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Home , Differentiable manifold, Immersion (mathematics), Local homeomorphism, Topological manifold

LECTURE 1. Differentiable , differentiable maps Def: topological m-. Locally euclidean, Hausdorff, second-countable . At each point we have a local chart. (U, h), where h : U → Rm is a topological ( onto its ) and h(U) is open in Rm. Def: differentiable structure. An of class Cr (r ≥ 1 or r = ∞ ) for a topological m-manifold M is a collection U of local charts (U, h) satisfying: (i) The domains U of the charts in U define an open of M; (ii) If two domains of charts (U, h), (V, k) in U overlap (U ∩V 6= ∅) , then the transition map:

k ◦ h−1 : h(U ∩ V ) → k(U ∩ V ) is a Cr diffeomorphism of open sets in Rm. (iii) The atlas U is maximal for property (ii). Def: differentiable map between differentiable manifolds. f : M m → N n (continuous) is differentiable at p if for any local charts (U, h), (V, k) at p, resp. f(p) with f(U) ⊂ V the map k ◦ f ◦ h−1 = F : h(U) → k(V ) is m differentiable at x0 = h(p) (as a map from an open subset of R to an open subset of Rn.) f : M → N (continuous) is of class Cr if for any charts (U, h), (V, k) on M resp. N with f(U) ⊂ V , the composition k ◦ f ◦ h−1 : h(U) → k(V ) is of class Cr (between open subsets of Rm, resp. Rn.) f : M → N of class Cr is an if, for any local charts (U, h), (V, k) as above (for M, resp. N), the composition F = k ◦ f ◦ h−1 satisfies ker[dF (x)] = {0} for all x ∈ h(U), where dF (x) ∈ L(Rm; Rn) is the dif- ferential of F . (In particular, m ≤ n necessarily.) f : M → N is a diffeomorphism if it is a homeomorphism onto N and, for any p ∈ M and local charts (U, h), (V, h) at p, f(p) (resp.) and F = k ◦ f ◦ h−1, the differential dF (x) ∈ L(Rm,Rn), x = h(p), is an . (In particular, m = n.) If a Cr map f : M → N is a diffeomorphism , the inverse f −1 is of class Cr.

1 LECTURE 2. at a point.

Recall a subset M ⊂ Rn is a m-dimensional of class Ck in Rn (m ≤ n) if for any p ∈ M we may find an open nbd. of p, W ⊂ Rn, and an immersion of class Ck, φ : U → Rn (U ⊂ Rm open) which is a homeomorphism onto its image: φ(U) = W ∩ M. n For surfaces in R , the tangent space TpM at p ∈ M is geometrically defined as the of all velocity vectors α0(0) at t = 0 of C1 α on M with α(0) = p. m Exercise 1. Show that TpM = dφ(x0)[R ], where φ(x0) = p, and that this subspace of Rn is independent of the choice of local parametrization φ of M near p. On a manifold, we use instead the idea of tangent space as the set of ‘directional ’ of differentiable functions, at p ∈ M. m r r Def. Let M be a C manifold, CM the of real-valued Cr functions on M; let p ∈ M.A at p is a linear map r Xp : CM → R satisfying: r (i)p: if f, g ∈ CM coincide in an open neighborhood of p, then Xp(f) = Xp(g). (ii)p: Xp(fg) = f(p)Xp(g) + g(p)Xp(f) (Leibniz rule.) It follows easily from these properties that Xp(const) = 0,Xp(f) = Xp(f + const). (Show this.)

The space TpM of tangent vectors at p is a vector space of m: m it admits a standard basis (∂xi |p)i=1 associated to a local chart (U, h) at p such that h(p) = 0:

∂(f ◦ h−1) ∂xi |p(f) = (0), where h(p) = 0, i = 1, . . . , m. ∂xi r Indeed, if Xp ∈ TpM and f ∈ CM with f(p) = 0 (as we may assume), we compute using (ii)p (see notes):

X i i i Xp(f) = v ∂xi |p(f), v = Xp(x ), i where xi : U → R is the ith component of the Cr map h : U → R. Def. The differential of a C1 map f : M m → N n at p ∈ M is the linear map df(p) ∈ L(TpM; Tf(p)N) given by:

r df(p)[Xp] = Yf(p),Yf(p)(g) = Xp(g ◦ f), g ∈ CN .

2 r Note that g ◦ f ∈ CM (check in a local chart.) r Exercise 2. The linear map Yf(p) ∈ L(CN ; R) satisfies the conditions (i),(ii) at f(p), required of an element of Tf(p)N. Reality check. If the manifold N is R (the ), the tangent space at t0 ∈ R is one-dimensional: dg ∂ (g) = (t ), for g ∈ C1( ); T N = {c∂ ; c ∈ }. t|t0 dt 0 R t0 t|t0 R We have an isomorphism: dg : T N → ; (Y ) = Y (id ) = c, where Y (g) = c (t ). It0 t0 R It0 t0 t0 R t0 dt 0 1 Now let f : M → N = R be C . Then:

1 df(p)[Xp] = Yf(p), where Yf(p)(g) = Xp(g ◦ f), for all g ∈ C (R).

In particular, if g = idR, we have Yf(p)(idR) = Xp(f). Thus, under the isomorphism If(p), which identifies Tf(p)N with R, we have:

If(p)(df(p)[Xp]) = (df(p)[Xp])(idR) = Yf(p)(idR) = Xp(f).

