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. We will give the ideas for these proofs: (i) We can show that any continuous map F : M → N between manifolds is homotopic to a smooth one. But this is painful to prove. (ii) What we really care about is maps σ : ∆p → M, and we can barycentrically subdivide the simplex so that it only lies in a single chart, and then it is easy to do smooth approximation. (iii) As Hp and H ∞ p have enough properties in common, in particular they both have Mayer-Vietoris and agree on convex sets, this implies they are the same. We will not spell out this proof, because we are going to do this to prove that de Rham cohomology is the same as singular cohomology Since we are working with R, we can cheat and deﬁne singular cohomology in a simple way: Deﬁnition (Singular cohomology). The singular cohomology of M is deﬁned as H p(M, R) = Hom(Hp(M, R), R). Similarly, the smooth singular cohomology is H p ∞(M, R) = Hom(H ∞ p (M, R), R). This is a bad deﬁnition in general! It just happens to work for singular cohomology with coeﬃcients in R, and we are not too bothered to write dowm the proper deﬁnition. Our goal is now to describe an isomorphism H p dR(M ) ∼= H p ∞(M, R). The map itself is given as follows: Suppose [w] ∈ H p dR(M ), so ω ∈ Ωp(M ) with dω = 0. Suppose that σ : ∆p → M is smooth with ∂σ = 0. We can then deﬁne I([ω]) = σ∗ω ∈ R. ∆p Note that we have not deﬁned what ∆p means, because ∆p is not even a manifold with boundary — it has corners. We can develop an analogous theory 68 7 De Rham’s theorem* III Diﬀerential Geometry of integration on manifolds with corners, but we can also be lazy, and just integrate over �
�× p = ∆p \ {codimension 2 faces}. p does not have compact support, but has the property that it is the Now ω\|∆∗ restriction of a (unique) p-form on ∆p, so in particular it is bounded. So the integral is ﬁnite. Now in general, if τ = aiσi ∈ Cp(M ), we deﬁne I([ω])(τ ) = ai ∆p i ω ∈ R. σ∗ Now Stokes theorem tell us ω = dω. ∂σ σ So we have Lemma. I is a well-deﬁned map H p dR(M ) → H p ∞(M, R). Proof. If [ω] = [ω], then ω − ω = dα. Then let σ ∈ H p ∞(M, R). Then σ (ω − ω) = σ dα = ∂σ α = 0, since ∂σ = 0. On the other hand, if [σ] = [σ], then σ − σ = ∂β for some β. Then we have ω = ω = dω = 0. σ−σ ∂β β So this is well-deﬁned. Lemma. I is functorial and commutes with the boundary map of Mayer-Vietoris. In other words, if F : M → N is smooth, then the diagram H p dR(M ) I H p ∞(M ) F ∗ F ∗ H p dR(N ). I H p ∞(N ) And if M = U ∪ V and U, V are open, then the diagram H p dR(U ∩ V ) I ∞(U ∩ V, R) H p δ δ H p+1 dR (U ∪ V ) I H p(U ∪ V, R) also commutes. Note that the other parts of the Mayer-Vietoris sequence commute because they are induced by maps of manifolds. Proof. Trace through the deﬁnitions. 69 7 De Rham’s theorem* III Diﬀerential Geometry Proposition. Let U ⊆ Rn is convex, then U : H
p dR(U ) → H p ∞(U, R) is an isomorphism for all p. Proof. If p > 0, then both sides vanish. Otherwise, we check manually that I : H 0 ∞(U, R) is an isomorphism. dR(U ) → H 0 These two are enough to prove that the two cohomologies agree — we can cover any manifold by convex subsets of Rn, and then use Mayer-Vietoris to patch them up. We make the following deﬁnition: Deﬁnition (de Rham). (i) We say a manifold M is de Rham if I is an isomorphism. (ii) We say an open cover {Uα} of M is de Rham if Uα1 ∩ · · · ∩ Uαp is de Rham for all α1, · · ·, αp. (iii) A de Rham basis is a de Rham cover that is a basis for the topology on M. Our plan is to inductively show that everything is indeed de Rham. We have already seen that if U ⊆ Rn is convex, then it is de Rham, and a countable disjoint union of de Rham manifolds is de Rham. The key proposition is the following: Proposition. Suppose {U, V } is a de Rham cover of U ∪ V. Then U ∪ V is de Rham. Proof. We use the ﬁve lemma! We write the Mayer-Vietoris sequence that is impossible to ﬁt within the margins: H p dR(U ) ⊕ H p dR(V ) H p dR(U ∪ V ) H p+1 dR (U ∩ V ) H p dR(U ) ⊕ H p+1 dR (V ) H p+1 dR (U ∪ V ) I⊕I I I I⊕I I H p ∞(U ) ⊕ H p ∞(V ) H p ∞(U ∪ V ) H p+1 ∞ (U ∩ V ) H p ∞(U ) ⊕ H p+1 ∞ (V ) H p+1 ∞ (U ∪ V ) This huge thing commutes, and
all but the middle map are isomorphisms. So by the ﬁve lemma, the middle map is also an isomorphism. So done. Corollary. If U1, · · ·, Uk is a ﬁnite de Rham cover of U1 ∪ · · · ∪ Uk = N, then M is de Rham. Proof. By induction on k. Proposition. The disjoint union of de Rham spaces is de Rham. Proof. Let Ai be de Rham. Then we have H p dR ∼= Ai H p dR(Ai) ∼= H p ∞(Ai) ∼= H p ∞. Ai Lemma. Let M be a manifold. If it has a de Rham basis, then it is de Rham. 70 7 De Rham’s theorem* III Diﬀerential Geometry Proof sketch. Let f : M → R be an “exhaustion function”, i.e. f −1([−∞, c]) for all c ∈ R. This is guaranteed to exist for any manifold. We let Am = {q ∈ M : f (q) ∈ [m, m + 1]}. We let A m = q ∈ M : f (q Given any q ∈ Am, there is some Uα(q) ⊆ A m in the de Rham basis containing q. As Am is compact, we can cover it by a ﬁnite number of such Uαi, with each Uαi ⊆ A m. Let Bm = Uα1 ∪ · · · ∪ Uαr. Since Bm has a ﬁnite de Rham cover, so it is de Rham. Observe that if Bm ∩ B ˜m = ∅, then ˜M ∈ {m, m − 1, m + 1}. We let U = m even Bm, V = Bm. m odd Then this is a countable union of de Rham spaces, and is thus de Rham. Similarly, U ∩ V is de Rham. So M = U ∪ V is de Rham. Theorem. Any manifold has a de Rham basis. Proof. If U ⊆ Rn is open, then it is de Rham, since there is a basis of convex
sets {Uα} (e.g. take open balls). So they form a de Rham basis. Finally, M has a basis of subsets diﬀeomorphic to open subsets of Rn. So it is de Rham. 71 8 Connections III Diﬀerential Geometry 8 Connections 8.1 Basic properties of connections Imagine we are moving in a manifold M along a path γ : I → M. We already know what “velocity” means. We simply have to take the derivative of the path γ (and pick the canonical tangent vector 1 ∈ TpI) to obtain a path γ : I → T M. Can we make sense of our acceleration? We can certainly iterate the procedure, treating T M as just any other manifold, and obtain a path γ : I → T T M. But this is not too satisfactory, because T T M is a rather complicated thing. We would want to use the extra structure we know about T M, namely each ﬁber is a vector space, to obtain something nicer, perhaps an acceleration that again lives in T M. We could try the naive deﬁnition d dt = lim h→0 γ(t + h) − γ(t) h, but this doesn’t make sense, because γ(t + h) and γ(t) live in diﬀerent vector spaces. The problem is solved by the notion of a connection. There are (at least) two ways we can think of a connection — on the one hand, it is a speciﬁcation of how we can take derivatives of sections, so by deﬁnition this solves our problem. On the other hand, we can view it as telling us how to compare inﬁnitesimally close vectors. Here we will deﬁne it the ﬁrst way. Notation. Let E be a vector bundle on M. Then we write Ωp(E) = Ω0(E ⊗ Λp(T ∗M )). So an element in Ωp(E) takes in p tangent vectors and outputs a vector in E. Deﬁnition (Connection). Let E be a vector bundle on M. A connection on E is a linear map such that dE : Ω
0(E) → Ω1(E) dE(f s) = df ⊗ s + f dEs for all f ∈ C∞(M ) and s ∈ Ω0(E). A connection on T M is called a linear or Koszul connection. Given a connection dE on a vector bundle, we can use it to take derivatives of sections. Let s ∈ Ω0(E) be a section of E, and X ∈ Vect(M ). We want to use the connection to deﬁne the derivative of s in the direction of X. This is easy. We deﬁne ∇X : Ω0(E) → Ω0(E) by ∇X (s) = dE(s), X ∈ Ω0(E), where the brackets denote applying dE(s) : T M → E to X. Often we just call ∇X the connection. Proposition. For any X, ∇X is linear in s over R, and linear in X over C∞(M ). Moreover, ∇X (f s) = f ∇X (s) + X(f )s for f ∈ C∞(M ) and s ∈ Ω0(E). 72 8 Connections III Diﬀerential Geometry This doesn’t really solve our problem, though. The above lets us diﬀerentiate sections of the whole bundle E along an everywhere-deﬁned vector ﬁeld. However, what we want is to diﬀerentiate a path in E along a vector ﬁeld deﬁned on that path only. Deﬁnition (Vector ﬁeld along curve). Let γ : I → M be a curve. A vector ﬁeld along γ is a smooth V : I → T M such that for all t ∈ I. We write V (t) ∈ Tγ(t)M J(γ) = {vector ﬁelds along γ}. The next thing we want to prove is that we can indeed diﬀerentiate these things. Lemma. Given a linear connection ∇ and a path γ : I → M, there exists a unique map Dt : J(
γ) → J(γ) such that (i) Dt(f V ) = ˙f V + f DtV for all f ∈ C∞(I) (ii) If U is an open neighbourhood of im(γ) and ˜V is a vector ﬁeld on U such that ˜V \|γ(t) = Vt for all t ∈ I, then Dt(V )\|t = ∇ ˙γ(0) ˜V. We call Dt the covariant derivative along γ. In general, when we have some notion on Rn that involves derivatives and we want to transfer to general manifolds with connection, all we do is to replace the usual derivative with the covariant derivative, and we will usually get the right generalization, because this is the only way we can diﬀerentiate things on a manifold. Before we prove the lemma, we need to prove something about the locality of connections: Lemma. Given a connection ∇ and vector ﬁelds X, Y ∈ Vect(M ), the quantity ∇X Y \|p depends only on the values of Y near p and the value of X at p. Proof. It is clear from deﬁnition that this only depends on the value of X at p. To show that it only depends on the values of Y near p, by linearity, we just have to show that if Y = 0 in a neighbourhood U of p, then ∇X Y \|p = 0. To do so, we pick a bump function χ that is identically 1 near p, then supp(X) ⊆ U. Then χY = 0. So we have 0 = ∇X (χY ) = χ∇X (Y ) + X(χ)Y. Evaluating at p, we ﬁnd that X(χ)Y vanishes since χ is constant near p. So ∇X (Y ) = 0. We now prove the existence and uniqueness of the covariant derivative. 73 8 Connections III Diﬀerential Geometry Proof of previous lemma. We ﬁrst prove uniqueness. By a similar bump function argument, we know that DtV \|t0 depends only on values of V (t) near t0. Suppose that locally on a chart, we have
V (t) = j Vj(t) ∂ ∂xj γ(t) for some Vj : I → R. Then we must have DtV \|t0 = ˙Vj(t) j ∂ ∂xj γ(t0) + j Vj(t0)∇ ˙γ(t0) ∂ ∂xj by the Leibniz rule and the second property. But every term above is uniquely determined. So it follows that DtV must be given by this formula. To show existence, note that the above formula works locally, and then they patch because of uniqueness. Proposition. Any vector bundle admits a connection. Proof. Cover M by Uα such that E\|Uα is trivial. This is easy to do locally, and then we can patch them up with partitions of unity. Note that a connection is not a tensor, since it is not linear over C∞(M ). However, if dE and ˜dE are connections, then (dE − ˜dE)(f s) = df ⊗ s + f dEs − (df ⊗ s + f ˜dES) = f (dE − ˜dE)(s). So the diﬀerence is linear. Recall from sheet 2 that if E, E are vector bundles and is a map such that α : Ω0(E) → Ω0(E) α(f s) = f α(s) for all s ∈ Ω0(E) and f ∈ C∞(M ), then there exists a unique bundle morphism ξ : E → E such that α(s)\|p = ξ(s(p)). Applying this to α = dE − ˜dE : Ω0(E) → Ω1(E) = Ω0(E ⊗ T ∗M ), we know there is a unique bundle map such that ξ : E → E ⊗ T ∗M dE(s)\|p = ˜dE(s)\|p + ξ(s(p)). So we can think of dE − ˜dE as a bundle morphism E → E ⊗ T ∗M. In other words, we have dE − ˜dE ∈ �
��0(E ⊗ E∗ ⊗ T ∗M ) = Ω1(End(E)). The conclusion is that the set of all connections on E is an aﬃne space modelled on Ω1(End(E)). 74 8 Connections III Diﬀerential Geometry Induced connections In many cases, having a connection on a vector bundle allows us to diﬀerentiate many more things. Here we will note a few. Proposition. The map dE extends uniquely to dE : Ωp(E) → Ωp+1(E) such that dE is linear and dE(w ⊗ s) = dω ⊗ s + (−1)pω ∧ dEs, for s ∈ Ω0(E) and ω ∈ Ωp(M ). Here ω ∧ dEs means we take the wedge on the form part of dEs. More generally, we have a wedge product Ωp(M ) × Ωq(E) → Ωp+q(E) (α, β ⊗ s) → (α ∧ β) ⊗ s. More generally, the extension satisﬁes dE(ω ∧ ξ) = dω ∧ ξ + (−1)qω ∧ dEξ, where ξ ∈ Ωp(E) and ω ∈ Ωq(M ). Proof. The formula given already uniquely speciﬁes the extension, since every form is locally a sum of things of the form ω ⊗ s. To see this is well-deﬁned, we need to check that dE((f ω) ⊗ s) = dE(ω ⊗ (f s)), and this follows from just writing the terms out using the Leibniz rule. The second part follows similarly by writing things out for ξ = η ⊗ s. Deﬁnition (Induced connection on tensor product). Let E, F be vector bundles with connections dE, dF respectively. The induced connection is the connection dE⊗F on E ⊗ F given by dE⊗F (s ⊗ t) = dEs �
�� t + s ⊗ dF t for s ∈ Ω0(E) and t ∈ Ω0(F ), and then extending linearly. Deﬁnition (Induced connection on dual bundle). Let E be a vector bundle with connection dE. Then there is an induced connection dE∗ on E∗ given by requiring ds, ξ = dEs, ξ + s, dE∗ ξ, for s ∈ Ω0(E) and ξ ∈ Ω0(E∗). Here ·, · denotes the natural pairing Ω0(E) × Ω0(E∗) → C∞(M, R). So once we have a connection on E, we have an induced connection on all tensor products of it. Christoﬀel symbols We also have a local description of the connection, known as the Christoﬀel symbols. 75 8 Connections III Diﬀerential Geometry Say we have a frame e1, · · ·, er for E over U ⊆ M. Then any section s ∈ Ω0(E\|U ) is uniquely of the form s = siei, where si ∈ C∞(U, R) and we have implicit summation over repeated indices (as we will in the whole section). Given a connection dE, we write dEei = Θj i ⊗ ej, where Θj i ∈ Ω1(U ). Then we have dEs = dEsiei = dsi ⊗ ei + sidEei = (dsj + Θj i si) ⊗ ej. We can write s = (s1, · · ·, sr). Then we have dEs = ds + Θs, where the ﬁnal multiplication is matrix multiplication. It is common to write dE = d + Θ, where Θ is a matrix of 1-forms. It is a good idea to just view this just as a formal equation, rather than something that actually makes mathematical sense. Now in particular, if we have a linear connection ∇ on T M and coordinates x1, · · ·, xn on U ⊆ M, then we have a frame for T M \|U given by �
�1, · · ·, ∂n. So we again have i ⊗ ∂k. i ∈ Ω1(U ). But we don’t just have a frame for the tangent bundle, but where Θk also the cotangent bundle. So in these coordinates, we can write dE∂i = Θk Θk i = Γk i dx, where Γk i ∈ C∞(U ). These Γk i are known as the Christoﬀel symbols. In this notation, we have ∇∂j ∂i = dE∂i, ∂j = Γk = Γk i dx ⊗ ∂k, ∂j ji∂k. 8.2 Geodesics and parallel transport One thing we can do with a connection is to deﬁne a geodesic as a path with “no acceleration”. Deﬁnition (Geodesic). Let M be a manifold with a linear connection ∇. We say that γ : I → M is a geodesic if Dt ˙γ(t) = 0. 76 8 Connections III Diﬀerential Geometry A natural question to ask is if geodesics exist. This is a local problem, so we work in local coordinates. We try to come up with some ordinary diﬀerential equations that uniquely specify a geodesic, and then existence and uniqueness will be immediate. If we have a vector ﬁeld V ∈ J(γ), we can write it locally as V = V j∂j, then we have DtV = ˙V j∂j + V j∇ ˙γ(t0)∂j. We now want to write this in terms of Christoﬀel symbols. We put γ = (γ1, · · ·, γn). Then using the chain rule, we have DtV = ˙V k∂k + V j ˙γi∇∂i∂j = ( ˙V k + V j ˙γiΓk ij)∂k. Recall that γ is a geodesic if Dt ˙γ = 0 on I. This holds i�
� we locally have ¨γk + ˙γi ˙γjΓk ij = 0. As this is just a second-order ODE in γ, we get unique solutions locally given initial conditions. Theorem. Let ∇ be a linear connection on M, and let W ∈ TpM. Then there exists a geodesic γ : (−ε, ε) → M for some ε > 0 such that ˙γ(0) = W. Any two such geodesics agree on their common domain. More generally, we can talk about parallel vector ﬁelds. Deﬁnition (Parallel vector ﬁeld). Let ∇ be a linear connection on M, and γ : I → M be a path. We say a vector ﬁeld V ∈ J(γ) along γ is parallel if DtV (t) ≡ 0 for all t ∈ I. What does this say? If we think of Dt as just the usual derivative, this tells us that the vector ﬁeld V is “constant” along γ (of course it is not literally constant, since each V (t) lives in a diﬀerent vector space). Example. A path γ is a geodesic iﬀ ˙γ is parallel. The important result is the following: Lemma (Parallel transport). Let t0 ∈ I and ξ ∈ Tγ(t0)M. Then there exists a unique parallel vector ﬁeld V ∈ J(γ) such that V (t0) = ξ. We call V the parallel transport of ξ along γ. Proof. Suppose ﬁrst that γ(I) ⊆ U for some coordinate chart U with coordinates x1, · · ·, xn. Then V ∈ J(γ) is parallel iﬀ DtV = 0. We put Then we need V = V j(t) ∂ ∂xj. ˙V k + V j ˙γiΓk ij = 0. This is a ﬁrst-order linear ODE in V with initial condition given by V (t0) = ξ, which has a unique solution. The general result then follows by patching
, since by compactness, the image of γ can be covered by ﬁnitely many charts. 77 8 Connections III Diﬀerential Geometry Given this, we can deﬁne a map given by parallel transport: Deﬁnition (Parallel transport). Let γ : I → M be a curve. For t0, t1, we deﬁne the parallel transport map Pt0t1 : Tγ(t0)M → Tγ(t1)M given by ξ → Vξ(t1). It is easy to see that this is indeed a linear map, since the equations for parallel transport are linear, and this has an inverse Pt1t0 given by the inverse path. So the connection ∇ “connects” Tγ(t0)M and Tγ(t1)M. Note that this connection depends on the curve γ chosen! This problem is in general unﬁxable. Later, we will see that there is a special kind of connections known as ﬂat connections such that the parallel transport map only depends on the homotopy class of the curve, which is an improvement. 8.3 Riemannian connections Now suppose our manifold M has a Riemannian metric g. It is then natural to ask if there is a “natural” connection compatible with g. The requirement of “compatibility” in some sense says the product rule is satisﬁed by the connection. Note that saying this does require the existence of a metric, since we need one to talk about the product of two vectors. Deﬁnition (Metric connection). A linear connection ∇ is compatible with g (or is a metric connection) if for all X, Y, Z ∈ Vect(M ), ∇X g(Y, Z) = g(∇X Y, Z) + g(Y, ∇X Z). Note that the ﬁrst term is just X(g(Y, Z)). We should view this formula as specifying that the product rule holds. We can alternatively formulate this in terms of the covariant derivative. Lemma. Let ∇ be a connection. Then ∇ is compatible with g if and only if for all γ : I → M and V, W ∈ J(γ), we have d dt g(V
