Hungry, Hungry Homology - PDF Free Download

September 27, 2017

Motiving Problem: Algebra Problem (Preliminary Version) Given two groups A, C, does there exist a group E so that A E and E /A = C? If such an group exists, we call E an extension of C by A. Clearly, the answer is yes as if we take E = A C, then A E and E /A = C.

Motiving Problem: Algebra Problem Given groups A, C, how many extensions E exist?

Motiving Problem: Topology Problem How do we tell spaces apart?

Motiving Problem: Topology C = {e 2πin : n [0, 1)} D = {(r cos θ, r sin θ): r [0, 1], θ [0, 2π)} T = (2 + 2 cos θ) cos φ, (2 + 2 cos θ) sin φ, 2 sin θ): 0 θ, φ 2π

Movie Time https://www.youtube.com/watch?v=pydjnefh6zw

Motiving Problem: Analysis Problem Find a set X 3 such that... V := {F : 3 \ X 3 : F = 0} W := {F : F = g} dim(v /W ) = 8

Algebra

Theorem (First Isomorphism Theorem) If φ : G H is a homomorphism of groups, then ker φ G and G/ ker φ = φ(g). Corollary If φ : G H is a surjective homomorphism of groups, then G/ ker φ = H.

Definition (Exact) A pair of R-homomorphisms M f N g P is called exact (at N) if im f = ker g. An infinite sequence of homomorphisms is exact if it is exact in each spot. A sequence of the form is called a short exact sequence. 0 M f N g P 0

Example (i) The sequence 0 M f N is exact if and only if f is injective. (ii) The sequence N g P 0 is exact if and only if g is surjective.

Example (i) The sequence 0 M f N is exact if and only if f is injective. (ii) The sequence N g P 0 is exact if and only if g is surjective. (iii) The sequence 0 M f N 0 is exact if and only if f is an isomorphism. (iv) The sequence 0 M f N g P 0 is exact if and only if f is injective, g is surjective, and im f = ker g.

Example (v) The sequence 0 ker f M f N coker f 0 is an exact sequence. (vi) If P N M is a tower of submodules, there is an exact sequence 0 N/P M/N M/P 0.

Example (v) The sequence 0 ker f M f N coker f 0 is an exact sequence. (vi) If P N M is a tower of submodules, there is an exact sequence 0 N/P M/N M/P 0. i (vii) The sequence 0 M M N π N 0 is an exact sequence. Therefore, an extension of N by M always exists. To see an explicit example, take the following: 0 i ( /n ) π /n 0

Definition (Homomorphism of Exact Sequence) If (M n, f n ) n and (P n, g n ) n are exact sequences, then a homomorphism of exact sequences is a collection of maps (β n ) n, β n : M n P n such that for all n and m M n, g n β n (m) = β n 1 f n (m); that is, the following diagram commutes f n+1 M n+1 M n M n 1 f n f n 1 β n+1 β n β n 1 g n+1 g n g n 1 P n+1 P n P n 1

Example (i) Suppose m, n + and n m. Let k = m/n. Then (abusing notation by letting π to denote the obvious projection maps in each case) the following diagram commutes. 0 /n 0 π n n 0 /k /m /n 0 π π π 1

Example (ii) The following is a non-canonical morphism of exact sequences: 0 /2 /2 /2 /2 0 1 i 1 0 /2 /2 /2 /2 0 s π 2 i 1 π 1 where i j is the inclusion into the jth component, π i is the projection onto the ith component, and s swaps components. Note defining maps the obvious way would not give a commutative diagram. 1

Proposition (The Short 5 Lemma) If the rows in the following commutative diagram are exact and α, γ are isomorphisms, then β is an isomorphism. 0 A B C 0 α f r 0 M N P 0 β g s γ

Proposition (The 5 Lemma) Consider the following commutative diagram with exact rows: f 1 f 2 M 1 M 2 M 3 M 4 M 5 f 3 f 4 h 1 h 2 h 3 h 4 h 5 g 1 g 2 g 3 g 4 N 1 N 2 N 3 N 4 N 5 (a) if h 2, h 4 are surjective and h 5 is injective, then h 3 is surjective. (b) if h 2, h 4 are injective and h 1 is surjective, then h 3 is injective. (c) if h 1, h 2, h 4, and h 5 are isomorphisms, then h 3 is an isomorphism.

