Symmetric spaces
Claudio Gorodski September 13, 2010
Contents
1 Basic structure 5
1.1 Basic definitions . . . 5
1.2 The Cartan-Ambrose theorem . . . 7
1.3 Transvections . . . 10
1.4 Local characterization of symmetric spaces . . . 13
1.5 Globally symmetric spaces and OIL-algebras . . . 14
1.6 Exercises . . . 16
2 Classification 19 2.1 Semisimple Lie algebras . . . 19
2.2 Structure of OIL-algebras . . . 22
2.3 Exercises . . . 26
C H A P T E R 1
Basic structure
1.1 Basic definitions
Let(M, g)be a Riemannian manifold. Any pointx∈M is known to admit anormal neighborhood, that is, a neighborhoodU which is the diffeomorphic image of a neighborhoodU0of the origin0x
in the tangent spaceTxM under the exponential mapexpx :TxM →M. It is clear thatU0 can be taken to be of the form of an open ballB(0x, ǫ)for someǫ >0. Now it makes sense to define the geodesic symmetryatxto be the mapsx:U →U takingexpx(v)toexpx(−v)for everyv∈B(0x, ǫ).
Note thatsxreverses geodesics emanating fromx.
1.1.1 Proposition Fix a pointxinM. Then the following assertions are equivalent:
a. The geodesic symmetrysxatxis a local isometry.
b. There exists a local isometry sof M defined on a neighborhood of x such thats(x) = x and the differentialdsx=−id.
c. There exists an involutive local isometry ofM defined on a neighborhood of x which has x as an isolated fixed point.
Proof. a. is equivalent to b. Assume the geodesic symmetry at x is a local isometry. We havesx(0) = sx(expx(0x)) = expx(0x) = x. Moreover, let v ∈ TxM and consider the geodesic γ(t) = expx(tv)for smallt. Thendsx(v) = dtd|t=0sx(expx(tv)) = dtd|t=0expx(−tv) =−v. Hence we can takes=sx. Conversely, ifsis as in b., thens(expx(v)) = exps(x)(dsx(v)) = expx(−v)forcings to be the geodesic symmetry.
b. implies c. The fixed point set of a local isometry satxis a totally geodesic submanifold S ofM through xwhose tangent space atx is precisely the fixed point set ofdsx inTxM. Since dsx =−id, we haveTxS ={0}and henceS is discrete atx. Moreover,s2is a local isometry with s2(x) =xandd(s2)x= id, hence it must be the identity on a neighborhood ofx.
c. implies b. Supposesis an involutive local isometry as in c. Then(dsx)2 = id. Sincedsxis an orthogonal transformation ofTxM, this implies that its eigenvalues are±1. However, owing to the fact thatxis an isolated fixed point ofs, the eigenvalues must be all−1by the same argument
as above. This finishes the proof.
A Riemannian manifold (M, g)is called a locally symmetric space if the assertions of Proposi- tion 1.1.1 hold at every point of M. Furthermore, (M, g) is called a globally symmetric space or, simply, asymmetric spaceif the geodesic symmetrysxis globally defined onM for everyx∈M. 1.1.2 Examples a. Euclidean space Rn is a symmetric space. The geodesic symmetry at the
origin is s0(y) = −y for q ∈ Rn. More generally, the geodesic symmetry at x ∈ Rn is
sx(y) = 2x−y. Note that sx anss0 are conjugate by the translation τx : q 7→ q+x. This suggests the following example.
b. A Riemannian manifold is calledhomogeneous if it admits a transitive group of isometries.
A homogeneous Riemannian manifold is also called aRiemannian homogeneous space. A Rie- mannian manifold is called locally homogeneous if given two points x, y in M, there exist neighborhoodsU,V ofx,y, respectively, and an isometryf :U →V such thatf(x) = q. If (M, g)is (locally) homogeneous, the (resp., local) symmetries at different points are all con- jugate by (resp., local) isometries among themselves. Therefore it suffices to check that the geodesic symmetry at one single point is a (resp., local) isometry in order to show thatM is (resp., locally) symmetric.
c. The canonical metric on the sphereSnis realized as the induced metric from its embedding as the unit sphere in Euclidean space of one dimension more. It is then clear that Sn is homogeneous under the groupO(n+ 1)of orthogonal transformations ofRn+1. Anyway, for anyx∈Sn, the reflection on the lineRxis an orthogonal transformation ofRn+1whose restriction toSnis the geodesic symmetrysx. HenceSnis a symmetric space.
d. The real hyperbolic spaceRHn+1 can be realized as the upper sheet of the a two-sheeted hyperboloid in Lorentzian space. Namely, consider the Lorentzian inner product inRn+1 given by
hx, yi=−x0y0+x1y1+· · ·+xnyn,
wherex = (x0, . . . , xn), y = (y0, . . . , yn) ∈ Rn+1. We will writeR1,n to denoteRn+1 with such a Lorentzian inner product. Note that if x ∈ R1,n is such that hx, xi < 0, then the restriction ofh,ito the orthogonal complementhxi⊥is positive-definite. Note also that the equationhx, xi=−1defines a two-sheeted hyperboloid inR1,n. Now we can define thereal hyperbolic spaceas the following submanifold ofR1,n,
RHn={x∈R1,n| hx, xi=−1 and x0>0},
equipped with a Riemannian metricg given by the restriction of h,i to the tangent spaces at its points. Since the tangent space of the hyperboloid at a pont xis given by hxi⊥, the Riemannian metricgturns out to be well defined. Actually, this submanifold is sometimes called the hyperboloid model of RPn. Of course, as a smooth manifold, RHn is diffeomor- phic toRn. It is not difficult to see thatRHn is homogeneous under the groupO(1, n) of Lorentzian transformations ofR1,n. Moreover,RHnis a symmetric space, for the geodesic symmetry at a pointx is induced by the reflection along the lineRx, similar to the case of the sphere.
e. Let G be a compact Lie group. It is known that G admits a bi-invariant metric. (If you don’t know this, don’t worry: we’ll come back to this point later.) This means that there exists a Riemannian metric ong such that the left translations Lg : G → G, Lg(x) = gx, and right translationsRg : G → G,Rg(x) = xg, are isometries for allg ∈ G. In particular, Gis homogeneous. Moreover we claimGis also symmetric. Indeed, let us check that the inversion mapι(x) =x−1satisfies the conditions in part b. of Proposition 1.1.1 at the identity 1. Plainly,ι(1) = 1, anddι1(X) = dtd|t=0ιexpG(tX) = dtd|t=0expG(−tX) =−XforX∈T1G, whereexpGdenotes the Lie group exponential map. In particular,dι1is a linear isometry. In order to see thatdιg is a linear isometry for allg∈G, we apply the chain rule to the identity ι =Rg−1 ◦ι◦Lg−1 to getdιg = (dRg−1)1 ◦dι1 ◦(dLg−1)g and note thatdιg = (dRg−1)1 and (dLg−1)g are linear isometries. Henceιis an isometry.
f. A Riemannian manifold locally isometric to a symmetric space is locally symmetric. In par- ticular, ifM˜ → M is a Riemannian covering andM˜ is a symmetric space, thenM is locally
symmetric. The manifoldM does not have to be globally symmetric; examples are given by a surface of genusg≥2(covered byRH2) the lens spaces (covered by spheres).