(t), W (t)) = g(DtV (t), W (t)) + g(V (t), DtW (t)). (∗) Proof. Write it out explicitly in local coordinates. We have some immediate corollaries, where the connection is always assumed to be compatible. Corollary. If V, W are parallel along γ, then g(V (t), W (t)) is constant with respect to t. Corollary. If γ is a geodesic, then \| ˙γ\| is constant. Corollary. Parallel transport is an isometry. In general, on a Riemannian manifold, there can be many metric conditions. To ensure that it is actually unique, we need to introduce a new constraint, known as the torsion. The deﬁnition itself is pretty confusing, but we will chat about it afterwards to explain why this is a useful deﬁnition. 78 8 Connections III Diﬀerential Geometry Deﬁnition (Torsion of linear connection). Let ∇ be a linear connection on M. The torsion of ∇ is deﬁned by τ (X, Y ) = ∇X Y − ∇Y X − [X, Y ] for X, Y ∈ Vect(M ). Deﬁnition (Symmetric/torsion free connection). A linear connection is symmetric or torsion-free if τ (X, Y ) = 0 for all X, Y. Proposition. τ is a tensor of type (2, 1). Proof. We have τ (f X, Y ) = ∇f X Y − ∇Y (f X) − [f X, Y ] = f ∇X Y − Y (f )X − f ∇Y X − f XY + Y (f X) = f (∇X Y − ∇Y X − [X, Y ]) = f τ (X, Y ). So it is linear. We also have τ (X, Y ) = −τ (Y, X) by inspection. What does being symmetric tell us? Consider the Christoﬀel symbols in some coordinate system x1, · · ·, xn. We then have So we have ∂ ∂xi, ∂ ∂xj = 0. τ ∂
∂xi, ∂ ∂xj = ∇i∂j − ∇j∂i = Γk ij∂k − Γk ji∂k. So we know a connection is symmetric iﬀ the Christoﬀel symbol is symmetric, i.e. Now the theorem is this: ij = Γk Γk ji. Theorem. Let M be a manifold with Riemannian metric g. Then there exists a unique torsion-free linear connection ∇ compatible with g. The actual proof is unenlightening. Proof. In local coordinates, we write g = gij dxi ⊗ dxj. Then the connection is explicitly given by Γk ij = 1 2 gk(∂igj + ∂jgi − ∂gij), where gk is the inverse of gij. We then check that it works. 79 8 Connections III Diﬀerential Geometry Deﬁnition (Riemannian/Levi-Civita connection). The unique torsion-free metric connection on a Riemannian manifold is called the Riemannian connection or Levi-Civita connection. Example. Consider the really boring manifold Rn with the usual metric. We also know that T Rn ∼= Rn → Rn is trivial, so we can give a trivial connection dRn f In the ∇ notation, we have ∂ ∂xi = df ⊗ ∂ ∂xi. ∇X f ∂ ∂xi = X(f ) ∂ ∂xi. It is easy to see that this is a connection, and also that it is compatible with the metric. So this is the Riemannian connection on Rn. This is not too exciting. Example. Suppose φ : M ⊆ Rn is an embedded submanifold. This gives us a Riemannian metric on M by pulling back g = φ∗gRn on M. We also get a connection on M as follows: suppose X, Y ∈ Vect(M ). Locally, we know X, Y extend to vector ﬁelds ˜X, ˜Y on Rn. We set ∇X Y = π( ¯∇ ˜X ˜Y ), where π is
the orthogonal projection Tp(Rn) → TpM. It is an exercise to check that this is a torsion-free metric connection on M. It is a (diﬃcult) theorem by Nash that every manifold can be embedded in Rn such that the metric is the induced metric. So all metrics arise this way. 8.4 Curvature The ﬁnal topic to discuss is the curvature of a connection. We all know that Rn is ﬂat, while Sn is curved. How do we make this precise? We can consider doing some parallel transports on Sn along the loop coun- terclockwise: We see that after the parallel transport around the loop, we get a diﬀerent vector. It is in this sense that the connection on S2 is not ﬂat. Thus, what we want is the following notion: 80 8 Connections III Diﬀerential Geometry Deﬁnition (Parallel vector ﬁeld). We say a vector ﬁeld V ∈ Vect(M ) is parallel if V is parallel along any curve in M. Example. In R2, we can pick ξ ∈ T0R2 ∼= R2. Then setting V (p) = ξ ∈ TpR2 ∼= R2 gives a parallel vector ﬁeld with V (0) = ξ. However, we cannot ﬁnd a non-trivial parallel vector ﬁeld on S2. This motivates the question — given a manifold M and a ξ ∈ TpM non-zero, does there exist a parallel vector ﬁeld V on some neighbourhood of p with V (p) = ξ? Naively, we would try to construct it as follows. Say dim M = 2 with coordinates x, y. Put p = (0, 0). Then we can transport ξ along the line {y = 0} to deﬁne V (x, 0). Then for each α, we parallel transport V (α, 0) along {x = α}. So this determines V (x, y). Now if we want this to work, then V has to be parallel along any curve, and in particular for lines {y = β} for β = 0. If we stare at it long enough, we ﬁgure
out a necessary condition is ∇ ∂ ∂xi ∇ ∂ ∂xj = ∇ ∂ ∂xj ∇ ∂ ∂xi. So the failure of these to commute tells us the curvature. This deﬁnition in fact works for any vector bundle. The actual deﬁnition we will state will be slightly funny, but we will soon show afterwards that this is what we think it is. Deﬁnition (Curvature). The curvature of a connection dE : Ω0(E) → Ω1(E) is the map FE = dE ◦ dE : Ω0(E) → Ω2(E). Lemma. FE is a tensor. In particular, FE ∈ Ω2(End(E)). Proof. We have to show that FE is linear over C∞(M ). We let f ∈ C∞(M ) and s ∈ Ω0(E). Then we have FE(f s) = dEdE(f s) = dE(df ⊗ s + f dEs) = d2f ⊗ s − df ∧ dEs + df ∧ dEs + f d2 = f FE(s) Es How do we think about this? Given X, Y ∈ Vect(M ), consider FE(X, Y ) : Ω0(E) → Ω0(E) FE(X, Y )(s) = (FE(s))(X, Y ) Lemma. We have FE(X, Y )(s) = ∇X ∇Y s − ∇Y ∇X s − ∇[X,Y ]s. In other words, we have FE(X, Y ) = [∇X, ∇Y ] − ∇[X,Y ]. 81 8 Connections III Diﬀerential Geometry This is what we were talking about, except that we have an extra term ∇[X,Y ], commute in which vanishes in our previous motivating case, since ∂ ∂xi general. and ∂ ∂xj Proof. We claim that if µ ∈ Ω1(E), then we have (dEµ)(X, Y ) = ∇X (µ(Y
)) − ∇Y (µ(X)) − µ([X, Y ]). To see this, we let µ = ω ⊗ s, where ω ∈ Ω1(M ) and s ∈ Ω0(E). Then we have dEµ = dω ⊗ s − ω ∧ dEs. So we know (dEµ)(X, Y ) = dω(X, Y ) ⊗ s − (ω ∧ dEs)(X, Y ) By a result in the example sheet, this is equal to = (Xω(Y ) − Y ω(X) − ω([X, Y ])) ⊗ s − ω(X)∇Y (s) + ω(Y )∇X (s) = Xω(Y ) ⊗ s + ω(Y )∇X s − (Y ω(X) ⊗ s + ω(X)∇Y s) − ω([X, Y ]) ⊗ s Then the claim follows, since µ([X, Y ]) = ω([X, Y ]) ⊗ s ∇X (µ(Y )) = ∇X (ω(Y )s) = Xω(Y ) ⊗ s + ω(Y )∇X s. Now to prove the lemma, we have (FEs)(X, Y ) = dE(dEs)(X, Y ) = ∇X ((dEs)(Y )) − ∇Y ((dEs)(X)) − (dEs)([X, Y ]) = ∇X ∇Y s − ∇Y ∇X s − ∇[X,Y ]s. Deﬁnition (Flat connection). A connection dE is ﬂat if FE = 0. Specializing to the Riemannian case, we have Deﬁnition (Curvature of metric). Let (M, g) be a Riemannian manifold with metric g. The curvature of g is the curvature of the Levi-Civita connection, denoted by Fg ∈ Ω2(End(T M )) = Ω0(Λ2T ∗M ⊗ T M ⊗ T ∗M ).
Deﬁnition (Flat metric). A Riemannian manifold (M, g) is ﬂat if Fg = 0. Since ﬂatness is a local property, it is clear that if a manifold is locally isometric to Rn, then it is ﬂat. What we want to prove is the converse — if you are ﬂat, then you are locally isometric to Rn. For completeness, let’s deﬁne what an isometry is. 82 8 Connections III Diﬀerential Geometry Deﬁnition (Isometry). Let (M, g) and (N, g) be Riemannian manifolds. We say G ∈ C∞(M, N ) is an isometry if G is a diﬀeomorphism and G∗g = g, i.e. DG\|p : TpM → TG(p)N is an isometry for all p ∈ M. Deﬁnition (Locally isometric). A manifold M is locally isometric to N if for all p ∈ M, there is a neighbourhood U of p and a V ⊆ N and an isometry G : U → V. Example. The ﬂat torus obtained by the quotient of R2 by Z2 is locally isometric to R2, but is not diﬀeomorphic (since it is not even homeomorphic). Our goal is to prove the following result. Theorem. Let M be a manifold with Riemannian metric g.Then M is ﬂat iﬀ it is locally isometric to Rn. One direction is obvious. Since ﬂatness is a local property, we know that if M is locally isometric to Rn, then it is ﬂat. To prove the remaining of the theorem, we need some preparation. Proposition. Let dim M = n and U ⊆ M open. Let V1, · · ·, Vn ∈ Vect(U ) be such that (i) For all p ∈ U, we know V1(p), · · ·, Vn(p) is a basis for TpM, i.e. the Vi are a frame. (ii) [Vi, Vj] = 0, i.e. the Vi
form a frame that pairwise commutes. Then for all p ∈ U, there exists coordinates x1, · · ·, xn on a chart p ∈ Up such that Vi = ∂ ∂xi. Suppose that g is a Riemannian metric on M and the Vi are orthonormal in TpM. Then the map deﬁned above is an isometry. Proof. We ﬁx p ∈ U. Let Θi be the ﬂow of Vi. From example sheet 2, we know that since the Lie brackets vanish, the Θi commute. Recall that (Θi)t(q) = γ(t), where γ is the maximal integral curve of Vi through q. Consider α(t1, · · ·, tn) = (Θn)tn ◦ (Θn−1)tn−1 ◦ · · · ◦ (Θ1)t1(p). Now since each of Θi is deﬁned on some small neighbourhood of p, so if we just move a bit in each direction, we know that α will be deﬁned for (t0, · · ·, tn) ∈ B = {\|ti\| < ε} for some small ε. Our next claim is that Dα ∂ ∂ti = Vi 83 8 Connections III Diﬀerential Geometry whenever this is deﬁned. Indeed, for t ∈ B and f ∈ C∞(M, R). Then we have Dα ∂ ∂ti t (f ) = f (α(t1, · · ·, tn)) ∂ ∂ti ∂ ∂ti t t = Vi\|α(t)(f ). = f ((Θi)t ◦ (Θn)tn ◦ · · · ◦ (Θi)ti ◦ · · · ◦ (Θ1)t1(p)) So done. In particular, we have Dα\|0 ∂ ∂ti 0 = Vi(p), and this is a basis for TpM. So Dα\|0 : T0Rn → TpM is an isomorphism. By the inverse function theorem, this is a local diﬀeomorphism, and in this chart, the claim tells us that
Vi = ∂ ∂xi. The second part with a Riemannian metric is clear. We now actually prove the theorem Proof of theorem. Let (M, g) be a ﬂat manifold. We ﬁx p ∈ M. We let x1, · · ·, xn be coordinates centered at p1, say deﬁned for \|xi\| < 1. We need to construct orthonormal vector ﬁelds. To do this, we pick an orthonormal basis at a point, and parallel transport it around. We let e1, · · ·, en be an orthonormal basis for TpM. We construct vector ﬁelds E1, · · ·, En ∈ Vect(U ) by parallel transport. We ﬁrst parallel transport along (x1, 0, · · ·, 0) which deﬁnes Ei(x1, 0, · · ·, 0), then parallel transport along the x2 direction to cover all Ei(x1, x2, 0, · · ·, 0) etc, until we deﬁne on all of U. By construction, we have ∇kEi = 0 (∗) on {xk+1 = · · · = xn = 0}. We will show that the {Ei} are orthonormal and [Ei, Ej] = 0 for all i, j. We claim that each Ei is parallel, i.e. for any curve γ, we have It is suﬃcient to prove that for all i, j. By induction on k, we show DγEi = 0. ∇jEi = 0 ∇jEi = 0 for j ≤ k on {xk+1 = · · · = xn = 0}. The statement for k = 1 is already given by (∗). We assume the statement for k, so ∇jEi = 0 (A) 84 8 Connections III Diﬀerential Geometry for j ≤ k and {xk+1 = · · · = xn = 0}. For j = k + 1, we know that ∇k+1Ei = 0 on {xk+2 = · · · = xn = 0} by (∗). So the only problem we have is
for j = k and {xk+2 = · · · = xn = 0}. By ﬂatness of the Levi-Civita connection, we have [∇k+1, ∇k] = ∇[∂k+1,∂k] = 0. So we know ∇k+1∇kEi = ∇k∇k+1Ei = 0 (B) on {xk+2 = · · · = xn = 0}. Now at xk+1 = 0, we know ∇kEi vanishes. So it follows from parallel transport that ∇kEi vanishes on {xk+2 = · · · = xn = 0}. As the Levi-Civita connection is compatible with g, we know that parallel transport is an isometry. So the inner product product g(Ei, Ej) = g(ei, ej) = δij. So this gives an orthonormal frame at all points. Finally, since the torsion vanishes, we know as the Ei are parallel. So we are done by the proposition. [Ei, Ej] = ∇EiEj − ∇Ej Ei = 0, What does the curvature mean when it is non-zero? There are many answers to this, and we will only give one. Deﬁnition (Holonomy). Consider a piecewise smooth curve γ : [0, 1] → M with γ(0) = γ(1) = p. Say we have a linear connection ∇. Then we have a notion of parallel transport along γ. The holonomy of ∇ around γ: is the map given by H : TpM → TpM H(ξ) = V (1), where V is the parallel transport of ξ along γ. Example. If ∇ is compatible with a Riemannian metric g, then H is an isometry. Example. Consider Rn with the usual connection. Then if ξ ∈ T0Rn, then H(ξ) = ξ for any such path γ. So the holonomy is trivial. Example. Say (M, g) is ﬂat, and p ∈ M. We have seen that there exists a neighbourhood of p such that (U, g\|
U ) is isometric to Rn. So if γ([0, 1]) ∈ U, then H = id. The curvature measures the extent to which this does not happen. Suppose we have coordinates x1, · · ·, xn on some (M, g). Consider γ as follows: (0, t, · · ·, 0) (s, t, · · ·, 0) 0 (s, 0, · · ·, 0) Then we can Taylor expand to ﬁnd H = id +Fg ∂ ∂x1 p, ∂ ∂x2 p st + O(s2t, st2). 85 Index Index C∞, 4 C∞(M ), 8 C∞(M, N ), 8 C∞(U, E), 39 F -related, 21 F ∗(Y ), 27 F ∗g, 27 Lg, 29 T ∗M, 41 T k M, 42 Der(C∞(M )), 21 Γk i, 76 Ωp(E), 72 Ωp(M ), 44 Θt(p), 25 Vect(U ), 19 VectL(G), 29 dxi, 41 ∂ ∂xi dVg, 61 ∂M, 62 p-form, 41, 44, 9, 11 atlas, 5 equivalence, 6 with boundary, 61 base space, 38 bilinear map, 34 boundary point, 62 bundle morphism, 42 over M, 42 center, 4 chain rule, 11 chart, 4 with boundary, 61 Christoﬀel symbols, 76 closed form, 50 cochain complex, 53 cochain map, 54 cocycle conditions, 39 cohomology, 54 complete vector ﬁeld, 25 connection, 72 cotangent bundle, 41 covariant derivative, 73 III Diﬀerential Geometry covariant tensor, 36 curvature, 81 of metric, 82 curve, 8 length, 42 de Rham cohomology, 50 deformation invariance, 51 homotopy invariance, 51 de Rham complex, 53 de Rham’s theorem, 67 deformation invariance of de Rham cohomology, 51 derivation, 9 derivative, 10, 11 diﬀeomorphism, 8 orientation preserving, 58 diﬀerentiable structure, 6 diﬀerential, 12
smooth function, 4, 8, 59 smooth homotopy, 51 smooth homotopy equivalence, 53 smooth singular homology, 67 snake lemma, 54 star-shaped, 52 Stokes’ theorem, 64 symmetric connection, 79 symplectic form, 48 tangent bundle, 19 tangent space, 9 tensor, 36 tensor product, 34 87 Index III Diﬀerential Geometry vector bundle, 40 vector bundle, 38 tensors on manifolds, 42 torsion of linear connection, 79 torsion-free connection, 79 total space, 38 transition function, 39 trivialization, 38 direct sum, 40 dual, 41 morphism, 42 tensor product, 40 vector ﬁeld, 19 along curve, 73 left invariant, 29 88olds of Euclidean space as in Examples 1.2.3 and 1.2.5. This is because we can visualize curves and surfaces in R3. However, there are a few topics in the later chapters which require the more abstract Deﬁnition 1.4.2 even to say interesting things about extrinsic geometry. There is a generalization to these manifolds involving a structure called a Riemannian metric. We will call this generalization intrinsic diﬀerential geometry. Examples 1.2.6 and 1.2.7 ﬁt into this more general deﬁnition so intrinsic diﬀerential geometry can be used to study them. Since an open set in Euclidean space is a smooth manifold the deﬁnition of a submanifold of Euclidean space (see §2.1 below) is mutatis mutandis the same as the deﬁnition of a submanifold of a manifold. The deﬁnitions in Chapter 2 are worded in such a way that it is easy to read them either extrinsically or intrinsically and the subsequent chapters are mostly (but not entirely) extrinsic. Those sections which require intrinsic diﬀerential geometry (or which translate extrinsic concepts into intrinsic ones) are marked with a *. 14 CHAPTER 1. WHAT IS DIFFERENTIAL GEOMETRY? Chapter 2 Foundations This chapter introduces various fundamental concepts that are central to the ﬁelds of diﬀerential geometry and diﬀerential topology. Both ﬁelds concern the study of
smooth manifolds and their diﬀeomorphisms. The chapter begins with an introduction to submanifolds of Euclidean space and smooth maps (§2.1), to tangent spaces and derivatives (§2.2), and to submanifolds and embeddings (§2.3). In §2.4 we move on to vector ﬁelds and ﬂows and introduce the Lie bracket of two vector ﬁelds. Lie groups and their Lie algebras, in the extrinsic setting, are the subject of §2.5, which includes a proof of the Closed Subgroup Theorem. In §2.6 we introduce vector bundles over a manifold as subbundles of a trivial bundle and in §2.7 we prove the theorem of Frobenius. The last two sections of this chapter are concerned with carrying over all these concepts from the extrinsic to the intrinsic setting and can be skipped at ﬁrst reading (§2.8 and §2.9). 2.1 Submanifolds of Euclidean Space To carry out the Master Plan §1.5 we must (as was done in [50]) extend the deﬁnition of smooth map to maps f : X → Y between subsets X ⊂ Rk and Y ⊂ R which are not necessarily open. In this case a map f : X → Y is called smooth iﬀ for each x0 ∈ X there exists an open neighborhood U ⊂ Rk of x0 and a smooth map F : U → R that agrees with f on U ∩ X. A map f : X → Y is called a diﬀeomorphism iﬀ f is bijective and f and f −1 are smooth. When there exists a diﬀeomorphism f : X → Y then X and Y are called diﬀeomorphic. When X and Y are open these deﬁnitions coincide with the usage in §1.4. 15 16 CHAPTER 2. FOUNDATIONS Exercise 2.1.1 (Chain rule). Let X ⊂ Rk, Y ⊂ R, Z ⊂ Rm be arbitrary subsets. If f : X → Y and g : Y → Z are smooth maps, then so is the composition g ◦ f : X → Z.
