A (Riemannian) symmetric space is a Riemannian manifold M such that every x ∈ M is the center of a “reflection”, in the sense that one can find an isometry sx of M such that sx(x) = x and d(sx)x : TxM → TxM is −id. It follows immediately from exercise 15 of chapter 3 that sx is involutive, in the sense that s2x = id. Note also that, if γv is the geodesic with γv(0) = x and γv′(0) =v for somev∈TxM, then sx◦γv =γ−v, sosx restricts to the geodesic reflection on some normal neighborhood of x.
Transvections
LetM be a symmetric space, fix x∈M and a geodesicγ withγ(0) =x. For each tin the domain of γ, denote sγ(t)=:st; the isometrypt:=st/2◦s0 is a called atransvection along γ (at x).
8.7.1 Proposition Let γ : (−ǫ, ǫ)→M be a geodesic through p=γ(0).
a. The transvectionpt induces translation along the curveγ, that is, pt(γ(t0)) =γ(t+t0). More generally, pt induces parallel transport on vectors along γ, in the sense that if v ∈ Tγ(t0)M thenX(t) = (dpt−t0)γ(t0)(v) is a parallel vector field along γ.
b. The transvections{pt}alongγ form a local one-parameter group of isometries ofM, namely, pt+t′ =ptpt′. whenever both hand sides are defined.
c. The transvection pt depends only on γ but not on the chosen initial point x=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 are going to use the fact that isometries act on vector fields by push-forward taking parallel vector fields to parallel vector fields. The assertion follows from the fact that if an isometry maps a geodesic to itself, up to a translation in the parameter, then it maps a parallel vector field along that geodesic to itself, up to a translation in the parameter. More formally, assume that t0 = 0 for simplicity of notation. We want to show that X(t) = (dpt)x(v) is parallel along γ. Let Y denote the parallel vector field along γ such that Y(0) = v. Fix t1. Then Z(t) = (dpt1)γ(t−t1)(Y(t−t1)) ∈ Tγ(t)M is a parallel vector field along γ. Note that Z(t1) = (dpt1)x(v) =X(t1). Since t1 is arbitrary, this completes the proof of a.
b. An isometry is 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. It is clear that for a isometryg, the conjugation gsxg−1 =sgx. Applying this tog=s0 =g−1 yields that s0s−t0s0 =st0, as desired.
It follows from Proposition 8.7.1 that each geodesic determines a unique local one-parameter group of transvections along it.
8.7.2 Proposition A connected symmetric space must be homogeneous and complete.
Proof. Define an equivalence relation on M by declaring that two points are equivalent in case there is an isometry of M mapping one point to the other. The existence of transvections implies that the equivalence class of a point contains a normal neighborhood of it, and hence equivalence classes are open. By connectedness ofM, there must be exactly one equivalence class, which means thatM is homogeneous.
Now every homogeneous manifold is complete (cf. exercise 13 in chapter 3).
Recall that the Myers-Steenrod Theorem states that isometry groupGof a Riemannian manifold M, 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 algebra g as the Lie algebra of Killing vector fields on M. It is also worth recalling that convergence of a sequence of isometries in G in the compact-open topology is equivalent to pointwise convergence in M. Finally, the isotropy group Gx at a point x∈M is compact.
Suppose now that M is a symmetric space. Let G be the identity component of the isometry group of G. Fix a base-point x in M, and let K denote the isotropy group of G at x. Then M = G/K. For every one-parameter group of transvections {pt} originating at x, there is a corresponding Killing vector fieldY whose value at y ∈M isY(y) = dtd
t=0pty; such a Y is called an infinitesimal transvection atx.
8.7.3 Proposition A Killing 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 at x.
Proof. Let {pt} be the transvection one-parameter group at x that Y generates and take any curveη(s) passing throughxats= 0. For the first assertion, it suffices to prove that∇(Yds◦η)|s=0 = 0.