We have seen that the class of symmetric spaces is a simultaneous generalization of the classes of spaces of constant curvature and compact Lie groups equipped with bi-invariant metrics. The relation of symmetric spaces to spaces of constant curvature will be made more explicit when we discuss the curvature of symmetric spaces; indeed, there is a local chaacterization of symmetric spaces in terms of curvature. On the other hand, the relation of symmetric spaces to Lie groups is related to the fact that symmetric spaces are a special type of homogeneous spaces, and this is the basis of the structure and classification results that we shall study.
1.2 The Cartan-Ambrose theorem
Probably the main application of the Cartan-Ambrose theorem is in the basic theory of symmetric spaces. Conversely, we think this theorem is so fundamental to the understanding of symmetric spaces that we give a complete proof of it. The proof is inspired by [CE75].
We first introduce some notation. LetM andM˜ be two Riemannian manifolds and letx∈M and x˜ ∈ M˜. Let I : TxM → Tx˜M˜ denote a linear isometry and letV ⊂ M denote a normal coordinate neighborhood around x such thatexpx˜ is defined in I(exp−x1(V)). We define a map ϕ:V →M˜ by setting
(1.2.1) ϕ(y) = expx˜◦I◦exp−1x (y).
The purpose of Cartan’s Theorem is to give a condition implying thatϕis a local isometry. For y∈V, letPγ:TxM →TqMdenote the parallel transport along the unique geodesicγ : [0,1]→M inV fromxtoyand letP˜γ :Tx˜M˜ →Tq˜M, where˜ y˜=ϕ(y), be the parallel transport alongγ˜=ϕ◦γ.
Finally we define
(1.2.2) Iγ :TyM →Ty˜M˜
by setting
Iγ(u) =P˜γ◦I◦Pγ−1(u).
1.2.3 Theorem (Cartan) If for everyy∈V and everyu, v, winTqM we have Iγ(R(u, v)w) = ˜R(Iγ(u), Iγ(v))Iγ(w)
whereRandR˜ are the curvature tensors ofM andM˜ respectively, thenϕis a local isometry. Moreover dϕy =Iγfor everyq∈V.
Proof. Letybe a point inV and letvbe a vector inTqM. We would like to show thatkdϕy(v)k= kvk.
There is a unique Jacobi fieldJ along the geodesicγ : [0,1]→V ⊂Mjoiningxandysuch that J(0) = 0andJ(1) =vsinceV is a normal neighborhood. It is known thatJ(t) =d(expx)tγ(0)˙ (tw) wherew=J′(0)and hence thatv=d(expx)γ(0)˙ (w). By the chain rule we have
dϕy(v) =d(expx˜)γ(0)˙˜ ◦I◦d(expx)−1γ(0)˙ (v) and hence
dϕy(v) =d(expx˜)γ(0)˙˜ (I(w)).
Thereforedϕy(v) = ˜J(1)whereJ˜is the Jacobi field along˜γsatisfyingJ˜(0) = 0andJ˜′(0) =I(w) = I(J′(0)).
LetE1, . . . , En−1 be an orthonormal set of parallel vector fields alongγ which form a basis of
˙
γ(t)⊥for everyt. We write
J =
n−1
X
j=1
αjEj
whereαjare real valued functions. The Jacobi equation now implies that the functionsαj are the unique solutions of the system of ordinary differential equations
α′′j +
n−1
X
k=1
fkjαk= 0
wherefkj = hR(Ek,γ) ˙˙ γ, Eji, satisfying the initial conditionαj(0) = 0andα′j(0) = wj wherewj
are the coefficents ofw=J′(0)with respect to the basisE1(0), . . . , En−1(0), i.e.,w=P
jwjEj(0).
It is clear thatkvk2=P
jα2j(1).
Now let E˜1, . . . ,E˜n−1 denote the parallel vector fields along ˜γ such that E˜j(0) = I(Ej(0)).
Notice thatϕγ(t)(Ej(t)) = ˜Ej(t)andϕγ(t)( ˙γ(t)) = ˙˜γ(t). Hence hR( ˜˜ Ek,γ) ˙˜˙˜ γ,E˜ki=hR(Ek,γ˙) ˙γ, Eki Using the Jacobi equation as above we therefore get
J˜=
n−1
X
j=1
αjE˜j,
i.e., the coefficients ofJandJ˜are the same with respect to the two bases. Here we have used that J˜satisfies the initial conditionJ˜(0) = 0andJ˜′(0) =P
jwjE˜j(0). Hencekdϕy(v)k2 =P
jα2j(1) = kvk2. Note also thatdϕy(v) = ˜J(1) =Iγ(J(1)) =Iγ(v). This finishes the proof.
There is a global version of Cartan’s theorem due to Ambrose. We start with some preliminar- ies. Abroken geodesicin a Riemannian manifoldM is a continuous curveγ : [0,1]→M such that there is a partitiont0= 0< t1 <· · ·< tk< tk+1 = 1with the property thatγ|[ti−1, ti]is a geodesic fori= 1, . . . , k+ 1. Let complete Riemannian manifoldsM andM˜ be given and letx ∈ M and
˜
x ∈ M˜. LetI : TxM → T˜xM˜ be a linear isometry. Then we associate to a broken geodesicγ in M starting atx a broken geodesicγ˜ inM˜ starting at x˜as follows: We first define ˜γ on [0, t1]by settingγ˜(t) = exp˜xtI( ˙γ(0)). Assume now that we have definedγ˜on[0, ti]and thatti <1. Then we define˜γon[ti, ti+1]by setting˜γ(t) = expγ(t˜ i)(t−ti)Iiγ( ˙γ(ti+)), whereIiγ= ˜Pi˜γ◦I ◦Pi−1γ , and PiγandPi˜γare the parallel translations alongiγ =γ|[0, ti]andiγ˜= ˜γ|[0, ti]respectively. Now we can more generally defineIη :Tη(1)M →Tη(1)˜ Mfor any broken geodesicη: [0,1]→M starting at xby settingIη =Pη˜◦I◦Pη−1.