The identity map id : X → X is smooth. Exercise 2.1.2. Let E ⊂ Rk be an m-dimensional linear subspace and let v1,..., vm be a basis of E. Then the map f : Rm → E deﬁned by f (x) := m Deﬁnition 2.1.3. Let k, m ∈ N0. A subset M ⊂ Rk is called a smooth m-dimensional submanifold of Rk iﬀ every point p ∈ M has an open neighborhood U ⊂ Rk such that U ∩ M is diﬀeomorphic to an open subset Ω ⊂ Rm. A diﬀeomorphism i=1 xivi is a diﬀeomorphism. φ : U ∩ M → Ω is called a coordinate chart of M and its inverse ψ := φ−1 : Ω → U ∩ M is called a (smooth) parametrization of U ∩ M (see Figure 2.1). Figure 2.1: A coordinate chart φ : U ∩ M → Ω. In Deﬁnition 2.1.3 we have used the fact that the domain of a smooth map can be an arbitrary subset of Euclidean space and need not be open. The term m-manifold in Rk is short for m-dimensional submanifold of Rk. In keeping with the Master Plan §1.5 we will sometimes say manifold rather than submanifold of Rk to indicate that the context holds in both the intrinsic and extrinsic settings. Lemma 2.1.4. If M ⊂ Rk is a nonempty smooth m-manifold, then m ≤ k. Proof. Fix an element p0 ∈ M, choose a coordinate chart φ : U ∩ M → Ω with p0 ∈ U and values in an open subset Ω ⊂ Rm, and denote its inverse by ψ := φ−1 : Ω → U ∩ M. Shrinking U, if necessary, we may assume that φ extends to a smooth map Φ : U → Rm. This extension satisﬁes Φ(ψ(x)) =
φ(ψ(x)) = x and hence dΦ(ψ(x))dψ(x) = id : Rm → Rm for all x ∈ Ω, by the chain rule. Hence the derivative dψ(x) : Rm → Rk is injective for all x ∈ Ω, and hence m ≤ k because Ω is nonempty. This proves Lemma 2.1.4. ψφMUMΩ 2.1. SUBMANIFOLDS OF EUCLIDEAN SPACE 17 Example 2.1.5. Consider the 2-sphere M := S2 = (x, y, z) ∈ R3 \| x2 + y2 + z2 = 1 depicted in Figure 2.2 and let U ⊂ R3 and Ω ⊂ R2 be the open sets U := (x, y, z) ∈ R3 \| z > 0, Ω := (x, y) ∈ R2 \| x2 + y2 < 1. The map φ : U ∩ M → Ω given by φ(x, y, z) := (x, y) is bijective and its inverse ψ := φ−1 : Ω → U ∩ M is given by ψ(x, y) = (x, y, 1 − x2 − y2). Since both φ and ψ are smooth, the map φ is a coordinate chart on S2. Similarly, we can use the open sets z < 0, y > 0, y < 0, x > 0, x < 0 to cover S2 by six coordinate charts. Hence S2 is a manifold. A similar argument shows that the unit sphere Sm ⊂ Rm+1 (see Example 2.1.14 below) is a manifold for every integer m ≥ 0. Figure 2.2: The 2-sphere and the 2-torus. Example 2.1.6. Let Ω ⊂ Rm be an open set and h : Ω → Rk−m be a smooth map. Then the graph of h is a smooth submanifold of Rm × Rk−m = Rk: M := graph(h) := {(x, y) \| x ∈ Ω, y = h(
x)}. It can be covered by a single coordinate chart φ : U ∩ M → Ω, where U := Ω × Rk−m, φ is the projection onto Ω, and ψ := φ−1 : Ω → U is given by ψ(x) = (x, h(x)) for x ∈ Ω. Exercise 2.1.7 (The case m = 0). Show that a subset M ⊂ Rk is a 0dimensional submanifold if and only if M is discrete, i.e. for every p ∈ M there exists an open set U ⊂ Rk such that U ∩ M = {p}. 18 CHAPTER 2. FOUNDATIONS Exercise 2.1.8 (The case m = k). Show that a subset M ⊂ Rm is an m-dimensional submanifold if and only if M is open. Exercise 2.1.9 (Products). If Mi ⊂ Rki is an mi-manifold for i = 1, 2, show that M1 × M2 is an (m1 + m2)-dimensional submanifold of Rk1+k2. Prove that the m-torus Tm := (S1)m is a smooth submanifold of Cm. The next theorem characterizes smooth submanifolds of Euclidean space. In particular condition (iii) will be useful in many cases for verifying the manifold condition. We emphasize that the sets U0 := U ∩ M that appear in Deﬁnition 2.1.3 are open subsets of M with respect to the relative topology that M inherits from the ambient space Rk and that such relatively open sets are also called M -open (see §1.3). Theorem 2.1.10 (Manifolds). Let m and k be integers with 0 ≤ m ≤ k. Let M ⊂ Rk be a set and p0 ∈ M. Then the following are equivalent. (i) There exists an M -open neighborhood U0 ⊂ M of p0 and a diﬀeomorphism φ0 : U0 → Ω0 onto an open set Ω0 ⊂ Rm. (ii) There exist open sets U, Ω ⊂ Rk and a di
ﬀeomorphism φ : U → Ω such that p0 ∈ U and φ(U ∩ M ) = Ω ∩ (Rm × {0}) (see Figure 2.3). (iii) There exists an open set U ⊂ Rk and a smooth map f : U → Rk−m such that p0 ∈ U, the derivative df (p) : Rk → Rk−m is surjective for every point p ∈ U ∩ M, and U ∩ M = f −1(0) = {p ∈ U \| f (p) = 0}. Moreover, if (i) holds, then the diﬀeomorphism φ : U → Ω in (ii) can be chosen such that U ∩ M ⊂ U0 and φ(p) = (φ0(p), 0) for every p ∈ U ∩ M. Figure 2.3: Submanifolds of Euclidean space. pUMφΩ0 2.1. SUBMANIFOLDS OF EUCLIDEAN SPACE 19 Proof. First assume (ii) and denote the diﬀeomorphism in (ii) by φ = (φ1, φ2,..., φk) : U → Ω ⊂ Rk. Then part (i) holds with U0 := U ∩ M, Ω0 := {x ∈ Rm \| (x, 0) ∈ Ω}, and φ0 := (φ1,..., φm)\|U0 : U0 → Ω0, and part (iii) holds with f := (φm+1,..., φk) : U → Rk−m. This shows that part (ii) implies both (i) and (iii). We prove that (i) implies (ii). Let φ0 : U0 → Ω0 be the coordinate chart in part (i), let ψ0 := φ−1 : Ω0 → U0 be its inverse, and let x0 := φ0(p0) ∈ Ω0. 0 Then the derivative dψ0(x0) : Rm → Rk is injective by Lem
ma 2.1.4. Hence there exists a matrix B ∈ Rk×(k−m) such that det([dψ0(x0) B]) = 0. Deﬁne the map ψ : Ω0 × Rk−m → Rk by ψ(x, y) := ψ0(x) + By. Then the k × k-matrix dψ(x0, 0) = [dψ0(x0) B] ∈ Rk×k is nonsingular, by choice of B. Hence, by the Inverse Function Theorem A.2.2, there exists an open neighborhood Ω ⊂ Ω0 × Rk−m of (x0, 0) such that U := ψ(Ω) ⊂ Rk is open and ψ\| Ω : Ω → U is a diﬀeomorphism. In particular, the restriction of ψ to Ω is injective. Now the set {x ∈ Ω0 \| (x, 0) ∈ Ω} is open and contains x0. Hence the set U0 := ψ0(x) x ∈ Ω0, (x, 0) ∈ Ω = p ∈ U0 (φ0(p), 0) ∈ Ω ⊂ M is M -open and contains p0. Hence, by the deﬁnition of the relative topology, there exists an open set W ⊂ Rk such that U0 = W ∩ M. Deﬁne U := U ∩ W, Ω := Ω ∩ ψ−1(W ). Then U ∩ M = U0 and ψ restricts to a diﬀeomorphism from Ω to U. Now let (x, y) ∈ Ω. We claim that ψ(x, y) ∈ M ⇐⇒ y = 0. (2.1.1) If y = 0, then obviously ψ(x, y) = ψ0(x) ∈ M. Conversely, let (x, y) ∈ Ω and suppose that p := ψ(x, y) ∈ M. Then = U0 �
�� U0 and hence (φ0(p), 0) ∈ Ω, by deﬁnition of U0. This implies ψ(φ0(p), 0) = ψ0(φ0(p)) = p = ψ(x, y). Since the pairs (x, y) and (φ0(p), 0) both belong to the set Ω and the restriction of ψ to Ω is injective we obtain x = φ0(p) and y = 0. This proves (2.1.1). It follows from (2.1.1) that the map φ := (ψ\|Ω)−1 : U → Ω satisﬁes φ(U ∩ M ) = {(x, y) ∈ Ω \| ψ(x, y) ∈ M } = Ω ∩ (Rm × {0}). Thus we have proved that (i) implies (ii). 20 CHAPTER 2. FOUNDATIONS We prove that (iii) implies (ii). Let f : U → Rk−m be as in part (iii). Then p0 ∈ U and the derivative df (p0) : Rk → Rk−m is a surjective linear map. Hence there exists a matrix A ∈ Rm×k such that det A df (p0) = 0. Deﬁne the map φ : U → Rk by φ(p) := Ap f (p) for p ∈ U. Then det(dφ(p0)) = 0. Hence, by the Inverse Function Theorem A.2.2, there exists an open neighborhood U ⊂ U of p0 such that Ω := φ(U ) is an open subset of Rk and the restriction φ := φ\|U : U → Ω is a diﬀeomorphism. In particular, the restriction φ\|U is injective. Moreover, it follows from the assumptions on f and the deﬁnition of φ that U ∩ M = p ∈ U f (p) = 0 = p ∈ U and so φ(U ∩M ) = Ω ∩Rm ×{0}. Hence the di
ﬀeomorphism φ : U → Ω satisﬁes the requirements of part (ii). This proves Theorem 2.1.10. φ(p) ∈ Rm × {0} The next corollary relates the notion of a smooth map on a smooth submanifold as deﬁned in the beginning of §2.1 to the standard notion of smoothness in local coordinates used in the intrinsic setting of §2.8 below. Corollary 2.1.11. Let M ⊂ Rk be a smooth m-dimensional submanifold and let f : M → R be a map. Then the following are equivalent. (i) For every p0 ∈ M there exists an open neighborhood U ⊂ Rk of p0 and a smooth map F : U → R that agrees with f on U ∩ M. (ii) If U0 ⊂ M is an M -open set and φ0 : U0 → Ω0 is a diﬀeomorhism onto an open set Ω0 ⊂ Rm, then the composition f ◦ φ−1 0 : Ω0 → R is smooth. Proof. Assume (ii), let p0 ∈ M, and choose φ = (φ1,..., φk) : U → Ω ⊂ Rk as in part (ii) of Theorem 2.1.10. Shrinking U, if necessary, we may assume that Ω = Ω0 × Ω1, where Ω0 ⊂ Rm is an open set and Ω1 ⊂ Rk−m is an open neighborhood of the origin. Then the map Ω0 → R : x → f ◦ φ−1(x, 0) is smooth by part (ii). Deﬁne F (p) := f ◦ φ−1(φ1(p),..., φm(p), 0,..., 0) for p ∈ U. Then the map F : U → R is smooth and agrees with f on U ∩M. Thus f satisﬁes (i). That (i) implies (ii) follows from Exercise 2.1.1 and this
proves Corollary 2.1.11. 2.1. SUBMANIFOLDS OF EUCLIDEAN SPACE 21 Deﬁnition 2.1.12 (Regular value). Let U ⊂ Rk be an open set and let f : U → R be a smooth map. An element c ∈ R is called a regular value of f iﬀ, for all p ∈ U, we have f (p) = c =⇒ df (p) : Rk → R is surjective. Otherwise c is called a singular value of f. Theorem 2.1.10 asserts that, if c is a regular value of f, then the preimage M := f −1(c) = p ∈ U f (p) = c is a smooth (k − )-dimensional submanifold of Rk. Examples and Exercises Example 2.1.13. Let A = AT ∈ Rk×k be a symmetric matrix and deﬁne the function f : Rk → R by f (x) := xTAx. Then df (x)ξ = 2xTAξ for x, ξ ∈ Rk and hence the linear map df (x) : Rk → R is surjective if and only if Ax = 0. Thus c = 0 is the only singular value of f and hence, for every element c ∈ R \ {0}, the set M := f −1(c) = {x ∈ Rk \| xTAx = c} is a smooth manifold of dimension m = k − 1. Example 2.1.14 (The sphere). As a special case of Example 2.1.13 consider the case k = m + 1, A = 1l, and c = 1. Then f (x) = \|x\|2 and so we have another proof that the unit sphere Sm = x ∈ Rm+1 \|x\|2 = 1 in Rm+1 is a smooth m-manifold. (See Examples 1.2.4 and 2.1.5.) Example 2.1.15. Deﬁne the map f : R3 × R3 → R by f (x, y) := \|x − y\|2. This is another special case of Example 2.1.13 and so, for every r > 0, the set M :=
{(x, y) ∈ R3 × R3 \| \|x − y\| = r} is a smooth 5-manifold. Example 2.1.16 (The 2-torus). Let 0 < r < 1 and deﬁne f : R3 → R by f (x, y, z) := (x2 + y2 + r2 − z2 − 1)2 − 4(x2 + y2)(r2 − z2). This map has zero as a regular value and M := f −1(0) is diﬀeomorphic to the 2-torus T2 = S1 × S1. An explicit diﬀeomorphism is given by (eis, eit) → (1 + r cos(s)) cos(t), (1 + r cos(s)) sin(t), r sin(s). This example corresponds to the second surface in Figure 2.2. Exercise: Show that f (x, y, z) = 0 if and only if (x2 + y2 − 1)2 + z2 = r2. Verify that zero is a regular value of f. 22 CHAPTER 2. FOUNDATIONS Example 2.1.17. The set M := (x2, y2, z2, yz, zx, xy) \| x, y, z ∈ R, x2 + y2 + z2 = 1 is a smooth 2-manifold in R6. To see this, deﬁne an equivalence relation on the unit sphere S2 ⊂ R3 by p ∼ q iﬀ q = ±p. The quotient space (the set of equivalence classes) is called the real projective plane and is denoted by RP2 := S2/{±1}. (See Example 1.2.6.) It is equipped with the quotient topology, i.e. a subset U ⊂ RP2 is open, by deﬁnition, iﬀ its preimage under the obvious projection S2 → RP2 is an open subset of S2. Now the map f : S2 → R6 deﬁned by f (x, y, z) := (x2, y2, z2, yz, zx, xy) descends to a homeomorphism from RP2 onto
M. The submanifold M is covered by the local smooth parameterizations Ω → M : (x, y) → f (x, y, Ω → M : (x, z) → f (x, Ω → M : (y, z) → f ( 1 − x2 − y2), 1 − x2 − z2, z), 1 − y2 − z2, y, z), deﬁned on the open unit disc Ω ⊂ R2. We remark the following. (a) M is not the preimage of a regular value under a smooth map R6 → R4. (b) M is not diﬀeomorphic to a submanifold of R3. (c) The projection Σ := (yz, zx, xy) \| x, y, z ∈ R, x2 + y2 + z2 = 1 of M onto the last three coordinates is called the Roman surface and was discovered by Jakob Steiner. The Roman surface can also be represented as the set of solutions (ξ, η, ζ) ∈ R3 of the equation η2ζ 2 + ζ 2ξ2 + ξ2η2 = ξηζ. It is not a submanifold of R3. Exercise: Prove this. Show that M is diﬀeomorphic to a submanifold of R4. Show that M is diﬀeomorphic to RP2 as deﬁned in Example 1.2.6. Exercise 2.1.18. Let V : Rn → R be a smooth function and deﬁne the Hamiltonian function H : Rn × Rn → R (kinetic plus potential energy) by H(x, y) := 1 2 \|y\|2 + V (x). Prove that c is a regular value of H if and only if it is a regular value of V. 2.1. SUBMANIFOLDS OF EUCLIDEAN SPACE 23 Exercise 2.1.19. Consider the general linear group GL(n, R) = g ∈ Rn×n \| det(g) = 0 Prove that the derivative of the function f = det : Rn×n → R is given by df (g)v = det(g) trace(g
−1v) for all g ∈ GL(n, R) and v ∈ Rn×n. Deduce that the special linear group SL(n, R) := {g ∈ GL(n, R) \| det(g) = 1} is a smooth submanifold of Rn×n. Example 2.1.20. The orthogonal group O(n) := g ∈ Rn×n \| gTg = 1l is a smooth submanifold of Rn×n. To see this, denote by Sn := S ∈ Rn×n \| ST = S the vector space of symmetric matrices and deﬁne f : Rn×n → Sn by f (g) := gTg. Its derivative df (g) : Rn×n → Sn is given by df (g)v = gTv + vTg. This map is surjective for every g ∈ O(n): then the matrix v := 1 if gTg = 1l and S = ST ∈ Sn, 2 gS satisﬁes 1 2 gTgS + 1 2 df (g)v = (gS)Tg = 1 2 S + 1 2 ST = S. Hence 1l is a regular value of f and so O(n) is a smooth manifold. It has the dimension dim O(n) = n2 − dim Sn = n2 − n(n + 1) 2 = n(n − 1) 2. Exercise 2.1.21. Prove that the set M := (x, y) ∈ R2 \| xy = 0 is not a submanifold of R2. Hint: If U ⊂ R2 is a neighborhood of the origin and f : U → R is a smooth map such that U ∩ M = f −1(0), then df (0, 0) = 0. 24 CHAPTER 2. FOUNDATIONS 2.2 Tangent Spaces and Derivatives The main reason for ﬁrst discussing the extrinsic notion of embedded manifolds in Euclidean space as explained in the Master Plan §1.5 is that the concept of a tangent vector is much easier to digest in the embedded case: it is simply the derivative of a curve in M, understood as a vector in the ambient Euclidean space in which M is
embedded. 2.2.1 Tangent Space Deﬁnition 2.2.1 (Tangent vector). Let M ⊂ Rk be a smooth m-dimensional manifold and ﬁx a point p ∈ M. A vector v ∈ Rk is called a tangent vector of M at p iﬀ there exists a smooth curve γ : R → M such that γ(0) = p, ˙γ(0) = v. The set TpM := { ˙γ(0) \| γ : R → M is smooth, γ(0) = p} of tangent vectors of M at p is called the tangent space of M at p. Theorem 2.2.3 below shows that TpM is a linear subspace of Rk. As does any linear subspace it contains the origin; it need not actually intersect M. Its translate p + TpM touches M at p; this is what you should visualize for TpM (see Figure 2.4). Figure 2.4: The tangent space TpM and the translated tangent space p+TpM. Remark 2.2.2. Let p ∈ M be as in Deﬁnition 2.2.1 and let v ∈ Rk. Then v ∈ TpM ⇐⇒ ∃ε > 0 ∃γ : (−ε, ε) → M such that γ is smooth, γ(0) = p, ˙γ(0) = v. To see this suppose that γ : (−ε, ε) → M is a smooth curve with γ(0) = p and ˙γ(0) = v. Deﬁne γ : R → M by εt ε2 + t2 γ(t) := γ t ∈ R. √, Then γ is smooth and satisﬁes γ(0) = p and ˙ γ(0) = v. Hence v ∈ TpM. ppMTMp +vMpT0 2.2. TANGENT SPACES AND DERIVATIVES 25 Theorem 2.2.3 (Tangent spaces). Let M ⊂ Rk be a smooth m-dimensional manifold and ﬁx a point p
