Math 6520 Differentiable Manifolds

Taught by Allen Knutson Notes by Aaron Lou Fall 2019

1 Contents

1 Aug 29 5 1.1 Introduction ...... 5 1.2 Various Flavors of Manifolds ...... 5

2 September 3 6 2.1 Tangent Bundle ...... 6 2.2 Inverse Function Theorem ...... 7 2.3 Immersions ...... 7

3 Sep 5 9 3.1 Submersions ...... 9 3.2 Transversality ...... 11

4 September 10 12 4.1 Homotopies and stable/open properties ...... 12

5 September 12 13 5.1 Critical and Regular values ...... 13 5.2 Linear Algebra Around Critical Points ...... 14 5.3 Some Morse Theory ...... 14

6 September 12 15 6.1 Recall ...... 15 6.2 Morse Theorem ...... 15

7 September 19 17 7.1 Whitney Embedding Theorem ...... 17 7.2 Manifolds with Boundary ...... 18

8 September 24 19 8.1 Manifolds w/Boundary Revisited ...... 19 8.2 Transversality is stable ...... 19

9 September 26 20 9.1 Connected ...... 20 9.2 Transversality ...... 21 9.3 Orientation ...... 22

2 10 Oct 1 22 10.1 More Orientation ...... 22 10.2 Return to Intersection ...... 23

11 Oct 3 24 11.1 Review from last time ...... 24 11.2 Euler Characteristic based off of Orientation ...... 24 11.3 Lefschetz Number ...... 25

12 Oct 8 26 12.1 Vector Bundles ...... 26

13 Oct 10 28 13.1 Lefchetz Number Continued ...... 28 13.2 Computing Euler Numbers ...... 29 13.3 Integration ...... 29

14 October 17 30 14.1 (0, p)-tensors ...... 30 14.2 Multivariable Change of Variable ...... 31

15 Oct 22 32 15.1 Integration on Manifolds ...... 32 15.2 Exterior Derivative ...... 33

16 Oct 24 33 16.1 Stoke’s Theorem ...... 33 16.2 De Rham Cohomology ...... 35

17 Oct 29 35 17.1 Operations on Forms ...... 35 17.2 Nullity and Rank Thm ...... 35 17.3 Goals for De Rham Cohomology ...... 36 17.4 A Tool : Partition of Unity ...... 37

18 Oct 31 37 18.1 Mayer-Vietoris ...... 37

19 Nov 5 39 19.1 Geodesics ...... 39 19.2 More on forms and integration ...... 39 ? 19.3 Mayer-Vietoris for Hc ...... 40

3 20 Nov 7 41 20.1 Integration and Homotopy ...... 41 20.2 M-V in H? ...... 41 20.3 Kunneth ...... 42

21 Nov 12 43 21.1 Revisiting Vector Bundles ...... 43

22 Nov 14 44 22.1 Lefchetz Numbers and Cohomology ...... 44

23 Nov 19 45 23.1 More Lefchetz ...... 45 23.2 Lie Groups ...... 46

24 Nov 21 47 24.1 More Lie Groups ...... 47 24.2 Some Representation Theory ...... 48 24.3 Quotient Manifold Theorem ...... 48

25 Nov 26 49 25.1 Culture ...... 49 n 25.2 H?(CP )...... 49 25.3 More Bundle Stuff ...... 50

26 Dec 3 51 26.1 Various Geometries ...... 51

27 Dec 5 54 27.1 Abstract Manifolds ...... 54

28 Dec 10 56 28.1 Review of Forms ...... 56 28.2 Intersection Numbers ...... 57

4 1 Aug 29

1.1 Introduction

ϕ ψ p Recall the chain rule: u → v → R we have Du(ψ ◦ ϕ) = Dϕ(u)ψ ◦ (Duϕ).

Def. ϕ : U → Rm,U ⊂ Rm open is an immersion if ϕ is differentiable at all u ∈ U and Tuϕ is 1 : 1, where Tu is the derivative at u. Ex.

(1) R → R2 given by θ → (cos θ, sin θ). (2) Not: (x, y) → 0.

(3) Open segment in R1 can be mapped to a circle with a open segment jutting out.

(4) Not: x → x3 because the derivative is not injective.

Def (Inductive Definition of Manifold Embedded in Rn).

(1) Rn is a n-manifold (2) Open subsets of n-manifolds are submanifolds

(3) M ⊆ Rn (not necessarily open) and ∀m ∈ M, there exists a open U ⊆ M,U 3 m (openness defined by subspace topology) and an open V ⊆ Rk then there exists an immersion

n ϕ : V → U,→ R

then M is a k-submanifold in Rn. These are also manifolds. Def. Chart: (U, V, ϕ) on M around m.

1.2 Various Flavors of Manifolds Def. Each ϕ is ∞ differentiable then it is a smooth manifold. If ϕ is con- tinuous topological manifolds. Twice differentiable would be C2 manifold.

Ex. Not: a closed interval of the real line.

5 Def. If M ⊆ Rn is a k-submanifold and (U, V, ϕ) is a chart around m ∈ M. We can reorient (through translation) s.t. ϕ(0) = m.

TmU := im(ηU ◦ D*ϕ) 0 and this is the tangent space at m where ηU is an inclusion mapping. Note that the dimension is k as ϕ is an immersion and also the tangent space is isomorphic with Rk.

Prop 1.1. The tangent space TmM is independent of our choice of chart.

Proof. Given two charts (U1,V1, ϕ1), (U2,V2, ϕ2) note that U1 ∩ U2 is a nbhd 0 −1 0 −1 of m. Let V1 = ϕ1 (U1 ∩ U2) and V2 = ϕ2 (U1 ∩ U2). Note that by homeo- 0 0 0 0 morphism there exists a homeomorphism b : V1 → V2 s.t. ϕ1 = ϕ2 ◦ b. Now we have that

0 0 0 D*ϕ = D * ϕ ◦ D*b = D*ϕ ◦ D*b 0 1 b(0) 2 0 0 2 0 0 0 Note that that D*ϕ and D*ϕ are linear and 1:1, which means that D*b is 0 1 0 2 0 0 0 0 linear and 1:1, so it is onto, so D*b(V ) = V . Similarly, we can see that V 0 1 2 2 0 0 0 and V1 are also bijective and it follows that ϕ1 and ϕ2 have the same image, which means that the tangent spaces are the same. Def. Let M,N be manifolds. A function ψ : M → N is smooth (∞- diff) if there exists charts (UM ,VM , ϕM ), (UN ,VN , ϕN ) s.t. m ∈ UM ⊆ M −1 m n ψ(m) ∈ UN ⊆ N and ϕN ◦ ψ ◦ ϕM : VM → VN is smooth from R → R . Def. We can define the category of manifolds. Objects are (smooth) man- ifolds, morphisms are smooth functions, and invertible morphisms are diffeo- morphisms (homeomorphisms but infinitely differentiable both ways). Ex (Milner ’60s). There exist 28 smooth manifolds that are homeomorphic to the 7-sphere S7 ⊆ R8.

2 September 3

2.1 Tangent Bundle

Def. Let M ⊆ RN , then the tangent bundle of TM = {(x, y): x ∈ M, y ∈ TxM}. If dim(M) = k, then the claim is that TM is a smooth manifold with dim(TM) = 2k.

Proof. It is M × Rk. Products of manifolds are manifolds with dimension +2.

6 Theorem 2.1. Let Mfld be the category of smooth manifolds. T : Mfld → Mfld is a functor.

Proof.

• First, note that if M is a manifold, TM is a manifold.

k • f : M → N,M,N ⊆ R , T f : TM → TN then T f(x, y) = (f(x), Dfx(y)). • If I have a map from f : N → K, g : M → N, smooth manifolds, T (f ◦ g) = T f ◦ T g. The idea is that D(f ◦ g) = Df ◦ Dg.

• T (idM ) = idTM . So T is a functor from smooth manifolds to smooth manifolds.

2.2 Inverse Function Theorem

Prop 2.1. f : X → Y manifolds of same dimension, if Dfx = TxX → Tf(x)T is an isomorphism, f is a local diffeomorphism at x.

Proof.

We see that since Dfx is an isomorphism, then dim X = dim Y and further- more

f X y

ϕ1 ϕ2 g 0 ∈ U 0 ∈ V

k k Then we see that Dg0 : R → R is a linear map and in particular has an inverse (by IFT for Euclidean open subsets) if det Dg0 6= 0. But this only happens in particular when it is an isomorphism.

Remark. f : R1 → S1 by x → eix, we see that this is a local diffeomorphism but not global.

2.3 Immersions

Def. f : X → Y manifolds, f is an immersion at x if Dfx is injective.

k l Ex. f : R → R by (x1, x2, . . . , xk) → (x1, . . . , xk, 0, 0,... 0). This is the canonical immersion.

7 Theorem 2.2. f : X → Y is an immersion at x. Then after some local parametrization around x, f is a canonical immersion.

Proof.

f X y

ϕ1 ϕ2 g 0 ∈ U 0 ∈ V

k l For f : X → Y immersion, we want to find ϕ1, ϕ2 s.t. g : R → R , since g is k l an immersion at 0 (local parametrization is also immersion), Dg0 : R → R and has rank k.

WTS after some change of coordinates, then g is canonical immersion. WLOG, I  Dg = k with change of basis in l. Define G : U × l−k → l,G(x, z) = 0 0 R R R g(x)+(0, z). g = G◦c, where c and this is the canonical immersion x → (x, 0). In particular DG0 = Il by IFT, G is a local diffeomorphism, so we can parametrize around y with ψ ◦ G. Shrinking U and V to U,˜ V˜ we have

f X y

ϕ1 ϕ2◦G 0 ∈ U c 0 ∈ V and this is our desired canonical immersion.

Remark. Image of immersion may not be a submanifold. For example, f : R1 → a figure 8 and this is not injective at the center. Another example is f : R1 → a figure 8 where we curve around the ±∞ converge to the intersection point (like an S). This is injective and onto but it is not a manifold since the preimage of a nbhd around the intersection point has infinitely many points.

Def. f : X → Y continuous from topological spaces X,Y . we say f is proper if f −1(K) is compact in X for compact K.

Ex. Let f : R1 → R1/Z × R1/Z, x → (x, ax) (torus) where a∈ / Q where we mod by 1. The preimage of a point in this has infinitely many, so not proper.

Def. f : X → Y manifolds. f is an embedding if f : X → f(X) is a diffeomorphism. Therefore, f(x) is a submanifold of Y .

8 Theorem 2.3. f : X → Y is an immersion. If f is injective and proper, f is an embedding. Proof. • f : X → f(X) is an open map. For x ∈ X and nbhd U, then want to show that f(U) is open in f(X). Suppose f(U) is not open, then there exists (yn) ⊆ f(X) s.t. yn ∈/ f(U) but yn → y ∈ f(U). ∃!x s.t. f(x) = y and ∃!xn s.t. f(xn) = yn. Consider {x, x1, x2,... }. −1 Note that f ({y, y1,... }) which is compact, so (xn) has a convergent 0 0 subsequence xni → x which means that f(xni ) = yni → f(x ) = y = 0 f(x) and since it is injective x = x , which implies that f(Xnk ) = ynk ∈ f(U) for some nk, which is our contradiction. • Given x ∈ X, using local immersion Theorem, f is local diffeomor- phism. By IFT, there exists f −1 : f(x) → x that is smooth around x.

3 Sep 5

3.1 Submersions

Def. f : X → Y smooth map, manifolds, f is a submersion at x if Dfx : TxX → TyY is onto where y = f(x). k ` Ex. f : R → R , k > ` where f(x1, . . . , x`, . . . , xk) = (x1, x2 . . . , x`). This is the canonical submersion. Theorem 3.1. f : X → Y smooth map between manifolds, f submersion at x, then ∃ local parametrization s.t. f is canonical submersion. Proof.

f X Y

ϕ1 ϕ2 g 0 ∈ U ⊆ Rk 0 ∈ V ⊆ R` where ϕ1(0) = x, ϕ2(0) = y = f(x) and k > `. Dg0 : T0U → T0V has rank `.   k We can rewrite Dg0 as I` | 0 after a change of basis of R .

k ` k−` Introduce G : U → R = R ×R . G(a) = (g(a), a`+1, . . . , ak). Can see that g = (canonical submersion)◦G and DG0 = Ik, so G is a local diffeomorphism by IFT. If we denote the canonical submersion as s, we have

9 f x y

ϕ1 ϕ2 u G G(u) s V

−1 ˜ and setting ϕ3 = ϕ1 ◦ G on an open set U we have

f X Y

ϕ3 ϕ1

U˜ s V˜

Def. f : X → Y smooth map between manifolds. We say y ∈ Y is a regular value if ∀x ∈ f −1(y), f is a submersion at x.

Def. y is a critical value if y is not regular.

Theorem 3.2 (Preimage Thm). f : X → Y . If y is a regular value, then L = {x ∈ X : f(x) = y} is a submanifold of X. X ⊆ Rk,Y ⊆ R`. ∼ Proof. WTS ∀x ∈ L, ∃U ⊆ X s.t. U ∩ L −→ V ⊆ Rdim L.

f X Y

ϕ1 ϕ2 g 0 ∈ U 0 ∈ V

−1 where ϕ1(0) = x, ϕ2(0) = f(x) = y, g(x1, . . . , xk) = (x1, . . . , x`). ϕ1 (L) ∩ −1 −1 ∼ k−` U = g (0) ∩ U, g (0) = {(0,..., 0, xk+1, . . . , x`)} = R .