Informally, this identifies Xp(f) ∼ df(p)[Xp]: Xp(f) is a ‘directional deriva- tive’ of f, as expected.

LECTURE 3. Tangent ; locally trivial vector bundles

Def. The of a Cr manifold M m is the set:

r TM = {(p, Xp) ∈ M × L(CM ; R); Xp ∈ TpM}, that is, Xp satisfies (i)p, (ii)p. We have the standard projection map π : TM → M, π(p, (Xp)) = p. (Intuitively, we think of elements of TM as ‘vectors with basepoint’–only vectors with the same basepoint can be added.) We endow TM with a topology by imposing two conditions: (a) π is a continuous map: π−1(U) is open in TM if U ⊂ M is an ; (b) Let (U, h) be a local chart for M, h(p) = 0. Define h˜ : π−1(U) → U × Rm by h˜(p, Xp) = (h(p), v),

3 P i where Xp = i v ∂i|p, the components of Xp in the basis ∂i|p of TpM asso- ciated with the chart h. We require h˜ to be a homeomorphism. Exercise 3. Show that (a),(b) define a topology on TM making it a (Hausdorff, second countable, locally euclidean.) De- scribe a local basis of neighborhoods at a point (p, Xp) of TM. Def. Differentiable structure on TM. Let (U, ϕ), (V, ψ) be overlapping charts for M, U ∩ V 6= ∅. So F = ψ ◦ϕ−1 : ϕ(U ∩V ) → ψ(U ∩V ) is a Cr diffeomorphism. Then the associated charts for TM also overlap, with the transition diffeomorphism (of class Cr−1): F˜ = ψ˜ ◦ ϕ˜−1 : ϕ(U ∩ V ) × Rm → ψ(U ∩ V ) × Rm. (x, α) 7→ (F (x), dF (x)[α]). Thus the Cr atlas on M induces a Cr−1 atlas on TM. The tangent bundle is a special case of an important class of manifolds. Def. A locally trivial with typical fiber V is a triple (E, M, π), where E and M are differentiable manifolds and π : E → M a differentiable map, satisfying: (i) Any p ∈ M admits a neighborhood U ⊂ M and a diffeomorphism −1 (‘local trivialization’) ΦU : π (U) → U × V commuting with the identity on U: π(e) = p1(ΦU (e)), where p1 : U × V → U is the first projection. (ii) If two trivializing domains overlap (U ∩ W 6= ∅), the transition map has the form:

−1 ΦW ◦ ΦU : U × V → W × V, (x, v) → (x, T (x)v), where T : U ∩ W → GL(V) is a differentiable map. E is trivial if there is a global diffeomorphism Φ : E → M × V, π(e) = −1 p1(Φ(e)), so that Φ|Ep ∈ L(Ep, V), Ep = π (p). Def. A of E is a differentiable map X : M → E such that π ◦ σ = idM , i.e. X(p) ∈ Ep.A vector field on M is a section of TM: X(p) ∈ TpM.

Exercise 4: E is trivial ⇔ ∃ a differentiable basis of sections X1,...,Xn (n = dim V), i.e. {Xi(p)} is a basis of Ep, for all p ∈ M. Example 1: TS1 is trivial: V (x, y) = (−y, x) is a nonvanishing tangent vector field onS1.

4 Example 2: TS2 is not trivial: there is no non-vanishing tangent vector field on S2. This follows from the following facts: (i) If f, g, ∈ C(Sn,Sn) and f(x) + g(x) 6= 0∀x ∈ Sn ⊂ Rn+1, then f is homotopic to g. (ii) In particular, if f ∈ C(S2,S2) has no fixed points, f is homotopic to α, the antipodal map α. (iii) The identity map of S2 is not homotopic to the antipodal map (this is true for even-dimensional ). The proof uses the invariance of the degree of a map. (iv) Suppose V were a nonvanishing vector field on S2. Consider the x+V (x) map: f(x) = ||x+V (x)|| . Then f is homotopic to the identity (replace V by tV, t ∈ [0, 1]). On the other hand, f has no fixed points, hence is homotopic to α. Contradiction! Exercise 5. Prove (i), (ii) and (iv) in detail. In fact, prove the more general version of (i): if f, g satisfy |f(x) + g(x)| < 2 ∀x ∈ Sn, then f is homotopic to g. Exercise 6. (i) Generalizing Example 1, show that any odd-dimensional admits a non-vanishing vector field. (Hint. Consider S2n+1 as the n unit sphere in C , complex n-dimensional space: 2n+1 X 2 S = {(z1, . . . , zn); |zi| = 1, zj = ui + ivj.}. j (ii) Show that the 3-sphere is parallelizable (i.e., the tangent bundle TS3 is trivial). Hint: Recall S3 can be thought of as the set of unit (denote the algebra by H). Use quaternion algebra and trial and error to find three l.i. tangent vector fields at a point x + yi + zj + wk ∈ 3 S ⊂ H. LECTURE 4. , immersions and . Recall the Implicit Theorem in : Theorem. Let U ⊂ Rm open, f : U → Rm be a Cr map (r ≥ 1). Suppose df(x) ∈ L(Rm) is an isomorphism, for some x ∈ U. Then there exist open neighborhoods U of x, V of f(x) in Rm, such that: (i) f is a homeomorphism from U to V ; (ii) f −1 : V → U is of class Cr, with df −1(f(x)) = (df(x))−1 ∈ L(Rm). Thus if df(x) is an isomorphism for all x ∈ U, then f is a Cr local diffeomorphism (a Cr with Cr local inverses.)

