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of another product called cross product. However, to define the cross product from scratch takes some work, so we will proceed in the opposite order, first giving an elementary definition of cup product by an explicit formula with simplices, then afterwards defining cross product in terms of cup product. The other approac... |
··, vk+ℓ+1] When we add these two expressions, the last term of the first sum cancels the first term of the second sum, and the remaining terms are exactly δ(ϕ ` ψ)(σ ) = (ϕ ` ψ)(∂σ ) ⊔⊓ since ∂σ = (−1)iσ || [v0, ···, vi, ···, vk+ℓ+1]. k+ℓ+1 i=0 b P b Cup Product Section 3.2 207 From the formula δ(ϕ ` ψ) = δϕ ` ψ ± ϕ ` δ... |
) is formed by the edges ai and bi, as we showed in Example 2.36 when we computed the homology of M using cellular homology. We have H 1(M) ≈ Hom(H1(M), Z) by cellular cohomology or the universal coefficient theorem. A basis for H1(M) determines a dual basis for Hom(H1(M), Z), so dual to ai is the cohomology class αi ass... |
, so ` ψ1)(c) is a generator it represents a nonzero element of H2(M). The fact that (ϕ1 ` ψ1 of Z implies both that c represents a generator of H2(M) ≈ Z and that ϕ1 represents the dual generator γ of H 2(M) ≈ Hom(H2(M), Z) ≈ Z. Thus α1 ` β1 = γ. In similar fashion one computes: αi ` βj = γ, i = j 0, i ≠ j = −... |
H 1(RP2; Z2) with itself is a generator of H 2(RP2; Z2). ∆ The remarks in the paragraph preceding this example apply here also, but with the following difference: When one tries to deform a second copy of the loop αi in the present example to be disjoint from the original copy, the best one can do is make ` αi is now n... |
a generator of H2(X; Zm) and there are 2 cocycles taking the value 1 on i Ti, for example the P cocycle taking the value 1 on one Ti and 0 on all the others. The cocycle ϕ ` ϕ takes i Ti, hence represents 0 + 1 + ··· + (m − 1) times the value 0 + 1 + ··· + (m − 1) on a generator β of H 2(X; Zm). In Zm the sum 0 + 1 + ... |
X, B; R)→C k+ℓ(X, A + B; R) where C n(X, A + B; R) is the subgroup of C n(X; R) consisting of cochains vanishing on sums of chains in A and chains in If A and B are open in X, the inclusions C n(X, A ∪ B; R) ֓ C n(X, A + B; R) B. induce isomorphisms on cohomology, via the five-lemma and the fact that the restriction map... |
vk, ···, vk+ℓ] f σ ||[vk, ···, vk+ℓ] ⊔⊓ The natural question of whether the cup product is commutative is answered by the following: Theorem 3.11. The identity α ` β = (−1)kℓβ ` α holds for all α ∈ H k(X, A; R) and β ∈ H ℓ(X, A; R), when R is commutative. Taking α = β, this implies in particular that if α is an element... |
vn−i). This reversal of vertices is the product of n + (n − 1) + ··· + 1 = n(n + 1)/2 transpositions of adjacent vertices, each of which reverses orientation of the n simplex since it is a reflection across an (n − 1) dimensional hyperplane. So to take orientations into account we would expect that a sign εn = (−1)n(n+1... |
σ ||[vn, ···, vn−i, ···, v0] Xi ρ∂(σ ) = ρ (−1)iσ ||[v0, ···, b vi, ···, vn] vn−i, ···, v0] b (−1)n−iσ ||[vn, ···, Xi = εn−1 Xi so the result follows from the easily checked identity εn = (−1)nεn−1. b To define a chain homotopy between ρ and the identity we are motivated by the construction of the prism operator P in th... |
··, vi, wn, ···, b wj, ···, wi] The j = i terms in these two sums give Xj≥i c εn[wn, ···, w0] + εn−i[v0, ···, vi−1, wn, ···, wi] Xi>0 + (−1)n+i+1εn−i[v0, ···, vi, wn, ···, wi+1] − [v0, ···, vn] Xi<n 212 Chapter 3 Cohomology In this expression the two summation terms cancel since replacing i by i − 1 in the second sum p... |
. This is easy to do if we simply define H ∗(X; R) to be the direct sum of the groups H n(X; R). Elements of H ∗(X; R) are finite sums i αi with αi ∈ H i(X; R), and the product of It is routine to check two such sums is defined to be that this makes H ∗(X; R) into a ring, with identity if R has an identity. Similarly, P H... |