L is a submanifold with dimension k − ` = dim X − dim Y .

−1 Corrollary. Z = f (y), f : X → Y y regular, then ∀x ∈ Z,TxZ = ker(Dfx).

−1 Proof. Since Z = f (y),TxZ ⊆ ker(Dfx). By dim condition TxZ = ker(Dfx).

Ex. Ξ = {(x, y, z) ∈ R3, x3 + y3 + z3 + 3xyz = 1} is a smooth manifold. If f : R3 → R by f(x, y, z) = x3 + y3 + z3 + 3xyz, then ∇f = (3x2 + 3xyz, 3y2 + 3xz, 3z2 + 3xy). If ∇f = 0, then we need that all are 0, but this is not inside the set. Therefore Ξ is a smooth manifold of dim = 3 − 1 = 2.

10 T Ex. n ∈ N, n ≥ 2, O(n) = {A ∈ M(n), AA = In}. WTS that O(n) is a smooth manifold of M(n). Define f : M(n) → S(n) by f(A) = AAT . We −1 see that O(n) = f (In) and WTS that In is a regular value of f, or that DfA : TAM(N) → Tf (A)S(n) is surjective. We calculate

f(A + sB) − f(A) DfA(B) = lim s→0 s (A + sb)(A + sB)T − AAT = lim s→0 s = BAT + ABT

Now we need to show DfA is onto. Note C ∈ Tf(A)s(n) = s(n) and TAM(n) = C+CT T C CA M(n) since they are Euclidean. C = 2 and want BA = 2 ,B = 2 =⇒ DfA is onto. So A is a regular value. By Preimage Thm, O(n) is a subman- n(n−1) ifold of M(n) with dim O(n) = dim M(n) − dim S(n) = 2 .

Remark. M manifold cut by g1, g2, . . . , gr, gi : M → R. We say g1, . . . , gr are independent functions on M around x ∈ M if (Dgi)x are independent vectors of TxM. Theorem 3.3. If y is a regular value f : X → Y, codim f −1(y) = `. Then −1 −1 f (y) is actually a cut by ` functions g1, g2, . . . , g` : X → R with f (y) = T` −1 i=1 gi (0) * ∼ ˜ ` −1 −1 T` −1 Proof. Pick h : V 3 y −→ V ⊆ R , y → 0.f (y) = (h◦f )(0) = i=1 gi (0) where h ◦ f = (g1, g2, . . . , g`).

3.2 Transversality We have shown that f −1(y) is a submanifold when y is regular. If Z is a submanifold of Y , what condition make f −1(Z) a submanifold of X. − Def. f : X → Y,Z ⊆ Y is a submanifold. f is transversal to Z f t T if −1 im(Dfx) + TyZ = TyY for all x ∈ f (Z).

− −1 Theorem 3.4. If f t Z then f (Z) is a submanifold. Proof. Ask Knutson. − − Ex. X,Y ⊆ Z, i : X → Z inclusion map. We say X t Y if i t Y .

11 4 September 10

4.1 Homotopies and stable/open properties

Def. A continuous function f : X × [0, 1] → Y is a homotopy between f0 and f1 where ft(x) = f(x, y).

Def. g, h : X → Y are homotopic if there exists a homotopy s.t. f0 = g, f1 = h. Ex.

• More generally, we can consider a family of maps fs : X → Y given by a f : X × S → Y by f(x, s) = fs(x). • Non example: if X is one point, Y two points and g(x) 6= h(x) then no homotopy between the two. Def. A property p of maps is called stable or open in families if

1. For all homotopic f : X × I → Y if f0 is p, then ft is p for all small enough t.

2. For all families over S of maps f : X × S → Y {s ∈ S|fs is p} is open in S. We can see that 2 implies 1 as we take S = I and s = 0. For 1 to 2, we take a path P ⊂ S and parametrize f : I → P and take s = f(0). Theorem 4.1. The following p are stable properties of maps from compact X → Y . (a) Local diffeomorphisms (including immersions and submersions). (b) Immersions (c) Submersions. (d) Transverse maps to some Z ⊆ Y . (e) Embeddings (injective immersions) (f) Diffeomorphisms

Ex. Non example: R → R2 by x → x + t. Def (transverse). For f : W → M, f is transverse to z if ∀ m ∈ M s.t. f(w) = z, TzZ + im(Twf) = TzM Proof. Ask knutson about this

12 5 September 12

5.1 Critical and Regular values

Recall that f : M → N smooth map between manifolds and Tpf : TpM → Tf(p)N.

Def. p is a regular point if Tpf is onto and is a critical point if it is not. The regular condition is an open coordination, and being critical is a closed condition (since determinants are 0 for closed and open is non zero) ( 0 x ≤ 0 Ex. A1(x) = . Everything on the positive is regular, but non- e−1/x x > 0 positive are critical.

What about {f(p)}.

Def. q ∈ N is a critical value if ∃ a critical point p : f(p) = q and a regular value if there doesn’t exist such a p.

Theorem 5.1. If q is a regular value, then f −1(g) ⊆ M is a submanifold of codimension dim N.

Ex (can have manifolds not preimages). Consider a Mobius strip M 2 in R3. Can’t get M 2 = f −1(0) a regular value f : U → R. Where U is an open set M ⊆ U 3 ⊆ R3 since U has a positive and negative side but the Mobius strip does not.

Ex (critical values might not be closed set). A2(x) = A1(x)A1(1 − x) is a P∞ bump function supported in [0, 1] and smooth. If f(x) = i=1 A2(x − 2i) ∗ 1 −i −i ( 2 + 2 ). Critical values are A2(0) ∗ (1/2 + 2 ) Theorem 5.2 (Sard). The set of critical values is measure 0 in N.

Def. X ⊆ Rk is measure zero if for all  there is a countable open cover s.t. the volume of the balls is less than . This definition is invariant under diffeomorphism from Rk → Rk. Corrollary. The set of of regular values of f : M → N is dense. Note that this uses the fact that f is smooth, unlike“space-filling curves”.

Remark. Sard’s theorem has a counterpart in algebraic geometry called “generic freeness” but only holds over fields of characteristic 0.

Ex. f : C → C, f(x) = x2, f −1(t) is 2 pts for t 6= 0, 1 pt for t = 0.

13 −1 Ex. F2 → F2 has f (t) is 1 point. T∞ Prop 5.1. If {fi : Mi → N} is a countable sequence, then i=1(regular values of fi) is dense.

Proof. Union of measure 0 sets is measure 0, and their complement is our desired.

5.2 Linear Algebra Around Critical Points

Def. Let p be a critical point of f : M → R. In coordinates on M near p, the second derivatives give a symmetric n × n matrix, the Hessian. Without picking coordinates we can say its defining a symmetric bilinear form on TpM.

In one set of coordinates, get H matrix and can use change of coordinates H → AHAT . A is the derivative of the map from one open chart’s open set to another’s.

Theorem 5.3 (Sylvester’s law of inertia). For all symmetric H, exists an invertible A s.t. AHAT is a diagonal matrix with several +1, −1, 0 in that order and the number of +1, −1, 0 is invariant.

Prop 5.2. If there are no 0 ie H is invertible then the number of 1 and −1 is stable in H. Ie if we perturb H by a small symmetric perturbation then the numbers remain the same.

Ex. M = s2 and map using the height. There are 2 critical values.

Ex. M is torus T 2 (stood up on it’s side) and use height function. There are 4 critical values at the top and bottom of the circles. The bottom point has two +1 in it’s hessian, as it looks like x2 + y2 and the second derivative is both positive. The middle two points are +−, and the top point is −−.

Ex. T 2 but laid on it’s side. There are two critical values corresponding to circles. The bottom circle is +0 and the top is −0 since is is flat on the x but increasing/decreasing on the y.

5.3 Some Morse Theory

Def. A critical point p ∈ M → R in f is Morse if the Hessian at p is “nondegenerate” (invertible, only +1, -1).

14 Def. A function f : M → R is a Morse function if all critical points are Morse.

Ex. In our two torus examples, the top is Morse but the bottom is not Morse. Note that in the top example in the z-axis there is one symmetry (a flip) while in the second the flat torus has infinitely many symmetries (rotation in z-axis).

Lemma (Morse Lemma). If p is a Morse crit point of f : M → R then ∃ c f local coordinates c : U → M taking 0 → p, U open in Rk s.t. U −→ M −→ Rk then (f ◦ c)(x , . . . , x ) = Pr+ x2 − Pk x2 1 k i=1 i j=r++1 j

6 September 12

6.1 Recall

Let f : M → R a smooth function. A point p ∈ M is critical if Tpf = 0 (not onto since dimension of R is 0). A critical point p is is Morse if the hessian at p is nondegenerate, TpM × TpM → R.

6.2 Morse Theorem

Def. Let U,→ M ⊆ Rn where M is k-dimensional be open. Cal u adapted n S to coordinates if ∃k-subset S ⊆ {1, . . . , n} so U,→ R  R is diffeomor- phic to its image.

Ex. On the circle, if u1 is an arc in quadrant I it is diffeomorphic to the projection on the x-axis and y axis (adapted to 1 or 2). If u2 is an arc that intersects (1, 0), it is diffeomorphic to the projection on x-axis (adapted to 2). If u3 is a curve that intersects (0, 1), (0, −1) then it is not adapted to either.

Remark. Note that this means that we can create the chart at Rs and get the chart to Rn. In 2230-2240 we can define a manifold through this process through out system of charts.

Def. V ? is the linear maps from V to R.

k n k n ? Theorem 6.1. Let M ,→ R . Let f0 : M → R. Define f : M ×(R ) → R * D* E n by (m, a → f0(m) + a, m ) (note we can just use R and use dot product). * * * For a.e. a, f a = f(m, a): M → R is Morse.

15 Remark. This lets us define things about M by using Morse functions since they exist.

Def. If f : M → R is Morse, M compact. Let χ(M, f) be the sum of the determinants Sylvester normal form of the Hessian of f at p for all critical points f at p. Note that this sum is finite by compactness. This is the Euler Characteristic of M

Ex. If we consider a sphere and take projection to z-axis. The Euler charac- teristic is 2 as the two critical points are the top with SNF [−1, −1] and the bottom with SNF [+1, +1].

If the indent the top point by pushing it down, then we have 4 critical points [−1, −1], [−1, −1] for the two newly created points, the top points with [+1, −1] and the bottom points is [+1, +1]. The Euler Characteristic is 2 as well.

insertdiagramshere

k k ? * Proof. Part 1 If M = U,→ R is open set. Define g a : U → (R ) essentially * * * as D(f a ). Note that Df a is a linear map from U → R as f a : U → R. * * g a = Df a in this case.

D* E * * k ? * * f a (m) = f0(m) + a, m so g a = Df~a = Df0 + a,. Note that a ∈ (R ) .

* * 1. If f a is critical at p then g(p) = −a, where g(p) = Tpf0. k k ? * * * 2. Hessian of f a = Dg a , where Tpg a : R → (R ) as th

* k ? * If −a is a regular value of g : U → (R ) (using Sard) and f a is critical value at p, then this occurs iff Tpg is onto, which means that Tpg is invertible so the Hessian is nondegenerate and f is Morse at p.

Part 2 Cover M with open sets U k adapted coordinates. Since it is sec- ond countable, we only need countably many. Assume some particular U is * n−k * adapted to {1, 2, . . . , k}. For each c ∈ R apply part 1 to f(ok, c ) to find * * k k a.e. b ∈ has f * * . If we say that this b is S* ⊆ where S* is of R ( b , c ) c R c measure 0. Then Z 1* = 1 * * * * c S { b , c , b ∈U} c ∈Rn−k Uk And this is the characteristic of a set of measure 0.

16 k Remark. Note that Dfa(p) = Tpfa is a linear functional on R . Knutson substitutes T instead of D but I’ll use T since it’s easier to parse.

Ex. Under our Morse function projection to height axis, χ(T 2) = 0. χ({0}) = 1. χ(S1) = 0 since bottom is + top is −. χ(genus) = 2 − 2g where the genus is a bunch of connected torus’s (g connected). Euler characteristic of Klein bottle is 0. The real project plane S2/± has characteristic 1.

7 September 19

7.1 Whitney Embedding Theorem Remark. We have shown that surfaces like the Klein bottle and Boy surface can be immersed in R3 but not embedded. The question is for M k ,→ Rn an embedding what is good value of n (project to linear space).

Theorem 7.1 (Whitney embedding). A.e. linear projection Rn → R2k+1 k n 2k+1 k gives an embedding M ,→ R  R when M is compact. Proof. It is enough to project Rn ,→ Rn−1 when n > 2k + 1 and then induct. Recall that if M,→ Rn then the tangent bundle TM,→ Rn × Rn, TM = * * n n {(m, v : v ∈ TmM)} is a 2k submanifold of R × R .

Take h : M × M × R → Rn by (x, y, t) → t(f(x) − f(y)) with g : TM → π n ∼ n n 2 n * * * n T R = R × R  R which is (m, v ) → T f(v ). Let a ∈ R be in neither orthogonal ⊥ * * f n * n * image by Sard. Now claim M ,→ R  a := { b ∈ R : b ⊥a} is an * embedding. Note that the kernel of this orthogonal map π is Ra.

* • Injectivity:If (π◦f)(x) = (π◦f)(y), then this occurs iff f(x)−f(y) = ta 1 * for some t ∈ R not 0. This implies that h(x, y, , t ) = a, which is a contradiction.