5 Exercise 7. If f : U → Rm, U ⊂ Rm open, is a C1 immersion, show that V = f(U) is open in Rm, and that if f is injective, then f is a C1 diffeomorphism from U onto V (that is, a homeomorphism of class C1, with C1 inverse.) Here we recall: Def. A Cr map f : M → N of Cr manifolds is an immersion if df(p) has m = dim M, for all p ∈ M (in particular, m ≤ n.) Up to taking local charts on the domain and range, immersions have a very simple form: Theorem: local form of immersions. Let f : M → N be a Ck map, let p ∈ M be such that df(p) has rank m = dim M. Then there exist a local chart (V, ϕ) at p, a real vector space F of dimension n − m, a neighborhood Z of f(p) in N with f(V ) ⊂ Z and a Ck diffeomorphism onto its image ψ : Z → Rm × F so that:

Φ = ψ ◦ f ◦ ϕ−1 : ϕ(V ) → Rm × F satisfies Φ(x) = (x, 0) ∀x ∈ ϕ(V ).

Proof. (Outline.) We consider the euclidean case, since the manifolds statement is easily reduced to it. So f : U → N, with U ⊂ Rm open and N an n-dimensional real vector space. The idea is to reduce the result to the . So let E = df(p)[Rm] ⊂ N (an m-dimensional subspace) and let F ⊂ N be any subspace complementary to E, so N = E ⊕ F . Now define:

h : U × F → N, h(x, y) = f(x) + y, h(p, 0) = f(p),

dh(p, 0) ∈ L(Rm ⊕ F ; E ⊕ F ) invertible: dh(p, 0)[u ⊕ v] = df(p)[u] ⊕ v. By the IFT, h is a diffeomorphism from a neighborhood V × W ⊂ U × F of (p, 0) (p ∈ V, 0 ∈ W ) to a neighborhood Z = h(V × W ) ⊂ N of f(p). (In particular, f(V ) = h(V × {0}) ⊂ Z.) Let ψ = h−1 : Z → V × W . Then since h(x, 0) = f(x), we have, for all x ∈ V :

Φ(x) = (ψ ◦ f)(x) = h−1(f(x)) = (x, 0), as claimed. Exercise 8. Show that (with V as in the statement of the theorem) the restriction f|V : V → f(V ) ⊂ Z is injective, and an open map (i.e. an embedding of V into Z ⊂ N), where f(V ) has the induced topology from Z. (But note in general f(V ) is not open in f(M), unless f : M → N is globally an embedding.)

6 Thus, informally, the topological content of the ‘local form of immersions’ is ‘any immersion is locally an embedding’. Def: . Let M ⊂ N be Cr manifolds, of dimension m, n resp. (m ≤ n.) M is a submanifold of N if for each p ∈ M one may find a chart m n−m m (V, ψ) for N at p, ψ : V → R ×R , such that ψ(M ∩V ) ⊂ R ×{0n−m}, r and (V ∩ M, ψ|V ∩M ) is a chart in the C atlas of M. In particular, M is a topological manifold, in the topology induced from N. Exercise 9: A Cr surface in Rn (see definition in an earlier lecture) is a submanifold of Rn; and conversely, a Cr submanifold of Rn satisfies the definition of a Cr surface in Rn. We have the following natural question: if f : M → N is an injective em- bedding, when is f(M) a submanifold of N? Simple examples (embeddings from R to R2) show this is not always the case. Proposition. Let f : M → N be a Ck immersion. If f : M → f(M) is an open map (where f(M) has the topology induced from N) then f(M) is a submanifold of M. This follows from the local form of immersions (see class notes for a proof.) And conversely, if f : M → N is a differentiable injective immersion and f(M) ⊂ N is a submanifold, then f is an embedding (since f is then an open map onto f(M), by a previous problem.) As a corollary, if an injective differentiable immersion f is a topological embedding (i.e. a homeomorphism onto its image, with the subspace topol- ogy), then f(M) is a submanifold of N. This is always the case if M is compact, or more generally if f is proper, since proper maps are closed (and hence also open, when bijective). Exercise 10. Recall a continuous map is proper if the preimage of compact sets is compact. Show that if M,N are differentiable manifolds and f : M → N a differentiable map, then f proper ⇒ f closed. However, it is easy to give examples of embeddings with image a sub- manifold, and which are not proper maps (for instance, the inclusion map of (−1, 1) into R). LECTURE 5. Regular values, submersions theorem. -euclidean space Let f : U → Rn be a Ck map, U ⊂ Rm open. Suppose p ∈ U is such that m n n −1 df(p) ∈ L(R ; R ) is surjective (so m ≥ n). Let c = f(p) ∈ R ,Lc = f (c).

7 k Then in a neighborhood of p, Lc is a C surface of dimension m − n, and TpLc = ker(df(p)). Specifically, let Rm = E ⊕ F be a direct sum decomposition, where n df(p)|F ∈ L(F ; R ) is an isomorphism: so F is n-dimensional and E is any m complement of F in R . Write p = (pE, pF ) ∈ E × F . Then there exist m n open neighborhoods Z ⊂ R of p, V ⊂ E of pE, W ⊂ R of f(p), and a diffeomorphism ϕ : V × W → Z such that (f ◦ ϕ)(x, w) = w.(local form of submersions.) Proof. Consider h : E × F → E × Rn, h(x, y) = (x, f(x, y)). So h(p) = (pE, f(p)) and it is easy to see the differential of h is an isomorphism. By m n the IFT, we find neighborhoods Z ⊂ R , V ⊂ E, W ⊂ R of p, pE, f(p) (resp.) so that h is a diffeomorphism from Z to V × W . Let ϕ = h−1. Then (x, w) = h(ϕ(x, w)) = (x, f(ϕ(x, w)), so (f ◦ ϕ)(x, w) = w.

