Some Generalites about Isometries Definition. Let M be a Riemannian manifold. Φ:M → M is said to be a local ∞ isometry if Φ ∈ C , and for all p ∈ M, ξ, η ∈ TpM, we have

hΦ∗ξ,Φ∗ηi = hξ,ηi. Definition. Let M be a Riemannian manifold. Φ:M → M is said to be a global isometry if in addtion to being a local isometry, Φ preserves the distance function, that is d(Φ(p), Φ(q)) = d(p, q), ∀p, q ∈ M. • If Φ is a local isometry, Φ preserves the Levi-Civita connection; in particular, one has that Φ preserves geodesics, that is,

Φ(exp ξ) = exp(Φ∗ξ), for all ξ in the domain of exp in TM. — Also, Φ preserves sectional curvature; that is, if p ∈ M, ξ, η ∈ TpM are linearly independent, then K(Φ∗ξ,Φ∗η)=K(ξ,η).

• If M1 is a submanifold of M, and Φ is an isometry of M satisfying Φ(M1) ⊂ M1,

then Φ is an isometry. M1 Proposition (Naturality of the Riemannian Connections). Suppose ϕ :(M,g) → (M,g) is an isometry. f (a) ϕ takes the Riemannian connection D of g to the Riemannian connection D of e g, in the sense that e ϕ∗(D Y )=D (ϕ∗Y ). X eϕ∗X (b) If γ is a curve in M and V is a vector field along γ, then

ϕ∗D V = D (ϕ∗V ). t et (c) ϕ takes geodesics to geodesics: if γ is the geodesic in M with initial point p and initial velocity V , then ϕ ◦ γ is the geodesic in M with initial point ϕ(p) and f initial velocity ϕ∗V . Proof. (a) Define a map ϕ∗D :Γ(TM) × Γ(TM) → Γ(TM) e by ∗ −1 (ϕ D) Y = ϕ (D (ϕ∗Y )). e X ∗ eϕ∗X Then ϕ∗D is a connection on M (called the pullback connection), and it is e symmetric and compatible; therefore ϕ∗D = D e by uniqueness of the Riemannian connection. (b) Define an operator ϕ∗D : T (γ) →T(γ) by a similar formula which is equal to et Dt .

• The naturality of the Riemannian connection and uniqueness of geodesics trans- late into the following important naturality property of the exponential map:

Typeset by AMS-TEX

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Proposition (Naturality of Exponential Map). Suppose that ϕ :(M,g) → (M,g) is an isometry. Then, for any p ∈ M, the following diagram f commutes: e ϕ T M −−−−→∗ T M p ϕ(p) f exp  exp p  ϕ(p) y ϕ y M −−−−→ M f Proposition. The Riemann curvature endomorphism and curvature tensor are local isometry invariants. More precisely, if ϕ :(M,g) → (M,g) is a local isometry, f then e ϕ∗Rm = Rm; g R(ϕ∗X, ϕ∗Y )ϕ∗Z = ϕ∗(R(X, Y )Z). e

Isometries between Riemannian Manifolds Proposition 2. Let M, N be a Riemannian manifolds, φ : M → N a local isometry. Then, for any p ∈ M, ∃ε>0 such that

φ : Bε(p) → Bε(φ(p)) Bε(p) is an isometry.

Proof. Let εp and εφ(p) satisfy

exp : Bεp (0p) → Bε(p) B (0 ) εp p

→ exp : Bεφ(p) (0p) Bε(φ(p)) B (0 ) εp φ(p) be diffeomorphisms, and choose ε = min{εp,εφ(p)}.

— The (Euclidean) disks Bε(0p) and Bε(0 ) are isometric under φ∗ , and φ(p) p

−1 φ = exp ◦ φ∗ ◦{exp } B (p) B (0 ) p B (0 ) ε εp φ(p) εp p is, therefore, a diffeomorphism, which implies the Proposition.  3

Theorem 1. Suppose we have two isometries ϕ, ψ :(M n,g) → (N n,h).If

Dϕp = Dψp.

Then ϕ ≡ ψ provided that M is connected. In other words, an isometry is uniquely determined by its differential at just one point. Proof. We use one of the standard closed-open arguments. Let

A = {p ∈ M : Dϕp = Dψp}.

Continuity of Dϕ and Dψ implies that A is closed. Hence we only need to claim: A is open. — Note that A =6 ∅ by assumption. — Fix q ∈ A and choose ε>0 such that

expq :Bε(0p) → Bε(q)

expϕ(q)=ψ(q) :Bε(0p) → Bε(ϕ(q))

are diffeomorphisms. (i) We will claim that

−1 ϕ(x) = expϕ(q) ◦Dϕq ◦ expq on Bε(ϕ(q)).