* * • If T (π ◦ f) = (T π ◦ T f)(v ) = 0. Note that since π is linear it’s * * derivative is itself, so we have that π(Txf(v )) = 0 which means that * * a * 1 * T f(v ) ∈ R so it is ta for some t 6= 0. Now g(x, t ) = a.

Remark. An Abstract k manifold M k not in any Rn is a second count- able Hausdorff space where every point is locally homeomorphic to Rk. In

17 −1 particular it has an atlas {ϕi : Ui ,→ M} s.t. ϕi that ϕi(Ui) ∩ ϕj(Uj) is open in Uj and Ui. This means it is countable. There exists a Whitney Embedding Theorem for these (can find in Munkres I think).

Ex. Non example: The long line is multiple [0, 1] glued together, one for each ordinal n < ω1 which is the uncountable ordinal.

7.2 Manifolds with Boundary

Def. A manifold with boundary M k ⊆ Rn has charts modeled on open sets k−1 in R × R≥0. Ex. An example is in the ball Bn for |x| ≤ 1. In the middle it is Rk and at k−1 k the boundary it is R × R≥0 (open set of R intersected with R≥0). Remark. A technical issue at the boundary is that at the boundary the half space demands an extension.

Theorem 7.2. If (M k, ∂M) is a manifold with boundary, then ∂M is a (k − 1) manifold without boundary.

Corrollary. ∂2 = 0.

Remark. Note that ∂(M × N) = (M × ∂N) ∪ (∂M × N), M,N manifolds with boundary. This is like Leibniz’s rule.

Theorem 7.3. If M is a compact connected 1-manifold with boundary (note that it might have empty boundary). Then M ∼= [0, 1] or S1. Proof. Perturb the constant function f = 0 to a (linear) Morse functionf : M → R. Make a graph w/ vertices are the connected components of regular points. Then edges correspond to connections on critical points. These give points of degree 1 and degree 2. We need to prove that M connected implies that the graph Γ is connected.

Corrollary. If M is compact 1 manifold with boundary, then there are two divides the total number of points in ∂M.

Def. A retraction of M to N ⊆ M a submanifold, is a map r : M → N where r |N = idN . Theorem 7.4. If M compact with boundary, then there doesn’t exit a retrac- tion r : M → ∂M.

18 Proof. Let n be a regular value of r. If n is a regular value of r, then r−1(n) is a 1-manifold with boundary (dimension is same as codimension of ∂M) that is also compact. ∂r−1(n) = ∂M ∩ r−1(n) = {n}, where we invoke sard.

Corrollary (Brouwer). If f : Bn → Bn is smooth then f has a fixed point.

Proof. If not we construct a retraction. But those don’t exist. The retraction is r(x) = f(x) → x → ∂Bn where we find the line containing f(x) and x and map to the boundary starting from f(x).

8 September 24

8.1 Manifolds w/Boundary Revisited

f Theorem 8.1 (Sard’s Thm for Manifold with Boundary). Let (M, ∂M) −→ Y . Then a.e. y ∈ Y is a regular value for f |M\∂M and ∂f = f |∂M and regular for the open M+ ⊇ M ∪ ∂M which we use to define the region for f being smooth. (IE the extension of M to another manifold). If y ∈ Y is a regular value, then f −1(y) is a submanifold with boundary and ∂f −1(y) = f −1(y) ∩ ∂M.

8.2 Transversality is stable Theorem 8.2. The F : X × S → Y where X is a manifold with boundary, S manifold, Y manifold and Z,→ Y a submanifold s.t. F is transverse to Z. If F , ∂F are transverse to z, then a.e. s ∈ S,Fx : X → Y and ∂Fx are transverse to z.

Proof. Pick s a regular value of F −1(z) = W → X × S → S. We want if

Fs(x) = z ∈ Z ⊆ Y that TFs Tx(X) + TzZ = TzY . We know TF (Tx,s(x × s)) + T π TzZ = TzY . Then we see that Fx(s, ) = Fs(x) = z so T(x,z)W  Ts(S) taking * * * * * * * * * * (u, e) → e. Then TF(x,s)((w, e) − (u, e)) − a ∈ TzZ since T(x,s)F (u, e) ∈ TzZ. Remark. There are two more technical results.

1. Let f : X → Y ⊇ Z and ∂Y = ∂Z = ∅. Then there exists perturbations of f (homotopies F w/ F0 = f) to make f transverse to Z. − 2. If f is already transversal on ∂X then we can perturb f (to t z) without changing ∂f.

19 Def. If f : X → Y ⊇ Z and dim X + dim Z = dim Y and f transverse to Z. Note that an equivalent of transversality is codim(X ∩Z) = dim Y −dim(X ∩ −1 Z) = codim X + codim Z. Let I2(f, z) be the number of points in f (z) mod 2.

Theorem 8.3. Let F be a homotopy from f to f . ie F : X × [0, 1] → Y . − 0 1 f0, f1 t Z,→ Y , X,Y,Z compact. Then I2(f0, z) = I2(f1, z).

Proof. Perturb F without changing f0, f1 so that it is transverse to Z. Now F −1(Z) is a 1 manifold with boundary ⊆ X × [0, 1]. ∂F −1(z) = F −1(z) ∩ −1 −1 −1 ∂(X × [0, 1]) = F (z) ∩ (X × {0, 1}). Then 0 ≡ #∂F (z) = #f0 (z) + −1 #f1 (z). Ex. Let X,→ T 2 ←-Z.

Def. If X,→ Y,→ n and TX,→ TY,→ Y × n submanifold. The normal R* * R bundle NX Y = {(y, v ) ∈ TY : y ∈ X, v ⊥ TyX}. Remark. If there exists a manifold T ?M, the cotangent bundle to M is ? * * ? T M = {(m, v ), m ∈ M, v ∈ (TmM) } which is a real morphism from TmM to R.

9 September 26

9.1 Connected Remark. Recall our two versions of “connected”. 1 is that every continuous function X → {±1} has exactly one value and version 2 is that for all points exists a continuous function [0, 1] → X that maps 0 → x, 1 → y. Note that this is path connected condition.

Def (One-connected). M manifold and connected is simply connected if (equivalently)

1. for all x, y and γ : [0, 1] → M with 0 → x, 1 → y and γ is unique up to homotopy relative to γ.

2. If γ : S1 → M is a curve then there exists an extension to D2.

Ex. The sphere any two points are simply connected (and path connected). On the empty torus, any two points around a circle in the torus are not simply connected (since they are not homotopic since you can’t go between them as it is hollow.)

20 Theorem 9.1. If γ : S1 → M and Z codimension 1 in M submanifold with no boundary and I2(γ, Z) 6= 0, then M is not simply connected. Proof. The contrapositive is that if M is simply connected we can use ho- motopy to take γ to a point, missing Z.

9.2 Transversality

Theorem 9.2 (-nbhd thm). If γ ⊆ Rm, γ compact, then ∃ > 0 s.t. γ := {m ∈ Rm, d(m, Y ) < } has a map γ → γ to the nearest point that is well defined, smooth, and submersion.

m * * Proof. Recall the normal bundle NY R = {(y, v ), v ⊥ TyY }. Map y to * m y + v with h. h is regular along Y,→ NyR so T h is an isomorphism. m h is injective on Y,→ NyR . This implies h is a local diffeomorphism in some U open ⊇ Y . For each  > 0 let v := {y ∈ Y : U ⊇ B(y)} and S Y = >0 V and since Y is compact we can find an finite cover V1 ∪ · · · ∪ Vk and  = mink k. Corrollary. If f : X → Y is a map and Z is a submanifold of Y with − − ∂Z = ∅, then f is homotopic to g : x → y s.t. g t Z, ∂g t Z. Proof. Let S be open in Rm around 0, X × S → Y given by (x, z) → f(x) + s to the nearest point in Y if s is small enough. This composite is a submersion since we can vary S to give us the perturbations of Y (definition − − of submersion). For a.e. s ∈ SFs t Z, ∂Fs t Z and as shown last time this implies F0 = f. − − Remark. A better version is that if W,→ X is closed, f |W t Z and ∂f |W t Z then can insist that g = f on W .

Theorem 9.3 (Jordan curve theorem). Let M be connected and simply con- nected and H ⊆ M closed with codimension 1 and ∂H 6= ∅. Then M \ H has ≥ 2 components each of which has H on the boundary of closures.

Proof. We need a function s : M \ H → [0, 1] that is continuous and onto. Pick m ∈ M \H. For any n ∈ M \H, pick γ : [0, 1] → M with 0 → m, 1 → n using the fact that M is connected. Let s = I2(γ, H) and we can perturb γ to be transverse to H (γ has dimension 1). Note that this is well defined since M is simply connected.

21 9.3 Orientation Def. An orientation of a finite dimensional real vector space V is a func- tion O

O : {ordered bases of V} → {±1} s.t. if T : V → V invertible and [b1, . . . , bn] a basis then O(T (b1) ...T (bn)) = sign(det T )O(b1, . . . , bn).

* Remark. If V = {0}. Then the ordered bases are {∅}.

Given V,W vector spaces and V ⊕ W := V × W with coordinate wise vector space operations. Call orientations OV , OW , OV ⊕W compatible if for bases bi of V , cj of w then

OV (b)OW (c) = OV ⊕W (b, c)

10 Oct 1

10.1 More Orientation Remark. Orientations of any two V,WV ⊕ W determine one on the third * * * * V ⊕ W → W ⊕ V, (v , w) → (w, v ) when is this orientation preserving? Note that this is the determinant of a matrix that is the determinant of   0 I * dim v = (−1)dim V dim W . * Idim w 0 Def. If M is an oriented manifold-w-boundary then ∂M is oriented by the * * * following recipe. O∂M ( b 1,..., bdim M−1) where b is a basis of Tm(∂M). Note that we must extend it by one dimension, so we extend it orthogonally to the tangent space with another vector of norm 1. The book defines it as * * OM (outward basis, b 1 ..., b dim M−1). Remark. Recall that ∂2 = 0. We should look at the ker/image.

Def. Ci = {formal linear combinations of diffeomorphism classes of compact ∂ ∂ i-mwb and oriented}. We see that C0 ←− C1 ←− ... . We see that C0 = {m[+] + n[−]}, C1 is the sum of n times the arrow from between two points (· → ·) and m times the circular arrow.

22 − + Remark. ∂ of our · → · is · · . We see that H0 = Z where H0 = ker / im at Ci. H1 = 0,H2 = 0,H3 = 0 since each manifold is not the boundary of 2 any manifold (???). H4 = Z in particular has an example of the CP or the 5 complex projective plane and is the ∂N . H5 = Z/2 since there is a manifold that is not the boundary but two of them is the boundary.

10.2 Return to Intersection

Remark. Recall that I2(f : X → Y,Z ⊆ Y ) X compact, dim X = codim Z −1 − and this is the number of points in f (Z) after jiggling f to be t Z mod 2. Def. I(f : compact X → Y,Z ⊆ Y ) oriented manifolds with dim X = P codim Z. This is defined as p∈f −1(Z) sign(T f(TpX) ⊕ Tf(p)Z versus Tf(p)Y . Ex. If we have curve (x + 2)(x − 2)x with standard orientation then at −2 it is negatively oriented (since the x vector is following the curve and the y vector is orthogonal), at 0 it is positive, and at 2 it is negative again.

Def. A coorientation of Z ⊆ Y is an orientation on the vector spaces * * in NZ Y = {(z, v ∈ TzY ): v ⊥ TzZ}. We always have that (NZ Y ) |z ⊕TzZ = TzY . To visualize, note that in a sin curve the orientation vectors are perpendicular, since we want to know which side is + which is −. The coorientation is orthogonal to the curve, and gives us which side “comes off” the manifold.

fW Prop 10.1. If f : X → Y and this factors through fW ie X,→ W −−→ where X = ∂W and is compact oriented and Z,→ Y is closed cooriented. Then I(f, Z) = 0.

− −1 Proof. Jiggle fW to be t Z also not changing fW |X from f. Note that fW (z) −1 −1 is an oriented 1 mwb since W is a mwb and ∂fW (z) = fW (z) ∩ ∂W . Note that it is 1-mwb since we dim W = codim Z and X is just one dimension −1 −1 less. We can also simplify ∂W = X and so ∂fW (z) = f (z) is an oriented 0 manifold. Recall that 1-manifolds are the arrow or circular arrow and 0 manifolds are some number of positive and negative points. We have that P sign(pts) = 0 and that is what we were computing.

Corrollary. If f0, f1 : X → Y are homotopic and X is oriented compact and Z cooriented in Y . Then I(f0,Z) = I(f, Z). Proof. Apply the previous to W = [0, 1] × X so ∂w = −X t +X (where the signs indicated orientation). We see that 0 = I(∂W, Z)−I(f0, z)+I(f, z).

23 Def. f : X → Y ←- {pt} = Z where X is compact oriented and Y connected and oriented. Then deg f = I(f, pt).

Theorem 10.1 (Fundamental Thm of Algebra). Any complex polynomial p(z) of deg > 0 has a root.

( x deg p p( t )t t 6= 0 Proof. Let pt(x) := . Assume p has no roots, then pt6=0 has xdeg p t = 0 1 1 no roots. Let ϕt : S → S = {z : |z| = 1} which takes z → pt(z)/|pt(z)|. We see that deg(ϕ1) = deg(ϕt > 0) = deg(ϕ0) = deg p 6= 0. So ϕ1 doesn’t extend 2 p(z) to D → C (disc). But we had an extension z → |p(z)| , contradiction.