In fact ϕ(x, w) = (x, ϕ2(x, w)). So define ζ : V → F by ζ(x) = ϕ2(x, c), where c = f(p). Then it is easy to see Lc ∩ Z = graph(ζ): Lc in Z is the graph of a Ck function defined on V ⊂ E, hence an (m − n)-dimensional surface of class Ck. On manifolds, we have the same statement, proved from the euclidean one via local charts. Def. A Ck map f : M → N is a if df(p) is surjective, for all p ∈ M. From the above, one easily shows: Prop. Let f : M m → N n be a Ck submersion of Ck manifolds. Let −1 k c = f(p) ∈ N,Lc = f (c). Then Lc is a C submanifold of M, of dimension m − n, with tangent space TpLc = ker(df(p)). Exercise 11. Any submersion is an open map. n2 Example 1. Let Mn = R be the space of n×n real matrices, det : M → R the determinant function. Then for any 0 6= c ∈ R, the level set {X ∈ Mn; det(X) = c} is a surface in Mn, of 1. In particular, the matrix SLn (corresponding to c = 1) is a codimension 1 submanifold of Rn2 . (It is enough to show any c 6= 0 is a regular value of det.) T Example 2. The orthogonal group O(n) = {X ∈ Mn; XX = In} is a compact surface of codimension n(n + 1)/2 in Mn. This follows from the T fact the identity matrix In is a regular value of the map X 7→ XX from Mn to Symn, the linear space of symmetric n × n matrices. Exercise 12. The set of m × n matrices of rank r is a submanifold of

8 Rmn of codimension (m − r)(n − r). ([G-P, p. 27])

LECTURE 6. Transversality

Def. Map transversal to a submanifold. A Cr map f : M → N is transversal at p ∈ M to a submanifold S ⊂ N if:

df(p)[TpM] + TqS = TqN, q = f(p). Notation: f tp S. f is transversal to S if this holds for all p ∈ f −1S (or if f −1(S) = ∅). (Notation: f t S.) Remark: Clearly if f is a submersion, f is transversal to any submanifold of N.

Transversal submanifolds. Two submanifolds S1,S2 ⊂ N are transversal (S1 t S2) if: TqS1 + TqS2 = TqN, ∀q ∈ S1 ∩ S2.

Equivalently, the inclusion map i : S1 → N is transversal to S2 at q (or the other way around.) Transversality may be characterized in terms of regular values. Proposition. Let f(p) = q ∈ S, ψ : V → Rs × Rn−s be a submanifold s chart for N at q (meaning ψ(S ∩ V ) ⊂ R × {0n−s}. ) Let U ⊂ M be a neighborhood of p so that f(U) ⊂ V . Then:

f|U t S ⇔ 0n−s is a regular value of π ◦ ψ ◦ f|U , where π : Rs × Rn−s → Rn−s is projection onto the second factor. −1 Corollary. If f t S, the preimage MS = f (S) is a submanifold of M, with codimension codimM (MS) = codimN (S). The tangent space at p ∈ MS is the preimage under the differential:

−1 TpMS = (df(p)) TqS, f(p) = q.

This follows from TpMS = Ker(d(π ◦ψ ◦f)(p)) and TqS = Ker(d(π ◦ψ)(q)), via the and the ‘linear algebra fact’:

Ker(ST ) = T −1Ker(S),T ∈ L(E; F ),S ∈ L(F ; W ).

Specializing to transversal submanifolds S1,S2 ⊂ N: codimN (S1∩S2) = codimN (S1)+codimN (S2),Tq(S1∩S2) = TqS1∩TqS2, q ∈ S1∩S2.

9 In particular if dim(S1) + dim(S2) = dim(N), we have dim(S1 ∩ S2) = 0: the intersection consists of isolated points in N. Def. Transversality of maps. Let f : M → P, g : N → P be differentiable maps, p ∈ M, q ∈ N such that f(p) = g(q) = r ∈ P . Then f and g are transversal at p, q (denoted f tp,q g) if:

df(p)[TpM] + dg(q)[TqN] = TrP.

We say f t g if this is true at all p, q with f(p) = g(q). Remark: If f is a submersion, clearly f t g for any g. Proposition. Let ∆ ⊂ P × P be the diagonal submanifold. Given f : M → P, g : N → P , consider the ‘cartesian prodict map’ f × g : M × N → P × P ,(f × g)(p, q) = (f(p), g(q)). Then if f(p) = f(q) = r:

f tp,q g ⇔ f × g t(p,q) ∆.

This follows from another linear algebra fact. Let A, B be subspaces of a vector space E. Then:

A + B = E ⇔ A × B + ∆ = E × E.

(Recall E × E has the ‘direct sum’ vector space structure.) Here ∆ is the diagonal subspace of E × E. Exercise 13. (i) Prove this linear algebra fact. (ii) Show how the proposition follows from it. (You may use ‘obvious’ facts about tangent spaces of cartesian product manifolds.) In the ‘ case’ of two differentiable (say smooth) functions f, g : R → R, f t g is not quite the same as transversality of their graphs Γf , Γg 2 at their points of intersection (as submanifolds of R .) Exercise 14. Draw examples showing these two notions of transversal- ity are not the same. Analyze how they are related: does either one imply the other? Corollary. Let f : M → P, g : N → P be transversal maps. Then the ‘coincidence set’

Q = {(p, q) ∈ M × N; f(p) = g(q)}

10 is a submanifold of M × N, of codimension equal to the dimension of P . (Since Q = (f × g)−1(∆)).