Since the Levi-Civit`a connection is torsionless,
(8.7.4) ∇
ds d
dtptη(s) = ∇ dt
d
dsptη(s) = ∇
dt(dpt)η(s)η′(s).
By Proposition 8.7.1(a), the vector field (dpt)xη′(0) is parallel alongγ(t) =pt(x). Since Y(η(s)) =
d
dt|t=0ptη(s), evaluating eqn. (8.7.4) at s=t= 0 yields one direction of the claim.
Conversely, assume (∇Y)x = 0, take γ to be the geodesic with γ(0) = x, γ′(0) = Yx, and consider the infinitesimal transvectionZ atx alongγ. ThenY =Z, due to exercise 6 of chapter 5.
The last assertion follows from [Y1, Y2] =∇Y1Y2− ∇Y2Y1. Denote the Lie algebras ofGandKbygandk, respectively, and denote the space of infinitesimal transvections atx by p.
8.7.5 Proposition There is a vector space direct sumg=k+p, where [k,k]⊂k, [k,p]⊂p, [p,p]⊂k.
Proof. Every Killing field Z on M decomposes as a sum of Killing fields X+Y, where Y ∈p is the infinitesimal transvection such thatYx=Zx, andX =Z−Y vanishes atx and thusX ∈k.
Further, a Killing fieldZ withZx = (∇Z)x= 0 is identically zero. This proves the existence of the direct sum. The first inclusion is just the statement thatkis a Lie subalgebra ofg. In order to see the second inclusion, letk∈Kand letptbe a transvection alongγatx, withY = dtd|t=0pt. Thenkptk−1 is clearly the transvection along k◦γ atx. Sincedk(Yk−1y) =dk dtd|t=0ptk−1y
= dtd|t=0kptk−1y, we see thatk∗Y =dk◦Y ◦k−1 is an infinitesimal transvection atx. ForX∈k, take ku = expuX so that
[X, Y]y = d
du
u=0AdkuY
y
= ∇ du
d dt u=0
t=0
kuptk−u1y
= d du
u=0dku(Yk−1 u y).
This shows [X, Y] = dud|u=0(ku)∗Y ∈p. The last inclusion is proved in Proposition 8.7.3.
Curvature
The calculation of the curvature of symmetric spaces was already known to Cartan. Let M be a symmetric space with Levi-Civit`a connection∇and curvature tensor R.
8.7.6 Lemma Let M be a locally symmetric space and fix x ∈ M. Let X be an infinitesinal transvection at x and let Y be any vector field defined on a neighborhood of x. Then (∇XY)x = (LXY)x.
Proof. Let {pt} denote the local one-parameter group of local transvection generated by X.
Since pt induces parallel transport of vectors along the geodesicγ(t) =pt(x), we have (LXY)x = d
dt
t=0dp−t(Yγ(t))
= d
dt
t=0P0,tγ (Yγ(t))
= (∇XY)x,
wherePtγ1,t0 denotes parallel transport along γ fromt0 tot1.
8.7.7 Proposition Let M be a locally symmetric space, x∈M andX, Y, Z ∈p. Then (8.7.8) Rx(X, Y)Z =−[[X, Y], Z]x,
where p∼=TxM is the space of infinitesimal tyransvections at x.
Proof. Let γ be the geodesic with initial speed Xx ∈ TxM. Since Y is a Killing field, its restriction along γ is a Jacobi field. Therefore, using dt∇ to denote the covariant derivative along γ,
R(X, Y)X = ∇
dt
∇ dtY
t=0
= ∇
dt[X, Y]γ(t)
t=0
(by Lemma 8.7.6)
= [X,[X, Y]]γ(0) (idem)
= −[[X, Y], X]x.
The proof is finished by recalling that the sectional curvature determines the curvature tensor and that the right-hand side of (8.7.8) has the symmetries of the curvature tensor.