Theorem (Mitchell Embedding Theorem) If is an abelian category, there is an associative unital ring R and a fully faithful exact functor F : Mod(R).

Definition (Chain Complex) A sequence of R-modules, (C n, d n ) n, with maps (called boundary maps) d n : C n C n 1 is said to be a chain complex if im d n+1 ker d n for all n. dn+3 d n+2 d n+1 d n d n 1 d n 2 C n+2 C n+1 C n C n 1 C n 2

Definition (Homology) Given a sequence of R-modules (C n, δ n ) n, the nth homology group of the complex is H n := ker δ n / im δ n+1. Definition (Chain Map) Given two chain complexes (C n, δ n ) n and (C n, δ n ) n, a map of chain complexes is a family of R-homomorphisms (f n : C n C n ) n so that δ n f n = f n 1 δ n for all n ; that is, that the diagram below commutes. δ n+1 C n+1 C n C n 1 δ n δ n 1 f n+1 f n f n 1 δ n+1 δ n δ n 1 C n+1 C n C n 1

Lemma (Snake Lemma) Given the following commutative diagram of R-modules with exact rows f M N P 0 α 0 M N P there is an exact sequence β g f g ker α f ker β g ker γ δ coker α f coker β g coker γ γ

0 0 0 ker α ker β ker γ M f N g P 0 α β γ 0 M N P f g coker α coker β coker γ 0 0 0

0 0 0 ker α ker β ker γ δ M f N g P 0 α β γ 0 M N P f g coker α coker β coker γ 0 0 0

Lemma (9 Lemma or 3 3 Lemma) If the following is a commutative diagram with exact columns and the bottom two rows exact, then the top row is exact. 0 0 0 f 0 A B C 0 α β γ f 0 A B C 0 0 A B C 0 g α β γ g f g 0 0 0

We can recognize all group extensions as a cohomology group using delooping from Topoi Theory. H 2 Grp (C, A) = H2 (BC, A) = H(BC, B 2 A) Or better yet using Ext groups. 0 A E C 0

We can recognize all group extensions as a cohomology group using delooping from Topoi Theory. H 2 Grp (C, A) = H2 (BC, A) = H(BC, B 2 A) Or better yet using Ext groups. 0 A E C 0 F 2 F 1 F 0 C 0

We can recognize all group extensions as a cohomology group using delooping from Topoi Theory. H 2 Grp (C, A) = H2 (BC, A) = H(BC, B 2 A) Or better yet using Ext groups. 0 A E C 0 F 2 F 1 F 0 C 0 0 Hom R (F 0, A) Hom R (F 1, A) Hom R (F 2, A) There is a bijective correspondence between the equivalence class of extensions of C by A and the elements of Ext 1 R (C, A).

Topology

Definition (Path Homotopy) A homotopy of paths f 0, f 1 (from x 0 to x 1 ) in a space X is a family f t : I X, 0 t 1 such that (i) f t (0) = x 0 and f t (1) = x 1 are independent of t. (ii) the associated map F : I I X defined by F (s, t) = f t (s) is continuous. If f 0 and f 1 are connected by a homotopy f t, we say that f 0 and f 1 are homotopic and denote this f 0 f 1. That is, f 0 and f 1 are path homotopic if we can continuously deform the path f 0 into the path f 1, as in the figure below.

Definition (Fundamental Group) The fundamental group of a space X at a point x 0 X is the set of equivalence classes of path homotopic loops at x 0, denoted π 1 (X, x 0 ).