Examples. (i) Any f : M → N is transversal to the identity idN : N → N (since idN is a submersion.) Thus the graph of f:Γf = {(p, q) ∈ M ×N; q = f(p)} is a submanifold of M × N, of dimension dim M. (This also follows from the fact that the map from M to M ×N, p 7→ (p, f(p)) is an embedding– why?) (ii) If f : M → N is a submersion, the graph of the induced by f is a submanifold of M × M:

G = {(p, q) ∈ M × M; f(p) = f(q)}.

Its dimension is 2 dim M − dim N. (Why is this true?) Exercise 15. (From [G-P p.33]). Let V be a real vector space, T ∈ L(V ), ∆ ⊂ V × V the diagonal subspace, ΓT ⊂ V × V the graph subspace of T . Then: ΓT t ∆ ⇔ 1 is not an eigenvalue of T.

In this case, what is the dimension of the intersection subspace ΓT ∩ ∆? What is its dimension if 1 is an eigenvalue of T ? Exercise 16. If M is a compact manifold and f : M → M is a Lefschetz map, then f has only finitely many fixed points ([G-P p.33]). Exercise 17. Let X be a vector field on M (a section of the tangent bun- dle π : TM → M.) A singularity of X is a point p ∈ M such that X(p) = 0.

A simple singularity is a singularity p at which dX(p) ∈ L(TpM,T0p TM) has rank m. Show that all singularities of X are simple iff X t Σ0, where Σ0 = {(p, 0p); p ∈ M} is the ‘zero section’ of TM. Show that simple singu- larities are isolated.

LECTURE 7. Paracompactness and differentiable partitions of unity

Definitions. A collection {Aα} of subsets of a space M is locally finite if for any p ∈ M there exists an open neighborhood U intersecting only finitely many Aα. A space M is paracompact if any open cover of M admits a locally finite refinement. Theorems (see [Munkres 1, no. 41]) 1. Every paracompact Hausdorff space is . 2. Closed subspaces of paracompact spaces are paracompact.

11 3. Every is paracompact. 4. Every regular Lindel¨ofspace (in particular every regular second- countable space) is paracompact. Proposition 1. X locally compact Hausdorff, second countable (for in- stance, a topological manifold. Then: (i) X is σ-compact: there exist precompact open subsets Gi, i ≥ 1, so that: [ X = Gi, Gi compact , Gi ⊂ Gi+1. i≥1 (ii) X is paracompact. Proof. Let {Uα} be an open cover. Choose a finite subcover of the open cover {Uα ∩ (Gi+1 \ Gi−2)} of the compact Gi \ Gi−1. Then choose a finite subcover of the open cover {Uα ∩G3} of G2. The collection of such open sets is a countable, locally finite refinement of {Uα}, consisting of precompact open sets. In fact, one can do more. For a chart h : U → B(3) ⊂ Rm with h(U) = B(3) (the open at 0 of radius 3), let W ⊂ V ⊂ U be the preimages under h of the open balls B(1) ⊂ B(2) ⊂ B(3). Proposition 2. Let C be an open cover of M. Then C has a countable, locally finite refinement {U1,U2,...}, domains of charts hi : Ui → B(3) (onto), so that the corresponding Wi define a locally finite cover of M. This is proved like Prop. 1. Note this is trivial if M is compact. k Definition. A set of C functions {ϕβ}β∈B : M → [0, 1] is a subordinate to an open cover A = {Uα}α∈A of M if: (i) the family {(ϕβ)}β∈B is locally finite, covers M, and is a refinement of A; P (ii) The sum β∈B ϕβ is identically 1 on M.

A partition of unity is strictly subordinate to the open cover A = {Uα}α∈A of M if the indexing set is the same and support(ϕα) ⊂ Uα, for each α ∈ A. There are two main existence results:

k Proposition 3. Given an open cover A = {Uα}α∈A of M, there exists a C countable partition of unity {ψi}i≥1 subordinate to A, where the supports support(ψi) are compact (and define a locally finite, countable cover of M refining A).

Proposition 4. Let A = {Uα}α∈A be an open cover of M. Then there ex-

12 k ists a C partition of unity {ϕα}α∈A strictly subordinate to A: support(ϕα) ⊂ Uα. Remark. In general one cannot simultaneously require a partition of unity to be strictly subordinate to a given cover, and to have compact sup- ports (as simple examples show.)