avoids potential confusion with the L degree of a polynomial. Cup Product Section 3.2 213 A graded ring satisfying the commutativity property of Theorem 3.11, ab = (−1)|a||b|ba, is usually called simply commutative in the context of algebraic topology, in spite of the potential for misunderstanding. In the older liter... |
graded R[α1, ···, αn] over a commutative ring R with identity. ring is the exterior algebra, i1 < ··· < ik, with This is the free R module with basis the finite products αi1 Λ associative, distributive multiplication defined by the rules αiαj = −αjαi for i ≠ j and α2 i = 0. The empty product of αi ’s is allowed, and pro... |
sometimes be used to rule out splittings of a space as a wedge sum up to homotopy equivalence. For example, consider CP2, which is S 2 with a cell e4 attached by a certain map f : S 3→S 2. Using homology or just the additive structure of cohomology it is impossible to conclude that CP2 is not homotopy equivalent to S ... |
product, or external cup product as it is sometimes called. This is the map H ∗(X; R) × H ∗(Y ; R) ×-----------------------→ H ∗(X × Y ; R) given by a× b = p∗ 2 (b) where p1 and p2 are the projections of X × Y onto X and Y. Since cup product is distributive, the cross product is bilinear, that is, linear 1 (a) ` p∗ in... |
� A. (5) Zn ⊗ A ≈ A/nA. (6) A pair of homomorphisms f : A→A′ and g : B→B′ induces a homomorphism f ⊗ g : A ⊗ B→A′ ⊗ B′ via (f ⊗ g)(a ⊗ b) = f (a) ⊗ g(b). (7) A bilinear map ϕ : A× B→C induces a homomorphism A ⊗ B→C sending a ⊗ b to ϕ(a, b). In (1)–(5) the isomorphisms are the obvious ones, for example a ⊗ b ֏ b ⊗ a in ... |
, 1 ⊗ q a ⊗ pb = 1 q a ⊗ b = 1 q qb =, √ 2 ⊗ 1, √ Property (7) of tensor products guarantees that the cross product as defined above √ √ √ √ gives rise to a homomorphism of R modules H ∗(X; R) ⊗R H ∗(Y ; R) ×-----------------→ H ∗(X × Y ; R), a ⊗ b ֏ a× b which we shall also call cross product. This map becomes a ring h... |
of a product space, are known as K¨unneth formulas. The hypothesis that X and Y are CW complexes will be shown to be unnecessary in §4.1 when we consider CW approximations to arbitrary spaces. On the other hand, the freeness hypothesis cannot always be dispensed with, as we shall see in §3.B when we obtain a completel... |
product, or a relative version of it, defines a map µ : hn(X, A)→kn(X, A) which we would like to show is an isomorphism when X is a CW complex and A = ∅. Cup Product Section 3.2 217 We will show: (1) h∗ and k∗ are cohomology theories on the category of CW pairs. (2) µ is a natural transformation: It commutes with induc... |
. The axiom for disjoint unions gives a further reduction to the case of the pair (Dn, ∂Dn). Finally, this case follows by applying the five-lemma to the long exact sequences of this pair, since Dn is contractible and hence is covered by the 0 dimensional case, and ∂Dn is (n − 1) dimensional. α, ∂Dn α (Dn ` ` Φ The case... |
), for a fixed n. This yields another exact sequence because H n(Y ; R) is a direct sum of copies of R, so the result of tensoring an exact sequence with this direct sum is simply to produce a direct sum of copies of the exact sequence, which is again an exact sequence. The second step is to let n vary, taking a direct ... |
�) ` p♯ p♯ 1 (ϕ) ` p♯ 1 (δϕ) ` p♯ 1 (δϕ) ` p♯ 1 (ϕ)` p♯ since p♯ = p♯ 1 (ϕ) ` p♯ 2 (ψ) 2 (ψ) It is sometimes important to have a relative version of the K¨unneth formula in Theorem 3.15. The relative cross product is H ∗(X, A; R) ⊗R H ∗(Y, B; R) ×-----------------→ H ∗(X × Y, A× Y ∪ X × B; R) for CW pairs (X, A) and (Y... |
retract of Y, the upper row of this diagram is a split short exact sequence. The lower row is the long exact sequence of a triple, and it too is a split short exact sequence since (X × y0, A× y0) is a retract of (X × Y, A× Y ). The middle and right cross product maps are isomorphisms by the case B = ∅ since H k(Y ; R)... |
2[α], where |α| = 1. In the complex case, H ∗(CPn; Z) ≈ Z[α]/(αn+1) and H ∗(CP∞; Z) ≈ Z[α] where |α| = 2. This turns out to be a quite important result, and it can be proved in a number of different ways. The proof we give here uses the geometry of projective spaces to reduce the result to a very special case of the K¨u... |
