, the Koszul formula and the adjoint invariance of a bi-invariant Riemannian metric give the displayed Levi-Civita connection. In particular ${\nabla }_{X}X=0$.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

Consider the orientation-preserving affine group of the real line, realized asThe coordinates $(\log a,b)$ identify it smoothly with ${\mathbb{R}}^2$, so this Lie group is connected. Choose the Lie algebra element $E$ and the group element $D$:Direct matrix multiplication gives ${\mathrm{Ad}}_{D}E=DE{D}^{-1}=2E$.

If a bi-invariant Riemannian metric existed, every conjugation map $x\mapsto Dx{D}^{-1}$ would be an isometry fixing the identity. Its differential there, the Adjoint representation of a Lie group, would preserve the positive-definite inner product at the identity. Thusa contradiction since $E\ne 0$. Therefore this connected Lie group admits no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric, an elementary alternative permitted by the question. The positive definite Riemannian metric interpretation matters; no claim about indefinite metrics is needed.

Take the orientation-preserving affine group of the real line, written asIt is a connected Lie group, since its underlying manifold is $(0,\infty )\times \mathbb{R}$. We use the permitted alternative to part (a), a direct obstruction from the Adjoint representation of a Lie group.

If a bi-invariant Riemannian metric existed, conjugation $h\mapsto gh{g}^{-1}$ would be an isometry, being a composition of a left and a right translation. Its differential at the identity would preserve the positive inner product on the Lie algebra. Let $T$ be the nonzero infinitesimal translation,Direct multiplication gives ${\mathrm{Ad}}_{g}T=gT{g}^{-1}=2T$. Metric invariance would then imply $\Vert T{\Vert }^2=\Vert 2T{\Vert }^2=4\Vert T{\Vert }^2$, impossible for a nonzero vector in a positive inner product. Therefore this connected affine Lie group has no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric; allowing an indefinite metric would be a different question.

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.

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.

Use the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. A Jacobi field is a smooth vector field $J$ along $\gamma$ satisfyingwhere $D_t$ is the covariant derivative along a curve for the Levi-Civita connection. Differentiating a geodesic variation gives this equation, and conversely every Jacobi field arises from a geodesic variation by varying its initial point and velocity.

Choose a parallel frame $E_1(t),\dots ,E_n(t)$ along $\gamma$ and write $J={\sum }_a{y}_a{E}_a$. The Jacobi equation becomes the linear systemThe existence and uniqueness theorem for linear ordinary differential equations says that each pair $(y(0),y^{\prime }(0))\in {\mathbb{R}}^n\oplus {\mathbb{R}}^n$ determines exactly one solution on $[0,1]$. Addition and scalar multiplication preserve the equation. Thus evaluation of initial position and covariant velocity is a linear isomorphismand the dimension is

For the convexity assertion, use the convex-normal meaning of a geodesically convex open set: its points are joined by a unique geodesic within the set, and the joining geodesic depends smoothly on its endpoints. Equivalently, the appropriate star-shaped restriction of ${\exp }_p$ is a diffeomorphism onto the set. The convex-normal-neighbourhood theorem supplies such sets around every point. Mere existence of some minimizing geodesic, without this uniqueness and normality, would not imply the assertion.

Let $p=\gamma (0)$ and $v=\dot{\gamma }(0)$. For a Jacobi field with $J(0)=0$ and $D_{t}J(0)=w$, differentiating ${\exp }_p(t(v+sw))$ gives the standard differential-of-the-exponential map identityIn a convex normal neighbourhood, $(d{\exp }_p{)}_v$ is invertible. Therefore a Jacobi field vanishing at both endpoints has $w=0$ and hence is identically zero. The difference of two fields with the same endpoint values consequently vanishes. More precisely, the mapis an isomorphism: it is injective and both spaces have dimension $2n$. The constant-geodesic case follows directly from $D_t^{2}J=0$. This uses the convex-normal-neighbourhood theorem, the differential-of-the-exponential map identity, and linear ordinary differential equation uniqueness.