LECTURE 8. Applications of partitions of unity. Application 1: differentiable Urysohn lemma. Let F,G be disjoint closed subsets of the Ck manifold M. Then there exists a Ck function f : M → [0, 1] which equals 0 in F and equals 1 in G. This follows from the existence of a partition of unity strictly subordinate to the open cover {M \ F,M \ G} of M. Application 2: Closed sets are level sets of differentiable functions. Let F ⊂ M be a closed subset of a Ck manifold. Then there exists a Ck −1 function f : M → R+ such that f (0) = F . (This is an application of the differentiable Urysohn lemma.) CASE 1: F = K compact, M = Rm. T m We have K = i Vi, where the Vi are the points in R with distance to K less than 1/i: a decreasing family of precompact open sets, contracting ∞ m to K. Then let fi be a C function in R which is zero on K, 1 on M \ Vi. The problem is fi could also vanish at other points. But suppose we find P m constants ci > 0 so that f = i cifi is a smooth function in R . Then if x is in the complement of K, x is in some Vi, so f(x) > 0. (j) Note that all derivatives fi (for j > 0) are continuous functions with compact support (since fi ≡ 1 outside the compact set Vi.) Thus, for each (j) m i ≥ 1, j ≥ 0, we may find Mij > 0 so that |fi | < Mij in R . (Set Mi0 = 1 for all i.) Now for each i ≥ 1 fixed, it is easy to find, inductively, constants 1 αij > 0 decreasing in j, α ≤ αij, with αij ≤ i . For each j fixed, we i(j+1) 2 Mij have the bound: X (j) X 1 α |f | < , ij i 2i i i thus the series on the left converges absolutely and uniformly in Rm and P (j) defines a for each fixed j, i αijfi . P Now let ci = αii, and note ci ≤ αij for i > j. Thus i cifi converges uni- m m P (j) formly on R to a function f ∈ C(R ), and the series i cifi also converge

13 m (p) P (p) uniformly in R . Thus for the derivatives of any order: f = i cifi , P ∞ showing f = i cifi defines a C function with the desired properties. P Remark. Note f takes the constant value ci in the complement of V1. Dividing by this number, we may assume f ≡ 1 outside of a compact set. Exercise 18. If C is any closed subset of Rk, there exists a submanifold X of Rk+1 such that X ∩ Rk = C (where Rk is embedded in Rk+1 as the vectors with last component zero.) [G-P, p.33] (This shows that, without the requirement of transversality, the intersection of two submanifolds can be arbitrarily bad.) CASE 2 (general). Take a countable, locally finite open cover of M by m domains Ui of local charts, hi : Ui → B(3) ⊂ R . Let Wi ⊂ Vi ⊂ Ui as before (preimages under hi of balls of smaller radii), and let Ki = Vi ∩ F . Note S k F = i Ki. From the first part (using the charts hi), we find fi ∈ C (M) −1 with fi (0) = Ki and fi ≡ 1 outside Vi.

Now define f : M → R as the product of all the Vi! Since the cover {Vi} of M is locally finite, near each p ∈ M only finitely many fi will be different k from 1, so f is well-defined and of class C . In , f(p) = 0 ⇔ fi(p) = 0 for some i ≥ 1 ⇔ p ∈ Ki for some i ≥ 1 ⇔ p ∈ F . This concludes the proof. Application 3. Extension of differentiable functions defined on arbitrary subsets of a manifold. Application 4. Differentiable Tietze extension theorem.

LECTURE 9. Embeddings in euclidean space.

Theorem 1. Let M be an m-dimensional compact manifold of class Cr, which can be covered by n domains of coordinate charts U1,...,Un, hi : m ∞ m Ui → B(3), the ball of radius 3 in R . Let φ ∈ C (R ) be a smooth ‘’: equal to 1 in B(1), equal to 0 in the complement of B(2). r Let ϕi = φ ◦ hi in Ui, extended to zero outside of Ui (so ϕi ∈ C (M).) Consider the map f : M → R(m+1)n = R × ... × R × Rm × ... × Rm (2n factors): f(x) = (ϕ1(x), . . . , ϕn(x), ϕ1h1(x), . . . , ϕnhn(x).). Then f is an injective immersion (and therefore an embedding, since M is compact.) To extend this result (and proof) to the general (noncompact) case, we need the covering dimension theory in the Appendix. It is used to prove the

14 following: Theorem 2. Let M be an m-dimensional topological manifold. Then M may be covered by m + 1 domains of coordinate charts (Ui, hi), i = 1, . . . , m + 1. Proof (outline). Let A be a locally finite open covering of M by finitely many domains of coordinate charts (which may assumed to have bounded m image in R –why?) Referring to the Appendix, let B = B0 ∪...∪Bm be the refinement of A given by the Corollary in the Appendix. Bi is a countable disjoint union of domains of coordinate charts:

m Bi = {Vi1,Vi2,...}, kij : Vij → Bij ⊂ R ,

m where the Bij = kij(Vij) are bounded open sets in R , but not necessarily disjoint (for fixed i and varying j.) To correct this, we replace the kij by k˜ij, so that: ˜ ˜ kij(Vij) ∩ kil(Vil) = ∅ if j 6= l.

Exercise 19. (i) Explain how the k˜ij are constructed. Then define coordi- m nate charts hi : Ui → R , where Ui is the disjoint union of the Vij over j and i = 1, . . . .m + 1, concluding the proof of Theorem 2. (ii) Show how one can cover the two-dimensional by three domains of coordinate charts to R2 (not necessarily connected.) Exercise 20. (i) Let M be an m-dimensional manifold of class Ck. Define an injective Ck immersion:

2 f : M → R × ... × R × Rm × ... × Rm = R(m+1)

(m + 1 factors of type R, m + 1 of type Rm.) That is, show that the map you define is injective and an immersion. (ii) If M is noncompact, show that this map is proper (hence f is an embedding).

Appendix to Lecture 9: Covering dimension of manifolds. Definition. A X has covering dimension at most m (dim X ≤ m) if any open cover of X admits an open refinement with order at most m (any point is in at most m + 1 open sets of the refinement.) (i) If dim X ≤ m and Y ⊂ X is a closed subspace, then dim Y ≤ m. m (ii) If X ⊂ R (compact), then dim X ≤ m. (iii) If X = Y1 ∪ ... ∪ Yp, where each Yi is a closed subspace of X, then dim Yi ≤ m∀i ⇒ dim X ≤ m.