1 and the other n coordinates arbitrary, so U is homeomorphic to Rn, with p corresponding to 0 under this homeomorphism. We can write this Rn as Ri × Rj, with Ri as the coordinates x0, ···, xi−1 and Rj as the coordinates xi+1, ···, xn. In the figure P n is represented as a disk with antipodal points of its boundary sph... |
retracts onto P i−1 by the following argument. The subspace P n − P j ⊂ P n consists of points represented by vectors v = (x0, ···, xn) with at least one of the coordinates x0, ···, xi−1 nonzero. The formula ft(v) = (x0, ···, xi−1, txi, ···, txn) for t decreasing from 1 to 0 gives a well-defined deformation retraction ... |
fined in Example 4.47. The cup product structure in quaternionic projective spaces is just like that in complex pro- jective spaces, except that the generator is 4 dimensional: H ∗(HP∞; Z) ≈ Z[α] and H ∗(HPn; Z) ≈ Z[α]/(αn+1), with |α| = 4 The same proof as in the real and complex cases works here as well. The cup produ... |
true, as the generator α ∈ H 1(RP2k+1; Z2) has α2k+1 ≠ 0, while α2k+1 = 0 for the generator α ∈ H 1(RP2k ∨ S 2k+1; Z2). Example 3.20. Combining the calculation H ∗(RP∞; Z2) ≈ Z2[α] with the K¨unneth formula, we see that H ∗(RP∞ × RP∞; Z2) is isomorphic to Z2[α1] ⊗ Z2[α2], which is just the polynomial ring Z2[α1, α2]. ... |
, as one sees by composing with the projections of RPn−1 × RPn−1 onto its two factors. The fact that h restricts to a homeomorphism on the first factor then implies that k1 is nonzero. Similarly k2 is nonzero, so since these coefficients lie in Z2 we have h∗(α) = α1 + α2. 1, αn n k αk n k k P 2 ), so the coefficient Since α... |
over C. This is assuming we are talking about n k finite-dimensional division algebras. For infinite dimensions there is for example the field of rational functions C(x). We saw in Theorem 3.19 that RP∞, CP∞, and HP∞ have cohomology rings that are polynomial algebras. We will describe now a construction for enlarging S 2... |
where one coordinate is e. These glueings are by homeomorphisms taking cells onto cells, so Jm(X) inherits a CW structure from Xm. There are natural inclusions Jm(X) ⊂ Jm+1(X) as subcomplexes, and J(X) is the union of these subcomplexes, hence is also a CW complex. Proposition 3.22. For n > 0, H ∗ multiple of n. If n ... |
maps for Jm(S n) will be trivial if they are trivial for (S n)m, as indeed they are since H ∗ is free with basis in one-to-one correspondence with the cells, by Theorem 3.15. (S n)m; Z We can compute cup products in H ∗ q∗. Let xk denote the generator of H kn by the cellular cocycle assigning the value 1 to the kn cel... |
(b) x2x2m−2 = mx2m in H ∗ J2m(S n); Z For (a) we apply q∗ and compute in the exterior algebra.. Z[α1, ···, α2m+1] : q∗(x1x2m) = αi α1 ··· αi ··· α2m+1 Λ Xi Xi αiα1 ··· = Xi b b αi ··· α2m+1 = (−1)i−1α1 ··· α2m+1 Xi The coefficients in this last summation are +1, −1, ···, +1, so their sum is +1 and (a) follows. For (b) w... |
x2, ··· and multiplication defined by Γ xixj = R[x]. Q[x] is just Q[x]. However, for R = Zp with p prime When R = Q it is clear that xi+j. The preceding proposition implies that H ∗ J(S 2n); R i+j i Γ ≈ something quite different happens: There is an isomorphism [x] ≈ Zp[x1, xp, xp2, ···]/(x Zp p2, ···) = Zp[xpi ]/(x p p... |
& Atiyah 1966]. However there is one slightly anomalous case, R = Z2, d = 8, which must be treated by special arguments; see [Toda 1963]. It is an interesting fact that for each even d there exists a CW complex Xd which is simultaneously an example for all the admissible choices of coefficients R in the table. Moreover,... |
[α1, ···, αm, β1, ···, βn], the product of two spaces with polynomial cohomology is again a space with polynomial cohomology, assuming the number of polynomial generators is finite in each dimension. For example, the n fold product (CP∞)n has H ∗ ≈ Z[α1, ···, αn] with each αi 2 dimensional. Similarly, products of (CP∞)n... |