∈ M. Then the following holds. (i) Let U0 ⊂ M be an M -open set with p ∈ U0 and let φ0 : U0 → Ω0 be a diﬀeomorphism onto an open subset Ω0 ⊂ Rm. Let x0 := φ0(p) and let ψ0 := φ−1 0 : Ω0 → U0 be the inverse map. Then TpM = im dψ0(x0) : Rm → Rk. (ii) Let U, Ω ⊂ Rk be open sets and φ : U → Ω be a diﬀeomorphism such that p ∈ U and φ(U ∩ M ) = Ω ∩ (Rm × {0}). Then TpM = dφ(p)−1 (Rm × {0}). (iii) Let U ⊂ Rk be an open neighborhood of p and f : U → Rk−m be a smooth map such that 0 is a regular value of f and U ∩ M = f −1(0). Then TpM = ker df (p). (iv) TpM is an m-dimensional linear subspace of Rk. Proof. Let ψ0 : Ω0 → U0 and x0 ∈ Ω0 be as in (i) and let φ : U → Ω be as in (ii). We prove that im dψ0(x0) ⊂ TpM ⊂ dφ(p)−1 (Rm × {0}). (2.2.1) To prove the ﬁrst inclusion in (2.2.1), choose a constant r > 0 such that Br(x0) := {x ∈ Rm \| \|x − x0\| < r} ⊂ Ω0. Now let ξ ∈ Rm and choose ε > 0 so small that ε \|ξ\| ≤ r. Then x0 + tξ ∈ Ω0 for all t ∈ R with \|t\| < ε. Deﬁne γ : (−ε, ε) → M by γ(t) := ψ0(x0
+ tξ) for − ε < t < ε. Then γ is a smooth curve in M satisfying γ(0) = ψ0(x0) = p, ˙γ(0) = d dt t=0 ψ0(x0 + tξ) = dψ0(x0)ξ. Hence it follows from Remark 2.2.2 that dψ0(x0)ξ ∈ TpM, as claimed. 26 CHAPTER 2. FOUNDATIONS To prove the second inclusion in (2.2.1) we ﬁx a vector v ∈ TpM. Then, by deﬁnition of the tangent space, there exists a smooth curve γ : R → M such that γ(0) = p and ˙γ(0) = v. Let U ⊂ Rk be as in (ii) and choose ε > 0 so small that γ(t) ∈ U for \|t\| < ε. Then φ(γ(t)) ∈ φ(U ∩ M ) ⊂ Rm × {0} for \|t\| < ε and hence dφ(p)v = dφ(γ(0)) ˙γ(0) = d dt t=0 φ(γ(t)) ∈ Rm × {0}. This shows that v ∈ dφ(p)−1 (Rm × {0}) and thus we have proved (2.2.1). Now the sets im dψ0(x0) and dφ(p)−1 (Rm × {0}) are both m-dimensional linear subspaces of Rk. Hence it follows from (2.2.1) that these subspaces agree and that they both agree with TpM. Thus we have proved assertions (i), (ii), and (iv). We prove (iii). Let v ∈ TpM. Then there is a smooth curve γ : R → M such that γ(0) = p and ˙γ(0) = v. For t suﬃciently small we have γ(t) ∈ U, where U ⊂ Rk is the open set in (iii), and f (γ(t
)) = 0. Hence df (p)v = df (γ(0)) ˙γ(0) = d dt t=0 f (γ(t)) = 0. This implies TpM ⊂ ker df (p). Since TpM and the kernel of df (p) are both m-dimensional linear subspaces of Rk, we deduce that TpM = ker df (p). This proves part (iii) and Theorem 2.2.3. Exercise 2.2.4. Let M ⊂ Rk be a smooth m-dimensional manifold and let pi ∈ M be a sequence that converges to a point p ∈ M. Let τi be a sequence of nonzero real numbers and let v ∈ Rk such that lim i→∞ τi = 0, lim i→0 pi − p τi = v. Prove that v ∈ TpM. Hint: Use part (iii) of Theorem 2.2.3. Example 2.2.5. Let A = AT ∈ Rk×k be a nonzero matrix as in Example 2.1.13 and let c = 0. Then part (iii) of Theorem 2.2.3 asserts that the tangent space of the manifold M = x ∈ Rk xTAx = c at a point x ∈ M is the (k − 1)-dimensional linear subspace TxM = ξ ∈ Rk xTAξ = 0. 2.2. TANGENT SPACES AND DERIVATIVES 27 Example 2.2.6. As a special case of Example 2.2.5 with A = 1l and c = 1 we ﬁnd that the tangent space of the unit sphere Sm ⊂ Rm+1 at a point x ∈ Sm is the orthogonal complement of x. i.e. TxSm = x⊥ = ξ ∈ Rm+1 \| x, ξ = 0. Here x, ξ = m i=0 xiξi denotes the standard inner product on Rm+1. Exercise 2.2.7. What is the tangent space of the 5-manifold M := (x, y) ∈ R3 × R3 \| \|x − y\| = r at a point (x, y) ∈ M
? (See Exercise 2.1.15.) Example 2.2.8. Let H(x, y) := 1 let c be a regular value of H. If (x, y) ∈ M := H −1(c), then 2 \|y\|2 + V (x) be as in Exercise 2.1.18 and T(x,y)M = {(ξ, η) ∈ Rn × Rn \| y, η + ∇V (x), ξ = 0}. Here ∇V := (∂V /∂x1,..., ∂V /∂xn) : Rn → Rn denotes the gradient of V. Exercise 2.2.9. The tangent space of SL(n, R) at the identity matrix is the space sl(n, R) := T1lSL(n, R) = ξ ∈ Rn×n \| trace(ξ) = 0 of traceless matrices. (Prove this, using Exercise 2.1.19.) Example 2.2.10. The tangent space of O(n) at g is TgO(n) = v ∈ Rn×n \| gTv + vTg = 0. In particular, the tangent space of O(n) at the identity matrix is the space of skew-symmetric matrices o(n) := T1lO(n) = ξ ∈ Rn×n \| ξT + ξ = 0 To see this, choose a smooth curve R → O(n) : t → g(t). Then g(t)Tg(t) = 1l for all t ∈ R and, diﬀerentiating this identity with respect to t, we obtain g(t)T ˙g(t) + ˙g(t)Tg(t) = 0 for every t. Hence every matrix v ∈ TgO(n) satisﬁes the equation gTv + vTg = 0. With this understood, the claim follows from the fact that gTv + vTg = 0 if and only if the matrix ξ := g−1v is skew-symmetric and that the space of skew-symmetric matrices in Rn×n has dimension n(n
− 1)/2. Exercise 2.2.11. Let Ω ⊂ Rm be an open set and h : Ω → Rk−m be a smooth map. Prove that the tangent space of the graph of h at a point (x, h(x)) is the graph of the derivative dh(x) : Rm → Rk−m: M = {(x, h(x)) \| x ∈ Ω}, T(x,h(x))M = {(ξ, dh(x)ξ) \| ξ ∈ Rm}. 28 CHAPTER 2. FOUNDATIONS Exercise 2.2.12 (Monge coordinates). Let M be a smooth m-manifold in Rk and suppose that p ∈ M is such that the projection TpM → Rm × {0} is invertible. Prove that there exists an open set Ω ⊂ Rm and a smooth map h : Ω → Rk−m such that the graph of h is an M -open neighborhood of p (see Example 2.1.6). Of course, the projection TpM → Rm × {0} need not be invertible, but it must be invertible for at least one of the k choices m of the m-dimensional coordinate plane. Hence every point of M has an M open neighborhood which may be expressed as a graph of a function of some of the coordinates in terms of the others as in e.g. Example 2.1.5. 2.2.2 Derivative A key purpose behind the concept of a smooth manifold is to carry over the notion of a smooth map and its derivatives from the realm of ﬁrst year analysis to the present geometric setting. Here is the basic deﬁnition. It appeals to the notion of a smooth map between arbitrary subsets of Euclidean spaces as introduced in the beginning of §2.1. Deﬁnition 2.2.13 (Derivative). Let M ⊂ Rk be an m-dimensional smooth manifold and let f : M → R be a smooth map. The derivative of f at a point p ∈ M is the map df (p) : TpM → R deﬁned as follows. Given a tangent vector v ∈ TpM, choose a smooth curve
satisfying γ : R → M γ(0) = p, ˙γ(0) = v, and deﬁne the vector df (p)v ∈ R by df (p)v := d dt t=0 f (γ(t)) = lim h→0 f (γ(h)) − f (p) h. (2.2.2) That the limit on the right in equation (2.2.2) exists follows from our assumptions. We must prove, however, that the derivative is well deﬁned, i.e. that the right hand side of (2.2.2) depends only on the tangent vector v and not on the choice of the curve γ used in the deﬁnition. This is the content of the ﬁrst assertion in the next theorem. 2.2. TANGENT SPACES AND DERIVATIVES 29 Theorem 2.2.14 (Derivatives). Let M ⊂ Rk be an m-dimensional smooth manifold and f : M → R be a smooth map. Fix a point p ∈ M. Then the following holds. (i) The right hand side of (2.2.2) is independent of γ. (ii) The map df (p) : TpM → R is linear. (iii) If N ⊂ R is a smooth n-manifold and f (M ) ⊂ N, then df (p)TpM ⊂ Tf (p)N. (iv) (Chain Rule) Let N be as in (iii), suppose that f (M ) ⊂ N, and let g : N → Rd be a smooth map. Then d(g ◦ f )(p) = dg(f (p)) ◦ df (p) : TpM → Rd. (v) If f = id : M → M, then df (p) = id : TpM → TpM. Proof. We prove (i). Let v ∈ TpM and γ : R → M be as in Deﬁnition 2.2.13. By deﬁnition there is an open neighborhood U ⊂ Rk of p and a smooth map F : U → R such that F (p) = f (p) for
all p ∈ U ∩ M. Let dF (p) ∈ R×k denote the Jacobian matrix (i.e. the matrix of all ﬁrst partial derivatives) of F at p. Then, since γ(t) ∈ U ∩ M for t suﬃciently small, we have dF (p)v = dF (γ(0)) ˙γ(0) = = d dt d dt t=0 t=0 F (γ(t)) f (γ(t)). The right hand side of this identity is independent of the choice of F while the left hand side is independent of the choice of γ. Hence the right hand side is also independent of the choice of γ and this proves (i). Assertion (ii) follows immediately from the identity df (p)v = dF (p)v just established. 30 CHAPTER 2. FOUNDATIONS Assertion (iii) follows directly from the deﬁnitions. Namely, if γ is as in Deﬁnition 2.2.13, then β := f ◦ γ : R → N is a smooth curve in N satisfying β(0) = f (γ(0)) = f (p) =: q, ˙β(0) = df (p)v =: w. Hence w ∈ TqN. Assertion (iv) also follows directly from the deﬁnitions. If g : N → Rd is a smooth map and β, q, w are as above, then d(g ◦ f )(p)v = d dt d dt t=0 t=0 = dg(q)w = g(f (γ(t))) g(β(t)) = dg(f (p))df (p)v. This proves (iv). Assertion (v) follows again directly from the deﬁnitions and this proves Theorem 2.2.14. Corollary 2.2.15 (Diﬀeomorphisms). Let M ⊂ Rk be a smooth m-manifold and N ⊂ R be a smooth n-manifold and let f : M → N be a diﬀeomorphism. Assume M and N are nonempty. Then m =
n and the derivative df (p) : TpM → Tf (p)N is a vector space isomorphism with inverse df (p)−1 = df −1(f (p)) : Tf (p)N → TpM for all p ∈ M. Proof. Deﬁne g := f −1 : N → M so that g ◦ f = idM and f ◦ g = idN. Then it follows from Theorem 2.2.14 that, for p ∈ M and q := f (p) ∈ N, we have dg(q) ◦ df (p) = id : TpM → TpM and df (p) ◦ dg(q) = id : TqN → TqN. Hence df (p) : TpM → TqN is a vector space isomorphism and its inverse is given by dg(q) = df (p)−1 : TqN → TpM. Hence m = n and this proves Corollary 2.2.15. Exercise 2.2.16. Let M ⊂ Rk be a smooth manifold and let f : M → R be a smooth map. Let pi ∈ M be a sequence that converges to a point p ∈ M, let τi be a sequence of nonzero real numbers, and let v ∈ TpM such that lim i→∞ τi = 0, lim i→∞ pi − p τi = v. (See Exercise 2.2.4.) Prove that f (pi) − f (p) τi Hint: Use the local extension F of f in the proof of Theorem 2.2.14. = df (p)v. lim i→∞ 2.2. TANGENT SPACES AND DERIVATIVES 31 2.2.3 The Inverse Function Theorem Corollary 2.2.15 is analogous to the corresponding assertion for smooth maps between open subsets of Euclidean space. Likewise, the inverse function theorem for manifolds is a partial converse of Corollary 2.2.15. Figure 2.5: The Inverse Function Theorem. Theorem 2.2.17 (Inverse Function Theorem). Assume that M ⊂ Rk and N ⊂ R are smooth m-manifolds
and f : M → N is a smooth map. Let p0 ∈ M and suppose that the derivative df (p0) : Tp0M → Tf (p0)N is a vector space isomorphism. Then there exists an M -open neighborhood U ⊂ M of p0 such that V := f (U ) ⊂ N is an N -open subset of N and the restriction f \|U : U → V is a diﬀeomorphism. Proof. Choose coordinate charts φ0 : U0 → U0, deﬁned on an M -open neighborhood U0 ⊂ M of p0 onto an open set U0 ⊂ Rm, and ψ0 : V0 → V0, deﬁned on an N -open neighborhood V0 ⊂ N of f (p0) onto an open set V0 ⊂ Rm. Shrinking U0, if necessary, we may assume that f (U0) ⊂ V0. Then the map f := ψ0 ◦ f ◦ φ−1 0 : U0 → V0 (see Figure 2.5) is smooth and its derivative d f (x0) : Rm → Rm is bijective at x0 := φ0(p0). Hence the Inverse Function Theorem A.2.2 asserts that there exists an open neighborhood U ⊂ U0 of x0 such that V := f ( U ) is an open subset of V0 and the restriction of f to U is a diﬀeomorphism from U to V. Hence the assertion holds with U := φ−1 0 ( V ). This proves Theorem 2.2.17. 0 ( U ) and V := ψ−1 Mφψq V~RU~x~ffU00pV0000000NRmm 32 CHAPTER 2. FOUNDATIONS Regular Values Deﬁnition 2.2.18 (Regular value). Let M ⊂ Rk be a smooth m-manifold, let N ⊂ R be a smooth n-manifold, and let f : M → N be a smooth map. An element q ∈ N is called a regular value of f iﬀ, for every p ∈ M with f (p) = q, the derivative df (p)
: TpM → Tf (p)N is surjective. Theorem 2.2.19 (Regular values). Let f : M → N be as in Deﬁnition 2.2.18 and let q ∈ N be a regular value of f. Then the set P := f −1(q) = {p ∈ M \| f (p) = q} is a smooth submanifold of Rk of dimension m − n and, for each point p ∈ P, its tangent space at p is given by TpP = ker df (p) = {v ∈ TpM \| df (p)v = 0}. Proof 1. Fix a point p0 ∈ P and choose a linear map A : Rk → Rm−n such that the restriction of A to the (m − n)-dimensional linear subspace ker df (p0) = {v ∈ Tp0M \| df (p0) = 0} ⊂ Tp0M ⊂ Rk is a vector space isomorphism. Deﬁne the map F : M → N × Rm−n by F (p) := (f (p), Ap) for p ∈ M. The derivative of F at p0 is given by dF (p0)v = (df (p0)v, Av) for v ∈ Tp0M and is a vector space isomorphism. Hence, by Theorem 2.2.17 there exists an M -open neighborhood U ⊂ M of p0 such that V := F (U ) is an open subset of N × Rm−n in the relative topology and the restriction F \|U : U → V is a diﬀeomorphism. Hence the P -open set U ∩ P is diﬀeomorphic to the open set Ω := {y ∈ Rm−n \| (q, y) ∈ V } ⊂ Rm−n by the diﬀeomorphism φ : U ∩ P → Ω, deﬁned by φ(p) := Ap for p ∈ U ∩ P, whose inverse is the smooth map ψ : Ω → U ∩ P given by ψ(y) = (F \|U )−1(q
, y) for y ∈ Ω. This shows that P is a smooth (m − n)-manifold in Rk. Now let p ∈ P and v ∈ TpP. Then there exists a smooth curve γ : R → P such that γ(0) = p and ˙γ(0) = v. Since f (γ(t)) = q for all t, we have df (p)v = d dt t=0 f (γ(t)) = 0 and so v ∈ ker df (p). Hence TpP ⊂ ker df (p) and equality holds because both TpP and ker df (p) are (m − n)-dimensional linear subspaces of Rk. This proves Theorem 2.2.19. 2.2. TANGENT SPACES AND DERIVATIVES 33 Proof 2. Here is another proof of Theorem 2.2.19 in local coordinates. Fix a point p0 ∈ P and choose a coordinate chart φ0 : U0 → φ0(U0) ⊂ Rm on an M -open neighborhood U0 ⊂ M of p0. Likewise, choose a coordinate chart ψ0 : V0 → ψ0(V0) ⊂ Rn on an N -open neighborhood V0 ⊂ N of q. Shrinking U0, if necessary, we may assume that f (U0) ⊂ V0. Then the point c0 := ψ0(q) is a regular value of the map f0 := ψ0 ◦ f ◦ φ−1 0 : φ0(U0) → Rn. Namely, if x ∈ φ0(U0) satisﬁes f0(x) = c0, then p := φ−1 0 (x) ∈ U0 ∩ P, so the maps dφ−1 0 (x) : Rm → TpM, df (p) : TpM → TqN, and dψ0(q) : TqN → Rn are all surjective, hence so is their composition, and by the chain rule this composition is the derivative df0(x) : Rm → Rn. With this understood, it follows from Theorem 2.1.10