11 Oct 3

11.1 Review from last time Remark. We are working with f : X → Y ←-Z where dim X = codim Z, X,Z compact X oriented Z cooriented (Y,Z both oriented). − P ∼ Def. If f t Z,I(f, Z) := x∈X,f(x)∈Z sign(T f(TxX) ⊕ Tf (x)Z −→ Tf(x)Y )

− 0 0 Theorem 11.1. If f t6 Z can jiggle it to f and I(f ,Z) doesn’t depend on f 0.

11.2 Euler Characteristic based off of Orientation Def. If X is compact oriented, then I(4, 4) = χ(X) is the Euler character- istic. Note that 4(x) = (x, x).

Def. Paint(M) := {(m ∈ M, o ∈ orientations of TmM)}. We can project using π1 : Paint(M) → M and this is a 2:1 map. The idea behind the paint is that if we paint the manifold and peel it off, for the sphere we have two spheres and for the Mobius strip we have one big strip of twice the size. Note that this paint is an abstract manifold, since we can double the charts on M and glue them together. So this paint comes with a canonical orientation.

We define χ(M) := χ(Paint(M))/2.

2 2 Ex. χ(RP ) = 1 since the paint of RP is S2. This is because for each pair 2 of opposite points in the sphere is mapped to a line in RP . χ(S2) = 2 (use the one from Morse theory but have not yet established I(4, triangle) = χ).

24 11.3 Lefschetz Number Remark. One way to get another copy of X insides X ×X which is hopefully − t 4 is as a graph of f which is {(a, f(a))} Def. The Lefschetz number of f : X → X compact manifolds is L(f) := I(graphf, 4).

Ex. L(idX ) = χ(X). − Def. f : X → X is Lefschetz if graph(f) t 4 inside X × X. Remark. When is this true?. If for all (a, b) ∈ graph(f) ∩ 4 the tangent spaces add up. We need to consider the a : f(a) = a. So we have that T(a,a) graph(f) + 4TaX = T(a,a)(X × X) or equivalently that they intersect at 0 (by dimension). Note that T(a,f(a)) graph(f) = im Tag which is g : a → * * * (a, f(a)). Note that Tag : TaX → T(a,a)X × X is given by v → (v , Taf(v )). * * * * In particular we want that {(v , T f(v ))} ∩ {v , v } = 0. f is Lefschetz if at fixed point a of f Taf doesn’t have 1 as an e-value. So in particular Taf − I is invertible.

P Note that L(f) = a:f(a)=a sign det(Taf − I)

1 Ex. X = RP = R ∪ ∞ = {[x, y] 6= [0, 0]} up to scale. f : X → X by 1 f(z) = 1.01z is homotopic to the identity on RP . To compute the Lefschetz
number note that the only fixed point 0 and that at that point T0f −I = 0.01 which has determinant of sign +. So the Lefschetz number is 1.

1 iθ n Ex. S = {e ∈ C}. Take fn(z) = z , n ∈ Z \{1}. The L(fn) = P sign det(n
− I) = n − 1 since for the positive case it is n − 1 roots µn−1 of unity (each contributing +1) and for n < 1 each term contributes −1 and there are 1 − n.

Corrollary. No fi is homotopic to fj in the above case. This is because Lefschetz number if invariant under homotopy.

Remark. If A, B ,→ M with dim A = codimM B. If all oriented then I(A, B) = (−1)dim A dim BI(B,A). if B = A and dim A ≡ 1 (mod 2) then I(A, A) = 0.

Ex. On the Mobius strip I2(A, A) = 1 since M isn’t orientable. Theorem 11.2. If M compact oriented odd-dimensional then χ(M) = I(4, 4) = −I(4, 4) = 0.

25 12 Oct 8

12.1 Vector Bundles

Def. A vector bundle over M ⊆ Rn is a submanifold B of M × RN s.t. B ∩ (m × RN ) ≤ RN (linear subspace) for allm ∈ M and is also of same dimension for each m.

Ex. TM = {(m, dx): dx ∈ TmM}. M,→ P then NM P is the normal bundle.

Def. A section of B → M is a map M → B s.t. M → B → M is identity. In particular we map m → (m, ·).

Ex. M × R is a vector bundle over M. A section over this is isomorphic a function f : M → R with m → (m, f(m)).

N1 N2 Def. If B1 ⊆ M × R ,B2 × R are vector bundles then B1 ⊕ B2 = N1 N2 Ni {(m, v1, v2) ∈ M × R × R } where (m, vi) ∈ M × R .

Def. A map of vector bundles B1 → B2 is a smooth map s.t.

f B1 B2

M commutes and is linear for each m ∈ M. (ie it maps m → m and each vector linearly)

Theorem 12.1 (Tubular nbhd thm). If Z,→ Y and is closed, then ∃ an ∼ open U ⊇ Z open in Y and a map NzY −→ U is local diffeomorphism around z

Ex. For Y a torus and Z a circle below the main one in the donut hole, the normal bundle is

N closest point Proof. NzY → R −−−−−−−→ Y works only locally (near Y )

Remark. Recall that χ(M) := I(M4,M4) both inside M × M. Of course 0 − we need to jiggle M4 a little to make M t M4 to compute I. By the tubular nbhd theorem, we note that this is instead equal to I(Z,Z) inside

NM4 (M × M) where Z is the zero section. Note that the zero section is M → M × {0}.

26 ∼ Theorem 12.2. NM4 (M × M) = TM as bundles over M. ∼ ⊕2 Proof. Note that (NM4 (M × M))m ,→ T(m,m)(M × M) = (TmM) . Note that the normal bundle is perpindicular to Tm×mM4 which is congruent TmM hookrightarrow ⊕2 ⊥ which is embedded in (TmM) . Then we see that V 4 V ×V ←-V and by projection to the first factor we see that V ⊥ −→∼ V Corrollary (Poincar´e-HopfThm). χ(M) = I(Z,Z) where Z is the zero section inside TM. Proof. Can perturb using a section of TM ie a vector field σ : M → TM. So this is equal to the intersection number of the image of σ and the zero section ((m, σ(m)) and (m, 0)) in TM and this is the equals to the sum over the zeros in σ of (−1)#negative eigenvalues of Tmσ

Near a 0 of σ,

σ : M TM

ball in Rn Rn

∼ So T σ : TmM → T*(TmM) = TmM. Let σ be generic enough that Tmσ is 0 1 : 1. ie the eigenvalues are non0. In particular we have that mσ is 1 and 2 block diagonalizable. Therefore the eigenvalues are > 0, < 0 or in pairs z, z ∈ C \ R. Corrollary (Hairy Ball Thm). There doesn’t exists a vector field σ on S2 w no zeros because χ(S2) is not 0, so we must have a zero of σ. Theorem 12.3. If M has a nonvanishing vector field then χ(M) = 0. Remark. If M is connected, then the converse holds. IE χ(M) = 0 implies that M has a nonvanishing vector field.

Remark. Recall that if f : M → R was sometimes Morse.

M R f

Rn

Note that if ∇f is a vector field on Rn, in particular the gradient. Σ: M → TM can be defined by m → (∇f) |m.

27 y Ex. For S2 ,→ R2 −→ R. The gradient would be +1 everywhere. But σ takes the +1 on the tangent plane. To visualize at (±1, 0) it is still 1. At (0, ±1) it is 0. Everywhere else is positive in the y direction. We see that if f1 = f2 on M then σ1 = σ2. Theorem 12.4. This σ is nondegenerate in the Poincare-Hopf sense (iso- lated and T σ is 1 : 1 at the zeros) if f is Morse. In particular the zeros of σ are the critical points of f. Remark. The set of 1 dimensional vector bundles with nonvanishing section over M is isomorphic to the set of 1-dim bundles ∼= to trivial line bundle M × R. 2 3 3 ∼ 2 Ex. The radius bundle over S is NS2 R . Note that (T R ) |S2 = S × R and 3 ∼ 2 3 2 NS2 R = S × R . Note that TS is neither trivial nor a line bundle. ∼ dim M k 0 1 3 7 Remark. When is TM trivial (= M × R ). For M = S . S ,S ,S ,S are the only ones. In these cases we identify each to Si+1 to the reals, complex, quaternions, and octonions. Then we imagine that each vector bundle is identified by some multiplication in the number system.

13 Oct 10

13.1 Lefchetz Number Continued

Remark. Recall that L(f) = I(M4 graph(f)) inside M × M. Of course, this could be infinite, so we perturb f a little bit and get another graph and this is our Lefschetz number. Remark. Claim: Small perturbation of graph(f) to another submanifold of M × M are of the form graph(g). We simply perturb f a little to g. We − figured out when M4 t graph(f). In particular f is Lefschetz and isolated fixed points {p} and Tpf − I : TpM → TpM is invertible. Remark. If we only had that isolated fixed points, how do we get a local Lefschetz number?

1 1 Ex. f : CP → CP = C ∪ ∞ by f : z → zn where n 6= 1. The fixed points are 0 and ∞. Near 0, if we wiggle f then z → zn has n solutions instead of 1 solution (0) of order n. These n solutions have Lefschetz number 1, so the total should be n. But we haven’t figured this out yet!

Remark. Near p inside B(p) in local coordinates, f(z) = z only at p. For f(z)−z k−1 z ∈ ∂B(p) we see that |f(z)−z| ∈ S .

28 Def. Local Lefschetz number of f at p is the degree of the above map k−1 from ∂B(p) → S . Remark. Why is this homotopy invariant (the sum of p of the Local Lefschetz numbers) and why is this correct when f is Lefschetz? Well this is correct when f is Lefschetz by a linear algebra calculation. In particular f − I shows up in the above map as the numberator and denominator. We get that deg = sign det.

˜ For homotopy invariant, say f deforms to f Leftchetz p to p1, p2. Let p1, p2 ⊆ B(p). Since deg is homotopy invariant we see that the degree. Note that deg is the only homotopy invariant map (Hopf-index theorem).

13.2 Computing Euler Numbers Fdim M Def. A triangulation of M is a decomposition M = i=0 Mi. Mi is the union of open i-simplices s.t. for each connected component C of Mdim M , “facet”, ∃U ⊆ M open C ⊆ U s.t. we have standard simplex X and X ⊆ U 0 ⊆ Rm and U is diffeomorphic to u0 and C is diffeomorphic to X. The P standard simplex is {(x1, . . . , xm): xi ≥ 0, xi ≤ 1}. P i Theorem 13.1. χ(M) = i(−1) #face of dim i Theorem 13.2. ∃ a vector field on k-simplex pointing toward center of i- simplices when i is even and away from the center of i-simplicies when i is odd. Remark. Triangulations exist on smooth manifolds but not necessarily on topological manifolds. Note that on topological manifolds, we replace diffeo- morphism with homeomorphism.

13.3 Integration Remark. We want to show that for some smooth f : M → N we have R T f : TM → TN and want to find the integration M f. What does this mean? When should one integrate it? Let’s begin with smooth and compact f support. Perhaps U −→∼ M −→ . and R f := R f ◦ ϕ. But this is no ϕ R M U R 10 good since the reparametrization, this is bad. Think how 0 x1dx = 10 but if y = 2x we must reparametrize dx.

Maybe integrate vector fields? Well, this is also not as clear. In the above idea, our vector field is 1 for x and we have to change it to 2x but this becomes hard.

29 Remark. Recall: If V is a real vector space then V ? = {f : V → R} homomorphisms. Note that V → (V ?)? by v → f(v) for f ∈ V ?. Of course we could also compose this with a linear map. If V is finite dimensional, this map is invertible. Def. A (0, 1)-tensor on V is an element of V ?.A 1-form on M is a section ? S ? of T M = m∈M {m} × (TmM) (cotangent bundle) and this is easily given structure of an abstract manifold, but not as easy to embed like the tangent bundle. Note that functions are sections of the trivial bundle by M ×R → M. A one form eats vector fields and gives a function.

14 October 17

14.1 (0, p)-tensors

Def. A (0, p)-tensors on V is a multilinear α : V × V · · · × V → R where there are p V ’s. We are building it this way since we are trying to avoid the direct definition, but it is good enough for finite dimensional vector spaces. We won’t need (p, q) tensors.

P m * * **T Ex. n × n → matrices −−−→ij . (v , w) → v w → P (v , w ). Each R R R cij i j n choice C = (cij) gives a (0, 2) tensor on R , and these are all the (0, 2) tensors.

Ex. det : (Rn)n → R. This is alternating (antisymmetric) ie it vanishes if two inputs are equal. Def. If α is a (0, p)-tensor in V ,

* * 1 X Alt(α)(v ,..., v ) = (−1)`(σ)α(v , . . . , v ) 1 p p! σ(1) σ(p) σ∈Sp Where ` is the number of inversions. Note that this definition is not standard, 1 in fact we mostly don’t include p! since over a general field we can get division by 0. Def. If α is a (0, p) -tensor and β is a (0, q)-tensor then α ×β is a (0, p+q)- tensor by (α ⊗ β)(v1, . . . , vp, w1, . . . , wq) = α(v)β(q). Def. α ∧ β := Alt(α ⊗ β).

Theorem 14.1. The space of alternating (0, p)-tensors of Rn has the basis n
{dxi1 ∧ · · · ∧ dxip : 1 ≤ i1 < ··· < ip ≤ n} (note there are p choices) where n dxi : R → R is projection to the i-th coordinate.