cients, according to a theorem in [Andersen & Grodal 2008]. The situation for Zp coefficients is more complicated and will be discussed in §3.G. Polynomial algebras are examples of free graded commutative algebras, where ‘free’ means loosely ‘having no unnecessary relations’. In general, a free graded com- mutative algeb... |
≈ 3 and e1 Z[α1, α2, α3]/(α2α3). 1 × e1 2 × e1 2 × e1 Λ Λ 228 Chapter 3 Cohomology we get a simplex n−1 ∩ q−1(X). This is a subcomplex of How many different subcomplexes of T n are there? To each subcomplex X ⊂ T n we can associate a finite simplicial complex CX by the following procedure. View T n as the quotient of th... |
same dimension. Let us conclude this section with an example of a cohomology ring that is not too far removed from a polynomial ring. Example 3.24: Cohen–Macaulay Rings. Let X be the quotient space CP∞/CPn−1. The quotient map CP∞→X induces an injection H ∗(X; Z)→H ∗(CP∞; Z) embedding H ∗(X; Z) in Z[α] as the subring g... |
) Prove the Borsuk–Ulam theorem by the following argument. Suppose on the contrary that f : S n→Rn satisfies f (x) ≠ f (−x) for all x. Then define g : S n→S n−1 by g(x) = /|f (x) − f (−x)|, so g(−x) = −g(x) and g induces a map RPn→RPn−1. Show that part (a) applies to this map. 4. Apply the Lefschetz fixed point theorem to... |
4. Thus X and Y have the same 3 skeleton but differ in the way their 4 cells are attached. Show that X and Y have isomorphic cohomology rings with Z coefficients but not with Zp coefficients. 9. Show that if Hn(X; Z) is free for each n, then H ∗(X; Zp) and H ∗(X; Z) ⊗ Zp are isomorphic as rings, so in particular the ring s... |
st by a geometric argument that the restriction q : RP2→CP1 induces a surjection on H 2 and then appealing to cup product structures. Next, form a quotient space X of RP∞∐CPn by identifying each point x ∈ RP2n with q(x) ∈ CPn. Show there are ring isomorphisms H ∗(X; Z) ≈ Z[α]/(2αn+1) and H ∗(X; Z2) ≈ Z2[α, β]/(β2 − α2n... |
spaces that depend only on homotopy type, so local topological properties do not play much of a role. Digressing somewhat from this viewpoint, we study in this section a class of spaces whose most prominent feature is their local topology, namely manifolds, which are locally homeomorphic to Rn. It is somewhat miraculo... |
, ···, zn] ∈ CPn | zi = 1 }. The same reasoning applies also for quaternionic projective spaces. Further examples of closed manifolds can be generated from these using the obvious fact that the product of closed manifolds of dimensions m and n is a closed manifold of dimension m + n. Poincar´e duality in its most primi... |
in the relation. The reader might also check that Poincar´e n k = n k n n−k duality is consistent with our calculations of the homology of projective spaces and lens spaces, which are all orientable except for RPn with n even. 232 Chapter 3 Cohomology For many manifolds there is a very nice geometric proof of Poincar´... |
ations of cellular chain groups C ∗ 0 = C2, 1 = C1, and C ∗ C ∗ 2 = C0. If we use Z coefficients these identifications are not quite canonical since there is an ambiguity of sign for each cell, the choice of a generator for the corresponding Z summand of the cellular chain complex. We can avoid this ambiguity by consideri... |
lattice points Z3, and then the dual cell structure is obtained by translating this by the vector (1/2, 1/2, 1/2). Each edge in either cell structure then has a dual 2 cell which it pierces orthogonally, and each vertex lies in a dual 3 cell. All the manifolds one commonly meets, for example all differentiable manifold... |
have degree −1, we see that a rotation ρ of Rn fixing x takes a generator α of Hn(Rn, Rn − {x}) to itself, ρ∗(α) = α, while a reflection takes α to −α. Note that with this definition, an orientation of Rn at a point x determines an orientation at every other point y via the canonical isomorphisms Hn(Rn, Rn−{x}) ≈ Hn(Rn, ... |
� Hn(Rn || B) under the natural maps Hn(M || B)→Hn(M || y). If an orientation exists for M, then M is called orientable. Every manifold M has an orientable two-sheeted covering space M. For example, RP2 is covered by S 2, and the Klein bottle has the torus as a two-sheeted covering space. The general construction goes ... |