Locally symmetric spaces
A locally symmetric space is a Riemannian manifold M such that the geodesic reflection at any point x ∈M is an isometry defined on a normal neighborhood of x onto itself (cf. exercise 12 of chapter 6). From exercises 12, 13 and 14 of chapter 6, one deduces:
8.7.9 Proposition A Riemannian manifold is locally symmetric if and only if the curvature tensor of its Levi-Civit`a connection is parallel, ∇R= 0.
Every symmetric space is plainly a locally symmetric space. On the other hand, it is a conse-quence of the Cartan-Ambrose theorem [CE75] that the geodesic symmetries in a simply-connected complete locally symmetric space can be extended to global isometries, so that it becomes a sym-metric space. It follows that every complete locally symsym-metric space is isosym-metrically covered by a global symmetric space, namely, its universal Riemannian covering.
Symmetric pairs and involutive Lie algebras
LetM be a symmetric space. Owing to Proposition 8.7.2,M is a homogeneous spaceG/K, where G is the identity component of the isometry group of M and K is the isotropy group of a chosen basepoint x. Since s2x = idM, conjugation by the symmetry sx induces an involutive automorhism σ ofG, namely,σ(g) =sxgs−x1. Ifk∈K, thensxks−x1 is an isometry ofM that fixesxand has the same differential atx as k; hence sxks−1x =k. This implies that K is contained in the fixed point set Gσ. On the other hand, the Lie algebra ofGσ is the fixed point set of dσ on the Lie algebrag ofG. IfX ∈g is such thatdσ(X) =X, thenσ(exptX) = exptX for allt∈R, which implies that sx and exptX commute. Now (exptX)xis a fixed point of sx. Sincex is an isolated fixed point of sx, we deduce that (exptX)x=x for smallt > 0 and thusXx = 0. This proves that X ∈ k, and hence the identity component (Gσ)0⊂K.
Let Gbe a connected Lie group and let K be a closed subgroup. The pair (G, K) is a called a (Riemannian) symmetric pair if there exists an involutive automorphismσ ofGsuch that (Gσ)0⊂ K ⊂Gσ (equivalently, K is open in Gσ) and, in addition, the group AdG(K) is a compact group of linear transformations of g.
8.7.10 Proposition Let (G, K) be a symmetric pair. Then there is a G-invariant Riemannian metric on G/K such that it is a symmetric space.
Proof. We have (Gσ)0 ⊂ K ⊂ Gσ for some involution σ of G. Let g = k+p be the ± 1-eigenspace decomposition of dσ. Then k is clearly the Lie algebra of K. If k ∈ K and Y ∈ p, then dσAdkY = Adσ(k)dσY = Adk(−Y) = −AdkY, showing that p is AdG(K)-invariant. Since AdG(K) is a compact group of linear transformations ofg, there is an invariant inner product onp.
The projection π :G→ G/K yields an isomorphism π∗ :p∼=Tx(G/K), which is equivariant with respect to the adjoint action ofK onp and the isotropy representation ofK on Tx(G/K). Indeed, fork∈K andY ∈p,
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. Now the AdG(K)-invariant inner product onp induces aK-invariant inner product on Tx(G/K), which we use to define aG-invariant Riemannian metric on G/K.
Putx= 1K and definesx(gK) =σ(g)K. ForY ∈p, we have dsx(Yx) = d
dt|t=0sxexp(tY)x
= d
dt|t=0σ(exp(tY))x
= d
dt|t=0exp(tdσY)x
= d
dt|t=0exp(−tY)x
=−Yx.
This shows that d(sx)x = −id, and that d(sx)x is a linear isometry of Tx(G/K). Using that sx◦g=σ(g)◦sx andg,σ(g)∈Gare isometries ofG/K, we see thatsxis an isometry everywhere.
Since G/K is homogeneous, this already implies that G/K is a symmetric space.