Definition (Fundamental Group) The fundamental group of a space X at a point x 0 X is the set of equivalence classes of path homotopic loops at x 0, denoted π 1 (X, x 0 ). Theorem Every subgroup of a free group is free. Problem Find all index 2 subgroups of the free group F (2) = a, b.

ab 1 = b 1 a

ab 1 = b 1 a a b = ( b) + a

ab 1 = b 1 a a b = ( b) + a a b + c d = (a c) + (b d) = (a d) + (b c)

0 δ 2 C 1 = = a δ 1 C 0 = = v δ 0 0

0 δ 2 C 1 = 2 = a, b δ 1 C 0 = 2 = v, w δ 0 0

0 δ 3 C 2 = = U δ 2 C 1 = 2 = a, b δ 1 C 0 = 2 = v, w δ 0 0

Analysis

Problem If f = (f 1, f 2 ) : U 2 is a smooth function on an open set U 2, is there a smooth function F : U so that F x = f 1 and F y = f 2, i.e. F = (f 1, f 2 )?

Since F is smooth function, we have 2 F f 1 y = f 2 x. y x = 2 F x y and hence

Clearly, this condition is necessary but is it sufficient?

Let f : 2 \ {(0, 0)} 2 be given by f (x, y) = y x 2 + y 2, x x 2 + y 2 It is routine to check that f 1 y = f 2 x. However, there is no function F : 2 \ {(0, 0)} so that F = f.

If there were then 2π 0 d F (cos θ, sin θ) dθ = F (1, 0) F (1, 0) = 0 dθ However, the chain rule gives d F F (cos θ, sin θ) = dθ x ( sin θ) + F y cos θ = f 1 (cos θ, sin θ) sin θ + f 2 (cos θ, sin θ) cos θ = 1

If there were then 2π 0 d F (cos θ, sin θ) dθ = F (1, 0) F (1, 0) = 0 dθ However, the chain rule gives d F F (cos θ, sin θ) = dθ x ( sin θ) + F y cos θ = f 1 (cos θ, sin θ) sin θ + f 2 (cos θ, sin θ) cos θ = 1 Therefore, no such F can exist. Note that the open set U where f (x, y) is defined contains a hole namely (0, 0). However for a nice open region, the question we asked as an affirmative solution.

Definition (Star-Shaped) A subset X n is star-shaped with respect to x 0 X if the line segment {tx 0 + (1 t)x : t [0, 1]} is contained in X for all x X. Proposition If U 2 is an open star-shaped region, then any smooth function f = (f 1, f 2 ) : U 2 with f 1 y = f 2 x, there is a function F : U with F = f.

Let U 3 be an open set and C (U, n ) be the set of smooth functions φ : U n on U. Taking U 3, we define the following maps: Grad : C (U, ) C (U, 3 ); Curl : C (U, 3 ) C (U, 3 ); Div : C (U, 3 ) C (U, ); φ φ φ φ φ φ

Let U 3 be an open set and C (U, n ) be the set of smooth functions φ : U n on U. Taking U 3, we define the following maps: Grad : C (U, ) C (U, 3 ); Curl : C (U, 3 ) C (U, 3 ); Div : C (U, 3 ) C (U, ); φ φ φ φ φ φ Now Grad, Rot, and Div are linear operators, Curl Grad = 0 (gradient fields are irrotational), and Div Rot = 0 (curls are incompressible) so that im Grad is a subspace of ker Curl and im Rot is a subspace of ker Div.

Proposition An open set U k is connected if and only if H 0 (U) =. In fact, this can be extended to show that dimh 0 (U) is the number of connected components of U. By Proposition 4.1 above, we have H 1 (U) = 0 whenever U 2 is star-shaped. Furthermore from our previous work, we know that H 1 ( 2 \ {(0, 0)} 0. Thus for a star-shaped region, we have

Definition (Alternating Map) A k-linear map ω : V k is alternating if ω(v 1,..., v k ) = 0 whenever v i = v j for i j. The vector space of alternating k-linear maps is denoted Alt k (V ). Definition (Exterior Product) For ω 1 Alt p (V ) and ω 2 Alt q (V ), define (ω 1 ω 2 )(v 1, v 2,..., v p+q ) := sgn(σ) ω 1 (v σ(1),..., v σ(p) ) σ S(p,q) ω 2 (v σ(p+1),..., v σ(p+q) ) where S(p, q) is all the permutations of {1, 2,..., p + q} with σ(1) < < σ(p) and σ(p + 1) < < σ(p + q).