Poincar´e Duality Section 3.3 235 to a path in M connecting two distinct points with the same image in M. The existence of such paths is equivalent to M being connected. f Proof: If M is connected, covering space of M. If it has two components, they are each mapped homeomorphi- M has either one or two components since... |
x. f One can generalize the definition of orientation by replacing the coefficient group Z by any commutative ring R with identity. Then an R orientation of M assigns to each x ∈ M a generator of Hn(M || x; R) ≈ R, subject to the corresponding local consistency condition, where a ‘generator’ of R is an element u such tha... |
n(M || x; R) ≈ R is an isomorphism for all x ∈ M. (b) If M is not R orientable, the map Hn(M; R)→Hn(M || x; R) ≈ R is injective with image { r ∈ R | 2r = 0 } for all x ∈ M. (c) Hi(M; R) = 0 for i > n. In particular, Hn(M; Z) is Z or 0 depending on whether M is orientable or not, and in either case Hn(M; Z2) = Z2. An el... |
and (b) of the R(M) be the set of sections of MR→M. The sum of two sections is a R(M) is an R module. There section, and a scalar multiple of a section is a section, so is a homomorphism Hn(M; R)→ R(M) sending a class α to the section x ֏ αx, where αx is the image of α under the map Hn(M; R)→Hn(M || x; R). Part (a) of... |
∈ Hn(M || A ∩ B) having image αx for all x in A, B, or A ∩ B respectively. The images of αA and αB in Hn(M || A ∩ B) satisfy the defining property of αA∩B, hence must equal αA∩B. Exactness of the sequence then implies that (αA, −αB) = (αA∪B ) for some αA∪B ∈ Hn(M || A ∪ B). This means that αA∪B maps to αA and αB, so αA... |
is equivalent to Hi(Rn || A)→Hi(Rn || x) by excision, and the latter map is an isomorphism for any x ∈ A, as both Rn − A and Rn − {x} deformation retract onto a sphere centered at x. (4) For an arbitrary compact set A ⊂ Rn ⊂ M let α ∈ Hi(M || A) be represented by a relative cycle z in Rn with ∂z in Rn − A, and let C b... |
ology a fundamental class must be represented by some linear comi kiσi of the n simplices σi of M. The condition that the fundamental bination class maps to a generator of Hn(M || x; Z) for points x in the interiors of the σi ’s means that each coefficient ki must be ±1. The ki ’s must also be such that i kiσi is a cycle... |
stein homomorphism Hn(M; Z2)→Hn−1(M; Z) coming from the short exact sequence of coefficient groups 0→Z→Z→Z2→0. The structure of Hn(M; G) and Hn−1(M; G) for a closed connected n manifold M can be explained very nicely in terms of cellular homology when M has a CW structure with a single n cell, which is the case for a lar... |
by Lemma 3.27 since U ∪ V and V are the complements of compact sets in M. Hence Hi(U; R) = 0, so z is a boundary in U and therefore in M, and we conclude that Hi(M; R) = 0. When i = n, the class [z] ∈ Hn(M; R) defines a section x֏[z]x of MR. Since M is connected, this section is determined by its value at a single poin... |
��(∂σ a ϕ − σ a δϕ) which is checked by a calculation: ∂σ a ϕ = ℓ Xi = 0 ℓ+1 (−1)iϕ σ ||[v0, ···, k + (−1)iϕ Xi = ℓ+1 vi, ···, vℓ+1] b σ ||[v0, ···, vℓ] σ ||[vℓ+1, ···, vk] σ ||[vℓ, ···, vi, ···, vk] b σ a δϕ = (−1)iϕ Xi = 0 k σ ||[v0, ···, σ ||[vℓ+1, ···, vk] vi, ···, vℓ+1] b ∂(σ a ϕ) = (−1)i−ℓϕ Xi = ℓ σ ||[v0, ···, v... |
, A; R) a-----------------------→ Hk−ℓ(X; R) For example, in the second case the cap product Ck(X; R)× C ℓ(X; R)→Ck−ℓ(X; R) restricts to zero on the submodule Ck(A; R)× C ℓ(X, A; R), so there is an induced cap product Ck(X, A; R)× C ℓ(X, A; R)→Ck−ℓ(X; R). The formula for ∂(σ a ϕ) still holds, so we can pass to homology... |
Theorem 3.30 (Poincar´e Duality). If M is a closed R orientable n manifold with fundamental class [M] ∈ Hn(M; R), then the map D : H k(M; R) -→ Hn−k(M; R) defined by D(α) = [M] a α is an isomorphism for all k. Recall that a fundamental class for M is an element of Hn(M; R) whose image in Hn(M || x; R) is a generator fo... |
[M] a ϕi = bi and [M] a ψi = −ai since in both cases there is just one 2 simplex [v0, v1, v2] where ϕi or ψi is nonzero on the edge [v0, v1]. Thus bi is the Poincar´e dual of αi and −ai is the Poincar´e dual of βi. If we interpret Poincar´e duality entirely in terms of homology, identifying αi with its Hom-dual ai and... |