1.2.4 Theorem (Cartan-Ambrose) Assume thatM andM˜ are complete Riemannian manifolds, M is simply connected, and that
Iγ(R(u, v)w) = ˜R(Iγu, Iγv)Iγw
for allu, v, w ∈Tγ(t)M and all broken geodesicsγ starting inx. We define a mapΦ :M → M˜ by setting Φ(y) = expx˜(Iγ˙(0))whereγ : [0,1]→ M is a geodesic joiningxandy. ThenΦis well-defined, a local isometry, and a covering map. In particular,Φis an isometry ifM˜ is also simply connected.
We will need the following lemma in the proof of Theorem 1.2.4.
1.2.5 Lemma Let M be a complete Riemannian manifold and let γ0 and γ1 be broken geodesics in M which are defined on [0,1], join x andy, and are homotopic to each other. Then there is a homotopyΓ : [0,1]×[0,1] → M betweenγ0 andγ1 such thatγs|[ti, ti+1]is a geodesic fori = 0, . . . , k−1and alls whereγs(t) = Γ(s, t)andt0 = 0< t1<· · ·< tk = 1is a partition of[0,1].
Proof. Letκ : [0,1]×[0,1] → M be a continuous homotopy betweenγ0 andγ1, i.e., κ0 = γ0, κ1 =γ1,κ(s,0) =x, andκ(s,1) =q. Letℓdenote the radius of a closed ball aroundxcontaining the image of the homotopyκ. Letr > 0be the minimum of the injectivity radius on the closed ball Bℓ(x). Let t0 = 0 < t1 < · · ·tk = 1 be a partition with the property that γ0|[ti, ti+1]and γ1|[ti, ti+1]are geodesics for alli= 0, . . . , k, and such thatκs|[ti, ti+1]is contained inBr(κs(ti))for alli= 0, . . . , kand allsin[0,1]. Now we define the homotopyΓ : [0,1]×[0,1]→ M by setting γs|[ti, ti+1]equal to the unique shortest geodesic betweenκs(ti) andκs(ti+1)for all i = 0, . . . , k and allsin[0,1]. The homotopyΓhas by construction the properties asked for in the claim of the
lemma.
Proof of Theorem 1.2.4. We prove the following claim which implies thatΦis well defined: For all broken geodesicsγ0,γ1 : [0,1] → M starting atxwithγ0(1) = γ1(1), we haveγ˜0(1) = ˜γ1(1)and Iγ0 =Iγ1. We may assume thatγ0andγ1have common break points att1, . . . , tksimply by adding some of them if nececesary.
The first case is when γ0 andγ1 are both contained in a normal coordinate neighborhoodV aroundx. Then Theorem 1.2.3 implies thatϕ= expx˜◦I◦exp−x1|V is an isometry and, forj= 0,1, thatdϕγj(t1)=I1γj.
In this proof it will be convenient to consider iθj = γj|[ti, ti+1]and iθ˜j = ˜γj|[ti, ti+1], and to notice thatIi+1γj =P
iθ˜j◦Iiγj◦P−1
iθj. Sinceϕis an isometry,
ϕ(γj(t)) = exp˜xtI( ˙γj(0+)) = ˜γj(t), for0≤t≤t1. Proceeding by induction oni= 1, . . . , kwe now have
ϕ(γj(t)) = ϕ(expγj(ti)(t−ti) ˙γj(ti+))
= expγ˜j(ti)(t−ti)(dϕγj(ti)γ˙j(ti+))
= expγ˜j(ti)(t−ti)(Iiγjγ˙j(ti+))
= ˜γj(t), forti≤t≤ti+1; this impliesϕ(iθj) =iθ˜j and then
dϕγj(ti+1) = P
iθ˜j◦dϕγj(ti)◦P−1
iθj
= P
iθ˜j◦Iiγj◦P−1
iθj
= Ii+1γj,
and the induction step is complete. This proves thatϕ◦γj = ˜γj and henceγ˜0(1) = ϕ(γ0(1)) = ϕ(γ1(1)) = ˜γ1(1).
The second case we want to consider is when for alli= 0, . . . , k−1(of course we may assume thatk ≥ 1) there is a normal coordinate neighborhood ofγ0(ti)(resp.γ˜0(ti)) containingγ0(ti+1), γ1(ti+1) and γ1(ti+2) (resp. ˜γ0(ti+1), γ˜1(ti+1) and γ˜1(ti+2)) and the minimal geodesic segments betwen them.
Ifk= 1, thenγ0andγ1are contained in a normal coordinate neighborhood aroundxand the result follows from the first case. We proceed by induction on k. Suppose that γ0 andγ1 have k ≥ 2 breaks. We introduce the auxiliary minimal geodesicτ : [tk−1, tk] → M fromγ0(tk−1) to γ1(tk). By the induction hypothesis we have
(1.2.6) (k−1^γ0∪τ)(tk) =k˜γ1(tk) and Ik−1γ0∪τ =Ikγ1.
Notice that the isometryIk−1γ0 : Tγ0(tk−1)M → T˜γ0(tk−1)M˜ induces a correspondence η 7→ ηˆbe- tween geodesics inM starting atγ0(tk−1)and geodesics inM˜ starting at˜γ0(tk−1). Set
η0 =γ0|[tk−1,1] and η1 =τ∪γ1|[tk,1] =τ ∪kθ1. By the first case we have
ˆ
η0(1) = ˆη1(1) and Iη0 =Iη1, which is clearly equivalent to
˜
γ0(1) =(k−1^γ0∪η1)(1) and Iγ0 =Ik−1γ0∪η1, and by (1.2.6) we have that
(k−1γ0^∪τ ∪kθ1)(1) =(k−1^γ0∪η1)(1) = ˜γ1(1).
It follows thatγ˜0(1) = ˜γ1(1), as desired.
Finally assume that γ0 and γ1 are arbitrary broken geodesics in M starting at x such that γ0(1) = γ1(1). Since M is simply connected, γ0 and γ1 are homotopic to each other, so by Lemma 1.2.5 a homotopy Γ between γ0 and γ1 can be chosen such that theγs(t) = Γ(s, t) for s ∈ [0,1]are broken geodesics, with common break points at t1, . . . , tk. By refining the partition 0 =t0 < t1< . . . < tk < tk+1= 1we may assume thatk≥1and, fors∈[0,1]andi= 0, . . . , k−1, thatγs(ti+1)andγs(ti+2)belong to a normal coordinate neighborhood ofγs(ti).