This product has the following properties: (i) ω 1 ω 2 is a (p + q)-linear map. (ii) if ω 1 Alt p (V ) and ω 2 Alt q (V ), then ω 1 ω 2 Alt p+q (V ).

This product has the following properties: (i) ω 1 ω 2 is a (p + q)-linear map. (ii) if ω 1 Alt p (V ) and ω 2 Alt q (V ), then ω 1 ω 2 Alt p+q (V ). (iii) if ω 1 Alt p (V ) and ω 2 Alt q (V ), then ω 1 ω 2 = ( 1) pq ω 2 ω 1. (iv) if ω 1 Alt p (V ), ω 2 Alt q (V ), and ω 3 Alt r (V ), then ω 1 (ω 2 ω 3 ) = (ω 1 ω 2 ) ω 3.

Definition (Differential p-form) A differential p-form on U is a smooth map ω : U Alt p ( n ). The vector space of all such maps is denoted by Ω p (U). For example, if p = 0 then Alt 0 ( ) and Ω 0 (U) is just the vector space of smooth real-valued functions on U so that Ω 0 (U) = C (U, ). Let Dω denote the ordinary derivative of a smooth map ω : U Alt p ( n ) and its value at x by D x ω.

Definition (Exterior Differential) The exterior differential d p : Ω p (U) Ω p+1 (U) is the linear operator p+1 d p x ω(v 1,..., v p+1 ) = ( 1) i 1 D x ω(v i )(v 1,..., v i,..., v p+1 ) i=1 where D x ω(e j ) = ω I x (x)ε j I, j = 1,..., n, ε I = ε i1 ε ip, and I runs over all sequences with 1 i 1 < i 2 < < i p n. Now d p x ω Altp+1 ( n ) and the composition d 2, Ω p (U) Ω p+1 (U) Ω p+2 (U), is zero. This should make one think that we can define a homology on these spaces and this is indeed the case.

Definition (de Rham Cohomology) The pth de Rham cohomology group is the quotient vector space H p (U) = ker d p / im d p 1 Then in particular H 0 (U) is the kernel of d : C (U, ) Ω 1 (U) the vector space of maps f C (U, ) with vanishing derivative, i.e. the space of locally constant maps. The case where U is again a star-shaped region, not surprisingly, have rigid restrictions on their de Rham cohomology groups.

Theorem (Poincaré s Lemma) If U n is an open star-shaped region, then H p (U) = 0 for p > 0 and H 0 (U) =. Theorem If an open set U n is covered by convex open sets U 1,..., U r, then H p (U) is finitely generated.

Theorem If C n is a closed subset, then H p+1 ( n+1 \ C) = H p ( n \ C), p 1 H 1 ( n+1 \ C) = H 0 ( n \ C)/ H 0 ( n+1 \ C) = Theorem If U n is an open contractible set, then H p (U) = 0 for p > 0 and H 0 (U) =.

Theorem (Mayer-Vietoris) If U 1, U 2 n are open subsets, there is an exact sequence of cohomology of vector spaces H p (U 1 U 2 ) H p (U 1 ) H p (U 2 ) H p (U 1 U 2 ) δ H p+1 (U 1 U 2 )

We now return to the original problem: V = {F : 3 \ X 3 : F = 0} W = {F : F = g} dim(v /W ) = 8

Notice then we can phrase our original question as follows: find a set X so that dim H 1 ( 3 \ X ) = 8. Take X to be the set of 8 lines parallel to the z-axis, distributed equally along the unit circle. Observe that 3 \ X deformation retracts to the plane minus the 8 intersections of these lines with the plane. It is a routine exercise to show that, p = 0, 1 H p ( 2 \ {(0, 0)}) = 0, p 2 It is then a matter of induction using Mayer-Vietoris to show that H 1 ( 2 \ (X 2 )) = 8. But then we have H 1 ( 3 \ X ) = 8, as desired. Generally, the dimension of H 1 will give the number of 1-dimensional holes in the space and the dimension H 2 gives the number 0-dimensional holes in the space.