that the set 0 (c0) = x ∈ φ0(U0) \| f (φ−1 f −1 0 (x)) = q = φ0(U0 ∩ P ) is a manifold of of dimension m − n contained in the open set φ0(U0) ⊂ Rm. Using Deﬁnition 2.1.3 and shrinking U0 further, if necessary, we may assume that the set φ0(U0 ∩ P ) is diﬀeomorphic to an open subset of Rm−n. Composing this diﬀeomorphism with φ0 we ﬁnd that U0 ∩ P is diﬀeomorphic to the same open subset of Rm−n. Since the set U0 ⊂ M is M -open, there exists an open set U ⊂ Rk such that U ∩ M = U0, hence U ∩ P = U0 ∩ P, and so U0 ∩ P is a P -open neighborhood of p0. Thus we have proved that every element p0 ∈ P has a P -open neigborhood that is diﬀeomorphic to an open subset of Rm−n. Thus P ⊂ Rk is a manifold of dimension m − n (Deﬁnition 2.1.3). The proof that the tangent spaces of P are given by TpP = ker df (p) remains unchanged and this completes the second proof of Theorem 2.2.19. Deﬁnition 2.2.20. Let M ⊂ Rk and N ⊂ R be smooth m-manifolds. A smooth map f : M → N is called a local diﬀeomorphism iﬀ its derivative df (p) : TpM → Tf (p)N is a vector space isomorphism for every p ∈ M. Example 2.2.21. The inclusion of an M -open subset U ⊂ M into M and the map R → S1 : t → eit are examples of local diﬀeomorphisms. Exercise 2.2.22. Prove that the image of a local diﬀeomorphism is an open subset of the target manifold. Hint: Use the Inverse Function
Theorem. In the terminology introduced in §2.3 and §2.6.1 below, local diﬀeomorphisms are both immersions and submersions. In particular, if f : M → N is a local diﬀeomorphism, then every element q ∈ N is a regular value of f and its preimage f −1(q) is a discrete subset of M. 34 CHAPTER 2. FOUNDATIONS 2.3 Submanifolds and Embeddings This section deals with subsets of a manifold M that are themselves manifolds as in Deﬁnition 2.1.3. Such subsets are called submanifolds of M. Deﬁnition 2.3.1 (Submanifold). Let M ⊂ Rk be an m-dimensional manifold. A subset P ⊂ M is called a submanifold of M of dimension n, iﬀ P itself is an n-manifold. Deﬁnition 2.3.2 (Embedding). Let M ⊂ Rk be an m-dimensional manifold and N ⊂ R be an n-dimensional manifold. A smooth map f : N → M is called an immersion iﬀ its derivatve df (q) : TqN → Tf (q)M is injective for every q ∈ N. It is called proper iﬀ, for every compact subset K ⊂ f (N ), the preimage f −1(K) = {q ∈ N \| f (q) ∈ K} is compact. The map f is called an embedding iﬀ it is a proper injective immersion. Remark 2.3.3. In our deﬁnition of proper maps it is important that the compact set K is required to be contained in the image of f. The literature also contains a stronger deﬁnition of proper which requires that f −1(K) is a compact subset of N for every compact subset K ⊂ M, whether or not K is contained in the image of f. This holds if and only if the map f is proper in the sense of Deﬁnition 2.3.2 and has an M -closed image. (Exercise!) Figure 2.6: A coordinate chart adapted to a submanifold
. Theorem 2.3.4 (Submanifolds). Let M ⊂ Rk be an m-dimensional manifold and N ⊂ R be an n-dimensional manifold. (i) If f : N → M is an embedding, then f (N ) is a submanifold of M. (ii) If P ⊂ M is a submanifold, then the inclusion P → M is an embedding. (iii) A subset P ⊂ M is a submanifold of dimension n if and only if, for every p0 ∈ P, there exists a coordinate chart φ : U → Rm on an M -open neighborhood U of p0 such that φ(U ∩ P ) = φ(U ) ∩ (Rn × {0}) (Figure 2.6). (iv) A subset P ⊂ M is a submanifold of dimension n if and only if, for every p0 ∈ P, there exists an M -open neighborhood U ⊂ M of p0 and a smooth map g : U → Rm−n such that 0 is a regular value of g and U ∩ P = g−1(0). The proof is pased on the following lemma. M0PφpU(U)φ 2.3. SUBMANIFOLDS AND EMBEDDINGS 35 Lemma 2.3.5 (Embeddings). Let M and N be as in Theorem 2.3.4, let f : N → M be an embedding, let q0 ∈ N, and deﬁne P := f (N ), p0 := f (q0) ∈ P. Then there exists an M -open neighborhood U ⊂ M of p0, an N -open neighborhood V ⊂ N of q0, an open neighborhood W ⊂ Rm−n of the origin, and a diﬀeomorphism F : V × W → U such that, for all q ∈ V and all z ∈ W, F (q, 0) = f (q), F (q, z) ∈ P ⇐⇒ z = 0. (2.3.1) (2.3.2) Proof. Choose any coordinate chart φ0 : U0 → Rm on an M -open neighborhood U
0 ⊂ M of p0. Then d(φ0 ◦ f )(q0) = dφ0(f (q0)) ◦ df (q0) : Tq0N → Rm is injective. Hence there is a linear map B : Rm−n → Rm such that the map Tq0N × Rm−n → Rm : (w, ζ) → d(φ0 ◦ f )(q0)w + Bζ (2.3.3) is a vector space isomorphism. Deﬁne the set Ω := (q, z) ∈ N × Rm−n \| f (q) ∈ U0, φ0(f (q)) + Bz ∈ φ0(U0). This is an open subset of N × Rm−n and we deﬁne F : Ω → M by F (q, z) := φ−1 0 (φ0(f (q)) + Bz). This map is smooth, it satisﬁes F (q, 0) = f (q) for all q ∈ f −1(U0), and the derivative dF (q0, 0) : Tq0N × Rm−n → Tp0M is the composition of the map (2.3.3) with dφ0(p0)−1 : Rm → Tp0M and so is a vector space isomorphism. Thus the Inverse Function Theorem 2.2.17 asserts that there is an N -open neighborhood V0 ⊂ N of q0 and an open neighborhood W0 ⊂ Rm−n of the origin such that V0 × W0 ⊂ Ω, the set U0 := F (V0 × W0) is M -open, and the restriction of F to V0 × W0 is a diﬀeomorphism onto U0. Thus we have constructed a diﬀeomorphism F : V0 × W0 → U0 that satisﬁes (2.3.1). We claim that the restriction of F to the product V × W of suﬃciently small open neighborhoods V ⊂ N of q0 and W ⊂ R
m−n of the origin also satisﬁes (2.3.2). Otherwise, there exist sequences qi ∈ V0 converging to q0 and zi ∈ W0 \ {0} converging to zero such that F (qi, zi) ∈ P. Hence there exists a sequence q i). This sequence converges to f (q0). Since f is proper we may assume, passing to a suitable subsequence if necessary, that q i ∈ N such that F (qi, zi) = f (q 0 ∈ N. Then i converges to a point q f (q 0) = lim i→∞ i) = lim i→∞ f (q F (qi, zi) = f (q0). 36 CHAPTER 2. FOUNDATIONS Since f is injective, this implies q suﬃciently large and F (q i, 0) = f (q that the map F : V0 × W0 → M is injective, and proves Lemma 2.3.5. 0 = q0. Hence (q i, 0) ∈ V0 × W0 for i i) = F (qi, zi). This contradicts the fact Proof of Theorem 2.3.4. We prove (i). Let q0 ∈ N, denote p0 := f (q0) ∈ P, and choose a diﬀeomorphism F : V × W → U as in Lemma 2.3.5. Then the set V ⊂ N is diﬀeomorphic to an open subset of Rn (after schrinking V if necessary), the set U ∩ P is P -open because U ⊂ M is M -open, and we have U ∩ P = {F (q, 0) \| q ∈ V } = f (V ) by (2.3.1) and (2.3.2). Hence the map f : V → U ∩ P is a diﬀeomorphism whose inverse is the composition of the smooth maps F −1 : U ∩ P → V × W and V × W → V : (q, z) → q. Hence a P -open neighborhood of p0 is diﬀeomorphic to an open subset of Rn. Since p0 ∈ P was chosen arbitrary, this shows
that P is an n-dimensional submanifold of M. We prove (ii). The inclusion ι : P → M is obviously smooth and injective (it extends to the identity map on Rk). Moreover, TpP ⊂ TpM for every p ∈ P and the derivative dι(p) : TpP → TpM is the obvious inclusion for every p ∈ P. That ι is proper follows immediately from the deﬁnition. Hence ι is an embedding. We prove (iii). If a coordinate chart φ as in (iii) exists, then the set U ∩ P is P -open and is diﬀeomorphic to an open subset of Rn. Since p0 ∈ P was chosen arbitrary this proves that P is an n-dimensional submanifold of M. Conversely, suppose that P is an n-dimensional submanifold of M and let p0 ∈ P. Choose any coordinate chart φ0 : U0 → Rm of M deﬁned on an M -open neighborhood U0 ⊂ M of p0. Then φ0(U0 ∩ P ) is an n-dimensional submanifold of Rm. Hence Theorem 2.1.10 asserts that there are open sets V, W ⊂ Rm with p0 ∈ V ⊂ φ0(U0) and a diﬀeomorphism ψ : V → W such that φ0(p0) ∈ V, ψ(V ∩ φ0(U0 ∩ P )) = W ∩ (Rn × {0}). Now deﬁne U := φ−1 0 (V ) ⊂ U0. Then p0 ∈ U, the chart φ0 restricts to a diﬀeomorphism from U to V, the composition φ := ψ ◦ φ0\|U : U → W is a diﬀeomorphism, and φ(U ∩ P ) = ψ(V ∩ φ0(U0 ∩ P )) = W ∩ (Rn × {0}). We prove (iv). That the condition is suﬃcient follows directly from Theorem 2.2.19. To prove that it is necessary, assume that P �
�� M is a submanifold of dimension n, ﬁx an element p0 ∈ P, and choose a coordinate chart φ : U → Rm on an M -open neighborhood U ⊂ M of p0 as in part (iii). Deﬁne the map g : U → Rm−n by g(p) := (φn+1(p),..., φm(p)) for p ∈ U. Then 0 is a regular value of g and g−1(0) = U ∩ P. This proves Theorem 2.3.4. 2.3. SUBMANIFOLDS AND EMBEDDINGS 37 Exercise 2.3.6. Let M ⊂ Rk be a smooth m-manifold and ∅ = P ⊂ M. (i) If P is an n-dimensional submanifold of M, then 0 ≤ n ≤ m. (ii) P is a 0-dimensional submanifold of M if and only if P is discrete, i.e. every p ∈ P has an M -open neighborhood U such that U ∩ P = {p}. (iii) P is an m-dimensional submanifold of M if and only if P is M -open. Example 2.3.7. Let S1 ⊂ R2 ∼= C be the unit circle and consider the map f : S1 → R2 given by f (x, y) := (x, xy). This map is a proper immersion but is not injective (the points (0, 1) and (0, −1) have the same image under f ). The image f (S1) is a ﬁgure 8 in R2 and is not a submanifold (Figure 2.7). The restriction of f to the submanifold N := S1 \ {(0, −1)} is an injective immersion but it is not proper. It has the same image as before and hence f (N ) is not a manifold. Figure 2.7: A proper immersion. Example 2.3.8. The map f : R → R2 given by f (t) := (t2, t3) is proper and injective, but is not an embedding (its derivatuve at t = 0 is not injective). The image
of f is the set f (R) = C := (x, y) ∈ R2 \| x3 = y2 (see Figure 2.8) and is not a submanifold. (Prove this!) Figure 2.8: A proper injection. Example 2.3.9. Deﬁne the map f : R → R2 by f (t) := (cos(t), sin(t)). This map is an immersion, but it is neither injective nor proper. However, its image is the unit circle in R2 and hence is a submanifold of R2. The map R → R2 : t → f (t3) is not an immersion and is neither injective nor proper, but its image is still the unit circle. 38 CHAPTER 2. FOUNDATIONS 2.4 Vector Fields and Flows This section introduces vector ﬁelds on manifolds (§2.4.1), explains the ﬂow of a vector ﬁeld and the group of diﬀeomorphisms (§2.4.2), and deﬁnes the Lie bracket of two vector ﬁelds (§2.4.3). 2.4.1 Vector Fields Deﬁnition 2.4.1 (Vector ﬁeld). Let M ⊂ Rk be a smooth m-manifold. A (smooth) vector ﬁeld on M is a smooth map X : M → Rk such that X(p) ∈ TpM for every p ∈ M. The set of smooth vector ﬁelds on M will be denoted by Vect(M ) := X : M → Rk \| X is smooth, X(p) ∈ TpM for all p ∈ M. Exercise 2.4.2. Prove that the set of smooth vector ﬁelds on M is a real vector space. Example 2.4.3. Denote the standard cross product on R3 by x × y :=   x2y3 − x3y2 x3y1 − x1y3 x1y2 − x2y1   for x, y ∈ R3. Fix a vector ξ ∈ S2 and deﬁne the maps X
, Y : S2 → R3 by X(p) := ξ × p, Y (p) := (ξ × p) × p. These are vector ﬁelds with zeros ±ξ. Their integral curves (see Deﬁnition 2.4.6 below) are illustrated in Figure 2.9. Figure 2.9: Two vector ﬁelds on the 2-sphere. 2.4. VECTOR FIELDS AND FLOWS 39 Example 2.4.4. Let M := R2. A vector ﬁeld on M is then any smooth map X : R2 → R2. As an example consider the vector ﬁeld X(x, y) := (x, −y). This vector ﬁeld has a single zero at the origin and its integral curves are illustrated in Figure 2.10. Figure 2.10: A hyperbolic ﬁxed point.   Example 2.4.5. Every smooth function f : Rm → R determines a gradient vector ﬁeld X = ∇f :=                       : Rm → Rm. ∂f ∂x1 ∂f ∂x2... ∂f ∂xm Deﬁnition 2.4.6 (Integral curve). Let M ⊂ Rk be a smooth m-manifold, let X be a smooth vector ﬁeld on M, and let I ⊂ R be an open interval. A continuously diﬀerentiable curve γ : I → M is called an integral curve of X iﬀ it satisﬁes the equation ˙γ(t) = X(γ(t)) for every t ∈ I. Note that every integral curve of X is smooth. Theorem 2.4.7. Let M ⊂ Rk be a smooth m-manifold and X ∈ Vect(M ) be a smooth vector ﬁeld. Fix a point
p0 ∈ M. Then the following holds. (i) There exists an open interval I ⊂ R containing 0 and a smooth curve γ : I → M satisfying the equation ˙γ(t) = X(γ(t)), γ(0) = p0 (2.4.1) for every t ∈ I. (ii) If γ1 : I1 → M and γ2 : I2 → M are two solutions of (2.4.1) on open intervals I1 and I2 containing 0, then γ1(t) = γ2(t) for every t ∈ I1 ∩ I2. 40 CHAPTER 2. FOUNDATIONS Figure 2.11: Vector ﬁelds in local coordinates. Proof. We prove (i). Let φ0 : U0 → Rm be a coordinate chart on M, deﬁned on an M -open neighborhood U0 ⊂ M of p0. The image of φ0 is an open set Ω := φ0(U0) ⊂ Rm and we denote the inverse map by ψ0 := φ−1 : Ω → M 0 (see Figure 2.11). Then, by Theorem 2.2.3, the derivative dψ0(x) : Rm → Rk is injective and its image is the tangent space Tψ0(x)M for every x ∈ Ω. Deﬁne f : Ω → Rm by f (x) := dψ0(x)−1X(ψ0(x)) for x ∈ Ω. This map is smooth and hence, by the basic existence and uniqueness theorem for ordinary diﬀerential equations in Rm (see [63]), the equation ˙x(t) = f (x(t)), x(0) = x0 := φ0(p0), (2.4.2) has a solution x : I → Ω on some open interval I ⊂ R containing 0. Hence the function γ := ψ0 ◦ x : I → U0 ⊂ M is a smooth solution of (2.4.1). This proves (i). The local uniqueness theorem asserts that two solutions γi : Ii → M of
(2.4.1) for i = 1, 2 agree on the interval (−ε, ε) ⊂ I1 ∩ I2 for ε > 0 suﬃciently small. This follows immediately from the standard uniqueness theorem for the solutions of (2.4.2) in [63] and the fact that x : I → Ω is a solution of (2.4.2) if and only if γ := ψ0 ◦ x : I → U0 is a solution of (2.4.1). To prove (ii) we observe that the set I := I1 ∩ I2 is an open interval containing zero and hence is connected. Now consider the set A := {t ∈ I \| γ1(t) = γ2(t)}. This set is nonempty, because 0 ∈ A. It is closed, relative to I, because the maps γ1 : I → M and γ2 : I → M are continuous. Namely, if ti ∈ I is a sequence converging to t ∈ I, then γ1(ti) = γ2(ti) for every i and, taking the limit i → ∞, we obtain γ1(t) = γ2(t) and hence t ∈ A. The set A is also open by the local uniqueness theorem. Since I is connected it follows that A = I. This proves (ii) and Theorem 2.4.7. U0M0ψ0φfXp00xmRΩ 2.4. VECTOR FIELDS AND FLOWS 41 2.4.2 The Flow of a Vector Field Deﬁnition 2.4.8 (The ﬂow of a vector ﬁeld). Let M ⊂ Rk be a smooth m-manifold and X ∈ Vect(M ) be a smooth vector ﬁeld on M. For p0 ∈ M the maximal existence interval of p0 is the open interval I ⊂ R is an open interval containing 0 and there is a solution γ : I → M of (2.4.1) I(p0) := I. By Theorem 2.4.7 equation (2.4.1) has a solution γ : I(p0) → M.