Now if s0,s1 ∈ [0,1]are sufficiently close, thenγs0 andγs1 are seen to satisfy the conditions of the second case. It follows thatγ˜s0(1) = ˜γs1(1). This shows thatγs(1)is locally constant with respect tos ∈ [0,1], which obvioulsy implies thatγ˜s(1)is constant with respect tos ∈ [0,1]. In particular,γ˜0(1) = ˜γ1(1). This completes the proof of the claim.
The proof so far implies that Φ(y) = ˜γ(1) where γ is any broken geodesic defined on[0,1]
and joiningxtoy. LetV be a normal coordinate neighborhood ofyinM. Now it is clear thatΦ restricted toV coincides with the map
expq˜◦Iγ◦exp−q1|V,
which is an isometry by Proposition 1.2.3. ThereforeΦis a local isometry. It is well known that a local isometry fromM toM˜ is a covering map ifM˜ is complete. Hence we have proved thatΦis
a covering map which finishes the proof.
1.3 Transvections
Let (M, g) be a locally symmetric space. The existence of a local isometry reversing geodesics through any point ofM has very strong consequences for the geometry of M that we begin to discuss now.
As a starter, we can try to see if the geodesic symmetries generate some kind of group. To study this question more sistematically, we first fix a pointxinM and a geodesicγ : (−ǫ, ǫ)→M through x = γ(0). Fort ∈ (−ǫ, ǫ), the geodesic symmetrysγ(t), which we denote simply byst, is defined and a local isometry. Now the composite mappt := st
2s0 is locally defined and a local isometry ofM; this is called a(local) transvection alongγ. Sinceγ passes through x, we also say thatptis a transvection atx.
1.3.1 Proposition Letγ : (−ǫ, ǫ)→Mbe a geodesic throughx=γ(0).
a. The transvectionpt induces translation along the curve γ, that is, pt(γ(t0)) = γ(t+t0). More generally, pt induces parallel transport on vectors alongγ, in the sense that ifv ∈ Tγ(t0)M then X(t) = (dpt−t0)γ(t0)(v)is a parallel vector field alongγ.
b. The transvections{pt}along γ form a local one-parameter group of local isometries ofM, namely, pt+t′ =ptpt′ whenever both hand sides are defined.
c. The transvectionptdepends only onγbut not on the chosen initial pointx=γ(0). In other words, st
2s0 =st0+t
2st0.
Proof. a. We have pt(γ(t0)) = st
2s0(γ(t0)) = st
2(γ(−t0)) = γ(t+t0). In the proof of the second assertion, we assume that t0 = 0 for simplicity of notation. Now we want to prove that X(t) = (dpt)x(v) is parallel along γ. Let Y denote the parallel vector field along γ such that Y(0) =v. We are going to use the fact that isometries act on vector fields by push-forward taking parallel vector fields to parallel vector fields. Fixt1. ThenZ(t) = (dpt1)γ(t1−t)(Y(t−t1))∈Tγ(t)M is a parallel vector field alongγ. Note thatZ(t1) = (dpt1)x(v) = X(t1). Sincet1 is arbitrary, this completes the proof of a.
b. An isometry is locally determined by its differential at one point. Moreover, the composition of parallel transports along two adjacent segments of γ equals the parallel transport along the justaposed segment, so the result follows from part a.
c. Use part b. to write pt = pt+2t0p−2t0 = st
2+t0s0s−t0s0. We have already remarked that for a local isometry g, the conjugationgsxg−1 = sgx. Applying this to g = s0 = g−1 yields that
s0s−t0s0=st0, as desired.
It follows from Proposition 1.3.1 that each geodesic determines a unique local one-parameter group of transvections along it.
1.3.2 Proposition A connected locally symmetric space is locally homogeneous. A connected globally symmetric space is homogeneous and complete.
Proof. Suppose (M, g) is connected and locally symmetric. Declare two points of M to be equivalent if there exists a local isometry of M mapping the first point to the second one. It is enough to prove that the equivalence classes are open. Indeed, the existence of transvections implies that a normal neighborhood of a point is contained in its equivalence class.
If (M, g) is in addition globally symmetric, transvections are global isometries and by this argumentMis globally homogeneous. Completeness follows from homogeneity in general.
It follows from Proposition 1.3.2 and the Hopf-Rinow theorem that a globally symmetric space M is geodesically complete. In this case, the transvections along a geodesic form a full one- parameter group of (global) isometries ofM. Moreover, since any two points ofM can be joined by a geodesic arc, the product of any two geodesic symmetries ofM is a transvection, and the transvections generate a transitive subgroup of the isometry group ofM.
IfM is globally symmetric, the group generated by transvections at a fixed pointx is a con- nected subgroup of the isometry group ofM. We will see in section 2.2 that in fact this group coincides with the identity component of the full isometry group ofMin most cases.
1.3.3 Example Consider the unit sphereSn. We have already remarked that the geodesic symme- trysxat a pointxis the isometry induced by reflection ofRn+1on the lineRx. It follows that for y 6=±x, the transvectionsysxis the isometry which rotates the geodesic arcxy by an angle twice the angle betweenxandy, and which is the identity on the orthogonal complement of the plane spanned byx,y. (ify =−x, thensy =sx). It is more or less clear (e.g. ifn = 2, every element of SO(3)has an axis of rotation) then that the group generated by transvections atxis the full special orthogonal groupSO(n+ 1).
We start by looking at the local transvections at a fixed pointxfor a locally symmetric space.
First we need to recall some facts about Killing fields. AKilling fieldXon a Riemannian manifold M is the infinitesimal generator of a (local) one-parameter group of (local) isometries ofM, and it can be characterized by the equationLXg = 0, or, equivalently, that(∇X)x is a skew-symmetric endomorphism ofTxMfor everyx∈M.
1.3.4 Lemma A Killing vector fieldXon a Riemannian manifoldMis completely determined by the values ofXand∇Xatx.
Proof. Fix a point x ∈ M, and denote by g the vector space of Killing vector fields on M. The assertion is then equivalent to the linear map X ∈ g 7→ ((∇X)x, Xx) ∈ so(TxM) ⊕TxM being injective. So supposeXx = (∇X)x = 0. For any geodesicγ originating atx, the restriction J :=X◦γis clearly a Jacobi field alongγ, and the assumption onXimplies thatJ(0) =J′(0) = 0, henceJ ≡0. The manifoldM is not necessarily complete, but any point of it can be joined toxby broken geodesic, so that the argument above suffices to conclude thatX≡0. It follows from the identityL[X,Y]= [LX, LY]and Lemma 1.3.4 that the space of Killing fields on an-dimensional Riemannian manifold forms a finite-dimensional Lie algebra of dimension at most12n(n+ 1). In the case of a complete manifold, it is easily seen that Killing fields are complete (since they have constant length along their integral curves).