8.7.11 Remark (i) For the symmetric space G/K constructed in Proposition 8.7.10, the action ofGby left-multiplication onG/K need not be effective. The kernelZ of this action is the largest normal subgroup of G contained in K. One obtains an effective presentation by dividing by Z, namely,G/K =G′/K′ whereG′ =G/Z and K′ =K/Z.
(ii) If (G, K) is a symmetric pair and g = k+p as in Proposition 8.7.10, there is a bijective correspondence between G-invariant Riemannian metrics on the symmetric spaceG/K and AdK -invariant inner products onp.
A (Riemannian) symmetric pair (G, K) gives rise on the Lie algebra level to an (orthogonal) involutive Lie algebra: this is a pair (g, s), wheregis a Lie algebra,sis an involutive automorphism of g, and the fixed point setk of sis a compactly embedded subalgebra of g, in the sense that the group of inner automorphisms ofg generated byk is compact.
Let (g, s) be an involutive Lie algebra, and writeg=k+pfor the±1-eigenspace decomposition of s. By assumption there are adk-invariant inner products on p, and sometimes one specifies one such inner product B and call the triple (g, s, B) an orthogonal involutive Lie algebra, or OIL-algebra, for short.
Given an OIL-algebra (g, s, B), whereg=k+punders, one can construct a symmetric pair as follows. Let ˜G be the simply-connected Lie group with Lie algebrag, and let ˜K be the connected subgroup of ˜Gwith Lie algebrak. Then ( ˜G,K) is a symmetric pair, with automorphism˜ σspecified by dσ = s. If one assumes in the definition of OIL-algebra that k does not contain an ideal of g, then the largest normal subgroup ˜Z of ˜G which is contained in ˜K is discrete, and we obtain an effective presentation of a simply-connected symmetric space ˜G/K˜ =G′/K′ where G′ = ˜G/Z,˜ K′ = ˜K/Z˜ have resp. Lie algebrasg,k. Note thatB defines aK′-invariant inner product onTxM, wherex= 1K′, that extends to a G′-invariant Riemannian metric on M.
8.7.12 Example There is a symmetric pair (SU(n+ 1),S(U(1)×U(n))), where σ is given by conjugation by the matrix
−1 1 ...
1
!
. The associated symmetric space is complex projective space CPn = SU(n+ 1)/S(U(1)×U(n)). More generally, if we conjugate by
−Ip
Iq
, where p+q =n+ 1, we obtain the complex GrassmannianGrp(Cn+1) ofp-planes in Cn+1 as a symmetric space.
Types and duality
LetM =G/K be a symmetric space whereG is the connected isometry group ofM and (g, s, B) is the OIL-algebra atx∈M. Write g=k+p for the decomposition into±1-eigenspaces ofs.
The symmetric space M is called of Euclidean type is [p,p] = 0. Recall that a Lie algebra is calledsemisimple if its Killing form is nondegenerate. Ifg is semisimple and its Killing formBg is negative (resp. positive) definite on p, thenM is called ofcompact type (resp. noncompact type).
8.7.13 Lemma a. Bg is negative definite on k.
b. M is of Euclidean type if and only if Bg= 0 onp.
Proof. (a) Let X∈k. Then Bg(X, X) = tracek(ad2X) + tracep(ad2X). Since k is the Lie algebra of a compact Lie group, there is an ad-invariant inner product on k, with respect to which adX is a skew-symmetric emdomorphism of k and has thus purely imaginary eigenvalues; it follows that ad2X has negative eigenvalues on k. Similarly, B is adX-invariant and therefore ad2X has negative eigenvalues on p. Now Bg(X, X) ≤0. In addition, Bg(X, X) = 0 is and only if all eigenvalues of adX are zero, which is to say thatX lies in the center ofg. Then the multiples ofX form an ideal of gcontained in k, and henceX = 0.