30 Proof. Let α be an alternating (0, p)-tensor. We need to show ∃! expansion P α = [n] cI ∗ (dxi1 ∧ · · · ∧ dxip ). I∈( p )

Uniques We plug in eji vectors to α and get the massive sum X 1 X c (−1)`(σ)dx e . . . dx e I p! iσ(1) j1 iσ(p) jp I σ∈Sp

X 1 X cJ = c (−1)`(σ) = I p! p! I σ:iσ(k)=jk

Therefore cJ = p!α(ej1 , . . . , ejp ). To check that α is the same for any list of basis vectors, use multilinearity to reduce p-tuples of basis vectors then use alternating to reduce to j1 < j2 ··· < jp. Therefore, cJ is unique. Remark. Note that α ∧ β = (−1)deg α deg ββ ∧ α.

14.2 Multivariable Change of Variable

∼ 0 Ex. (0,R) × (0, 2π) −→ DR \ (x ≥ 0, 0) by (r, θ) → r(cos θ, sin θ). For ∼ a U −→ V −→ R we have that Z Z adx1 . . . dxk = (a ◦ f)| det T f|dx1 . . . dxK V U This kind of sucks since we need to take the absolute value over the deter- minant of the Jacobian. For manifolds, we use forms to orient so we don’t have to deal with this.

Def. A p-form on a manifold M (not necessarily of dim p) is a smoothly varying choice of (0, p)-tensor on each TmM. Ex. p = 0 we have functions on M. p = 1 1-forms are vector eater fields.

For f : M → N we can define f ? : 0 − forms on N → 0 − forms on M by a → a ◦ f is the pushforward.

? * * * * If α is a p-form on N then (f α)(v 1,..., v p) := α(T f(v 1), . . . , T f(v p)). Furthermore f ?(α ∧ β) = f ?(α) ∧ f ?(β) where we wedge forms defined point- wise.

31 15 Oct 22

15.1 Integration on Manifolds

k R Theorem 15.1. f : U → V where U, V open in R . Then gdx1 . . . dxk = R V U g ◦ f| det(df)|dy1 . . . dyk. Note that if f is orientation preserving then R R ? we don’t need the absolute value. In particular V ω = U f ω, where ω = ? Pk ∂fi gdx1dx2 . . . dxk and f ω = g ◦ fdf1 . . . dfk, where dfi = dyj and so j=1 ∂yj k Pk ∂fi the above value is g ◦ f ∧ ( dyj) = g ◦ f(det(df))dy1 ∧ · · · ∧ dyk. i=1 j=1 ∂yj Def. X is a k dimensional manifold and ω a k-form. We say that ω is compactly-supported if {x ∈ X : ω(x) 6= 0} is compact in X.

R R R ? Def. We define X ω = V ω = U ϕ ω, where the support of ω is contained in an open set V and ϕ : U → ϕ is orientation preserving chart. We can see ? ? that it is well defined by using a change of variables from ϕ2ω to ϕ1ω S We could also define it using a partition of unity. If supp(ω) ⊆ i Ui. If we pick a partition of unity {ρi} where each ρi has support contained in Ui, then R P∞ R we have X w = j=1 X ρiω.

For submanifolds Z and ω is an `-form and supp(ω) ∩ Z is compact in Z,, R R ? we can define Z ω = Z i ω. 3 Ex. ω = f1dx1 + f2dx2 + f3dx3 and γ : [0, 1] → R , x → (γ1(x), γ2(x), γ3(x)) and C = γ(i) then

Z Z Z 1 3 X dγi ω = γ?ω = f ◦ γ dx i i dx i C [0,1] 0 i=1 i

Ex. F : [0, 1]×[0, 1] → R where K = [0, 1]×[0, 1] and S = {(x1, x2,F (x1, x2))}. Then ω = f1dx1 ∧ dx2 + f2dx2 ∧ dx3 + f3dx3 ∧ dx1. Then Z Z ω = G?ω S K ? ∂F and we can calculate G ω = f1 ◦ Gdx1 ∧ dx2 + f2 ◦ G(− )dx1dx2 + (f3 ◦ ∂x1 * ∂F n * G)(− )dx1dx2 = f ◦ G, * |n|dx1dx2, where the form is dA or the area ∂x2 | n | form.

32 15.2 Exterior Derivative Theorem 15.2. There is a unique map d from p-forms to p + 1 forms s.t.

1. d(w1 + w2) = dw1 + dw2 2. d(ω ∧ θ) = dωθ + (−1)pω ∧ dθ

3. d(dω) = 0.

PN ∂f 4. df = dxi when f is a 0-form. i=1 ∂xi 5. g : X → Y smooth then g? ◦ d = d ◦ g?. P P In particular we define ω = i fidxi =⇒ dω = i dfi ∧ dxi where ω is a k-forms and we sum over I ⊆ [n] where |I| = k.

∂f1 ∂f2 ∂f2 ∂f3 Def. We define the curl as curl(f) = [− + ]dx1∧dx2+[− + ]dx2∧ ∂x2 ∂x1 ∂x3 ∂x2 ∂f1 ∂f3 dx3 + [− + ]dx1 ∧ dx3. ∂x3 ∂x1 Def. We can also define smooth map on manifolds.

∂f ∂f ∂f Ex. 0-form df = dx1 + dx2 + dx3. 1-form given by ω = f1dx1 + ∂x1 ∂x2 ∂x3 * f2dx2 +f3dx3 has dω = curl(F ). The 2-form ω = f1dx2 ∧dx3 +f2dx3 ∧dx1 + ∂f1 ∂f2 ∂f3 f3dx1 ∧ dx2 is given by dω = ( + + )dx1dx2dx3 is given by div(f). ∂x1 ∂x2 ∂x3

16 Oct 24

16.1 Stoke’s Theorem Theorem 16.1. IF X is k-dim and ω is a (k − 1)-form, then Z Z dω = ω X ∂X Proof. If supp(ω) ∩ ∂Hk = ∅ then both sides are 0.

k Pk i−1 If supp(ω) ∩ ∂H 6= ∅ where ω = i=1(−1) fidx1 ∧ ... dxci ∧ · · · ∧ dxk then

33 Z Z k ! X ∂fi dω = dx1 . . . dxk k ∂x X H i=1 i Z ∂fk = dx1 . . . dxk Hk ∂xk Z = − f(x1, x2, . . . , xk−1, 0)dx1 . . . dxk−1 Rk−1

Define ∂X = {(x1, . . . , xk−1, 0) : x ∈ R}. Given a unit normal to ∂X −ek = (0,..., 0, −1) then

Z Z k K X i−1 ω = (−1) (−1) fidx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxk ∂X ∂X i=1 For i 6= k, we note that Z i−1 (−1) fidx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxk = 0 Rk Since the k-coordinate is always 0. So we see that

Z Z k k−1 ω = (−1) (−1) fkdx1 . . . dxk ∂X ∂X Z Z = fK dx1 . . . dxk−1 = xdω ∂X and we can use a partition of unity in the general case. Ex (Fundamental Theorem of Calculus). f is a 0-form on X = [a, b] and R R ∂X = {a, b}. Then X df = ∂X f = f(b) − f(a). 2 Ex (Green’s Theorem). X = S ⊆ R and ∂X = C. If ω = f1dx1 + f2dx2 and so we ahve that

Z Z Z   Z ∂f2 ∂f1 dω = ω =⇒ − dx1dx2 = f1dx1 + f2dx2 S C S ∂x1 ∂x2 C 3 Ex. X ⊆ and ω = f3dx − 1 ∧ dx2 + f1dx2 ∧ dx3 + f2dx3 ∧ dx1 and say R * ∂X = S. Then dω = curl(|F ) · udx1dx2dx3 then

Z * Z curl(|F ) · udx1dx2dx3 = ω X S

34 16.2 De Rham Cohomology Def. X smoooth manifold then we have exterior derivative from k-forms to k + 1-forms. A k-form ω is closed if dω = 0 and is exact if ω = dθ for some k − 1 form θ. The k-thm dim De Rham cohomology group Hk(X) = {closed k forms}/{exact k-forms}. It has a ring structure ???. Since ω1 + dθ1 + ω2 + dθ2 = (ω1 + ω2) + (dθ1 + dθ2) and ω1, ω2 → ω1 ∧ ω2

−y x Ex. ω = x2+y2 dx = x2+y2 dy is closed but not exact. Def. f : X → Y given by f # : Hk(Y ) → Hk(X) given by f #ω = f ?ω then f ? ◦ d = d ◦ f ?.

R ? R Theorem 16.2. f : X → Y then X f ω = deg(f) Y ω. m m−1 Ex. f(z) = z + am−1z + ··· + a1z + a0 given by f : C → C. If arg(z) = θ + 2πn where z = reiθ, then d arg is a well defined 1-form on C \{0}. we f define p = |f| Z Z p?d arg = deg(f) d arg = 2π deg(p) C S1 which gives us the zeros inside our region Ω.

Proof.

17 Oct 29

17.1 Operations on Forms Remark. We have ∧ which takes α ∧ β and dα the exterior derivative. We also have f ?α where f : M → N. We note that, in fact we have some commutativity and distributivity guarantees, f ?d = df ?, f ?(α∧β) = f ?α∧f ?β and d(α ∧ β) = dα ∧ β + (−1)deg d deg αα ∧ dβ.

17.2 Nullity and Rank Thm Theorem 17.1 (Rank Nullity). If T : V → W is linear and V,W are finite I 0 dimensional vector spaces, then there are bases of v, w s.t. T = k . 0 0 Note that the rank is the number of 1’s the the nullity is the number of 0’s in the bottom left corner, and we recover that dim V is the height which is rank + nullity.

35 T1 Td−1 Theorem 17.2 (Gabriel’s Thm). For type A quivers ie let V1 −→ V2 ... −−−→ Ve be a sequence of finite dimensional vector spaces and maps. Then there 0 0 0 exists bases of all the Vi s.t. each Ti = 0 Ik 0. This means that 0 0 0 basis elements are mapped into other basis elements until they go into the kernel.

T1 Te−1 Remark. What about complexes? Complexes are V1 −→ ... −−→ Ve where i ? Ti+1 ◦ Ti = 0. We see that H (complex) = (ker Ti)/(im Ti−1), so H is the isolated bases vectors (not in the image and also in the kernel).

Pe i Pe i Corrollary. i=1(−1) dim(Vi) = i=1(−1) dim H . This is the Euler char- acteristic of complex/its cohomology.

17.3 Goals for De Rham Cohomology 1. We wish to prove computability of De Rham Cohomology (Mayer- Vietoris).

2. Using that we wish to show that De Rham Cohomology of compact manifolds is finite dimensional.

R ? dim M 3. If M is compact oriented we get a function : H (M)  H (M) → R M R which gives us a pairing hα, βi → M α ∧ β. Our goal is to show that this pairing is perfect (ie nothing kills everything else). ( i = 0, k Ex. We use H? practically. Hi(Sk) = R and Hi(Bk+1) = 0 otherwise ( i = 0 R . Brouwer asked if there exists a map Sk → Bk+1 → Sk 0 otherwise that is invertible. But if we apply the map to the cohomology then we are mappinHk(Sk) → Hk(Bk+1) that is invertible, which is from R → 0, so that is impossible.

Remark. Why is cohomology a ring? A × B  A, B using πA, πB. THen π? ? A ? ? ? H (A) −→ H (A×B) and similarly for πB which means we have a mapH (A)× 4? H?(B) → H?(A × B). If A = B then H?(A × A) −→ H?(A) given by A,→ A × A for map 4A.

36 17.4 A Tool : Partition of Unity P Def. A partition of unity on M is a countable or finite sum fi = 1 s.t. fi ≥ 0. This is a “locally finite sum” ie at any m ∈ M the number of i s.t. fi(m) 6= 0 is < ∞. Furthermore supp(fi) is compact. S Def. A partition of unity subordinate to an open cover M = Ui s.t. supp(fi) ⊆ Ui ie same indexing set or equivalently we can find i s.t. supp(fj) ⊆ Ui for all j. Theorem 17.3. There exists a partition of unity subordinate to any open cover.

? Theorem 17.4 (Homotopy Invariance of H ). If f0 : f1 : X → Y are ? ? ? homotopic then f0 = f1∗ : H (Y ) → H (X).

p ? ? Proof. On forms, we’re saying α ∈ ω (Y ) want f1 α = f0 α = dβ, β ∈ p−1 R 1 ? d Ω (x). Let β(v1, . . . , vp−1) = t=0 F α( dt , v1 . . . vp−1)dt. We will finish this next time.

18 Oct 31

18.1 Mayer-Vietoris

β Def. A short exact sequence (SES) of groups is 0 → A −→α B −→ C where ker(β) = im(α) and α is injective, β surjective.

d d Def. A complex of v space is a sequence 0 → A1 −→ A2 ... −→ An → 0 where d2 = 0.

Def. A morphism of complexes A → B is a sequence of fi that map each Ai → Bi. Theorem 18.1 (Mayer-Vietoris). A SES of complexes associated to M = U ∪ V open sets is

pullback difference of pullbacks 0 → Ωi(M) −−−−→ Ωi(U) ⊕ Ωi(V ) −−−−−−−−−−−−→ Ωi(U ∩ V ) → 0

To show Ω(U) ⊕ Ω(V ) → Ω(U ∩ V ) is onto, hits some α, pick a partition of unity subordinate to V,V where 1M = fu + fv and supp(fU ) ⊆ U and similar for fV . (αfi, −αfv) → α = α(fu) + α(fv). Remark. What do we get in H?.