Suppose now that M is a locally symmetric space. Fix a base-point x in M. For every one- parameter group of transvections {pt} originating at x, there is a corresponding Killing vector fieldY whose value atq ∈ M isY(y) = dtd|t=0ptq; such a Y is called an infinitesimal transvection atx.
1.3.5 Proposition A vector field Y is an infinitesimal transvection at x if and only if (∇Y)x = 0. It follows that the bracket of two infinitesimal transvections vanishes atx.
Proof. Let {pt} be the transvection at x that Y generates and take any curve η(s) passing through x at s = 0. For the first assertion, it suffices to prove that ∇(Yds◦η)|s=0 = 0. Since the Levi-Civit`a connection is torsionless,
(1.3.6) ∇
ds d
dtptη(s) = ∇ dt
d
dsptη(s) = ∇
dt(dpt)η(s)η(s).˙
By Proposition 1.3.1.a., the vector field(dpt)xη(0)˙ is parallel alongγ(t) = pt(x). SinceY(η(s)) =
d
dt|t=0ptη(s), evaluating eqn. (1.3.6) at s = t = 0 yields one direction of the claim. Also, this argument can be reversed to prove the other direction.
The last assertion follows from[Y1, Y2] =∇Y1Y2− ∇Y2Y1.
Let(M, g)be a locally symmetric space and fix a pointx ∈ M. Thelinear isotropy groupofM atx is defined to be the groupK of all linear isometries of the tangent spaceTxM that preserve the curvature tensor. ThenK has the structure of a Lie group because it is a closed subgroup of the orthogonal group ofTxM. By Theorem 1.2.3, each element ofK extends to an isometry of a normal neighborhoodUofx. NowKis a group of isometries ofU, called thelocal isotropy groupof M atx. Letkbe the Lie algebra ofK. Then the action ofKonU representskfaithfully as a finite- dimensional Lie algebra of vector fields on U. These vector fields vanish at x because K(x) = {x}. Denote bypthe set of infinitesimal transvections atx. It follows from Proposition 1.3.5 and Lemma 1.3.4 thatpis a vector space and the mapY ∈p7→Yx ∈TxMis a linear isomorphism.
1.3.7 Proposition We have
[k,k]⊂k, [k,p]⊂p, [p,p]⊂k.
In particular,g = k+p(direct sum of vector spaces) is a Lie algebra of vector fields defined on a normal neighborhood ofx. The geodesic symmetrysxinduces an involutive automorphismsofgsuch thatkandp are respectively the±1-eigenspaces.
Proof. The first inclusion is clear. For the second one, letk ∈ K andY ∈ p. Then Y gener- ates a one-parameter of transvections atx, which we denote byetY andAdkY is the Killing field
d
dt|t=0ketYk−1. Using that d(etY)x is parallel transport alongγ(t) = etY ·x, it is easy to see that dketY·xd(etY)xdk−1x is parallel transport alongk·γ. ThereforeketYk−1is the one-parameter group of transvections alongk·γand thusAdkY ∈p, implying the second inclusion. Finally, ifY1,Y2∈p, then[Y1, Y2]is a Killing field generating a one-parameter group{gt}of local isometries fixing x, by Proposition 1.3.5, and obviously preservingRx. Hencegt∈Kand[Y1, Y2] = dgdtt ∈k.
Finally, for Z ∈ g, set sZ = dtd|t=0sxetZs−x1. IfX ∈ k, thensxetXs−x1 is a local isometry fixing x with differential atx given by d(etX)x. HencesxetXs−x1 = etX andsX = X. If Y ∈ p, then sxetYs−1x is the local transvection atxalongsxetY ·x. ThereforesY =−Y. This finishes the proof.
To a point x in a locally symmetric space, there is now associated a triple (g, s, B), where g = k+p andsare as in Proposition 1.3.7, and B is the inner product of TxM lifted topunder the identificationp ∼= TxM,Y 7→ Yx. Note thatkis the Lie algebra of a compact Lie group. It is also clear from the above discussion thatkacts faithfully onp. An ideal ofg, contained ink, would be in the kernel of the action ofkonp. Hencekdoes not contain nontrivial ideals ofg. Finally,K acts by isometries on a normal neighborhood ofx, soB isAdK-invariant. The triple (g, s, B)is called theorthogonal involutive Lie algebra(or, for short,OIL-algebra) ofM atx. IfM is in addition connected, then the choice of pointxis unimportant by Proposition 1.3.2, and the OIL-algebra is associated toM.
1.4 Local characterization of symmetric spaces
LetM be a Riemannian manifold with Levi-Civit`a connection∇. The condition in Theorem 1.4.1 is clearly equivalent to the sectional curvature ofM being invariant under parallel transport of tangent2-planes. The theorem makes clear the degree of generalization that we get from space forms to symmetric spaces.
1.4.1 Theorem We have thatM is a locally symmetric space if and only if∇R= 0.
Proof. Letx ∈ M and consider the geodesic symmetrysx. IfM is locally symmetric, this is a local isometry atx. Since(dsx)x=−id, the equation
d(sx)x∇uR(v, w)z=∇d(sx)xuR(d(sx)xv, d(sx)xw)d(sx)xz foru,v,w,z∈TxM yields that∇R= 0.
Conversely, assume that ∇R = 0 and let x ∈ M. Take I = −idand define ϕand Iγ as in eqns. (1.2.1) and (1.2.2). Then φ is the geodesic symmetry sx. Since R is a tensor of degree 4, Rx is invariant under I. Since ∇R = 0, R is preserved by parallel transport. It follows from Theorem 1.2.3 thatsxis a local isometry. Sincex∈M is arbitrary,M is locally symmetric.
1.4.2 Proposition A complete simply-connected locally symmetricMspace is globally symmetric.
Proof. Letx ∈M. The local symmetrysx admits an extension to a global isometryΦofM by
Theorem 1.2.4. HenceM is globally symmetric.