15 (Proved in [Munkres 1, no. 50].) Exercise 21(i). Prove (using these facts) that any compact topological m-manifold M satisfies dim M ≤ m. In fact this is true for any topological m-manifold, but is harder to prove.

Remark. If C, U are open covers of X, the intersection C∩U = {C∩U; C ∈ C,U ∈ U} is also an open cover of X, refining both C and U. If C is an open cover of X and Y ⊂ X is a closed subspace, then the restriction of C to Y is the open cover of Y :

C|Y = {C ∩ Y ; C ∈ C}.

Theorem. Let M be a topological m-manifold. Then dim M ≤ m. Proof. (Outline–cp [Munkres 2] or problem 8, p.316 in [Munkres 1].) Recall M is σ-compact in a strong sense: [ M = Kn,Kn compact ,Kn ⊂ int(Kn+1). n≥1

Denote by An the compact ‘annular region’: An = Kn \ int(Kn−1)(n ≥ 2), A1 = K1. Then M is also the union of the An.

Exercise 21 (ii). dim(An) ≤ m for all n ≥ 1. (This is like the previous problem.)

Given an open cover of M, consider a refinement B0 with the property: if U ∈ B0 intersects Kn, then U ⊂ Kn+1. Thus each U is contained in at most two of the An. Such a refinement can be obtained by intersecting the given cover with the open cover {int(Kn+1) \ Kn−1}n≥2, int(K2) of M. Now inductively construct a sequence of refining covers of M:

B0 < B1 < B2 < . . . , Bn < Bn+1, as follows: Let B1 be a refinement of B0, so that the restriction of B1 to K1 has order at most m (possible due to (i)–(iii) above.)

Now suppose Bn is a refinement of Bn−1, whose restriction to Kn has order ≤ m. And let C be a refinement of Bn whose restriction to the compact annular region An+1 has order ≤ m (from the previous problem.)

16 The idea is to construct an open cover Bn+1 of M refining Bn, whose restriction to Kn+1 has order ≤ m, by ‘splicing’ Bn and C, as follows. Bn+1 consists of the following open sets:

(a) Open sets of Bn intersecting Kn−1 (and hence contained in Kn.) (b) Open sets of C which do not intersect Kn. (c) Open sets of the form

0 [ U = {V ∈ C; V ∈ Cn and f(V ) = U}, where U ∈ Bn, Cn is the set of V ∈ C intersecting Kn, but not Kn−1; and f : Cn → Bn is a fixed ‘choice function’ assigning to each such V an open set in Bn containing it (which exists since C refines Bn). 0 Note that Bn+1 refines Bn, since C does, and any set U of type (c) is contained in a set of Bn (namely U).

Exercise 22. Prove the claim: Bn+1 restricted to Kn+1 has order ≤ m.

Hint: Note Kn+1 = Kn ∪ An+1. We must show that any p ∈ Kn+1 is in at most m + 1 open sets of Bn+1. There are two cases to consider: (i) p ∈ An+1. Then any open set of Bn+1 containing p is of type (b) (a set in C) or type (c) (a union of sets in C). And C restricted to An+1 has order ≤ m.

(ii) p ∈ Kn. Then any open set of Bn+1 containing p is of type (a) (a 0 set in Bn) or type (c) (a union U of sets of C, contained in U ∈ Bn). We know that Bn restricted to Kn has order ≤ m, but something still needs to be said to conclude the proof of the claim.

Now let B be the collection of open sets which are in all Bn for n suffi- ciently large. (In particular B refines each Bn.) Claim: B is an open cover of M, with order ≤ m. The claim follows from the following two facts: (i) If B ∈ BN intersects KN−1 then B ∈ Bn for all n ≥ N (from the inductive construction), thus B ∈ B. Such B cover KN−1, so B covers M.

(ii) Let x ∈ M, B1,...,Bp open sets in B containing x. Then one may find N ∈ N so that all Bi ∈ BN . Taking a larger N if needed, we may assume x ∈ KN . Since BN restricted to KN has order ≤ m, it follows that p ≤ m+1. Thus any finite collection of open sets in B with a common point x has at most m + 1 elements. This shows B has order ≤ m.

17 Corollary. Let A be an open covering of M. There exists a locally finite open refinement B, union of B0,..., Bm (each Bi is a collection of open sets) so that the sets in each Bi are disjoint.

Proof outline (see [Munkres 2].) Let C = {Uj} be a refinement of A given by the result dim M ≤ m, and {Vj} be a locally finite open cover such that Vj ⊂ Uj∀j (from paracompactness.) Let {ϕj} be a partition of unity strictly dominated by {Vj}.

Given n ∈ {0, . . . , m}, define Bn as follows: for each set of n + 1 positive i0 < i1 < . . . < in, define:

W (i0, i1, . . . , in) = {x ∈ M; ∀i ∈ N\{i0, . . . , in}ϕi(x) < min{ϕi0 (x), . . . , ϕin (x)}}.

Let Bn be the collection of all such open sets W . It is not hard to show: (i) The sets in Bn are all disjoint; (ii) The union B = B0 ∪ ... ∪ Bm covers M. (iii) B is locally finite.