For the special unitary group example, use Jacobi fields from conjugation at a central endpoint. WriteThe endpoints of $\gamma (t)=\exp (tA)$ are $I$ and $zI$. The latter is central. For every $X\in \mathfrak{su}(3)$ considerConjugation is an isometry for a bi-invariant Riemannian metric, and geodesics of a bi-invariant metric are one-parameter subgroups. Thus $F_X$ is a geodesic variation. Its Jacobi field iswhich vanishes at $t=0$ and $t=1$ because both endpoints are central. Its initial covariant derivative is $[X,A]$.

The linear map $X\mapsto J_X$ has kernel equal to the centralizer of an element of a Lie algebra of $A$. Indeed, $J_X\equiv 0$ implies $[X,A]=0$ on differentiating at $t=0$, and that commutation conversely implies $J_X\equiv 0$. The repeated first two eigenvalues giveAs ${\dim }_{\mathbb{R}}\mathfrak{su}(3)=8$, rank-nullity theorem givesExplicit independent generators are the fields belonging to $E_{13}-E_{31}$, $i(E_{13}+E_{31})$, $E_{23}-E_{32}$ and $i(E_{23}+E_{32})$. Their initial derivatives $[X,A]$ are independent because the eigenvalue differences in the $13$ and $23$ entries are $2\pi i$. This also demonstrates directly why the endpoint-value conclusion fails along this geodesic.

For $v$ in the domain of the exponential map at $p$, let ${\gamma }_v$ be the unique geodesic with initial data $(p,v)$ and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists of initial velocities whose geodesics exist on $[0,1]$; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on $TM$, it gives smoothness of the exponential map. Rescaling the parameter gives ${\exp }_p(tv)={\gamma }_v(t)$ and hence $(d{\exp }_p{)}_0(v)=v$. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighbourhood of zero to one of $p$. Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.

The geodesic sphere of radius $r$ is $S_r(p)=\{x:d(p,x)=r\}$. For small $r>0$, it is ${\exp }_p(\{v:\Vert v\Vert =r\})$, a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.

For a piecewise smooth path $c$, put $L(c)=\int \Vert \dot{c}(t)\Vert dt$. The Riemannian distance is $d(x,y)=\inf L(c)$ over such paths from $x$ to $y$. Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.

Choose $r_0>0$ so that ${\exp }_p$ is a diffeomorphism on a neighbourhood of the closed tangent ball of radius $2r_0$. For $r<r_0$, radial geodesics realize the distance from $p$ on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius-$2r_0$ ball already has length at least $2r_0$. Thus the radius-$r$ sphere is the compact exponential image described above.

Given $q\ne p$, take $0<\delta <\min \{r_0,d(p,q)\}$. The continuous function $z\mapsto d(z,q)$ attains its minimum at some $p_0\in S_{\delta }(p)$. Every path from $p$ to $q$ crosses that sphere, since the distance from $p$ is continuous. Splitting at a crossing gives length at least $\delta +{\min }_{S_{\delta }(p)}d(\cdot ,q)$. Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to $q$.

For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives $⟨[X,Y],Z⟩+⟨Y,[X,Z]⟩=0$. The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.

Let $\sigma$ be the integral curve through the identity of the left-invariant field with identity value $v$. Uniqueness and left translation show $\sigma (s+t)=\sigma (s)\sigma (t)$ wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus $\sigma$ exists on $\mathbb{R}$, and uniqueness now gives the group law for every $s,t$.

Since ${\nabla }_{X}X=0$, this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here ${\exp }_G$ is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.

For a bi-invariant Riemannian metric on a Lie group, the induced inner product on its Lie algebra makes every adjoint operator skew-adjoint. The Levi-Civita connection of a bi-invariant metric gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$. Thus $K(X,Y)=\frac{1}{4}\Vert [X,Y]{\Vert }^2$ for orthonormal $X,Y$, andThe nullspace of this quadratic form is exactly the center of a Lie algebra. A zero center therefore gives positive Ricci curvature, uniformly bounded below by a positive constant times the metric through left invariance and compactness of the unit sphere in the Lie algebra.