1.4.3 Corollary A complete simply-connected manifold is globally symmetric if and only if∇R= 0.
1.4.4 Corollary The universal covering of a complete locally symmetric space is a globally symmetric space.
We close this section with some interesting side remarks. The following result poses a topo- logical obstruction for a smooth manifold to admit the structure of a symmetric space.
1.4.5 Proposition The fundamental group of a symmetric space is Abelian.
Proof. LetM be a symmetric space and fixx ∈ M. By applying a standard curve-shortening process, one shows that any nontrivial element inπ1(M, x)can be represented by a closed geodesic through x, sayγ. The geodesic symmetrysx reverses geodesics throughx, sosx(γ(t)) = γ(−t).
Now the homomorphism induced bysxon the fundamental group level is group inversion. The result follows from noticing that group inversion is a homomorphism only if the underlying group
is Abelian.
1.4.6 Corollary A surface of genusg≥2does not admit a Riemannian metric with respect to which it is a symmetric space.
1.5 Globally symmetric spaces and OIL-algebras
Recall that the isometry groupGof a Riemannian manifoldM, equipped with the compact-open topology, has a natural structure of Lie group such that the action of G on M is smooth and represents its Lie algebragas the Lie algebra of Killing vector fields onM. It is also worth recalling that convergence of a sequence of isometries inGin the compact-open topology is equivalent to pointwise convergence inM. Finally, the isotropy groupGxat a pointx∈M is compact.
1.5.1 Proposition Let M be a simply-connected symmetric space with Lie group of isometries G. Fix x∈M and writeKbe the local isotropy group atx, andP for the set of local transvections atx. Then the elements ofKandP extend to global isometries ofM,K =GxandG=KP =P K.
Proof. The extension follows from Theorem 1.2.4. Ifg ∈ Gandy = g−1x, there isp ∈ P such that py = x. Nowgp−1x = x, so gp−1 ∈ K andg = (gp−1)p ∈ KP, provingG = KP. Finally,
P K=KP becauseKis a subgroup ofGandP−1=P.
The following proposition is clear. In particular, we see that symmetric spaces can be viewed as a very special class of homogeneous spaces.
1.5.2 Proposition LetM be a globally symmetric space, letGbe the Lie group of all isometries ofM, fix x∈M and writeK be the isotropy groupGx. Then:
a. M is represented as the coset spaceG/K;
b. the OIL-algebra associated toM is(g, s, B), wheregis the Lie algebra ofG,s= Adsx, the projection π : G→ M given byπ(g) =gxhas differentialπ∗ : p∼= TxM, andB is the pull back underπof the inner product onTxM.
c. ifk∈KandY ∈p, thenπ∗(AdkY) =dkx(π∗Y).
Proof. (a) is obvious. In (b), the Lie algebra ofGcoincides with the Lie algebra introduced in Proposition 1.3.7 by Proposition 1.5.1, but a direct argument is also easy: every Killing fieldZ on Mdecomposes as a sum of Killing fieldsX+Y, whereY ∈pis the infinitesimal transvection such thatYx=ZxandX=Z−Y ∈k. The assertions abouts,π,Bare clear. For (c), we compute
π∗(AdkY) = π∗ d
dt
t=0kexptY k−1
= d
dt
t=0π(kexptY k−1)
= d
dt
t=0kexptY ·x
= dkx(Yx)
= dkx(π∗Y),
as desired.
Anabstract OIL-algebrais defined to be a triple(g, s, B), wheregis a real finite-dimensional Lie algebra,sis an involutive automorphism ofg, the fixed point setkofsdoes not contain nontrivial ideals ofgandBis anadk-invariant inner product on the−1-eigenspacepofs.
Given an abstract OIL-algebra(g, s, B), we construct a simply-connected symmetric space as follows. Let G˜ be the simply-connected Lie group with Lie algebra g, and let K˜ ⊂ G˜ be the connected subgroup with Lie algebrak. There is an involutionσofG˜ such thatdσ =s, andK˜ is the identity component of the fixed point subgroupG˜σ; thusK˜ is closed inG. Now˜ M = ˜G/K˜ is a simply-connected homogeneous space, but in general G˜ does not act effectively on M; let Z˜ = {g ∈ G˜ :g :M → M is the identity} ⊂K˜ be the kernel of the action. Z˜ is discrete because k does not contain nontrivial ideals ofg. NowM = G′/K′ where G′ = ˜G/Z,˜ K′ = ˜K/Z˜ have resp. Lie algebrasg,k. The projectionπ : G′ → M yields π∗ : p ∼= TxM, wherex = 1K′, andB defines aK′-invariant inner product onTxMthat extends to aG′-invariant Riemannian metric on M. Finally,σ inducesψ : M → M by the ruleψ(gK′) = σ(g)K′, orψ(gx) = σ(g)x = πσ(g)for g ∈ G′. Note that ψis well defined and fixes x. Also,dψx(Yx) = π∗s(Y) = −Yx forY ∈ p, so dψx =−id. Sinceψ(gg1K′) =σ(g)σ(g1)K′ =σ(g)ψ(g1K′), we haveψ◦g=σ(g)◦ψ. By the chain ruledψgx = (d(σ(g)))x◦(dψ)x◦(dg−1)gxfor everyg ∈ G′. It follows thatψis an isometry ofM. HenceM is symmetric.
In particular, suppose that we apply the construction described in the previous paragraph to the OIL-algebra(g, s, B)associated to a given symmetric spaceM =G/Kas in Proposition 1.5.2 to obtainM˜ = G′/K′. Since G′ is simply-connected and has the same Lie algebra asG, there is a covering homomorphism p : G′ → G. Moreover, K′ andK also have the same Lie algebra, sop(K′) = K andp inducesp¯ : ˜M → M. Owing to the identificationsTx˜M˜ ∼= p ∼= TxM, d¯px˜
is the identity so, in particular, an isometry. The equivariance ofp¯now implies that it is a local isometry. Finally, the completeness ofM,M˜ implies thatp¯: ˜M → M is a Riemannian covering.
In particular, ifMis taken simply-connected, then it is isometric toM.˜
1.5.3 Example It is known that the group of isometries of the unit sphere isO(n+ 1). The isotropy group at, say, x = (1,0, . . . ,0)t is O(n). Now the associated OIL-algebra (g, s, B) is given by g=so(n+ 1),sis conjugation by the matrix
−1 0 0 In
,
whereInis an identity block of ordern, its eigenspaces are given by k=
0 0 0 A
:A∈so(n)
, p=
Y =
0 −vt v 0
:v ∈Rn
,
andB(Y, Y) = (const) tr(YtY) = 2||v||2.