The ﬂow of X is the map φ : D → M deﬁned by D := {(t, p0) \| p0 ∈ M, t ∈ I(p0)} and φ(t, p0) := γ(t), where γ : I(p0) → M is the unique solution of (2.4.1). Theorem 2.4.9. Let M ⊂ Rk be a smooth m-manifold and X ∈ Vect(M ) be a smooth vector ﬁeld on M. Let φ : D → M be the ﬂow of X. Then the following holds. (i) D is an open subset of R × M. (ii) The map φ : D → M is smooth. (iii) Let p0 ∈ M and s ∈ I(p0). Then and, for every t ∈ R with s + t ∈ I(p0), we have I(φ(s, p0)) = I(p0) − s φ(s + t, p0) = φ(t, φ(s, p0)). The proof is pased on the following lemma. (2.4.3) (2.4.4) Lemma 2.4.10. Let M, X, D, φ be as in Theorem 2.4.9 and let K ⊂ M be a compact set. Then there exists an M -open set U ⊂ M and an ε > 0 such that K ⊂ U, (−ε, ε) × U ⊂ D, and φ is smooth on (−ε, ε) × U. Proof. In the case where M = Ω is an open subset of Rm this is proved in [64, Satz 4.1.4 & Satz 4.3.1 & Satz 4.4.1]. Using local coordinates (as in the proof of Theorem 2.4.7) we deduce that, for every p ∈ M, there exists an M -open neighborhood Up ⊂ M of p and a constant εp > 0 such that (−εp, εp) × Up ⊂ D and the restriction of φ to (−εp, εp) × Up is
smooth. Using this observation for every p ∈ K (and the axiom of choice) we obtain an M -open cover K ⊂ p∈K Up. Since the set K is compact there exists a ﬁnite subcover K ⊂ Up1 ∪ · · · ∪ UpN =: U. Now deﬁne ε := min{εp1,..., εpN } to deduce that (−ε, ε) × U ⊂ D and φ is smooth on (−ε, ε) × U. This proves Lemma 2.4.10. 42 CHAPTER 2. FOUNDATIONS Proof of Theorem 2.4.9. We prove (iii). The map γ : I(p0) − s → M deﬁned by γ(t) := φ(s+t, p0) is a solution of the initial value problem ˙γ(t) = X(γ(t)) with γ(0) = φ(s, p0). Hence I(p0) − s ⊂ I(φ(s, p0)) and equation (2.4.4) holds for every t ∈ R with s + t ∈ I(p0). In particular, with t = −s we have p0 = φ(−s, φ(s, p0)). Thus we obtain equality in equation (2.4.3) by the same argument with the pair (s.p0) replaced by (−s, φ(s, p0)). We prove (i) and (ii). Let (t0, p0) ∈ D so that p0 ∈ M and t0 ∈ I(p0). Suppose t0 ≥ 0. Then K := {φ(t, p0) \| 0 ≤ t ≤ t0} is a compact subset of M. (It is the image of the compact interval [0, t0] under the unique solution γ : I(p0) → M of (2.4.1).) Hence, by Lemma 2.4.10, there is an M -open set U ⊂ M and an ε > 0 such that K ⊂ U, (−ε, ε) × U ⊂ D, and φ is smooth on (−
ε, ε) × U. Choose N so large that t0/N < ε. Deﬁne U0 := U and, for k = 1,..., N, deﬁne the sets Uk ⊂ M inductively by Uk := {p ∈ U \| φ(t0/N, p) ∈ Uk−1}. These sets are open in the relative topology of M. We prove by induction on k that (−ε, kt0/N + ε) × Uk ⊂ D and φ is smooth on (−ε, kt0/N + ε) × Uk. For k = 0 this holds by deﬁnition of ε and U. If k ∈ {1,..., N } and the assertion holds for k − 1, then we have p ∈ Uk =⇒ p ∈ U, φ(t0/N, p) ∈ Uk−1 =⇒ (−ε, ε) ⊂ I(p), (−ε, (k − 1)t0/N + ε) ⊂ I(φ(t0/N, p)) =⇒ (−ε, kt0/N + ε) ⊂ I(p). Here the last implication follows from (2.4.3). Moreover, for p ∈ Uk and t0/N − ε < t < kt0/N + ε, we have, by (2.4.4), that φ(t, p) = φ(t − t0/N, φ(t0/N, p)) Since φ(t0/N, p) ∈ Uk−1 for p ∈ Uk the right hand side is a smooth map on the open set (t0/N − ε, kt0/N + ε) × Uk. Since Uk ⊂ U, φ is also a smooth map on (−ε, ε) × Uk and hence on (−ε, kt0/N + ε) × Uk. This completes the induction. With k = N we have found an open neighborhood of (t0, p0) contained in D, namely the set (−ε, t0 + ε) × UN, on which φ is smooth. The case t
0 ≤ 0 is treated similarly. This proves (i) and (ii) and Theorem 2.4.9. Deﬁnition 2.4.11. A vector ﬁeld X ∈ Vect(M ) is called complete iﬀ, for each p0 ∈ M, there exists an integral curve γ : R → M of X with γ(0) = p0. 2.4. VECTOR FIELDS AND FLOWS 43 Lemma 2.4.12. Let M ⊂ Rk be a compact manifold. Then every vector ﬁeld on M is complete. Proof. Let X ∈ Vect(M ). It follows from Lemma 2.4.10 with K = M that there exists an ε > 0 such that (−ε, ε) ⊂ I(p) for all p ∈ M. By Theorem 2.4.9 this implies I(p) = R for all p ∈ M. Hence X is complete. Let M ⊂ Rk be a smooth manifold and X ∈ Vect(M ). Then X is complete ⇐⇒ I(p) = R ∀ p ∈ M ⇐⇒ D = R × M. Assume X is complete, let φ : R × M → M be the ﬂow of X, and deﬁne the map φt : M → M by φt(p) := φ(t, p) for t ∈ R and p ∈ M. Then Theorem 2.4.9 asserts that φt is smooth for every t ∈ R and that φs+t = φs ◦ φt, φ0 = id (2.4.5) for all s, t ∈ R. In particular, this implies that φt ◦ φ−t = φ−t ◦ φt = id. Hence φt is bijective and (φt)−1 = φ−t, so each φt is a diﬀeomorphism. Exercise 2.4.13. Let M ⊂ Rk be a smooth manifold. A vector ﬁeld X on M is said to have compact support iﬀ there exists a compact subset K ⊂
M such that X(p) = 0 for every p ∈ M \ K. Prove that every vector ﬁeld with compact support is complete. We close this subsection with an important observation about incomplete vector ﬁelds. The lemma asserts that an integral curve on a ﬁnite existence interval must leave every compact subset of M. Lemma 2.4.14. Let M ⊂ Rk be a smooth m-manifold, let X ∈ Vect(M ), let φ : D → M be the ﬂow of X, let K ⊂ M be a compact set, and let p0 ∈ M be an element such that I(p0) ∩ [0, ∞) = [0, b), 0 < b < ∞. Then there exists a number 0 < tK < b such that tK < t < b =⇒ φ(t, p0) ∈ M \ K Proof. By Lemma 2.4.10 there exists an ε > 0 such that (−ε, ε) ⊂ I(p) for every p ∈ K. Choose ε so small that ε < b and deﬁne tK := b − ε > 0. Choose a real number tK < t < b. Then I(φ(t, p0)) ∩ [0, ∞) = [0, b − t) by equation (2.4.3) in part (ii) of Theorem 2.4.9. Since 0 < b − t < b − tK = ε, this shows that (−ε, ε) ⊂ I(φ(t, p0)) and hence φ(t, p0) /∈ K. This proves Lemma 2.4.14. 44 CHAPTER 2. FOUNDATIONS The next corollary is an immediate consequence of Lemma 2.4.14. In this formulation the result will be used in §4.6 and in §7.3. Corollary 2.4.15. Let M ⊂ Rk be a smooth m-manifold, let X ∈ Vect(M ), and let γ : (0, T ) → M be an integral curve of X. If there exists a compact set K ⊂ M that contains
the image of γ, then γ extends to an integral curve of X on the interval (−ρ, T + ρ) for some ρ > 0. Proof. Here is another more direct proof that does not rely on Lemma 2.4.10. Since K is compact, there exists a constant c > 0 such that \|X(p)\| ≤ c for all p ∈ K. Since γ(t) ∈ K for 0 < t < T, this implies \|γ(t) − γ(s)\| = t s ˙γ(r) dr ≤ t s \| ˙γ(r)\| dr = t s \|X(γ(r))\| dr ≤ c(t − s) for 0 < s < t < T. Thus the limit p0 := limt0 γ(t) exists in Rk and, since K is a closed subset of Rk, we have p0 ∈ K ⊂ M. Deﬁne γ0 : [0, T ) → M by γ0(t) := p0, γ(t), for t = 0, for 0 < t < T. We prove that γ0 is diﬀerentiable at t = 0 and ˙γ0(0) = X(p0). To see this, ﬁx a constant ε > 0. Since the curve [0, T ) → Rk : t → X(γ(t)) is continuous, there exists a constant δ > 0 such that 0 < t ≤ δ =⇒ \|X(γ(t)) − X(p0)\| ≤ ε. Hence, for 0 < s < t ≤ δ, we have \|γ(t) − γ(s) − (t − s)X(p0)\| = = ≤ t s t s t ( ˙γ(r) − X(p0)) dr (X(γ(r)) − X(p0)) dr \|X(γ(r)) − X(p0)\| dr s ≤ (t − s)ε. Take the limit s → 0 to obtain γ(t) − p0 t − X(p0) = lim s→0 \|γ(t) − γ(s) − (t − s)X(p0)\| t − s ≤ ε for 0 <
t ≤ δ. Thus γ0 is diﬀerentiable at t = 0 with ˙γ0(0) = X(p0), as claimed. Hence γ extends to an integral curve γ : (−ρ, T ) → M of X for some ρ > 0 via γ(t) := φ(t, p0) for −ρ < t ≤ 0 and γ(t) := γ(t) for 0 < t < T. Here φ is the ﬂow of X. That γ also extends beyond t = T, follows by replacing γ(t) with γ(T −t) and X with −X. This proves Corollary 2.4.15. 2.4. VECTOR FIELDS AND FLOWS 45 The Group of Diﬀeomorphisms Let us denote the space of diﬀeomorphisms of M by Diﬀ(M ) := {φ : M → M \| φ is a diﬀeomorphism}. This is a group. The group operation is composition and the neutral element is the identity. Now equation (2.4.5) asserts that the ﬂow of a complete vector ﬁeld X ∈ Vect(M ) is a group homomorphism R → Diﬀ(M ) : t → φt. This homomorphism is smooth and is characterized by the equation d dt φt(p) = X(φt(p)), φ0(p) = p for all p ∈ M and t ∈ R. We will often abbreviate this equation in the form d dt φt = X ◦ φt, φ0 = id. (2.4.6) Exercise 2.4.16 (Isotopy). Let M ⊂ Rk be a compact manifold and I ⊂ R be an open interval containing 0. Let I × M → Rk : (t, p) → Xt(p) be a smooth map such that Xt ∈ Vect(M ) for every t. Prove that there is a smooth family of diﬀeomorphisms I × M → M : (t, p) → φt(p) satisfying d dt φt = Xt ◦ φt
, φ0 = id (2.4.7) for every t ∈ I. Such a family of diﬀeomorphisms I → Diﬀ(M ) : t → φt is called an isotopy of M. Conversely prove that every smooth isotopy I → Diﬀ(M ) : t → φt is generated (uniquely) by a smooth family of vector ﬁelds I → Vect(M ) : t → Xt. 46 CHAPTER 2. FOUNDATIONS 2.4.3 The Lie Bracket Let M ⊂ Rk and N ⊂ R be smooth m-manifolds and X ∈ Vect(M ) be smooth vector ﬁeld on M. If ψ : N → M is a diﬀeomorphism, then the pullback of X under ψ is the vector ﬁeld on N deﬁned by (ψ∗X)(q) := dψ(q)−1X(ψ(q)) (2.4.8) for q ∈ N. If φ : M → N is a diﬀeomorphism, then the pushforward of X under φ is the vector ﬁeld on N deﬁned by (φ∗X)(q) := dφ(φ−1(q))X(φ−1(q)) (2.4.9) for q ∈ N. Lemma 2.4.17. Let M ⊂ Rk, N ⊂ R, and P ⊂ Rn be smooth m-dimensional submanifolds, let φ : M → N and ψ : N → P be diﬀeomorphisms, and let X ∈ Vect(M ) and Z ∈ Vect(P ). Then φ∗X = (φ−1)∗X (2.4.10) and (ψ ◦ φ)∗X = ψ∗φ∗X, (ψ ◦ φ)∗Z = φ∗ψ∗Z. (2.4.11) Proof. Equation (2.4.10) follows from the fact that dφ−1(q) = d
φ(φ−1(q))−1 : TqN → Tφ−1(q)M for all q ∈ N (Corollary 2.2.15) and the equations in (2.4.11) follow directly from the chain rule (Theorem 2.2.14). This proves Lemma 2.4.17. We think of a vector ﬁeld on M as a smooth map X : M → Rk that satisﬁes the condition X(p) ∈ TpM for every p ∈ M. Ignoring this condition temporarily, we can diﬀerentiate X as a map from M to Rk and its derivative at p is then a linear map dX(p) : TpM → Rk. In general, this derivative will not take values in the tangent space TpM. However, if we have two vector ﬁelds X and Y on M, then the next lemma shows that the diﬀerence of the derivative of X in the direction Y and of Y in the direction X does take values in the tangent spaces of M. 2.4. VECTOR FIELDS AND FLOWS 47 Lemma 2.4.18 (Lie bracket). Let M ⊂ Rk be a smooth m-manifold and let X, Y ∈ Vect(M ) be complete vector ﬁelds. Denote by R → Diﬀ(M ) : t → φt, R → Diﬀ(M ) : t → ψt the ﬂows of X and Y, respectively. Fix a point p ∈ M and deﬁne the smooth map γ : R → M by γ(t) := φt ◦ ψt ◦ φ−t ◦ ψ−t(p). (2.4.12) Then ˙γ(0) = 0 and γ√ d dt t=0 t = = = ¨γ(0) 1 2 d ds d dt s=0 t=0 ((φs)∗ Y ) (p) ψt∗ X (p) (2.4.13) = dX(p)Y (p) − dY (p)X(p) ∈ T
pM. Exercise 2.4.19. Let γ : R → Rk be a C2-curve and assume ˙γ(0) = 0. t) = limt→0 t−2γ(t) − γ(0) = 1 Prove that d dt 2 ¨γ(0). t=0 γ( √ Proof of Lemma 2.4.18. Deﬁne the map β : R2 → M by β(s, t) := φs ◦ ψt ◦ φ−s ◦ ψ−t(p) for s, t ∈ R. Then γ(t) = β(t, t) and ∂β ∂s ∂β ∂t (0, t) = X(p) − dψt(ψ−t(p))X(ψ−t(p)), (s, 0) = dφs(φ−s(p))Y (φ−s(p)) − Y (p) (2.4.14) (2.4.15) for all s, t ∈ R. Hence ˙γ(0) = ∂β ∂s (0, 0) + ∂β ∂t (0, 0) = 0. This implies the ﬁrst equality in (2.4.13) by Exercise (2.4.19). To prove the remaining assertions, note that β(s, 0) = β(0, t) = p, hence the second derivatives ∂2β/∂s2 and ∂2β/∂t2 vanish at s = t = 0, and therefore ¨γ(0) = 2 ∂2β ∂s∂t (0, 0). (2.4.16) 48 CHAPTER 2. FOUNDATIONS Combining equations (2.4.15) and (2.4.16) we ﬁnd 1 2 ¨γ(0) = = ∂ ∂s d ds s=0 s=0 ∂β ∂t (s, 0) = d ds s=0 ((φs)∗ Y ) (p)). dφs(φ−s(p))Y (φ−s(p)) Likewise,
combining equations (2.4.14) and (2.4.16) we ﬁnd 1 2 ¨γ(0) = = = = ∂ ∂t d dt d dt d dt t=0 t=0 t=0 t=0 ∂β ∂s (0, t) = − d dt t=0 dψt(ψ−t(p))X(ψ−t(p)) dψ−t(ψt(p))X(ψt(p)) dψt(p)−1X(ψt(p)) ψt∗ X (p)). In both cases the right hand side is the derivative of a smooth curve in the tangent space TpM and so is itself an element of TpM. Moreover, we have 1 2 ¨γ(0) = = = = ∂ ∂s ∂ ∂s ∂ ∂t ∂ ∂t s=0 s=0 t=0 t=0 dφs(φ−s(p))Y (φ−s(p)) ∂ ∂t ∂ ∂s t=0 s=0 φs ◦ ψt ◦ φ−s(p) φs ◦ ψt ◦ φ−s(p) X(ψt(p)) − dψt(p)X(p) = dX(p)Y (p) − = dX(p)Y (p) − = dX(p)Y (p) − ∂ ∂t ∂ ∂s ∂ ∂s t=0 s=0 s=0 ∂ ∂s ∂ ∂t s=0 t=0 ψt ◦ φs(p) ψt ◦ φs(p) Y (φs(p)) = dX(p)Y (p) − dY (p)X(p). This proves Lemma 2.4.18. 2.4. VECTOR FIELDS AND FLOWS 49 Deﬁnition 2.4.20 (Lie bracket). Let M ⊂ Rk be a smooth manifold and let X, Y ∈ Vect(M ) be smooth vector ﬁeld
s on M. The Lie bracket of X and Y is the vector ﬁeld [X, Y ] ∈ Vect(M ) deﬁned by [X, Y ](p) := dX(p)Y (p) − dY (p)X(p). (2.4.17) Warning: In the literature on diﬀerential geometry the Lie bracket of two vector ﬁelds is often (but not always) deﬁned with the opposite sign. The rationale behind the present choice of the sign will be explained in §2.5.7. Lemma 2.4.21. Let M ⊂ Rk and N ⊂ R be smooth manifolds, let X, Y, Z be smooth vector ﬁelds on M, and let be a diﬀeomorphism. Then φ : N → M φ∗[X, Y ] = [φ∗X, φ∗Y ], [X, Y ] + [Y, X] = 0, [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0. (2.4.18) (2.4.19) (2.4.20) The last equation is called the Jacobi identity. Proof. Let R → Diﬀ(M ) : t → ψt be the ﬂow of Y. Then the map R → Diﬀ(N ) : t → φ−1 ◦ ψt ◦ φ is the ﬂow of the vector ﬁeld φ∗Y on N. Hence, by Lemma 2.4.17 and Lemma 2.4.18, we have [φ∗X, φ∗Y ] = φ−1 ◦ ψt ◦ φ∗ φ∗X d dt d dt t=0 t=0 = φ∗[X, Y ]. = φ∗ ψt∗ X This proves (2.4.18). Equation (2.4.19) is obvious. To prove (2.4.20), let φt be the ﬂow of X. Then by (2.4.18)
and (2.4.19) and Lemma 2.4.18 we have [[Y, Z], X] = = d dt d dt t=0 t=0 (φt)∗[Y, Z] [(φt)∗Y, (φt)∗Z] = [[Y, X], Z] + [Y, [Z, X]] = [Z, [X, Y ]] + [Y, [Z, X]]. This proves Lemma 2.4.21. 50 CHAPTER 2. FOUNDATIONS Deﬁnition 2.4.22. A Lie algebra is a real vector space g equipped with a skew-symmetric bilinear map g × g → g : (ξ, η) → [ξ, η] that satisﬁes the Jacobi identity [ξ, [η, ζ]] + [η, [ζ, ξ]] + [ζ, [ξ, η]] = 0 for all ξ, η, ζ ∈ g. Example 2.4.23. The Vector ﬁelds on a smooth manifold M ⊂ Rk form a Lie algebra with the Lie bracket (2.4.17). The space gl(n, R) = Rn×n of real n × n-matrices is a Lie algebra with the Lie bracket [ξ, η] := ξη − ηξ. It is also interesting to consider subspaces of gl(n, R) that are invariant under this Lie bracket. An example is the space o(n) := ξ ∈ gl(n, R) ξT + ξ = 0 of skew-symmetric n × n-matrices. It is a nontrivial fact that every ﬁnitedimensional Lie algebra is isomorphic to a Lie subalgebra of gl(n, R) for some n. For example, the cross product deﬁnes a Lie algebra structure on R3 and the resulting Lie algebra is isomorphic to o(3). Remark 2.4.24. There is a linear map Rm×m → Vect(Rm) : ξ → Xξ which assigns to a matrix ξ ∈ gl(m, R) the linear
vector ﬁeld Xξ : Rm → Rm given by Xξ(x) := ξx for x ∈ Rm. This map preserves the Lie bracket, i.e. [Xξ, Xη] = X[ξ,η], and hence is a Lie algebra homomorphism. To understand the Lie bracket geometrically, consider again the curve γ(t) := φt ◦ ψt ◦ φ−t ◦ ψ−t(p) in Lemma 2.4.18, where φt and ψt are the ﬂows of the vector ﬁelds X and Y, respectively. Since ˙γ(0) = 0, Exercise 2.4.19 asserts that [X, Y ](p) = 1 2 ¨γ(0) = d dt t=− √ t ◦ ψ− √ t(p). (2.4.21) Geometrically this means that by following ﬁrst the backward ﬂow of Y for time ε, then the backward ﬂow of X for time ε, then the forward ﬂow of Y for time ε, and ﬁnally the forward ﬂow of X for time ε, we will not, in general, get back to the original point p where we started but approximately obtain an “error” ε2[X, Y ](p). An example of this (which we learned from Donaldson) is the mathematical formulation of parking a car. 2.4. VECTOR FIELDS AND FLOWS 51 Example 2.4.25 (Parking a car). The conﬁguration space for driving a car in the plane is the manifold M := C × S1, where S1 ⊂ C denotes the unit circle. Thus a point in M is a pair p = (z, λ) ∈ C × C with \|λ\| = 1. The point z ∈ C represents the position of the car and the unit vector λ ∈ S1 represents the direction in which it is pointing. The left turn is represented by a vector ﬁeld X and the right turn by a vector ﬁeld Y on M. These vector ﬁeld are given