Choose γ(t) = expq(tv), v ∈ Bε(0), t ∈ [0, 1], which is a geodesic segment. — The curve σ(t)=ϕ ◦ γ(t), t ∈ [0, 1] must also be a geodesic segment. For otherwise, we could find a shorter curve σ :[0, 1] → N with e 1 1 L(ϕ−1 ◦ σ)=Z |Dϕ−1 ◦ σ0(t)| = Z |σ0(t)| = L(σ), e 0 e 0 e e since Dϕ−1 preserves length of vectors, and we will have found a curve from γ(0) to γ(t) of length L(σ)

0 ϕ ◦ γ(t) = expϕ(q)(t(ϕ ◦ γ) (0)) 0 = expϕ(q)(tDϕq · γ (0)) 0 = expϕ(q)(Dϕq (tγ (0))) −1 = expϕ(q)(Dϕq (expq (γ(t)))).

(ii) Similar reasoning shows that

−1 ϕ(x) = expψ(q) ◦Dψq ◦ expq on Bε(ϕ(q)).

Thus ϕ = ψ on Bε(q), as we assumed that Dψq = Dϕq.  4

• What about the inverse problem?

— Given any isometry L : TpM → TqN, is there an isometry ϕ : M → N such that

Dϕp = L?

— If we let M = N, this would in particular mean that if π is a 2-plane in TpM and p a 2-plane in TqM, then there should be an isometry ϕ : M → M such that e ϕ(π)=π. e But this would imply that M has constant sectional curvature. The above problem can therefore not be solved in general. n n n n — Let Sk be R if k =0,Sk ,ifk>0 and Hk ,ifk<0. If we go back and inspect n our knowledge of Iso(Sk ), we see that these spaces have enough isometries so that any linear isometry n n L : TpSk → TqSk n n can be extended to a global isometry ϕ : Sk → Sk with Dϕp = L. — In some sense these are the only spaces with this property. Theorem 2. Suppose (M,g) is a Riemannian manifold of dimension n and con- stant curvature k. n If M is simply connected and L : TpM → TqSk is a linear isometry, then there is a n unique isometric immersion called the monodromy map ϕ : M → Sk with

Dϕp = L.

Furthermore, this map is a diffeomorphism if (M,g) is complete.

We need the following. Theorem 3. Suppose M and M are connected Riemannian manifolds, with M f f complete, and π : M → M is a local isometry. Then M is complete and π is a f covering map.

Proof of Theorem 2. We know that M can be covered by sets Uα such that each n Uα admits a Riemannian embedding ϕ : Uα → Sk . – Furthermore, if p ∈ U , then we can choose ϕ such that Dϕ is any α0 α0 α0 p predetermined isometry. n – Also, each ϕα is well defined up an element in Iso(Sk ); in other words, if ϕα, n −1 n Uα → Sk are isometries then ϕα ◦ψα is the restriction of an element in Iso(Sk ). • The construction of ϕ now proceeds in the same way as one does analytic con- tinuation on simply connected domains. The geodesically complete situation is divided into two cases. (I) Case k ≤ 0. The uniqueness theorem implies that ϕ maps geodesics to geodesics. (i) Claim: ϕ is injective. 5

– Completeness⇒Any two points p, q ∈ M can be joined by a geodesic γ on M. n –Nowϕ ◦ γ is also a geodesic in Sk , k ≤ 0. n – Since geodesics in Sk never intersect themselves, we have ϕ(p) =6 ϕ(q). (ii) Claim: ϕ is onto. n – Choose p ∈ ϕ(M) and q ∈ Sk . b b n Then join p and q by a geodesic γ :[0, 1] → Sk . –InM, let γb:[0, 1]b→ M be the unitb geodesic with

−1 0 0 γ(0) = ϕ (p) and Dϕϕ−1(p)(γ (0)) = γ (0). b b This geodesic exists by geodesic completeness. – Now observe that ϕ ◦ γ is a geodesic with the initial values at γ. Thus ϕ ◦ γ = γ everyehere and ϕ(γ(1)) = q in particular. b b b (II) Case k>0. Surjectivity is established by the same method, but since n geodesics in Sk are closed curves, we cannot use the same proof for injectivity. — Using that M is geodesically complete, we know that ϕ is a covering map. n But then it is a diffeomorphism, since π1(Sk )=0.