1.5.4 Example LetH be a simply-connected compact connected semisimple Lie group, and de- note its Lie algebra byh. We can define an abstract OIL-algebra(g, s, B) by setting g = h⊕h, s(X, Y) = (Y, X) forX, Y ∈ h; then k is the diagonal of g and p = {(X,−X) : X ∈ h}; put B((X,−X),(Y,−Y)) =λβ(X, Y)forX,Y ∈h, whereβ <0andβdenotes the Killing form ofh.
The simply-connected symmetric space M associated to(g, s, B)isM = H×H/∆H, where
∆H ={(h, h) :h ∈H}. Note thatH×Hacts transitively onH by the rule(h1, h2)·x =h1xh−12 ; the isotropy at1∈His∆H, hence there is a diffemorphismH×H/∆H ∼=H,(h1, h2)∆H 7→h1h−21. Using this identificationM =H, the projectionπ:H×H→Hisπ(h1, h2) =h1h−21, the geodesic symmetry ψ : H → H at 1 is ψ(x) = ψ((x,1)∆H) = σ(x,1)∆H = (1, x)∆H = x−1. Finally π∗(X, Y) = X−Y forX, Y ∈ h, soπ∗(X,−X) = 2X and the metric on H isH×H-invariant (i.e. bi-invariant) with value at1given by a negative multiple of the Killing form ofh.
1.5.5 Example It is not difficult to show that the full group of isometries of Euclidean space is the semidirect productG=O(n)⋉Rn(cf. Exercise 4). The isotropy group at the origin isO(n). Now its associated OIL-algebra(g, s, B)isg =so(n) +Rn(semi-direct sum),s: g → gis+1onso(n) and−1onRn, andBis the inner product onRn. Note that[p,p] = 0in this example.
More generally, for any subalgebrak ⊂ so(n), the semi-direct sumg = k+Rnhas a similar structure of OIL-algebra, including the casek = 0. The associated simply-connected symmetric space is again Euclidean space.
1.6 Exercises
1 LetM be a complete connected Riemannian manifold with vanishing sectional curvature. De- duce from Cartan-Ambrose theorem that for everyx∈M,expx :TxM →M is a smooth covering.
2 LetM be a symmetric space andx ∈M. Prove that the geodesic symmetrysx normalizes the group generated by transvections atx.
3 Let(M, g)be a connected Riemannian manifold and consider the underlying metric space struc- ture(M, d). Prove that any isometryf of(M, g)isdistance-preserving, that is, it satisfies the condi- tion thatd(f(x), f(y)) =d(x, y)for everyx,y∈M.
4 Describe the isometry groupGofRn:
a. Show thatGis generated by orthogonal transformations and translations.
b. Show thatGis isomorphic to the semidirect productO(n)⋉Rn, where (B, w)·(A, v) = (BA, Bv+w)
forA,B ∈O(n)andv,w∈Rn.
(Hint: Use the result of the previous exercise.)
5 Prove that every isometry of the unit sphereSnof Euclidean spaceRn+1 is the restriction of a linear orthogonal transformation ofRn+1. Deduce that the isometry group ofSnis isomorphic to O(n+ 1). What is the isometry group ofRPn?
6 Prove that every isometry of the hyperboloid model ofRHnis the restriction of a linear Loret- zian orthochronous transformation ofR1,n. Deduce that the isometry group ofRHnis isomorphic toO0(1, n).
7 Let(g, s, B)and(g′, s′, B′)be two OIL-algebras such thatgis a subalgebra ofg′,s=s′|g,p=p′ andB = B′. Prove that the corresponding simply-connected symmetric spaces M andM′ are isometric. (There is essentially only one nontrivial concrete example; what is it?)
C H A P T E R 2
Classification
In this chapter, we explain how OIL-algebras can be used to classify simply-connected symmetric spaces.
2.1 Semisimple Lie algebras
In this section, we recall in more detail the structure of semisimple Lie algebras.
Let g be a Lie algebra. Then GL(g) is a Lie group with Lie algebra gl(g) consisting of all endomorphisms of the vector space underlyingg. The group of automorphisms ofg, denoted by Aut(g), is clearly a closed subgroup ofGL(g). Recall that a closed subgroup of a Lie group is a Lie subgroup with the subspace topology. HenceAut(g) is a Lie subgroup ofGL(g). Its Lie algebra consists of the endomorphismsD∈gl(g)such that
[exp(tD)·X,exp(tD)·Y] = exp(tD)·[X, Y] forX,Y ∈g,t∈R, or, equivalently,
(2.1.1) [DX, Y] + [X, DY] =D[X, Y]
forX,Y ∈g. The endomorphismsDsatisfying eqn. (2.1.1) are calledderivationsofg. Thus the Lie algebra ofAut(g)is the spaceDer(g)of all derivations ofg.
In particular, the Jacobi identity shows thatadX ∈Der(g)for allX∈g, so that the image of the adjoint representationad[g]is a subalgebra ofDer(g), again by Jacobi. Its elements are calledinner derivations. LetInt(g)be the connected subgroup ofAut(g) defined byad[g]. In accordance with the next proposition, this group is called theadjoint group of gand its elements are called inner automorphismsofg.
2.1.2 Proposition The adjoint group Int(g) is canonically isomorphic toG/Z(G), where G is any Lie group with Lie algebragandZ(G)denotes the center ofG.
Proof. The image of the adjoint representationAd : G → Aut(g) isInt(g), becauseAdexpX =
exp adX forX ∈g, and its kernel isZ(G).
Letgbe a Lie algebra. TheCartan-Killing formofgis the symmetric bilinear form β(X, Y) = tradXadY (trace)
whereX,Y ∈g.
2.1.3 Proposition a. Ifa ⊂ g is an ideal, then the Cartan-Killing form ofa is the restriction ofβ to a×a.
b. Ifs∈Aut(g), thenβ(sX, sY) =β(X, Y)forX,Y ∈g.
c. β(adXY, Z) +β(Y,adXZ) = 0forX,Y,Z ∈g (β isad-invariant).
Proof. (a) IfX,Y ∈athenadXadY mapsaintoa. (b) Ifs∈Aut(g)thenadsX =s◦adX ◦s−1.
(c) It follows from Jacobi.
In order to avoid unnecessary technicalities, we adopt the following nonstandard, but com- pletely equivalent definitions. We call a Lie algebrag: semisimpleif β is nondegenerate;simpleif it is semisimple and does not contain nontrivial ideals.1 Note that by thead-invariance ofβ, its kernel is always an ideal of the underlying Lie algebra.