LECTURE 10. Riemannian metrics Definition. A Riemannian g on a differentiable manifold M is an assignment of an inner product (positive-definite bilinear form) on each tangent space TpM. Notation hv, wip or hv, wig, for v, w ∈ TpM. To make sense of differentiability, consider a local chart h : U → Rm at p (h(p) = 0) and the associated basis (∂i) of TpM. We have functions:

gij : U → R, gij(p) = h∂i, ∂jip.

m n Let S+(R ) ⊂ L(R ) denote the set of positive-definite symmetric linear transformations, an open convex cone in the space S(Rm) of all symmetriic linear transformations. (Recall ‘symmetric’ means hT v, wi0 = hv, T wi0, for v, w ∈ Rm and denoting the standard inner product (‘dot product’) in Rm with the subscript 0.) We have the map from U to S+: h h −1 −1 m g : U → S+, g (p)(v, w) = hdh (0)[v], dh (0)[w]ip., v, w ∈ R .

k k We say g is of class C if, for each chart (U, h), the functions gij ∈ C (U); h k equivalently, if g : U → S+ is of class C . Example. If (N, h) is a (a Ck manifold with a Ck Riemannian metric h), and f : M → N is a Ck+1 immersion, we may induce a metric g = f ∗h (‘ metric’) on M by setting:

hv, wig = hdh(p)v, dh(p)wih, v, w ∈ TpM.

18 In particular, if M is an m-dimensional surface in Rn, the standard inner n n product in R induces in each subspace TpM ⊂ R (by restriction) an inner product on M. In general, any Ck manifold carries a Ck−1 Riemannian metric: cover M by countably many domains Ui of local charts (a loc. finite cover), introduce m a metric in each Ui via local charts (pulling back from R ) , then define a metric globally on M using a subordinate partition of unity. 1. of the differential. Recall two natural ways to define a norm in a space of linear transformations: (i) If E,F are normed vector spaces (finite-dimensional, over R), define the norm on L(E; F ):

||T || = sup{||T v||F ; ||v||E = 1}.

Then clearly ||T v||F ≤ ||T ||||v||E. The problem with this norm is that the function T 7→ ||T || is not differentiable on L(E; F )\{0} (as simple examples show.) It is only continuous in L(E; F ). (ii) If E,F are inner product spaces and T ∈ L(E; F ), the adjoint T ∗ ∈ L(F ; E) is defined by:

∗ hT v, wiF = hv, T wiE; v ∈ E, w ∈ F.

We may define an inner product in L(E; F ) via: hT,Si = tr(T ∗S), where T ∗S ∈ L(E) and tr means trace. Then define the Hilbert-Schmidt norm on L(E; F ) by:

||T ||2 = hT,T i.

Introducing orthonormal bases in E,F we obtain matrices aij, bij for A, B and an isomorphism L(E; F ) ∼ Rmn, with the natural inner product:

n m X X hA, Bi = aijbij, m = dim E, n = dim F. i=1 j=1

2 P 2 mn Then ||A|| = i,j aij is certainly smooth as a function of A ∈ R , and ||A|| is smooth away from the origin. Proposition 1. Let f : M m → N n be a Ck+1 map of Riemannian man- ifolds (M, g), (N, g). Then the function p 7→ ||df(p)||2 is in Ck(M) (where we use the Hilbert-Schmidt norm.)

19 Proof: It is enough to consider Riemannian metrics g resp. h in open sets U ⊂ Rm,V ⊂ Rn, and a map f : U → V of class Ck+1. Then we have k m n C maps G : U → S+(R ),H : V → S+(R ) so that:

m n hu, vig(p) = hG(p)u, vi0, p ∈ U, u, v ∈ R ; hw, zih(q) = hH(q)w, zi0, q ∈ V, w, z ∈ R . Recall the Hilbert-Schmidt norm is given by:

||df(p)||2 = tr(df(p)∗df(p)), where * is the adjoint with respect to the metrics g, h. Let t denote the adjoint with respect to the euclidean inner products in Rm,Rn. Exercise 23. Prove the linear algebra lemma:

G(p)df(p)∗ = df(p)tH(q), q = f(p).

(State and prove the corresponding general lemma, for inner product spaces (E, g), (F, h) and T ∈ L(E; F ).) Thus: ||df(p)||2 = tr(G(p)−1df(p)t(H ◦ f)(p)df(p)), which is clearly Ck as a function of p ∈ U, proving the proposition. 2. The metric structure of a Riemannian manifold. Let (M, g) be a connected Riemannian manifold (in particular, path connected.) Then given p, q ∈ M, there exists a piecewise C1 path α : R 1 0 [0, 1] → M from p to q (why?) with length L[α] = 0 ||a (t)||dt. Define: d(p, q) = inf{L[α]; α : [0, 1] → M is a p.w. C1 path from p to q }.

Proposition 2. d defines a metric on M. (The only condition that is not immediate is p 6= q ⇒ d(p, q) > 0.) Proposition 3. (M, d) is homeomorphic to M with its original topology. Proof. There are two things to show (see scanned notes): (1) Any nbd. V ⊂ M of p ∈ M contains a d-ball with center p; (2) Any metric ball Bd(p, ) contains a coordinate neighborhood of p. Remark. When d is a complete metric, we say M is a complete Rieman- nian manifold. In that case C1 paths joining two points and achieving the infimum d(p, q) (distance-minimizing curves) always exist, and (M, d) has the Heine-Borel property: closed bounded subsets are compact. The metric completion of a manifold is not always a manifold (e.g. remove the tangency point from a closed self-tangent at a point, bounding a connected re- gion in the plane.)

20