with a complex ∆ borhood that meets only finitely many simplices. Consider the subgroup X which is locally compact. This is equivalent to saying that every point has a neighi c(X; G) i(X; G) consisting of cochains that are compactly supported in the sense that they take nonzero values on only finitely many sim- of the s... |
However, what we really need is singular cohomology with compact supports for spaces without any simplicial or cellular structure. The quickest definition of this is the following. Let C i c(X; G) be the subgroup of C i(X; G) consisting of cochains ϕ : Ci(X)→G for which there exists a compact set K = Kϕ ⊂ X such that ϕ... |
varying K. This is indeed the case, and it is algebraic limits that provide the description. Suppose one has abelian groups Gα indexed by some partially ordered index set I having the property that for each pair α, β ∈ I there exists γ ∈ I with α ≤ γ and β ≤ γ. Such an I is called a directed set. Suppose also that for... |
[a]. A useful consequence of this is that if we have a subset J ⊂ I with the property that for each α ∈ I there exists a β ∈ J with α ≤ β, then lim--→Gα is the same whether we compute it with α varying over I or just over J. In particular, if I has a maximal element γ, we can take J = {γ} and then lim--→Gα = Gγ. P P i... |
K ⊂ L of compact sets we associate the natural homomorphism H i(X, X −K; G)→H i(X, X −L; G). The resulting limit group lim--→H i(X, X −K; G) is then equal to H i c(X; G) since each element of this limit group is represented by a cocycle in C i(X, X − K; G) for some compact K, and such a cocycle is zero in lim--→H i(X,... |
incar´e duality, however, we will need induced maps of a different sort going in the opposite direction from what is usual for cohomology, maps H i c(V ; G) associated to inclusions U ֓ V of open sets in the fixed manifold M. c(U; G)→H i The group H i(X, X−K; G) for K compact depends only on a neighborhood of K in X by e... |
�(x) for all x ∈ H k(M || K; R), so µK ax = µL Since H ∗ Poincar´e duality for closed manifolds: Theorem 3.35. The duality map DM : H k for all k whenever M is an R oriented n manifold. c (M; R)→Hn−k(M; R) is an isomorphism The proof will not be difficult once we establish a technical result stated in the next lemma, con... |
ycles. Less trivial is the third square, which we rewrite in the following way: (∗) Letting A = M − K and B = M − L, the map δ is the coboundary map in the Mayer– Vietoris sequence obtained from the short exact sequence of cochain complexes 0 -→ C ∗(M, A + B) -→ C ∗(M, A) ⊕ C ∗(M, B) -→ C ∗(M, A ∩ B) -→ 0 where C ∗(M, ... |
) ≈ Hn(U ∩ V || K ∩ L). Similarly, αU −L + αU ∩V represents µK. In the square (∗) let ϕ be a cocycle representing an element of H k(M || K ∪ L). Under δ this maps to the cohomology class of δϕA. Continuing on to Hn−k−1(U ∩ V ) a δϕA, which is in the same homology class as ∂αU ∩V we obtain αU ∩V a ϕA since ∂(αU ∩V a ϕA)... |
L + αU ∩V represents µK, and ϕA vanishes on chains in A = M − K. Thus the square (∗) commutes up to a sign depending only on k. ⊔⊓ Proof of Poincar´e Duality: There are two inductive steps, finite and infinite: (A) If M is the union of open sets U and V and if DU, DV, and DU ∩V are isomorphisms, then so is DM. Via the fiv... |
a maximal open set U for which the theorem holds. If U ≠ M, choose a point x ∈ M − U and an open neighborhood V of x homeomorphic to Rn. The theorem holds for V and U ∩ V by (1) and (2), and it holds for U by assumption, so by (A) it holds for U ∪V, contradicting ⊔⊓ the maximality of U. Poincar´e Duality Section 3.3 2... |
α ∈ Ck+ℓ(X; R), ϕ ∈ C k(X; R), and ψ ∈ C ℓ(X; R). This holds since for a singular (k + ℓ) simplex σ : k+ℓ→X we have ψ(σ a ϕ) = ψ ∆ = ϕ ϕ σ ||[v0, ···, vk] σ ||[v0, ···, vk] ψ σ ||[vk, ···, vk+ℓ] σ ||[vk, ···, vk+ℓ] = (ϕ ` ψ)(σ ) The formula (∗) says that the map ϕ` : C ℓ(X; R)→C k+ℓ(X; R) is equal to the map HomR(Cℓ(X... |