37 Theorem 18.2 (Homological Algebra). A SES of complexes a long exact sequence (LES) on H?. This LES given by → HiA → HiB → HiC → Hi+1 → Hi+1B → Hi+1C. Def. The Mayer-Vietoris LES is 0 → H0(M) → H0(U) ⊕ H0(V ) → H0(U ∩ V ) → H1(M) → H1(U) ⊕ H1(V ) → H1(U ∩ V ) → ... . Eventually we get to Hdim M (U ∩ V ) → 0. Ex. S1 = U ∪ V where U is (cos t, sin t) from t = π/4 → 7π/4 and V is from 5π/4 → 3π/4. The MV LES is 0 → H0(M) → H0(U) ⊕ H0(V ) → H0(U ∩ V ) → H1(M) → 0 → 0 → 0. H0(M) is 0-dimensional forms on M (connected components) so this sequence is 0 → R → R ⊕ R → R2 →? → 0 → 0 → 0. Ex. Sn = (Sn \ north) ∪ (Sn \ south). Note that their intersection, the fat hemisphere, is homeomorphic to Sn−1. If we examine the sequence 0 → 0 n 0 1 0 2 0 1 n H (S ) → H (D ) ⊕ H (D ) → H (D1 ∩ D2) → H (S ) → ... . Note that i 1 i 2 i−1 ∼ i n H (D ) ⊕ H (D ) = 0. This means that H (D1 ∩ D2) −→ H (S ) since the surrounding parts are 0 so it must be isomorphic. We can see this because all three parts are similar to Sn−1 and we know the ranks of the Cohomology groups of Sn−1. 2 Ex. RP (klein bottle). Let U be an open set in the center of the rectangle, V is the rectangle minus a small open set in the middle. Their intersection 1 is a ring. They are isomorphic to a point, RP and S1 respectively.

0 0 2 1 0 0 1 0 0 The sequence is 0 −→ H (RP ) −→ H (D1) ⊕ H (D2) −→ H (D1 ∩ D2) −→ 1 2 a 1 1 1−a 1 a H (RP ) −→ H (D1) ⊕ H (D2) −−→ H (D1 ∩ D2) −→ ... , where we use the fact that H1({p}) = 0 and examine the ranks (which are 0 → 1 → 1 + 1 → 1 → a → 0 + 1 → 1). So we need to examine one of these maps to get the other values. In particualr, D1 ∩ D2 ,→ D2 is the boundary of the mobius 1 n xdy−ydx 2 strip to the mobius strip. This is dθ ∈ Ω (S ), x2+y2 in R \ 0. In fact as 1 1 we map from H (D2) → H (D1 ∩ D2) is by θ → 2θ, so the rank of this map 1 2 is 1. We see that H1(RP ) = H2(RP ) = 0.

i i f Proof. take an α ∈ A and note that A −→ Bi by α → f(α). Note that if we take an ω ∈ Ai−1 then α + dω then it’s equal to f(a) + dω0 and we are modding out by dω so we are good. This define HiA → HiB.

Another thing is that we can eventually find a γ ∈ Ci s.t. γ → α. Then we define a βinBi. Honestly, this proof is very long and a lot of diagram chasing. The basic learning moments from this theorem is that it’s mostly just diagram chasing. Find a youtube “it’s my turn” snake lemma.

38 19 Nov 5

19.1 Geodesics S Def. A good cover {Ui} of M is a cover, = M s.t. for all finite T i∈I subsets J j∈J Uj is contractible (homotopic to a point) or empty. Remark. If M is compact and has a good cover, it has a finite good subcover. S Remark. Note that the set of J ⊆ I s.t. J uj 6= ∅ is closed under shrinkage. So this defines a simplical complex.

Def. A geodesic on M is a map ϕ :(a, b) → M,→ Rn that is as straight as possible over stretches in (a, b). This means that it is infimum and this is a min in this case.

Def. An open set U ⊆ M is geodesically convex if ∀u1, u2 ∈ U, there is a unique geodesic in M [0, 1] → [u1, u2] and that geodesic is in U. Remark.

1. Geodesically convex subsets are contractible.

2. A finite intersection of geodesically convex sets is geodesically convex.

Theorem 19.1. Around any m ∈ M submanifold of Rn, there is a geodesi- cally convex nbhd. * Proof. The geodesic spray is defined as TmM → M takes each v to a flow * time 1 along geodesic through m with derivative v . This maps an open subset of TmM to an open subset of M and this is our nbhd.dfc

19.2 More on forms and integration Remark. If M compact oriented manifold, H?(M) finite dimensional, we R dim M dim M have M : ω (m) ,→ H (M) → R. This gives us h·, ·i by hα, βi = R p dim M−p M α ∪ β where α ∈ H (M) and β ∈ H (M). Our goal is to show that this is a perfect pairing.

Remark. Distributions on Forms are called Currents. The cohomology of the currents is the same as the cohomology of manifolds.

p R Def. Ωc (M) is the compactly supported p-forms on M. Can define M : d im M p p+1 Ωc (M) → R. We also see that d :Ωc (M) → Ωc (M). We define ? Hc (M) as compactly supported cohomology.

39 ? 0 d 1 Ex. Hc (R). We see that 0 → Ωc (R) −→ Ωc (R) → 0. The kernel of the first d is constant functions. Since they are compact it follows that the only kernel 0 R t 1 elements is {0} so Hc (R) = 0. We examine t → −∞ α for α ∈ Ωc (R). R t We see that if we have some bump function the t → −∞ α is not finitely supported since for t → ∞ the value is constant. But with some work we can 1 ∼ R ∞ see that Hc (R) = R since any β we just define β − −∞ /cαc = df.

R p dim M−p dim M M Remark. Our new goal is to show that H (M)×Hc (M) → Hc (M) −−→ R is a perfect pairing for oriented M. Remark. Compactly supported cohomology is not homotopic invariant.

Remark. Ω: manifolds → supercommutative graded algebras given by f : ? M → N =⇒ f : Ω(N) → Ω(M). What about Ωc? It is covariant for open inclusions Ωp(u) 3 α → extend by 0 off of U ∈ Ωp(M). This only works because α is of compact support in U.

? 19.3 Mayer-Vietoris for Hc · Theorem 19.2. M = U ∪ V open. Then the sequence is Ωc(U ∩ V ) → · · difference · Ωc(U) ⊕ Ωc(V ) −−−−−−→ Ωc(M) → 0. We can see this is exact.

? 1 Ex. M − V for Hc (S ) where U and V are large arcs.

U ∩ VU ⊕ VM 0 Hc 0 0 ⊕ 0 ? 1 2 Hc R R ⊕ R ? 2 Hc 0 0 0

1 1 1 we can find a map from Hc (U ∩ V ) → Hc (U) ⊕ Hc (V ) that is of rank 1. Therefore, we have that both ? are of dimension 1 and we can see that they are the same. R ? n N p Remark. Still need base case Hc (R ). Covariance M ×N −→ M by Ωc (M × p−dim N N) → Ωc (M).

p ∼ p π Warmup H (M × R) = H (M). This is true because M × R −→ and η m −→ (m, 0) is the reverse. π? ◦ η? :Ω·(M × R) → Ω·(M × R) want 1 − π? ◦ η? to be d on closed forms. Need “homotopy operator” of degree −1.

40 20 Nov 7

20.1 Integration and Homotopy Remark. Our goal is to show that if M is oriented and non compact then R i dim M−i dim M H (M) × Hc (M) → Hc (M) −→ R is a perfect pairing.

A warmup is given an operator η : M → M ×R with inverse π : M timesR → M

Ωi(M) Ωi+1(M)

Ωi(M × R) Ωi+1(M × R)

The trick is that there exists a K “Homotopy Operator” s.t. 1 − η?π? = q−1 R t d H (dK − Kd) on q-forms. We define K(α)(x, y) := s=0 α( dt ,... ) where α is a q-form, x ∈ M, t ∈ R. i+1 ∼ i Lemma. Hc (M × R) = Hc(M). R i+1 i ? Proof. We note that Ωc (M × R) −→ Ωc(M) and with reverse α → (π α) ∧ (fixed bump form on R). We want a homotopy operator K s.t. 1 − (e ∧ π?) ◦ R = (−1)q−1(dK − Kd). We set K(α)(x, y) = R t α( d ,... ) − R −∞ dt R t R ∞ d ? ? q−1 −∞ e −∞ α( dt ,... ). K satisfies the equation 1−π ◦η = (−1) (dK −Kd) and so is a homotopy. ( i n R i = n Corrollary (Poincar´eDuality). Hc(R ) = . Recall that the orig- 0 i 6= n R i dim M−i ◦∧ inal statement was H (M) × Hc (M) −−→ R is perfect. Note that htis only occurs for M oriented with finite good cover.

20.2 M-V in H? Remark. Suppose Hk = U ∪ V and recall the long M-V sequence · · · → Hi(M) → Hi(U) ⊕ Hi(V ) → Hi(U ∩ V ) −→∂ Hi+1(M) → ... . We note k−i ? k−i ? k−i ? k−i ? that by duality that Hc (M) → Hc (U) ⊕ Hc (V ) → Hc (U ∩ V ) → k−i−1 ? Hc (M) . We also assume that this manifold has finite good cover and is oriented. In particular we have isomorphisms

41 Hi(M) Hi(U) ⊕ Hi(V ) Hi(U ∩ V ) ∂ Hi+1(M)

∼

k−i ? k−1 ? k−i ? k−i ? k−i−1 ? Hc (M) Hc (U) Hc (V ) Hc (U ∩ V ) Hc (M) and to show that all are isomorphism we just apply the Five Lemma.

Lemma (The Five Lemma). If we have a exact sequence morphism

A B C D E ∼ ∼ A0 B0 C0 D0 E0

Then C → C0 is also an isomorphism.

dim M ∼ Corrollary. If M is compact oriented connected, then Hc = R.

20.3 Kunneth

Def. R is a commutative ring with M,N modules. Then M ⊗R N is the free module on (m, n) quotient by (m1 + m2, n) = (m1, n) + (m2, n) and (rm, n) = r(m, n) = (m, rn). ∼ P Ex. Z6 ⊗Z Z4 = Z/2. Notice that the free module is sums finite rab(a, b) but since we quoteitenting out by stuff like 6(1, 1) = (6, 1) = (0, 1) = 0(1, 1).

∼ a b ∼ ab R⊗R = R and in general R ⊗R R = R Remark. Bilinear (M × N, ·) set of bilinear maps. We see that R-Modules −−−−−→Covariant Set by A → Bilinear(M × N → A) and if we compose both with ϕ then we map B → Bilinear(M × N → B) which gives us the functor aspect. In general the basic functors are Hom(T, ·) that maps A → HomR(T,A) and these are representable functors. The Tensor product gives us this information.

Theorem 20.1 (Kunneth). Let M,N be manifolds where M has a finite good cover. Then the natural map H?(M) × H?(N) → H?(M × N) by P P ? ? i αI ⊗ βi → i(π αi) ∨ (π αi). Proof. Involved (Refer to book).

42 21 Nov 12

21.1 Revisiting Vector Bundles Def. A vector bundle V on M is a map from M → V with scalar multi- plication and addition and a zero section satisfying linear axioms.

fA fB Def. The fiber product A ×M B of A −→ M,B −→ M is {(a, b) ∈ A × B : fA(a) = fB(b)}. Theorem 21.1 (Tubular Neighborhood). If m ,→ 0 then there exists U ⊇ ∼ M, U ⊆ 0 open with U = NM 0 the normal bundle. Def. An oriented vector bundle has an orientation on each fiber contin- uously.

Def. A Thom form on an oriented vector bundle V n+k −→π M n is a closed form β on V s.t. for all m ∈ M β |π−1(m) is a compactly supported bump form with integral equal to 1.

2 Ex. Take V = R ←- {(x0, ∗)} = {x0} × R where β = f(x, y)dx + g(x, y)dy. Then we see that on {x0} × R then f(x, y)dx vanishes so we only know that g(x, y)dy is a bump form for all x.

? ∼ ?+k Theorem 21.2 (Thom). Hc (M) = Hc (V ) where V is an oriented k-plane bundle over M. This is dual to H?(M) = H?(V ) if M is compact oriented.

Proof. The first part of this proof is that there exists a Thom form β on V constructed normally when M = Rn once and for all, and glued together using a partition of unity, subordinate to an open cover {U} on which V is ∼ k P a trivial vector bundle. V |U = U × R . Take the forms βU and β = fU βU .

For the second part, we take a map from Ω· (M) → Ω·+k(V ) by α → (π?α)∧β. R c c We can construct the inverse by fiber. Therefore there exists a homotopy operator that makes each forward and backwards map homotopic to the identity.

−1 −1 On V we can define a weird complex of forms ϕ each ϕ/π (m) ∈ Ωc(π (m)) · ·+k of “compactly supported in vertical directions” and we have Ω (M) → Ωvertical(v) by α → π?(α) ∧ β.[m] is the Poincar´eDual of M ⊆ O.

Ex.

43 * 1. M,→ V where m → (m, 0) an oriented vector bundle. We get the k −1 k Thom form β ∈ Ωvertical(V ) where σ (β) ∈ Ω (M) closed. Then [σ?(β)] = σ?[β] is the Euler class of V and is independent of β. R 2. M,→ O. M os compact and both are oriented. We get that M : dim M ∼ H (O) → R. We have the sequence M,→ NM O = U,→ O where U is the tubular nbhd of M. Thom form, extended by 0 represents [m]. We claim that

Z Z Z ? X η (β Thom form on NM O) ∧ α = α|M = (α + fW )|W O M W

Note that we find this since this is equal to

Z Z X Z β ∧ α = β ∧ α = β ∧ (αfw) −1 ∼ k U NM O W π (W )=W ×R and we just apply fubini to get our desired since β becomes nothing. We use a partition of unity M subordinate to an atlas on which NM O is trivial, to replace α by α · fw. − Theorem 21.3. If M1,M2 ,→ O all compact oriented and M1 t M2 then [M1][M2] = [M1 ∩ M2]. The product is cup product (wedge product on duals).