(b) If M is of Euclidean type then [p,p] = 0. For Y1, Y2 ∈ p, this implies that Bg(Y1, Y2) = tracek(adY1 ◦adY2) + tracep(adY1 ◦adY2) = 0. In fact the first term on the right hand-side is zero because [Y1,[Y2, X]]∈[Y2,p] = 0 forX ∈k, and the second term is clealy zero. Conversely, suppose thatBg= 0 on p, and let Y1,Y2 ∈p. Then Bg([Y1, Y2],[Y1, Y2]) =Bg(Y1,[Y2,[Y1, Y2]]) = 0 because Y1, [Y2,[Y1, Y2]]∈p. By part (a) we get that [Y1, Y2] = 0.
Cartan duality establishes a pairing between simply-connected symmetric spaces of compact and noncompact type. We first contruct the dual OIL-algebra to (g, s, B). Consider the complexification gc =g⊗RC. Note that g∗ := k+√
−1p is a real subalgebra of gc. There is a natural involution
s∗ of g∗, with ±1-eigenspaces k∗ = k and p∗ = √
−1p. We complete the definition of dual OIL-algebra (g, s, B)∗ = (g∗, s∗, B∗) by settingB∗(√
−1X,√
−1Y) =B(X, Y) forX,Y ∈p. Note that (g, s, B)∗∗= (g, s, B).
Now we say that simply-connected symmetric spaces aredual one to the other if they have dual OIL-algebras.
8.7.14 Proposition A simply-connected symmetric space M is of compact (resp. noncompact) type if and only if its dual M∗ is of noncompact type (resp. compact) type.
Proof. Let (g, s, B) be the OIL-algebra of M. Then g is semisimple, which is to say that Bg is nondegenerate. The Killing form of gc is the complexification of Bg, so, as a complex bilinear symmetric form, it has the same matrix as Bg, on a basis (over C) consisting of elements of g.
It follows that Bgc is also nondegenerate. Since gc is also the complexification of g∗, the same argument yields that g∗ is semisimple. Now we have just to note that
Bg∗(√
−1X,√
−1) =Bgc(√
−1X,√
−1Y)
=−Bgc(X, Y) (byC-bilinearity ofBgc)
=−Bg(X, Y)
forX,Y ∈p.
8.7.15 Example The Cartan dual ofSn isRHn. Irreducibility
LetM be a simply-connected symmetric space. By the de Rham decomposition theorem, there is a Riemannian product decomposition of M and the identity component of its isometry group:
(8.7.16) M =M0×M1× · · · ×Mk, Isom(M)0 ∼= Isom(M0)0× · · · ×Isom(Mk)0,
where M0 is an Euclidean space (with the flat metric), and the Mi for i = 1, . . . , k are simply-connected complete Riemannian manifolds which are irreducible, in the sense that the holonomy group Holxi(Mi) at a pointxi∈Miacts irreducibly onTxiMi.8 Every geodesicγ ofMemanating fromx= (x0, . . . , xk) is the productγ0× · · · ×γk of geodesicsγi emanating fromxi. It follows that the geodesic symmetry ofM atx is also decomposed as a productsx=sx0 × · · · ×sxk, and hence each factor Mi is a symmetric space.
We say a symmetric space is irreducible if it is holonomy irreducible. In the proof of Proposi-tion 8.7.19, we will use the following two results.
8.7.17 Theorem (Ambrose-Singer) Let M be a connected Riemannian manifold. Then the Lie algebra of the holonomy group of M at a point p is generated by Pγ◦R(Pγ(x), Pγ(y))◦(Pγ)−1, where γ is an arbitrary piecewise smooth curve emanating from p and x,y ∈TpM.
8.7.18 Lemma Let g be a semisimple Lie algebra, let a⊂g be an ideal and a⊥ ={X∈g:Bg(X,a) = 0}.
Then a⊥ is an ideal, a and a⊥ are semisimple and g=a⊕a⊥ (direct sum of ideals).
8The holonomy group Holx(M) of a Riemannian manifoldM at a point x∈M is defined to be the subgroup of O(TxM) generated by parallel transport along piecewise smooth loops atx. It is a Lie group.