by X(z, λ) := (λ, iλ) and Y (z, λ) := (λ, −iλ). Their Lie bracket is the vector ﬁeld [X, Y ](z, λ) = (−2iλ, 0). This vector ﬁeld represents a sideways move of the car to the right. And a sideways move by 2ε2 can be achieved by following a backward right turn for time ε, then a backward left turn for time ε, then a forward right turn for time ε, and ﬁnally a forward left turn for time ε. This example can be reformulated by identifying C with R2 via z = x+iy and representing a point in the unit circle by the angle θ ∈ R/2πZ via λ = eiθ. In this formulation the manifold is M = R2 × R/2πZ, a point in M is represented by a triple (x, y, θ) ∈ R3, the vector ﬁelds X and Y are X(x, y, θ) := (cos(θ), sin(θ), 1), Y (x, y, θ) := (cos(θ), sin(θ), −1), and their Lie bracket is [X, Y ](x, y, θ) = 2(sin(θ), − cos(θ), 0). Lemma 2.4.26. Let X, Y ∈ Vect(M ) be complete vector ﬁelds on a manifold M and φt, ψt ∈ Diﬀ(M ) be the ﬂows of X and Y, respectively. Then the Lie bracket [X, Y ] vanishes if and only if the ﬂows of X and Y commute, i.e. φs ◦ ψt = ψt ◦ φs for all s, t ∈ R. Proof. If the ﬂows of X and Y commute, then the Lie bracket [X, Y ] vanishes by Lemma 2.4.18. Conversely, suppose that [X, Y ] = 0. Then we have d ds for every s ∈ R and hence (φs)∗ Y = (φs)�
� d dr r=0 (φr)∗ Y = (φs)∗ [X, Y ] = 0 (φs)∗Y = Y. (2.4.22) Fix a real number s and deﬁne the curve γ : R → M by γ(t) := φs(ψt(p)) for t ∈ R. Then γ(0) = φs(p) and ˙γ(t) = dφs(ψt(p))Y (ψt(p)) = ((φs)∗ Y ) (γ(t)) = Y (γ(t)) for all t. Here the last equation follows from (2.4.22). Since ψt is the ﬂow of Y we obtain γ(t) = ψt(φs(p)) for all t ∈ R and this proves Lemma 2.4.26. Exercise 2.4.27. In the situation of Lemma 2.4.26 prove that {φt ◦ ψt}t∈R is the ﬂow of the vector ﬁeld X + Y. 52 CHAPTER 2. FOUNDATIONS 2.5 Lie Groups Combining the concept of a group and a manifold, it is interesting to consider groups which are also manifolds and have the property that the group operation and the inverse deﬁne smooth maps. We shall only consider groups of matrices. 2.5.1 Deﬁnition and Examples Deﬁnition 2.5.1 (Lie group). A nonempty subset G ⊂ Rn×n is called a Lie group iﬀ it is a submanifold of Rn×n and a subgroup of GL(n, R), i.e. g, h ∈ G =⇒ gh ∈ G (where gh denotes the product of the matrices g and h) and g ∈ G =⇒ det(g) = 0 and g−1 ∈ G. (Since G = ∅ it follows from these conditions that the identity matrix 1l is an element of G.) Example 2.5.2. The general linear group G = GL(n, R) is an open subset of Rn×n and hence
is a Lie group. By Exercise 2.1.19 the special linear group SL(n, R) = g ∈ GL(n, R) det(g) = 1 is a Lie group and, by Example 2.1.20, the special orthogonal group g ∈ GL(n, R) gTg = 1l, det(g) = 1 SO(n) := is a Lie group. so SO(n) is an open subset of O(n) (in the relative topology). In fact every orthogonal matrix has determinant ±1 and In a similar vein the group GL(n, C) := {g ∈ Cn×n \| det(g) = 0} of complex matrices with nonzero (complex) determinant is an open subset of Cn×n and hence is a Lie group. As in the real case, the subgroups SL(n, C) := g ∈ GL(n, C) U(n) := g ∈ GL(n, C) det(g) = 1, g∗g = 1l, g∗g = 1l, det(g) = 1 SU(n) := g ∈ GL(n, C) are submanifolds of GL(n, C) and hence are Lie groups. Here g∗ := ¯gT denotes the conjugate transpose of a complex matrix. Exercise 2.5.3. Prove that SL(n, C), U(n), and SU(n) are Lie groups. Prove that SO(n) is connected and that O(n) has two connected components. 2.5. LIE GROUPS 53 Exercise 2.5.4. Prove that GL(n, C) can be identiﬁed with the group G := {Φ ∈ GL(2n, R) \| ΦJ0 = J0Φ}, J0 := 0 −1l 0 1l. Hint: Use the isomorphism Rn × Rn → Cn : (x, y) → x + iy. Show that a matrix Φ ∈ R2n×2n commutes with J0 if and only if it has the form Φ = X −Y X Y, X, Y ∈ Rn×n. What is the relation between the real determinant of Φ and
the complex determinant of X + iY? Exercise 2.5.5. Let J0 be as in Exercise 2.5.4 and deﬁne ΨTJ0Ψ = J0 Ψ ∈ GL(2n, R) Sp(2n) :=. This is the symplectic linear group. Prove that Sp(2n) is a Lie group. Hint: See [49, Lemma 1.1.12]. Example 2.5.6 (Unit quaternions). The quaternions form a fourdimensional associative unital algebra H, equipped with a basis 1, i, j, k. The elements of H are vectors of the form x = x0 + ix1 + jx2 + kx3 (2.5.1) The product structure is the bilinear map H × H → H : (x, y) → xy, determined by the relations i2 = j2 = k2 = −1, jk = −kj = i, ki = −ik = j. ij = −ji = k, x0, x1, x2, x3 ∈ R. This product structure is associative but not commutative. The quaternions are equipped with an involution H → H : x → ¯x, which assigns to a quaternion x of the form (2.5.1) its conjugate ¯x := x0 − ix1 − jx2 − kx3. This involution satisﬁes the conditions x + y = ¯x + ¯y, for x, y ∈ H, where \|x\| := x2 of the quaternion (2.5.1). Thus the unit quaternions form a group Sp(1) := x ∈ H xy = ¯y¯x, 0 + x2 x¯x = \|x\|2, 2 + x2 \|xy\| = \|x\| \|y\| 2 + x2 \|x\| = 1 3 denotes the Euclidean norm with the inverse map x → ¯x. Note that the group Sp(1) is diﬀeomorphic to the 3-sphere S3 ⊂ R4 under the isomorphism H ∼= R4. Warning: The unit quaternions (a compact Lie group) are not to be confused with the symplectic linear group in Exercise 2
.5.5 (a noncompact Lie group) despite the similarity in notation. 54 CHAPTER 2. FOUNDATIONS Let G ⊂ GL(n, R) be a Lie group. Then the maps G × G → G : (g, h) → gh, G → G : g → g−1 are smooth (see [64]). Fixing an element h ∈ G we ﬁnd that the derivative of the map G → G : g → gh at g ∈ G is given by the linear map (2.5.2) TgG → TghG : g → gh. Here g and h are both matrices in Rn×n and gh denotes the matrix product. In fact, if g ∈ TgG, then, since G is a manifold, there exists a smooth curve γ : R → G with γ(0) = g and ˙γ(0) = g. Since G is a group we obtain a smooth curve β : R → G given by β(t) := γ(t)h. It satisﬁes β(0) = gh and so gh = ˙β(0) ∈ TghG. The linear map (2.5.2) is obviously a vector space isomorphism whose inverse is given by right multiplication with h−1. It is sometimes convenient to deﬁne the map Rh : G → G by Rh(g) := gh for g ∈ G (right multiplication by h). This is a diﬀeomorphism and the linear map (2.5.2) is the derivative of Rh at g, so dRh(g)g = gh for g ∈ TgG. Similarly, each element g ∈ G determines a diﬀeomorphism Lg : G → G, given by Lg(h) := gh for h ∈ G (left multiplication by g). Its derivative at h ∈ G is again given by matrix multiplication, i.e. the linear map dLg(h) : ThG → TghG is given by dLg(h)h = gh for h ∈ ThG. (2.5.3) Since Lg is a diﬀeomorphism its derivative dLg(h) : ThG → TghG is again a vector space isomorphism for every
h ∈ G. Exercise 2.5.7. Prove that the map G → G : g → g−1 is a diﬀeomorphism and that its derivative at g ∈ G is the vector space isomorphism TgG → Tg−1G : v → −g−1vg−1. Hint: Use [64] or any textbook on ﬁrst year analysis. 2.5. LIE GROUPS 55 2.5.2 The Lie Algebra of a Lie Group Let G ⊂ GL(n, R) be a Lie group. Its tangent space at the identity matrix 1l ∈ G is called the Lie algebra of G and will be denoted by g = Lie(G) := T1lG. This terminology is justiﬁed by the fact that g is in fact a Lie algebra, i.e. it is invariant under the standard Lie bracket operation [ξ, η] := ξη − ηξ on the space Rn×n of square matrices (see Lemma 2.5.9 below). The proof requires the notion of the exponential matrix. For ξ ∈ Rn×n and t ∈ R we deﬁne exp(tξ) :=. (2.5.4) ∞ k=0 tkξk k! A standard result in ﬁrst year analysis asserts that this series converges absolutely (and uniformly on compact t-intervals), that the map R → Rn×n : t → exp(tξ) is smooth and satisﬁes the diﬀerential equation d dt exp(tξ) = ξ exp(tξ) = exp(tξ)ξ, (2.5.5) and that exp((s + t)ξ) = exp(sξ) exp(tξ), exp(0ξ) = 1l (2.5.6) for all s, t ∈ R. This shows that the matrix exp(tξ) is invertible for each t and that the map R → GL(n, R) : t → exp(tξ) is a group homomorphism. Exercise 2.5.8. Prove the following analogue of (2.4.12). For
ξ, η ∈ g d dt t=0 √ exp( √ tξ) exp( √ tη) exp(− √ tξ) exp(− tη) = [ξ, η]. (2.5.7) In other words, the inﬁnitesimal Lie group commutator is the matrix commutator. (Compare Equations (2.5.7) and (2.4.21).) 56 CHAPTER 2. FOUNDATIONS Lemma 2.5.9. Let G ⊂ GL(n, R) be a Lie group and denote by g := Lie(G) its Lie algebra. Then the following holds. (i) If ξ ∈ g, then exp(tξ) ∈ G for every t ∈ R. (ii) If g ∈ G and η ∈ g, then gηg−1 ∈ g. (iii) If ξ, η ∈ g, then [ξ, η] = ξη − ηξ ∈ g. Proof. We prove (i). For every g ∈ G we have a vector space isomorphism g = T1lG → TgG : ξ → ξg as in (2.5.2). Hence each element ξ ∈ g determines a vector ﬁeld Xξ ∈ Vect(G), deﬁned by Xξ(g) := ξg ∈ TgG, g ∈ G. (2.5.8) By Theorem 2.4.7 there is an integral curve γ : (−ε, ε) → G satisfying ˙γ(t) = Xξ(γ(t)) = ξγ(t), γ(0) = 1l. By (2.5.5), the curve (−ε, ε) → Rn×n : t → exp(tξ) satisﬁes the same initial value problem and hence, by uniqueness, we have exp(tξ) = γ(t) ∈ G for all t ∈ R with \|t\| < ε. Now let t ∈ R and choose N ∈ N such that t < ε. N Then exp( t
N ξ) ∈ G and hence it follows from (2.5.6) that exp(tξ) = exp N t N ξ ∈ G. This proves (i). We prove (ii). Consider the smooth curve γ : R → Rn×n deﬁned by γ(t) := g exp(tη)g−1. By (i) we have γ(t) ∈ G for every t ∈ R. Since γ(0) = 1l we have gηg−1 = ˙γ(0) ∈ g. This proves (ii). We prove (iii). Deﬁne the smooth map η : R → Rn×n by η(t) := exp(tξ)η exp(−tξ). By (i) we have exp(tξ) ∈ G and, by (ii), we have η(t) ∈ g for every t ∈ R. Hence [ξ, η] = ˙η(0) ∈ g. This proves (iii) and Lemma 2.5.9. By Lemma 2.5.9 the curve γ : R → G deﬁned by γ(t) := exp(tξ)g is the integral curve of the vector ﬁeld Xξ in (2.5.8) with initial condition γ(0) = g. Thus Xξ is complete for every ξ ∈ g. 2.5. LIE GROUPS 57 Lemma 2.5.10. If ξ ∈ g and γ : R → G is a smooth curve satisfying γ(s + t) = γ(s)γ(t), γ(0) = 1l, ˙γ(0) = ξ, (2.5.9) then γ(t) = exp(tξ) for every t ∈ R. Proof. For every t ∈ R we have ˙γ(t) = d ds s=0 γ(s + t) = d ds s=0 γ(s)γ(t) = ˙γ(0)γ(t) = ξγ(t). Hence γ is the integral curve
of the vector ﬁeld Xξ in (2.5.8) with γ(0) = 1l. This implies γ(t) = exp(tξ) for every t ∈ R, as claimed. Example 2.5.11. Since the general linear group GL(n, R) is an open subset of Rn×n its Lie algebra is the space of all real n × n-matrices gl(n, R) := Lie(GL(n, R)) = Rn×n. The Lie algebra of the special linear group is sl(n, R) := Lie(SL(n, R)) = ξ ∈ gl(n, R) trace(ξ) = 0 (see Exercise 2.2.9) and the Lie algebra of the special orthogonal group is so(n) := Lie(SO(n)) = ξ ∈ gl(n, R) ξT + ξ = 0 = o(n) (see Example 2.2.10). Exercise 2.5.12. Prove that the Lie algebras of the general linear group over C, the special linear group over C, the unitary group, and the special unitary group are given by gl(n, C) := Lie(GL(n, C)) = Cn×n, sl(n, C) := Lie(SL(n, C)) = ξ ∈ gl(n, C) u(n) := Lie(U(n)) = ξ ∈ gl(n, R) trace(ξ) = 0, ξ∗ + ξ = 0, ξ∗ + ξ = 0, trace(ξ) = 0. su(n) := Lie(SU(n)) = ξ ∈ gl(n, C) These are vector spaces over the reals. Determine their real dimensions. Which of these are also complex vector spaces? Remark 2.5.13. Let G ⊂ GL(n, R) be a subgroup. In Theorem 2.5.27 below it is shown that G is a Lie group if and only if it is a closed subset of GL(n, R) in the relative topology. This observation can be used in many of the examples and exercises of the present section. 58 CHAPTER 2. FOUNDATIONS Exercise 2.
5.14. Let V be a ﬁnite-dimensional vector space. Prove that the vector space g := V × End(V ) is a Lie algebra with the Lie bracket [(u, A), (v, B)] := (Av − Bu, AB − BA) (2.5.10) for u, v ∈ V and A, B ∈ End(V ). Find the corresponding Lie group. Find an embedding of g into End(R × V ) as a Lie subalgebra. Exercise 2.5.15. Let (V, ω) be a 2n-dimensional symplectic vector space, so ω : V × V → R is a nondegenerate skew-symmetric bilinear form. The Heisenberg algebra of (V, ω) is the Lie algebra h := V × R with the Lie bracket of two elements (v, t), (v, t) ∈ V × R deﬁned by (v, t), (v, t) := 0, ω(v, v). (2.5.11) Find a corresponding Lie group structure on H = V × R. Embed H as a Lie subgroup into GL(n + 2, R) and ﬁnd a formula for the exponential map. Hint: Take V = Rn × Rn and ω(x, y), (x, y) = x, y − y, x and deﬁne (x, y, t) · (x, y, t) := (x + x, y + y, t + t + x, y). 2.5.3 Lie Group Homomorphisms Let G, H be Lie groups and g, h be Lie algebras. A Lie group homomorphism from G to H is a smooth map ρ : G → H that is a group homomorphism. A Lie group isomorphism is a bijective Lie group homomorphism whose inverse is also a Lie group homomorphism. A Lie group automorphism is a Lie group isomorphism from a Lie group to itself. A Lie algebra homomorphism from g to h is a linear map Φ : g → h that preserves the Lie bracket. A Lie algebra isomorphism is a bijective Lie algebra homomorphism whose inverse is also a Lie algebra homomorphism. A Lie algebra automorphism
is a Lie algebra isomorphism from a Lie algebra to itself. Lemma 2.5.16. Let G and H be Lie groups and denote their Lie algebras by g := Lie(G) and h := Lie(H). Let ρ : G → H be a Lie group homomorphism and denote its derivative at 1l ∈ G by ˙ρ := dρ(1l) : g → h. Then ˙ρ is a Lie algebra homomorphism. Moreover, ρ(exp(ξ)) = exp( ˙ρ(ξ)), ρ(gξg−1) = ρ(g) ˙ρ(ξ)ρ(g)−1 for all ξ ∈ g and all g ∈ G. 2.5. LIE GROUPS 59 Proof. The proof has three steps. Step 1. For all ξ ∈ g and t ∈ R we have ρ(exp(tξ)) = exp(t ˙ρ(ξ)). Fix an element ξ ∈ g. Then exp(tξ) ∈ G for every t ∈ R by Lemma 2.5.9. Thus we can deﬁne a curve γ : R → H by γ(t) := ρ(exp(tξ)). Since ρ is smooth, this is a smooth curve in H and, since ρ is a group homomorphism and the exponential map satisﬁes (2.5.6), our curve γ satisﬁes the conditions γ(s + t) = γ(s)γ(t), γ(0) = 1l, ˙γ(0) = dρ(1l)ξ = ˙ρ(ξ). Hence γ(t) = exp(t ˙ρ(ξ)) by Lemma 2.5.10. This proves Step 1. Step 2. For all g ∈ G and η ∈ g we have ˙ρ(gηg−1) = ρ(g) ˙ρ(η)ρ(g)−1. Deﬁne the smooth curve γ : R → G by γ(t) := g exp(tη)g−1. It
takes values in G by Lemma 2.5.9. By Step 1 we have ρ(γ(t)) = ρ(g)ρ(exp(tη))ρ(g)−1 = ρ(g) exp(t ˙ρ(η))ρ(g)−1 for every t. Since γ(0) = 1l and ˙γ(0) = gηg−1 we obtain ˙ρ(gηg−1) = dρ(γ(0)) ˙γ(0) d dt d dt t=0 t=0 = = ρ(γ(t)) ρ(g) exp(t ˙ρ(η))ρ(g)−1 This proves Step 2. = ρ(g) ˙ρ(η)ρ(g)−1. Step 3. For all ξ, η ∈ g we have ˙ρ([ξ, η]) = [ ˙ρ(ξ), ˙ρ(η)]. Deﬁne the curve η : R → g by η(t) := exp(tξ)η exp(−tξ) for t ∈ R. It takes values in the Lie algebra of G by Lemma 2.5.9 and ˙η(0) = [ξ, η]. Hence ˙ρ([ξ, η]) = ˙ρ (exp(tξ)η exp(−tξ)) = d dt d dt d dt t=0 t=0 t=0 = [ ˙ρ(ξ), ˙ρ(η)]. = ρ (exp(tξ)) ˙ρ(η)ρ (exp(−tξ)) exp (t ˙ρ(ξ)) ˙ρ(η) exp (−t ˙ρ(ξ)) Here the ﬁrst equality follows from the fact that ˙ρ is linear, the second equality follows from Step 2 with g = exp(tξ), and the third equality follows from Step 1. This proves Step 3 and Lemma 2.5.16. 60 CHAPTER 2. FOUNDATIONS Exercise 2.5
.17. A Lie group homomorphism ρ : G → H is uniquely determined by the Lie algebra homomorphism ˙ρ whenever G is connected. Hint: If ρ1, ρ2 : G → H are Lie group homomorphisms such that ˙ρ1 = ˙ρ2, prove that the set A := {g ∈ G \| ρ1(g) = ρ2(g)} is both open and closed. Exercise 2.5.18. If ˙ρ : g → h is a bijective Lie algebra homomorphism, then its inverse is also a Lie algebra homomorphism. Exercise 2.5.19. If ρ : G → H is a bijective Lie group homomorphism, then ρ−1 : H → G is smooth and hence ρ is a Lie group isomorphism. Hint: Use Lemma 2.5.16 to prove that ˙ρ : g → h is injective. If ˙ρ is not surjective, show that ρ has no regular value in contradiction to Sard’s theorem. Example 2.5.20. The complex determinant deﬁnes a Lie group homomorphism det : U(n) → S1. The associated Lie algebra homomorphism is trace = ˙det : u(n) → iR = Lie(S1). Example 2.5.21 (Unit quaternions and SU(2)). The Lie group SU(2) is diﬀeomorphic to the 3-sphere. Every matrix in SU(2) can be written as g = x0 + ix1 x2 + ix3 −x2 + ix3 x0 − ix1, 0 + x2 x2 1 + x2 2 + x2 3 = 1. (2.5.12) Here the xi are real numbers. They can be interpreted as the coordinates of a unit quaternion x = x0 + ix1 + jx2 + kx3 ∈ Sp(1) (see Example 2.5.6). The reader may verify that the map Sp(1) → SU(2) : x → g in (2.5.12) is a Lie group isomorphism. Exercise 2.5.22 (The double cover of SO(3)). Identify the imaginary part of
H with R3 and write a vector ξ ∈ R3 = Im(H) as a purely imaginary quaternion ξ = iξ1 + jξ1 + kξ3. Prove that if ξ ∈ Im(H) and x ∈ Sp(1), then xξ ¯x ∈ Im(H). Deﬁne the map ρ : Sp(1) → SO(3) by ρ(x)ξ := xξ ¯x for x ∈ Sp(1) and ξ ∈ Im(H). Prove that the linear map ρ(x) : R3 → R3 is represented by the 3 × 3-matrix ρ(x) =   0 + x2 2 − x2 1 − x2 x2 3 2(x1x2 + x0x3) 2(x1x3 − x0x2) 2(x1x2 − x0x3) 3 − x2 2 − x2 0 + x2 x2 1 2(x2x3 + x0x1) 2(x1x3 + x0x2) 2(x2x3 − x0x1) 1 − x2 3 − x2 0 + x2 x2 2  . Show that ρ is a Lie group homomorphism. Find a formula for the map ˙ρ := dρ(1l) : sp(1) → so(3) and show that it is a Lie algebra isomorphism. For x, y ∈ Sp(1) prove that ρ(x) = ρ(y) if and only if y = ±x. 2.5. LIE GROUPS 61 Example 2.5.23. Let g be a ﬁnite-dimensional Lie algebra. Then the set Aut(g) :=   Φ : g → g  Φ is a bijective linear map and Φ[ξ, η] = [Φξ, Φη] for all ξ, η ∈ g    (2.5.13) of Lie algebra automorphisms of g is a Lie group. Its Lie algebra is the space of derivations on g denoted by Der