Proof. Put a tubular nbhd around M1 ∩ M2 that is equivalent to NM1∩M2 O where NM1∩M2 Mi ⊆ NM1∩M2 O. The two NM1∩M2 Mi are transverse in NM1∩M2 O. ? ? Now fiber by fiber we’re looking for bump forms β1 β2 s.t. π1(β1) ∧ π2(β2) is a bump form on V1 × V2.

22 Nov 14

22.1 Lefschetz Numbers and Cohomology Remark. Recall that given f : M k → M compact oriented then the Lefschetz R number if L(f) = I(graph(f), 4) = M×M [graph(f)] ∨ [M4]. Theorem 22.1. Ha(M) ⊗ Hb(M) kunneth Ha+b(M × M) R not hom M4 trace end(H?(M)) R

44 ? ? R ? Def. The β-transform Φβ : H (M) → H (N) by α → M (πM (α) ∨ β) where β ∈ H?(M × N) and both M and N are compact oriented. * R * ih v ,wi Ex. Recall the Fourier transform . We can write it as fb(w) = V f(v )e and this is analogous to the β-transform. Note that as opposed to our normal requirement of analysis, since we make things compact then this is well defined always. Remark. Our goal is that if f : M → M which is compact oriented then Pdim M i ? i i L(f) = i=0 (−1) tr(f : H (M) → H (M)). The corollary is that χ(M) = P(−1)i dim Hi(M). Lemma. Given f : N → M we have that f ? : H?(M) → H?(N). We claim ? that f = Φ[graph f]T ⊆M×N . Proof. Z Z Z ? T Φ[(graph f)T ](α)β β πM (α) ∨ [(graph f) ] N N M Z Z Z = (α ⊗ β)[(graph f)T ] = α ⊗ β = f ?(α)β M×N (graph f)T N

? R Def. The Lefschetz formula for β ∈ H (M × M) is given by M×M β ∨ R ? Pdim M i [4(M)] = M 4 β where 4 : M → M×M. We define this as i=0 (−1) tr(Φβ : Hi(M) → Hi(M)). Both sides are linear in β. Enough to check for β = j αi ⊗ α the basis and dual basis elements. j R R j Ex. If we take β = αI ⊗ α then we get that M×M β ∨ [4(M)] = M αiα = [i = j]. However, for the right hand side of our above equation, we have that deg(α`) ` R ` j j P j this is α` (−1) Φαi⊗α (α ), α` = M (α ⊗ ·(αo ⊗ α )) = [i = `]α = [i = `][j = `] = (−1)deg(αi)[i = j]. We didn’t find where the −1 comes from or how to remove it. Remark. In our definition of L(f) and the Euler characteristic, we note that the supertrace we define does not need that M is compact oriented. Does this mean we can define it for non compact manifolds? How much do the Betti Numbers dim Hi affect our actual understanding?

23 Nov 19

23.1 More Lefschetz Remark. Recall that if M is compact oriented and f : M → M is Lefschetz (isolated fixed points enough) that we have

45 X X local Lefschetz number = (−1)itr(f ∗ : Hi(M) → Hi(M)) m:f(m)=m i

n Ex. We showcase an example on CP . We can calculate to get that the cohomologies are of dimension 1. In particular we get that R[h(2)]/ hhn+1i = R ⊕ Rh ⊕ · · · ⊕ Rhn.

23.2 Lie Groups Def. A Lie group G is a group that is the manifold.

Ex. GL(n, R) is an example of a Lie group. This is because the determinant of AB is the composition of the determinant of A and B. This is the general linear group.

Remark. If G is a closed subset of GL(n, R), then G is a Lie group. If G is connected and H ≤ G is discrete and normal, it is central if it commutes with all of G ie ghg−1 = h for all g and h ∈ H.

Ex. SL(n, R) special linear group, GL(n, C) ⊆ GL(2n, R) are good examples. So are O(n, R) the orthogonal matrices and SO(n, R) the special orthogonal ones. Another good example is GL(n, H) ⊆ GL(4n, R) and well as the unitary T group Un = {M ∈ GL(n, C): MM = I}. Remark. If G is a topological group that is compact and simple, then it is Lie.

Theorem 23.1. Let G0 be a connected component containing I ∈ G (an example of is O(n, R)0 = SO(n, R)). Then G0 is normal in G.

Proof. If h ∈ G0 there is a path γ : [0, 1] → G0 with 0 → I and 1 → h. If we −1 −1 look at t → g γ(t)g then 0 → I and 1 → g hg ∈ G0 by our definitions. Def. A Lie group action G : M → M where M is a manifold is a map G × M −→α M smooth (αM = M) s.t.

G × G × M Iα×α G × M

·G×Im α G × M α M

Remark. We want that when G acts on M that M/G is again a manifold. In particular we need the action to be free and proper.

46 Ex. Good examples of cases where if we don’t have free and proper that don’t 2 work are√ if R acts on T = {(w, z) ∈ C : |w| = |z| = 1} by s · (w, z) = (eisw, eis 2) is free but not proper. In particular it is a ray on the torus (in the square identification) and is dense but is not a manifold. This is because it is free but not proper.

If R× acts on R2 \ 0 by z · (x, y) = (zx, z−1y). (R2 \ 0)/R× is R with two origins, so this is not a manifold since it is not proper (it is free).

Def. An action is free if for all g 6= 1 then g · m 6= m for all m. In particular this means that the stabilizer of each m is IG. An action is proper if (g, m) → (m, g·m) is proper (X compact implies preimage is also compact).

Theorem 23.2. If G acts on M properly and freely then M/G is a manifold.

24 Nov 21

24.1 More Lie Groups Remark. Let G, H be Lie Groups (groups and manifolds). Examine ϕ : G → H. Then we note that T ϕ : TgG → Tϕ(g)H is the tangent map. For the tangent space TgG, we note that this is isomorphic to T1G by map g· : G → G (we use the tangent space map). As a corollary this parallelizes the tangent bundle to G and that S2n is not a group by hairy ball theorem. Let T1G = Lie(G).

Remark. If G = R and ϕ : R → H, then ϕ is determined by T1ϕ : R → Lie(H) or by T1ϕ(1) ∈ Lie(H). The H is the one parameter subgroup and T1ϕ(1) is the infinitesimal generator.

Given x ∈ Lie(H) there exists a unique ϕ : R → H by T1ϕ(1) = x. If G is connected and ϕ : G → H is a smooth homomorphism, then ϕ is uniquely determined by T1ϕ : Lie(G) → Lie(H). We define the lie exponential map exp : Lie(G) → G by x → ϕx(1).

We claim that any open U ⊆ G with 1 ∈ U 6= ∅,U = U −1 generates G. To do so we note that H = S U i and so any open subgroup in G is also closed, i∈N where U i are the cosets.

Remark. If ρ : Lie(G) → Lie(H) respects the Lie bracket and G is connected and simply connected then there exists a unique ϕ : G → H given by T1ϕ = ρ.

47 Ex. We note that if we have lie group homomorphism S1 → R then we note that since S1 is compact it’s image is compact. But the only compact subgroup of R is {0}. Similarly, when we map S1 → S1 we see that the map must be of the form z → zn for some integer n.

Many of the examples come from quantum mechanics. In particular, we note that for ϕ : SO(3) → U(H) then does this map come from so(3) → u(H) the Lie Algebras. We note that SO(3) is not simply connected so this only occurs half the time.

24.2 Some Representation Theory

n Def. If ϕ1, ϕ2 : G → GL(C ) then there exists M ∈ GL(C) s.t. ϕ1(g) = −1 Mϕ2(G)M iff for all g ∈ G Trϕ1(g) = Trϕ2(G) where this is the charac- ter of ϕ1. Now we examine two seemingly contradictory theorems. Theorem 24.1. If G is finite and H ≤ G and H 6= G then H misses some conjugation class of G ie there exists some g s.t. for all k kgk−1 ∈/ H.

Theorem 24.2. Unitary matrices are diagonalizable.

24.3 Quotient Manifold Theorem Theorem 24.3 (Quotient Manifold Thm). If G is a Lie group and G acts on M freely and properly ie for all m, g · m = m =⇒ g = 1 and (g, m) → (m, gm) is proper, then

1. Each orbit is a closed submanifold

2. M/G := M/ ∼ is a manifold s.t. M → M/G is a submersion.

/G Def. A principal bundle G when E −→ M is a

1. a space E with free proper G-action.

2. diffeomorphism ie E/G ∼= M. In particular each fiber is diffeomorphic to G as a G-space.

Ex. S1 = {eiθ} ⊆ C. ϕ : S1 → S1, z → zn then we note that Z/n acts on S1 2πi k 1 3 2 2 2 by k · z = (e n ) z. If we take S → S = {(z, w) ∈ C : |z| + |w| = 1} then we can construct the action eiθ · (z, w) = (eiθz, eiθw) satisfies our conditions. In this case G → G × M → M is a trivial principal G-bundle.

48 1 Ex. What are the Z/2-bundles on S ? We see that it is either ϕ1 or ϕ2 the trivial and nontrivial bundles.

k * * * Ex. M then the Frame(M) = {(m, v 1,..., v m):(v i) basis of TmM} ⊆ Qk i=1 TM. We can act on this by GL(R) by replacing our list of vectors, which makes the frame a principal GLk(R) bundle. Def. Let G → E → M be a principal G-bundle. Let G act on X smoothly. The associated X-bundle (X ×E)/Gdiagonalization is a nonprincipal bundle.

1 1 * Ex. Z/2 → S = E → S = M via ϕ2. Let Z2 act on R by I ie k · v = k* 1 1 (−1) v . What is the associated R bundle?. Well it is R → (R×S )/Z2 → S . Simply replace the Z2 we had with reals and this forms the Mobius band.

k k k Ex. M by GLk(R) acts on R then the associated bundle is given by (R × Frame(M))/GKk(R), which is just the tangent bundle. What is the map?

p k ? Ex. Examine GLk(R) acting on (Alt R ) and sections of associated bundle are Ωp(M).

−1 Ex. GLk(R) acts on R by M → | det M | are called top forms are better versions of

25 Nov 26

25.1 Culture Def. K0(M) is defined as the finite differences of isomorphism classes of C-vector bundles on M.

0 ∼ ? Theorem 25.1 (Chern Character). R ⊗Z K (M) −→ H (M).

n 25.2 H?(CP ) n Remark. Recall that CP = {z ∈ Cn, z 6= 0}/C× = {z : |z| = 1}/U(1) where n ∼ U(1) = {[eiθ]}. We see that CP −→ Hermitian matrices M with eigenvalues = 1, 0,..., 0 by z → z∗z where z∗ is the conjugate transpose.

n We define CP = {[z0, . . . , zn]: zn = 1}∪{[z0, . . . , zn], z0, . . . , zn−1 not all 0}. n n−1 n ∼ 2n−1 We see that U = C and V = CP with intersection C \{0} = S .

We construct the Mayer Vietoris sequence. Note that near the end it becomes

49 n>1 CP U q VU ∩ V H2n−3 ? 0 0 0 H2n−2 ? 0 1 0 H2n−1 ? 0 0 1 H2n ? 0 0 0 H2n+1 0 0 0 0

n We can fill up the above. To get that H∗(CP ) = R[h(2)]/ hhn+1i

n Taking h → the Poincar´edual of hyperplane gives us R[h] → H∗(CP ). Obviously hn+1 → 0 if hn → 0 (indeed hn → [hn hyperplanes] = [pt] 6= 0) n then each p(hi) 6= 0 so it generates our desired H2i(CP ). Def. There is something called the tautological line bundle which is * * * * * n * {[v ], wαv } where w is just scalings of v . We see that CP = {[v ]}. This is as opposed to the better line bundle, which you learn in Algebraic Geometry.

25.3 More Bundle Stuff

k Remark. Given M a manifold we have a principal GLk(R)-bundle Frame(m) → m and to each GLnR : X → X, associated X-bundle X → (X×Frame(m))/GLkR → M.

Ex.

k k 1. GLkR : R → R then the associated bundle is TM.

k k ? 2. GLnR :(R ⊗ · · · ⊗ R ) where these are antisymmetric tensors and p compositions. The associated bundle is the p-form bundle.

3. M → [1] is the trivial bundle.

−1 4. p = k by GLkR : R → R by M → [(det M) ]. This is the k-form bundle or the bundle of volume forms.

−1 5. GLkR : R → R where M → [|(det M) |] is the bundle of densities. These are like volume forms but better (since we can integrate without orientation as done previously).

Remark. Given ϕ : M −→∼ N we get that Frame(M) −→∼ Frame(N) and ∼ p k ? AssX (M) −→ AssX (N). Can define twisted p-forms GLkR :(Alt R ) ⊗ R with sign(det) representation.

Ex.

50 2 k ? k k ? 1. GLkR :(Sym R ) by (R )⊗(R ) . What does a section g ∈ Γ(Associated Bundle) * * do? gm(v1, v2) ∈ R. This is indefinite degenerate Riemannian Metrics.

2 k ∼ 1. GLk(R): Sym R = Symmetric k x k matrices ←- positive definite ma- trices with eigenvalue > 0 by A · M = AMAT .