2.1.4 Proposition Letgbe a semisimple Lie algebra, leta⊂gbe an ideal and a⊥={X∈g:β(X,a) = 0}.
Thena⊥is an ideal,aanda⊥are semisimple andg=a⊕a⊥(direct sum of ideals).
Proof. Thead-invariance ofβ implies thata⊥is an ideal ofg. Thena∩a⊥is an ideal ofg, and again byad-invariance ofβ,a∩a⊥is Abelian; in fact in fact, for everyZ∈gandX,Y ∈a∩a⊥,
β(Z,[X, Y]) =β([Z, X], Y) = 0,
so[X, Y] = 0by nondegeneracy ofβ. Fix now a complementary subspacebofa∩a⊥ing. Then, forX∈a∩a⊥andY ∈g, the linear mapadXadY has no diagonal elements and thusβ(X, Y) = 0 yielding X = 0. We have shown that a∩a⊥ = 0. The nondegeneracy of β also implies that dima+ dima⊥ = dimg whenceg = a⊕a⊥. Thataanda⊥ are semisimple is a consequence of
Proposition 2.1.3(a).
2.1.5 Corollary A semisimple Lie algebragdecomposes into a direct sum of simple idealsg=g1⊕· · ·⊕gr. In particular,[g,g] =g.
2.1.6 Proposition Ifgis semisimple thenad[g] = Der(g), i.e. every derivation is inner.
Proof. The centerzis always contained in the kernel of the Cartan-Killing form, soz= 0. Now ad : g → Der(g)is injective, soad[g]is semisimple. Denote byβ, β′ the Cartan-Killing forms of ad[g]andDer(g). LetD ∈ Der(g). ThenadDX = [D,adX]forX ∈ gso thatad[g]is an ideal of Der(g). Letadenote itsβ′-orthogonal complement inDer(g). Then alsoais an ideal ofDer(g). We have
β(ad[g],ad[g]∩a) =β′(ad[g],ad[g]∩a) = 0
so ad[g]∩a = 0 . Therefore adDX = [D,adX] ∈ ad[g]∩a = 0 for D ∈ a and X ∈ g. Since
ker ad =z= 0, this impliesa= 0, as desired.
2.1.7 Corollary Ifgis semisimple thenInt(g) = Aut(g)0.
Proof.Int(g)is connected and both hand sides have the same Lie algebra.
1The more usual definitions are that a Lie algebra is calledsemisimpleif it does not admit nontrivial solvable ideals, and it calledsimpleif it is not Abelian and does not admit nontrivial ideals. Cartan’s criterium for semisimplicity is that the Killing form be nondegenerate.
2.1.8 Proposition Letgbe a Lie algebra. The following assertions are equivalent:
a. gis the Lie algebra of a compact Lie group.
b. Int(g)is compact.
c. gadmits anad-invariant positive definite inner product.
d. g = z⊕[g,g]wherezis the center ofgand[g,g]is semisimple with Cartan-Killing form negative definite.
In this case, we say thatgis acompactLie algebra.
Proof. (c) implies (d). Leth,ibe anad-invariant positive definite inner product ong, h[X, Y], Zi+hY,[X, Zi= 0
forX,Y,Z ∈ g. The centerzisad-invarint, so that also itsh,i-orthogonal complementz⊥ isad- invariant. Nowg = z⊕z⊥, direct sum of ideals. The Cartan-Killing form of z⊥is the restriction of the Cartan-Killing formβofg. Owing to thead-invariance ofh,i,adX is skew-symmetric with respect toh,iforX∈g, thus it has purely imaginary eigenvalues. Thereforeβ(X, X) = trad2X ≤0 and equality holds if and only ifadX = 0if and only ifX∈z. This proves thatβ|z⊥×z⊥is negative definite and hencez⊥is semisimple and[g,g] = [z⊥,z⊥] =z⊥.
(d) implies (b). We haveInt(g) = Int(z)×Int([g,g]) = Int([g,g])sincezis Abelian thus, without loss of generality, we may assumegis semisimple with negative definite Cartan-Killing form. Let O(g)⊂GL(g)the compact subgroup ofβ-preserving transformations. ClearlyAut(g)is contained inO(g)as a closed, thus compact subgroup. Now Corollary 2.1.7 yields the result.
(b) implies (c). SinceInt(g) = Ad(G)is compact there exists anAd-invariant positive definite inner product ong. It is alsoad-invariant.
Now (b), (c) and (d) are equivalent. We next show that (b) and (d) imply (a). Int([g,g]) = Int(g) is compact and andInt([g,g]) has Lie algebra ad([g,g]) ∼= [g,g]since [g,g] is semisimple. Now g=z⊕[g,g]is the Lie algebra ofS1× · · ·S1×Int([g,g]).
Finally (a) implies (b). Sincegis the Lie algebra of a compact Lie groupG,Int(g)∼=G/Z(G)is
compact.
2.1.9 Theorem (Weyl) LetGbe a compact connected semisimple Lie group. Then the universal covering Lie groupG˜is also compact.
Proof. The universal coveringG˜ has a structure of Lie group so that the projection G˜ → G is a smooth homomorphism. Letg be the Lie algebra ofG. By Proposition 2.1.8, there exists an ad-invariant inner producth,i on g = T1G. We extend this to a Riemannian metric on Gusing the identification TgG ∼= T1Gvia left-translations. This endows Gwith a bi-invariant metric in which the geodesics through 1 coincide with the one-parameter groups. Now the Koszul for- mula yields the Levi-Civit`a connection∇XY = 12[X, Y]forX,Y ∈ g, from which one computes the sectional curvature to beK(X, Y) = 14||[X, Y]||2. Therefore the Ricci curvatureRic(X, X) =
1 4
Pn
i=1||[X, Xi]||2where{Xi}is an orthonormal basis ofg. Sincegis centerless,Ric(X, X)>0for X 6= 0. By compactness of the unit sphere,Ric≥agfor somea >0. Finally,Gis complete, so the
Bonnet-Myers theorem yields thatG˜ is compact.
We close this section with some remarks about the classification of compact Lie groups. Let Gbe a compact connected Lie group. Then its Lie algebrag = z+g′ where zis the center and g′ is compact semisimple. The universal coveringG˜′ofInt(g′)is a simply-connected compact Lie group with Lie algebrag′, by Weyl’s theorem, which decomposes into the direct product of finitely many simple Lie groups by Corollary 2.1.5. NowGis the quotient ofRk×G˜′ by a discrete central subgroup, where k = dimz. Hence the classification of compact connected Lie groups is equivalent