a function of each variable separately, are both isomorphisms. Proposition 3.38. The cup product pairing is nonsingular for closed R orientable manifolds when R is a field, or when R = Z and torsion in H ∗(M; Z) is factored out. Proof: Consider the composition H n−k(M; R) h-----→ HomR(Hn−k(M; R), R) D∗-----→ HomR(H k(M... |
α ` β generating H n(M; Z) is equivalent to having ϕ with ϕ(α) = ±1. The case of field coefficients is similar but easier. ⊔⊓ Example 3.40: Projective Spaces. The cup product structure of H ∗(CPn; Z) as a truncated polynomial ring Z[α]/(αn+1) with |α| = 2 can easily be deduced from this as follows. The inclusion CPn−1֓CP... |
since it is the pullback of a generator of H 2(S 2; Z) under the projection S 2 × S 4→S 2, and in H ∗(S 2; Z) the square of the generator of H 2 is zero. More generally, an exercise for §4.D describes closed 6 manifolds having the same cohomology groups as CP3 but where the square of the generator of H 2 is an arbitra... |
` kαβn−1 generates H 2n+1(L2n+1; Zm) for some integer k. We must have k relatively prime to m, otherwise the product β ` kαβn−1 = kαβn would have order less than m and so could not generate H 2n+1(L2n+1; Zm). Then since k is relatively prime to m, αβn is also a generator of H 2n+1(L2n+1; Zm). From this it follows that... |
the extra Z ≈ H 2n+1(L2n+1; Z). A different derivation of the cup product structure in lens spaces is given in Example 3E.2. Using the ad hoc notation H k f r ee(M) for H k(M) modulo its torsion subgroup, the preceding proposition implies that for a closed orientable manifold M of dimenf r ee(M)→Z is a sion 2n, the mid... |
with the same form if the square α ` α of some α ∈ H 2(M 4) is an odd multiple of a generator of H 4(M 4), for example for CP2, and otherwise the M 4 is unique, for example for S 4 or S 2 × S 2 ; see [Freedman & Quinn 1990]. In §4.C we take the first step in this direction by proving a classical result of J. H. C. Whit... |
with boundary, then ∂M has a collar neighborhood. Proof: Let M ′ be M with an external collar attached, the quotient of the disjoint union of M and ∂M × [0, 1] in which x ∈ ∂M is identified with (x, 0) ∈ ∂M × [0, 1]. It will suffice to construct a homeomorphism h : M→M ′ since ∂M ′ clearly has a collar neighborhood. Sinc... |
M ′, finishing the proof. ⊔⊓ More generally, collars can be constructed for the boundaries of paracompact manifolds in the same way. A compact manifold M with boundary is defined to be R orientable if M − ∂M is R orientable as a manifold without boundary. If ∂M × [0, 1) is a collar neighborhood of ∂M in M then Hi(M, ∂M;... |
. Via a collar neighc (M − ∂M; R), and there are obvious borhood of ∂M we see that H k(M, ∂M; R) ≈ H k isomorphisms Hn−k(M; R) ≈ Hn−k(M − ∂M; R). The general case reduces to the case B = ∅ by applying the five-lemma to the following diagram, where coefficients in R are implicit: For commutativity of the middle square one ... |
) ≈ lim--→H n−i(U). We will show that the natural map from this limit to H n−i(K) is an isomorphism. This is easy when K has a neighborhood that is a mapping cylinder of some map X→K, as in the ‘letter examples’ at the beginning of Chapter 0, since in this case we can compute the direct limit us- c ing neighborhoods U ... |
, then H n−i−1(K; Z) for all i. Hi(S n − K; Z) ≈ Proof: The long exact sequence of reduced homology for the pair (S n, S n − K) gives e Hi(S n −K; Z) ≈ Hi+1(S n, S n −K; Z) for most values of i. The exception isomorphisms e is when i = n − 1 and we have only a short exact sequence e 0 -→ Hn(S n; Z) -→ Hn(S n, S n − K; ... |
e Corollary 3.46. If X ⊂ Rn is compact and locally contractible then Hi(X; Z) is 0 for i ≥ n and torsionfree for i = n − 1 and n − 2. e For example, a closed nonorientable n manifold M cannot be embedded as a subspace of Rn+1 since Hn−1(M; Z) contains a Z2 subgroup, by Corollary 3.28. Thus the Klein bottle cannot be e... |
Vβ is contained in some Uα, then these inclusions induce a simplicial map Poincar´e Duality Section 3.3 257 N(V)→N(U) that is well-defined up to homotopy. We can then form the direct limit lim--→H i(N(U); G) with respect to finer and finer open covers U. This limit group is by definition the ˇCech cohomology group ˇH i(X;... |