(g) :=   δ : g → g  δ is a linear map and δ [ξ, η] = [δ ξ, η] + [ξ, δ η] for all ξ, η ∈ g   . (2.5.14) Now suppose that g = Lie(G) is the Lie algebra of a Lie group G. Then there is a map Ad : G → Aut(g) deﬁned by Ad(g)η := gηg−1 (2.5.15) for g ∈ G and η ∈ g. Part (ii) of Lemma 2.5.9 asserts that Ad(g) maps g to itself for every g ∈ G. It follows directly from the deﬁnitions that the map Ad(g) : g → g is a Lie algebra automorphism for every g ∈ G and that the map Ad : G → Aut(g) is a Lie group homomorphism. The associated Lie algebra homomorphism is the linear map ad : g → Der(g) deﬁned by ad(ξ)η := [ξ, η] (2.5.16) for ξ, η ∈ g. To verify the equation ad = ˙Ad we compute ˙Ad(ξ)η = d dt t=0 Ad(exp(tξ))η = d dt t=0 exp(tξ)η exp(−tξ) = [ξ, η]. Exercise 2.5.24. Let g be any Lie algebra. Deﬁne the map ad : g → End(g) by (2.5.16) and prove that the endomorphism ad(ξ) : g → g is a derivation for every ξ ∈ g. Prove that ad : g → Der(g) is a Lie algebra homomorphism. Exercise 2.5.25. Let g be any ﬁnite-dimensional Lie algebra. Prove that the group Aut(g) in (2.5.13) is a Lie subgroup of GL(g) with the Lie algebra Lie(Aut(g)) = Der(g). Hint
: Show that a linear map δ : g → g is a derivation if and only if the linear map exp(tδ) : g → g is a Lie algebra automorphism for every t ∈ R. Use the Closed Subgroup Theorem 2.5.27. 62 CHAPTER 2. FOUNDATIONS 2.5.4 Closed Subgroups This section deals with subgroups of a Lie group G that are also submanifolds of G. Such subgroups are called Lie subgroups. We assume throughout that G ⊂ GL(n, R) is a Lie group with the Lie algebra g := Lie(G) = T1lG. Deﬁnition 2.5.26 (Lie subgroup). A subset H ⊂ G is called a Lie subgroup of G iﬀ it is both a subgroup and a smooth submanifold of G. A useful general criterion is the Closed Subgroup Theorem which asserts that a subgroup H ⊂ G is a Lie subgroup if and only if it is a closed subset of G. This was ﬁrst proved in 1929 by John von Neumann [54] for the special case G = GL(n, R) and then in 1930 by ´Elie Cartan [15] in full generality. Theorem 2.5.27 (Closed Subgroup Theorem). Let H be a subgroup of G. Then the following are equivalent. (i) H is a smooth submanifold (and hence a Lie subgroup) of G. (ii) H is a closed subset of G. It (i) holds, then the Lie algebra of H is the space h := η ∈ g exp(tη) ∈ H for all t ∈ R. (2.5.17) Proof of Theorem 2.5.27 (i) =⇒ (ii) and (2.5.17). Assume that H is a Lie subgroup of G and let h ⊂ g be deﬁned by (2.5.17). We prove that h is the Lie algebra of H. Assume ﬁrst that η ∈ h. Then the curve γ : R → G deﬁned by γ(t) := exp(tη) for t ∈ R takes values in H and satis�
��es γ(0) = 1l and ˙γ(0) = η, and this implies η ∈ T1lH = Lie(H). Conversely, if η ∈ Lie(H), then Lemma 2.5.9 asserts that exp(tη) ∈ H for all t ∈ R and hence η ∈ h. This shows that h = Lie(H). Next we prove in three steps that H is a closed subset of G. Choose any inner product on g, denote by \|·\| the associated norm, and denote by h⊥ ⊂ g the orthogonal complement of h with respect to this inner product. Step 1. There exist open neighborhoods V ⊂ H of 1l and W ⊂ h⊥ of the origin such that the map φ : V × W → G, deﬁned by φ(h, ξ) := h exp(ξ) for h ∈ V and ξ ∈ W, is a diﬀeomorphism from V × W onto an open neighborhood U = φ(V × W ) ⊂ G of 1l. The derivative of the map H × h⊥ → G : (h, ξ) → h exp(ξ) at the point (1l, 0) is bijective. Hence Step 1 follows from the Inverse Function Theorem. 2.5. LIE GROUPS 63 Step 2. There exists a δ > 0 such that, if ξ, ξ ∈ h⊥ satisfy \|ξ\|, \|ξ\| < δ and exp(ξ) exp(−ξ) ∈ H, then ξ = ξ. Let V, W, φ be as in Step 1, choose an open neigborhood V ⊂ G of 1l such that V ∩ H = V, and choose a constant δ > 0 such that the following holds. (a) If ξ ∈ h⊥ satisﬁes \|ξ\| < δ, then ξ ∈ W. (b) If ξ, ξ ∈ g satisfy \|ξ\|, \|ξ\| < δ, then exp(ξ) exp(−ξ) ∈ V. Let ξ, ξ
∈ h⊥ such that \|ξ\|, \|ξ\| < δ and h := exp(ξ) exp(−ξ) ∈ H. Then we have ξ, ξ ∈ W by (a) and h ∈ V ∩ H = V by (b). Also φ(h, ξ) = φ(1l, ξ) and so ξ = ξ, because φ is injective on V × W. This proves Step 2. Step 3. Let hi be a sequence in H that converges to an element g ∈ G. Then g ∈ H. i g = h Let φ : V × W → U be as in Step 1 and let δ > 0 be as in Step 2. Since the sequence h−1 i g converges to 1l, there exists an i0 ∈ N such that h−1 i g ∈ U for all i ≥ i0. Hence, for each i ≥ i0, there exists a unique pair (h i, ξi) ∈ V × W such that h−1 i exp(ξi). This sequence satisﬁes limi→∞ ξi = 0. Hence there exists an integer i1 ≥ i0 such that \|ξi\| < δ for all i ≥ i1. Since i exp(ξi) = g = hjh i)−1hjh j exp(ξj), we also have exp(ξi) exp(−ξj) = (hih j ∈ H for all i, j ≥ i1. By Step 3, this implies ξi = ξj for all i, j ≥ i1. Hence ξi = limj→∞ ξj = 0 for all i ≥ i1 and so g = hih i ∈ H. This proves Step 3. hih By Step 3 the Lie subgroup H is a closed subset of G. Thus we have proved that (i) implies (ii) and (2.5.17) in Theorem 2.5.27. The proof of the converse implication requires three preparatory lemmas. Lemma 2.5.28. Let ξ ∈ g and let γ : R → G be a curve that is diﬀerentiable at t = 0 and sat
isﬁes γ(0) = 1l and ˙γ(0) = ξ. Then exp(tξ) = lim k→∞ γ(t/k)k (2.5.18) for every t ∈ R. Proof. Fix a nonzero real number t and deﬁne ξk := kγ(t/k) − 1l ∈ Rn×n for k ∈ N. Then lim k→∞ ξk = t lim k→∞ γ(t/k) − γ(0) t/k = t ˙γ(0) = tξ and hence = lim k→∞ (See [64, Satz 1.5.2].) This proves Lemma 2.5.28. exp(tξ) = lim k→∞ 1l + γ(t/k)k. k ξk k 64 CHAPTER 2. FOUNDATIONS Lemma 2.5.29. Let H ⊂ G be a closed subgroup. Then the set h in (2.5.17) is a Lie subalgebra of g Proof. Let ξ, η ∈ h and deﬁne the curve γ : R → H by γ(t) := exp(tξ) exp(tη) for t ∈ R. This curve is smooth and satisﬁes γ(0) = 1l and ˙γ(0) = ξ + η. Since H is closed, it follows from Lemma 2.5.28 that exp(t(ξ + η)) = lim k→∞ γ(t/k)k ∈ H for all t ∈ R and so ξ + η ∈ h by deﬁnition. Thus h is a linear subspace of g. Now ﬁx an element ξ ∈ h. If h ∈ H, then exp(sh−1ξh) = h−1 exp(sξ)h ∈ H for all s ∈ R and hence h−1ξh ∈ h by deﬁnition. Take h = exp(tη) with η ∈ h to obtain exp(−tη)ξ exp
(tη) ∈ h for all t ∈ R. Diﬀerentiating this curve at t = 0 gives [ξ, η] ∈ h and this proves Lemma 2.5.29. Lemma 2.5.30. Let H ⊂ G be a closed subgroup and let h ⊂ g be the Lie subalgebra in (2.5.17). Let ξ ∈ g, let (ξi)i∈N be a sequence in g, and let (τi)i∈N be a sequence of positive real numbers such that exp(ξi) ∈ H, ξi = 0 for all i ∈ N and lim i→∞ τi = 0, lim i→∞ ξi = 0, lim i→∞ ξi τi = ξ. Then ξ ∈ h. Proof. Fix a real number t. Then, for each i ∈ N, there exists a unique integer mi ∈ Z such that miτi ≤ t < (mi + 1)τi. The sequence mi satisﬁes lim i→∞ miτi = t, lim i→∞ miξi = lim i→∞ miτi ξi τi = tξ and hence exp(tξ) = lim i→∞ exp(miξi) = lim i→∞ exp(ξi)mi ∈ H. Thus exp(tξ) ∈ H for every t ∈ R and so ξ ∈ h by (2.5.17). This proves Lemma 2.5.30. 2.5. LIE GROUPS 65 Proof of Theorem 2.5.27 (ii) =⇒ (i). Choose any inner product on g. Let H ⊂ G be a closed subgroup of G and deﬁne the set h ⊂ g by (2.5.17). Then h is a Lie subalgebra of g by Lemma 2.5.29. Deﬁne k := dim(h), := dim(g) ≥ k, and choose a basis η1,..., η of g such that the vectors η1...., ηk form a basis of h and ην
∈ h⊥ for ν > k. Let h0 ∈ H and deﬁne the map Θ : R → G by Θ(t1,..., t) := h0 exp(t1η1 + · · · + tkηk) exp(tk+1ηk+1 + · · · + tη). Then Θ(0) = h0, Θ(Rk × {0}) ⊂ H, and the derivative dΘ(0) : R → Th0G is bijective. Hence the inverse function theorem asserts that Θ restricts to a diﬀeomorphism from an open neighborhood Ω ⊂ R of the origin to an open neighborhood U := Θ(Ω) ⊂ G of h0 that satisﬁes Θ(0) = h0, ΘΩ ∩ (Rk × {0}) ⊂ U ∩ H. We prove the following. Claim. There exists an open set Ω0 ⊂ R such that 0 ∈ Ω0 ⊂ Ω, ΘΩ0 ∩ (Rk × {0}) = U0 ∩ H, U0 := Θ(Ω0). (2.5.19) Assume, by contradiction, that such an open set Ω0 does not exist. Then there exists a sequence ti = (t1 i) ∈ R such that i,..., t lim i→∞ ti = 0, ti ∈ Ω \ (Rk × {0}), Θ(ti) ∈ H. Deﬁne hi := h0 exp tν i ην ∈ H, ξi := k ν=1 ν=k+1 i ην ∈ h⊥ \ {0}. tν Then hi exp(ξi) = Θ(ti) ∈ H and hence lim i→∞ ξi = 0, ξi = 0, exp(ξi) = h−1 i Θ(ti) ∈ H. Passing to a subsequence, if necessary, we may assume that the sequence ξi/\|ξi\| converges. Denote its limit by
ξ := limi→∞ ξi/\|ξi\|. Then ξ ∈ h by Lemma 2.5.30 and ξ ∈ h⊥ by deﬁnition. Since \|ξ\| = 1, this is a contradiction. This contradiction proves the Claim. Thus there does, after all, exist an open set Ω0 ⊂ R that satisﬁes (2.5.19), and the map Θ−1 : U0 → Ω0 is then a coordinate chart on G which satisﬁes Θ−1(U0 ∩ H) = Ω0 ∩ (Rk × {0}). Hence H is a submanifold of G and this proves Theorem 2.5.27. 66 CHAPTER 2. FOUNDATIONS Exercise 2.5.31. The subgroup {exp(it) \| t ∈ Q} ⊂ S1 is not closed. Exercise 2.5.32. Choose a nonzero vector (ω1,..., ωn) ∈ Rn such that at least one of the ratios ωi/ωj is irrational. Prove that the subgroup Sω := {(e2πitω1, e2πitω2,..., e2πitωn) \| t ∈ R} ⊂ (S1)n ∼= Tn of the torus is not closed. Similar examples exist in any Lie group that contains a torus of dimension at least two. Exercise 2.5.33. Let G0 and G1 be Lie subgroups of GL(n, R) with the Lie algebras g0 := Lie(G0) and g1 := Lie(G1). Prove that G := G0 ∩ G1 is a Lie subgroup of GL(n, R) with the Lie algebra g = g0 ∩ g1. Exercise 2.5.34 (Center). The center of a group G is the subgroup Z(G) := g ∈ G gh = hg for all h ∈ G. (2.5.20) Let G ⊂ GL(n, R) be a Lie group. Prove that its center Z(G) is a Lie subgroup of G. If G is connected, prove that the
Lie algebra of the center Z(G) is the center of the Lie algebra g = Lie(G), deﬁned by [ξ, η] = 0 for all η ∈ g. Z(g) := ξ ∈ g (2.5.21) Hint: If G is connected, prove that an element ξ ∈ g satisﬁes [ξ, η] = 0 for all η ∈ g if and only if exp(tξ)h = h exp(tξ) for all t ∈ R and all h ∈ G. Exercise 2.5.35. Let G ⊂ GL(n, R) be a compact Lie group with the Lie algebra g := Lie(G) and let ξ ∈ g. Prove that the set Tξ := exp(tξ) t ∈ R is a closed, connected, abelian subgroup of G and deduce that it is a Lie subgroup of G (called the torus generated by ξ). Exercise 2.5.36. Let G ⊂ GL(n, R) be a Lie group with the Lie algebra g and let ξ : R → g be a smooth function. Prove that the diﬀerential equation ˙γ(t) = ξ(t)γ(t), γ(0) = 1l, has a unique solution γ : R → G. Hint: Prove the existence of a solution γ : R → GL(n, R) and show that the set {t ∈ R \| γ(t) ∈ G} is open and closed. Remark 2.5.37 (Malcev’s Theorem). Let G ⊂ GL(n, R) be a Lie group with the Lie algebra g and let h ⊂ g be a Lie subalgebra. Then the set H :=   h(1)  h : [0, 1] → G is a smooth path such that h(0) = 1l and ˙h(t)h(t)−1 ∈ h for all t ∈ [0, 1]    (2.5.22) is a subgroup of G, called the integral subgroup of h. A
theorem by Anatolij Ivanovich Malcev [47] (see also [28, Corollary 13.4.6]) asserts that H is a Lie subgroup of G if and only if Tη := exp(tη) t ∈ R ⊂ H for all η ∈ h. 2.5. LIE GROUPS 67 2.5.5 Lie Groups and Diﬀeomorphisms There is a natural correspondence between Lie groups and Lie algebras on the one hand and diﬀeomorphisms and vector ﬁelds on the other hand. We summarize this correspondence in the following table. Lie groups G ⊂ GL(n, R) g = Lie(G) = T1lG exponential map t → exp(tξ) adjoint representation ξ → gξg−1 Lie bracket on g [ξ, η] = ξη − ηξ Diﬀeomorphisms Diﬀ(M ) Vect(M ) = TidDiﬀ(M ) ﬂow of a vector ﬁeld t → φt = “ exp(tX) pushforward X → φ∗X Lie bracket of vector ﬁelds [X, Y ] = dX · Y − dY · X To understand the correspondence between the exponential map and the ﬂow of a vector ﬁeld compare equation (2.4.6) with equation (2.5.5). To understand the correspondence between the adjoint representation and pushforward observe that φ∗Y = d dt t=0 φ ◦ ψt ◦ φ−1, gηg−1 = d dt t=0 g exp(tη)g−1, where ψt denotes the ﬂow of Y. To understand the correspondence between the Lie brackets recall that [X, Y ] = d dt t=0 (φt)∗Y, [ξ, η] = d dt t=0 exp(tξ)η exp(−tξ), where φt denotes the ﬂow of X. We emphasize that the analogy between Lie groups and Diﬀeomorphisms only works well when the manifold M
is compact so that every vector ﬁeld on M is complete. The next exercise gives another parallel between the Lie bracket on the Lie algebra of a Lie group and the Lie bracket of two vector ﬁelds. Exercise 2.5.38. Let G ⊂ GL(n, R) be a Lie group with Lie algebra g and let ξ, η ∈ g. Deﬁne the smooth curve γ : R → G by γ(t) := exp(tξ) exp(tη) exp(−tξ) exp(−tη). Show that ˙γ(0) = 0 and 1 2 ¨γ(0) = [ξ, η] (cf. Exercise 2.5.8 and Lemma 2.4.18). Exercise 2.5.39. Let G ⊂ GL(n, R) be a Lie group with Lie algebra g and let ξ, η ∈ g. Show that [ξ, η] = 0 if and only if the exponential maps commute, i.e. exp(sξ) exp(tη) = exp(tη) exp(sξ) = exp(sξ + tη) for all s, t ∈ R. How can this observation be deduced from Lemma 2.4.26? 68 CHAPTER 2. FOUNDATIONS Deﬁnition 2.5.40. Let M ⊂ Rk be a smooth manifold and let G ⊂ GL(n, R) be a Lie group. A (smooth) group action of G on M is a smooth map G × M → M : (g, p) → φg(p) (2.5.23) that for each pair g, h ∈ G satisﬁes the condition φg ◦ φh = φgh, φ1l = id. (2.5.24) If (2.5.23) is a smooth group action, then the inﬁnitesimal action of the Lie algebra g := Lie(G) on M is the map g → Vect(M ) : ξ → Xξ deﬁned by t=0 φexp(yξ)(p) Xξ(p)
:= (2.5.25) d dt for ξ ∈ g and p ∈ M. Exercise 2.5.41. Let (2.5.23) be a smooth group action of a Lie group G on a manifold M. Prove that Xg−1ξg = φ∗ gXξ, X[ξ,η] = [Xξ, Xη] (2.5.26) for all g ∈ G and all ξ, η ∈ g = Lie(G). Exercise 2.5.42. Show that the maps GL(m, R) × Rm → Rm : (gx) → gx, SO(m + 1) × Sm → Sm : (g, x) → gx, and R × S1 → S1 : (θ, z) → eiθz are smooth group actions. Verify the formulas in (2.5.26) in these examples. A smooth group action of a Lie group G on a manifold M can the thought of as a “Lie group homomorphism” G → Diﬀ(M ) : g → φg. (2.5.27) While the group Diﬀ(M ) is inﬁnite-dimensional, and so cannot cannot be a Lie group in the formal sense, it has many properties in common with Lie groups as explained above. For example, one can deﬁne what is meant by a smooth path in Diﬀ(M ) and extend formally the notion of a tangent vector (as the derivative of a path through a given element of Diﬀ(M )) to this setting. In particular, the tangent space of Diﬀ(M ) at the identity can then be identiﬁed with the space of vector ﬁelds TidDiﬀ(M ) = Vect(M ), and the inﬁnitesimal action in (2.5.25) is the Lie algebra homomorphism associated to the “Lie group homomorphism” (2.5.27). In fact, we have chosen the sign in the deﬁnition of the Lie bracket of vector ﬁelds so that the map ξ → Xξ is a Lie algebra homomorphism and not a
Lie algebra anti-homomorphism. This will be discussed further in §2.5.7. 2.5. LIE GROUPS 69 2.5.6 Smooth Maps and Algebra Homomorphisms Let M be a smooth submanifold of Rk. Denote by F (M ) := C∞(M, R) the space of smooth real valued functions f : M → R. Then F (M ) is a commutative unital algebra. Each p ∈ M determines a unital algebra homomorphism εp : F (M ) → R deﬁned by εp(f ) = f (p) for p ∈ M. Theorem 2.5.43. Every unital algebra homomorphism ε : F (M ) → R has the form ε = εp for some p ∈ M. Proof. Assume that ε : F (M ) → R is an algebra homomorphism. Claim. For all f, g ∈ F (M ) we have ε(g) = 0 =⇒ ε(f ) ∈ f (g−1(0)). Indeed, the function f − ε(f ) · 1 lies in the kernel of ε and so the function h := (f − ε(f ) · 1)2 + g2 also lies in the kernel of ε. There must be at least one point p ∈ M where h(p) = 0 for otherwise 1 = ε(h)ε(1/h) = 0. For this point p we have f (p) = ε(f ) and g(p) = 0, hence p ∈ g−1(0), and therefore ε(f ) = f (p) ∈ f (g−1(0)). This proves the claim. The theorem asserts that there exists a p ∈ M such that every f ∈ F (M ) satisﬁes ε(f ) = f (p). Assume, by contradiction, that this is false. Then for every p ∈ M there exists a function f ∈ F (M ) such that f (p) = ε(f ). Replace f by f − ε(f ) to obtain f (p) = 0 = ε(f ). Now use the axiom of choice to obtain a family of functions f

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