Remark. What is the use of a Riemannian metric g? Given a Morse func- tion f : M → R get df ∈ Γ(T ?M) now use g : TM → T ?M and use g positive definite (nondegenerate) to get g−1 : T ?M → TM. Note that posi- tive definite is better than nondegenerate because we can go to submanifolds using positive definite but nondegenerate.

df g−1 M −→ T ?M −−→ TM. This g−1 ◦ f is the gradient of f, ie ∇f.

k −1 k Ex. GLkR : End(R ) by A·M = AMA . A submanifold is {M ∈ End(R ): M 2 = −1}. Note that in this submanifold the eigenvalues of M are ±i in equal numbers (only occurs when 2 | k). We have k0 = k/2. Let J be a section of this associated bundle. M has an action J which is rotation by 90 degrees on the tangent space. This makes each real tangent space into a C-vector space. This is an almost-complex structure (J the rotation by 90 degrees). The question is that given a manifold could we give it an almost complex structure? A fundamental result is that we can give a Riemannian structure on all manifolds, but what about almost complex? Well we need that 2 | k.

Theorem 25.2. Given an almost-C structure J on M ie ∀m ∈ M,J : 2 TmM → TmM where J = −1 varies smoothly can use to construct an orien- ∼ 2k0 0 tation on M. We see that TmM = R and we pick k vectors v1, . . . , vk0 , Jv1 . . . , Jvk0 has orientation 1. We see that this is actually general with respect to how we pick our vi since the set is connected.

26 Dec 3

26.1 Various Geometries Def. An almost complex structures J on a manifold M is a smooth 2 section of {j ∈ TmM, j = −I}. Note that this is a bundle over M (not a vector bundle). Remark. We claim that a complex vector space has a natural orientation. * * To construct a complex vector-space V . Pick a complex basis v1,..., vk and * * * * demand that the orientation of O(v1, iv1,..., vk, ivk) = 1. The basis is unique

51 up to GL(k, C), but unlike GL(k, R) this group is connected so the bases are continuously related.f

−1 Remark. Recall that the gradient of f is g (df) given metric g : TmM → ? 2 ? TmM. Given a 2-form ω ∈ Ω (M) can also define ω : TM → T M which is not a metric because antisymmetric. Def. An almost symplectic structure on M is a nondegenerate 2-form ie ? each TmM → TmM invertible (which implies that the 2 | dim M). Symplectic Sylvesters law says all symplectic functions on TmM are equivalent.

Def. The symplectic gradient/Hamiltonian vector field of f : M → R iw ω−1(df) where ω : TM → T ?M. Ex. Some basic geometries are

1. Rn with usual metric g

2. Cn with usual almost complex form J

2n 3. R with usual ω given by dx1 ∧ dxn+1 + dx2 ∧ dxn+2 . . . dxn ∧ dx2n If M is locally ∼= to one of these, then we can describe it. If it is locally isomorphic to the Euclidean, then if complete (ie bounded vector fields have ∼ n flows, which is implied by compactness), then M = R /Γ where Γ is a discrete group acting freely.

If locally isomorphic to symplectic, then by Moser M, ω is a nondegenerate 2 form ie dω = 0. An example is T ?F over some field (?). Remark. Note that even if M is not compact then we can actually put a complete Riemannian metric on M. Ex. If M is compact oriented and Riemannian, we can use g to put a positive definite inner product on Ωp(M) = Γ(M; AltpT ?M) which we can use to define d? :Ωp+1(M) → Ωp(M). Compact oriented comes in because

p p ? ? Def. Ω (M)harmonic := {α ∈ Ω (M):(dd + d d)α = 0}. Theorem 26.1 (Hodge). We have that

1 0 2 0 ... Ω (M)harmonic Ω (M)harmonic ...

... Ω1(M) Ω2(M) ...

52 and this gives an isomorphism on H? ie every c ∈ Hp(M) has a unique harmonic representative. This is incredibly difficult.

D* *E D* *E Def. On a Hermitian vector space V , we have v , w = w, v and * * * * D* *E * sv , tw = st v , w . Note that mathematicians normally put s t instead. * * * * D* *E * * We take <(v , w) = g(v , w) and =( v , w ) = ω(v , w) and so this Hermitian inner product gives us the Riemannian and Symplectic structures. They are * * * * related by g(v , Jw) = ω(v , w).

n ∼ n+1 Ex. CP = U ·e11 where · is action by conjugation. The complex structure * * *T * is given by Cv → v · v /|v |2. So every tangent space is Hermitian. Def. A Kahler manifold is complex with choice of Hermitian structure on TM.

Theorem 26.2 (Hard Lefschetz). If M is compact and Kahler ie has ω ∈ Ω2(M), then

∧[ω]i 1. The map Hn−i(M) −−−→ Hn+i(M) is an isomorphism. Note that these two vector spaces are isomorphic since M is compact and Kahler implies oriented and we just use Poincar´eduality.

2. [∧ω, (∧ω)?] = (p − n)1 on Hp.

This means that the Betti numbers form a palindrome that is increasing then decreasing (it’s a mountain) ie something like 1, 2, 3, 4, 5, 4, 3, 2, 1.

Remark. There was a push to try to prove that Symplectic manifolds are Kahler. But someone found a manifold with Betti Numbers that aren’t a mountain, which disproved the assertion.

Theorem 26.3. If A, B ,→ M are all complex (so oriented) and A − B, t− then the orientations on all are compatible. The special case is that if A t B is a set of points, then all points have positive orientation.

Ex. Let V = C3. Then {(p, `): ` 3 p, [?00]} ,→ PV × PV ?, where PV are the points and PV ? are the liens. This is just projective geometry. note that given p we can find the unique line ` if p 6= [?00], but we must include the ` in our formulation since if p = [?00] we can pick the line. This means that mostly PV,→ PV but sometimes we must take it with the dual.

53 27 Dec 5

27.1 Abstract Manifolds

Def. An abstract manifold M is a union of open subsets Ui, i ∈ I s.t. for k Vi ⊆ R there is a bijection ϕi : Vi → Ui and for all i, j ∈ I we have that k −1 ϕi ϕj −1 k R ⊇ Vi ⊇ ϕi (Ui ∩ Uj) −→ Ui ∩ Uj ←− ϕj (Ui ∩ Uj) ⊆ Vj ⊆ R .

n n
Ex. For submanifolds of R , at each m ∈ M ask hat are of the k projections π be locally invertible and smooth from Rk → Rn. * k Def. A tangent space to an abstract manifold is TmM = {v ∈ T −1 } |varphii (m)R for each Ui 3 m. This is also equal to {γ :(−1, 1) → M}/ ∼ where the equivalence relation is if the have the same derivative at 0. Concretely, this is im(T (π−1)) and shown in our example. In algebraic geometry, the Zariski definition is the dual to the quotient of the ideal of smooth functions on open sets at m vanishing at m with products of two such functions.

Def. Tangent bundle is TM is defined by M × Rk. The derivative is linear defined for f : M → N by T f : TM → TN as a map TmM −−−→ Tf(m)N. Theorem 27.1 (Inverse function thm). If f : M → N is smooth and T f : TmM → Tf(n)N is invertible then

1. ∃U ⊆ M, U 3 m s.t. f : U → f(U) is a diffeomorphism 2. or there exists charts U, U 0 around m, f(m) that are diffeomorphic to the same V ⊆ Rk.

Theorem 27.2 (Local Immersion Thm). If f : M → N, Tmf is 1 : 1 then there exist coordinates ϕ : V −→∼ U ⊆ M around m and ϕ0 : V 0 → U 0 around
f(m) s.t. the map is the canonical immersion ie something like In 0 . In particular f : U → U 0 is an immersion. Ex.

1. A bad example is R → R2 given by looking like the figure 8 (except not completed 8 where the arrows are) and note that this is bad. 2. Even when injective if we make a loop s.t. as x → ∞ it approaches the original line 3. For the irrational flow on T 2, we note that while the image is dense in T 2. It is clearly not onto although it is dense (topologically onto), and this suffers from the same problem as above (it is not proper).

54 Def. f : M → N is an embedding if proper injective immersion. This implies im(f) is a submanifold.

Def. m ∈ M is a regular point if Tmf is onto n ∈ N is a regular value if f −1(n) is regular points. f : M → N is a submersion if all values are regular.

Proof of local immersion thereom. In charts, we have f : A → B, T0f is ⊥ k * * 1 : 1. We have f+ : A × (im T0f) → R by (a, v ) → f(a) + v . Applying inverse function theorem to f+.

Theorem 27.3 (Local Submersion Thm). If f : M → N, Tmf is onto, then j k f is a local submersion. ∃ coordinates s.t. f looks like R  R . Def. f : M → R fails to be a submersion at p ∈ R (a critical point) is Morse if its Hessian is nondegenerate.

n P 2 Ex. f : R → R by (x1, x2, . . . , xn) → ±xi . Theorem 27.4 (Morse Lemma). Around Morse Critical points, there exist coordinates looking like the basic example.

Theorem 27.5 (Preimage Theorem). If f : M → N has regular value p ∈ N, then f −1(P ) is a dim M − dim N-manifold.

* * Def. The normal bundle to M,→ L, NM L = {(`, v ) ∈ TL | ` ∈ M, v ⊥ * TmM}. Algebraic geometry would say v ∈ TmL/TmM instead because ⊥ is gross.

Theorem 27.6 (Tubular Nbhd Theorem). If M,→ N is a closed submani- ∼ fold then there exists an open set U : M ⊆ U ⊆ L s.t. U −→ NM L. Theorem 27.7 (Sard’s Theorem). Almost every value is regular.

Def. A, B ,→ M are transverse if TpA + TpB = TpM for all p ∈ A ∩ B. Def. A manifold with boundary is defined similarly to a manifold except with half space.

Def. Orientation on a real vector spaces O : {bases} → {±1} given by O(B)O(M · B) = sign(det M).

55 28 Dec 10

28.1 Review of Forms

p Def. Ω (M) is smooth p-forms ie antisymmetric multilinear maps from (m, TmM) → ∼ p ? R and this is = Γ(M; ∧ T M) which are p-th exterior powers on the cotan- gent bundle.

f ? We have operations on a f : M → N with Ω?(M) ←− Ωp(N) defined as the pullback, ∧ :Ωp(M) × Ωq(M) −−−−→bilinear Ωp+q(M), and d :Ωp(M) → Ωp+1(M).

For the algebra, we have that d2 = 0, (f ◦ g)? = g? ◦ f ?, f ?(α ∧ β) = f ?α ∧ f ?β, α∧β = (−1)deg α deg ββ∧α, (α∧β)∧γ = α∧(β∧γ), f ?(dα) = df ?(α), d(α∧ β) = (dα) ∧ β + (−1)deg αα ∧ dβ.

d d 2 i · Def. For any Complex −→ A1 −→ A − 2 ... with d = 0, we define H (A ) = d (ker Ai −→ Ai+1)/ im(Ai−1 → Ai). This forms the idea of SES : 0 → A → B → C → 0 which implies LES.

Ex. Mayer-Vietoris is defined as for M = U ∪ V we have

M←U∪ V )? difference 0 → Ω(M) −−−−−−−−−−→disjoint Ω(U) ⊕ Ω(V ) −−−−−−→ Ω(U ∩ V ) → 0

Def. A good cover of M is a collection of open sets s.t. any finite nontrivial intersection is ∅ or contractible.

Theorem 28.1. Based on Riemannian Geometry (geodesic nbhds), we have that there always exists a good cover on M.

Remark. In Riemannian Geometry, we note that for larger dimensions (ie > 4) because of magic with the fundamental group, this becomes trivial. How- ever, there are still people working on 3 and 4 manifolds (since 2 is trivial). ( R ? = 0 Lemma (Poincar´e). H?(Rn) = 0 ? > 0 R Def. Orientation on M is M compactly supported top forms on M. Def. Boundary Orientation R R Theorem 28.2 (Stokes). M dα = ∂M α for oriented manifolds with bound- ary.

56 R k dim M−k ∧ Theorem 28.3 (Poincar´ePairing). Ω (M) × Ωc −−→ R Theorem 28.4 (Poincar´eDuality). This descends to a perfect pairing Hk(M)× dim M−k Hc (M) → R

? ? ? Theorem 28.5. If f0, f1 : M → N are homotopic then f0 = f1 as H (N) → H?(M).

Def. If M,→ N both compact oriented, [M] ∈ HcodimN M (N) is the Poincar´e ? R e ι e M e ? ∼ c dual. We have that (H (N) −→ H (M) −−→ R) ∈ (H (N)) = H (N) where e = dim M and c = codimN M.

Ex. {0} ,→ R with [{0}] are bump forms on R ie compactly supported non- negative R = 1. More generally {0} ,→ V on an oriented vector space. Even more generally, we have M,→ V on the oriented vector bundle is a Thom Form.

Def. If M ← V oriented vector bundle a Thom form α has

(i) has degree = codim M = dim Fibers

(ii) is a bump form on every fiber

(iii) closed

An so α ∈ Ωdim Fiber(V ).

∼ If M,→ P are both oriented, then M,→ NM P = tubular nbhd extended by 0 ,→ P .

Theorem 28.6. [M] is the class of extension by 0. Thom form on NM P .

28.2 Intersection Numbers Def. χ(M) = I(4, 4) M is compact oriented

P i i Theorem 28.7 (Poincar´eHopf). This is equal to i(−1) dim H (M). Def. Lefschetz Number of f : M → NI(4, graph f) for Lefschetz func- tions f.

P i i i Theorem 28.8. This is equal to i(−1) tr(f : H (M) → H (M)).

57