. Show that every covering space of an orientable manifold is an orientable manifold. 4. Given a covering space action of a group G on an orientable manifold M by orientation-preserving homeomorphisms, show that M/G is also orientable. 5. Show that M × N is orientable iff M and N are both orientable. 6. Given two disjoi... |
where εi is +1 or −1 homeomorphically by f onto B. Show the degree of f is according to whether f : Bi→B preserves or reverses local orientations induced from given fundamental classes [M] and [N]. P 9. Show that a p sheeted covering space projection M→N has degree ±p, when M and N are connected closed orientable mani... |
, the subspace of R2 consisting of the circles of radius 1/n and center (1/n, 0) for n = 1, 2, ···. (a) If fn : I→X is the loop based at the origin winding once around the nth circle, show that the infinite product of commutators [f1, f2][f3, f4] ··· defines a loop in X that is nontrivial in H1(X). [Use Exercise 12.] (b)... |
limit of countable abelian groups over a countable indexing set is countable. Apply this to show that if X is an open set in Rn then Hi(X; Z) is countable for all i. c (X; G) = 0 if X is path-connected and noncompact. 20. Show that H 0 21. For a space X, let X + be the one-point compactification. If the added point, de... |
(M; Z) torsionfree, then Hk(M; Z) is also torsionfree. 26. Compute the cup product structure in H ∗(S 2 × S 8♯S 4 × S 6; Z), and in particular show that the only nontrivial cup products are those dictated by Poincar´e duality. [See Exercise 6. The result has an evident generalization to connected sums of S i × S n−i ’s... |
contractible n manifold then ∂M is a homology (n − 1) sphere, that is, Hi(∂M; Z) ≈ Hi(S n−1; Z) for all i. 34. For a compact manifold M verify that the following diagram relating Poincar´e duality for M and ∂M is commutative, up to sign at least: 35. If M is a noncompact R orientable n manifold with boundary ∂M having... |
(X, A; G) becomes the map ∂ ⊗ 11 : Cn(X, A) ⊗ G→Cn−1(X, A) ⊗ G where ∂ : Cn(X, A)→Cn−1(X, A) is the usual boundary map for Z coefficients. Thus we have the following algebraic problem: ∂n-----→ Cn−1 -→ ··· of free abelian groups Cn, Given a chain complex ··· -→ Cn is it possible to compute the homology groups Hn(C; G) of... |
sequence of homology groups has the form (iii) ··· -→ Bn ⊗ G -→ Zn ⊗ G -→ Hn(C; G) -→ Bn−1 ⊗ G -→ Zn−1 ⊗ G -→ ··· The ‘boundary’ maps Bn ⊗ G→Zn ⊗ G in this sequence are simply the maps in ⊗ 11 where in : Bn→Zn is the inclusion. This is evident from the definition of the boundary map in a long exact sequence of homology... |
⊗ g where j(b) = c. This ϕ is well-defined since if j(b) = j(b′) = c then b − b′ = i(a) for some a ∈ A by exactness, so b ⊗ g − b′ ⊗ g = (b − b′) ⊗ g = i(a) ⊗ g ∈ Im(i ⊗ 11). Since ϕ is a homomorphism in each variable separately, it induces a homomorphism C ⊗ G→B ⊗ G/ Im(i ⊗ 11). This is clearly an inverse to the map B... |
⊗ G and H ⊗ G, but to the left of these two terms it may not be exact. For the moment let us write Hn(F ⊗ G) for the homology group Ker(fn ⊗ 11)/ Im(fn+1 ⊗ 11). Lemma 3A.2. For any two free resolutions F and F ′ of H there are canonical isomorphisms Hn(F ⊗ G) ≈ Hn(F ′ ⊗ G) for all n. Proof: We will use Lemma 3.1(a). I... |
⊔⊓ The group Hn(F ⊗ G), which depends only on H and G, is denoted Torn(H, G). Since a free resolution 0→F1→F0→H→0 always exists, as noted in §3.1, it follows that Torn(H, G) = 0 for n > 1. Usually Tor1(H, G) is written simply as Tor(H, G). As we shall see later, Tor(H, G) provides a measure of the common torsion of H ... |
oring C with G gives another complex C ⊗ G whose homology groups are denoted Hn(C; G). The following result is known as the universal coefficient theorem for homology since it describes homology with arbitrary coefficients in terms of homology with the ‘universal’ coefficient group Z. Theorem 3A.3. If C is a chain complex of... |
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