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cycles in C, by the definition of p. Corollary 3A.4. For each pair of spaces (X, A) there are split exact sequences 0 -→ Hn(X, A) ⊗ G -→ Hn(X, A; G) -→ Tor(Hn−1(X, A), G) -→ 0 for all n, and these sequences are natural with respect to maps (X, A)→(Y, B). ⊔⊓ The splitting is not natural, for if it were, a map X→Y that i... |
free, it has a free resolution with Fn = 0 for n ≥ 1, so Tor(A, B) = 0 for all B. On the other hand, if B is free, then tensoring a free resolution of A with B preserves exactness, since tensoring a sequence with a direct sum of Z ’s produces just a direct sum of copies of the given sequence. So Tor(A, B) = 0 in this ... |
process. These lie in a finitely generated subgroup B0 ⊂ B, so i xi ⊗ bi lies in the kernel of ϕ ⊗ 11 : F1 ⊗ B0→F0 ⊗ B0. This kernel is zero since Tor(A, B0) = 0, as B0 is finitely generated and torsionfree, hence free. P P Finally, we can obtain statement (4) by applying (6) to the short exact sequence 0→T (A)→A→A/T (A... |
Z), (ii) a Zp summand for each Zpk summand in Hn(X; Z), k ≥ 1, (iii) a Zp summand for each Zpk summand in Hn−1(X; Z), k ≥ 1. ⊔⊓ Even in the case of nonfinitely generated homology groups, field coefficients still give good qualitative information: Hn(X; Z) = 0 for all n iff Corollary 3A.7. (a) all n and all primes p. e (b) ... |
� Q is injective, hence A = 0 if A ⊗ Q = 0. ⊔⊓ The algebra by means of which the Tor functor is derived from tensor products has a very natural generalization in which abelian groups are replaced by modules over a fixed ring R with identity, using the definition of tensor product of R modules given in §3.2. Free resoluti... |
for all n and all primes p, then 3. Show that if Hn(X; Z) = 0 for all n, and hence e e 4. Show that ⊗ and Tor commute with direct limits: (lim--→Aα) ⊗ B = lim--→(Aα ⊗ B) and e Tor(lim--→Aα, B) = lim--→ Tor(Aα, B). 5. From the fact that Tor(A, B) = 0 if A is free, deduce that Tor(A, B) = 0 if A H n(X; G) = 0 for all G ... |
��nition for singular homology, involving explicit simplicial formulas. More enlightening, however, is the definition in terms of cellular homology. This necessitates assuming X and Y are CW complexes, but this hypothesis can later be removed by the technique of CW approximation in §4.1. We shall focus therefore on the ... |
chain group Ci(X) is free with basis the i cells of X, but there is a sign ambiguity for the basis element corresponding to each cell ei, The General K¨unneth Formula Section 3.B 269 namely the choice of a generator for the Z summand of Hi(X i, X i−1) corresponding to ei. Only when i = 0 is this choice canonical. We r... |
can be continuously deformed to an orthonormal basis, by the Gram–Schmidt process, and two orthonormal bases are related either by a rotation or a rotation followed by a reflection, according to the sign of the determinant of the transformation taking one to the other.) With this in mind, we adopt the convention that a... |
2 × ··· × en, which has a minus sign in (∗). To check that this is right, consider the n simplex [v0, ···, vn] with v0 at the origin and vk the unit vector along the k th coordinate axis for k > 0. This simplex defines the ‘positive’ orientation of In as described earlier, and in the usual formula for its boundary the f... |
. The coefficient g∗(e mαγ is the degree of the composition fαγ : S i→X i/X i−1→Z i/Z i−1→S i where the first and third maps are induced by characteristic maps for the cells ei γ, and the middle map is induced by the cellular map f. With the natural choices of basepoints in these quotient spaces, fαγ is basepoint-preservi... |
= S i, we have the commutative diagram at the right, and from the induced commutative diagram of homology groups Hi+1 we deduce that Sf and f ∧ 11 have the same degree. Since suspension preserves degree by Proposition 2.33, we conclude that deg(f ∧ 11) = deg(f ). The 11 in this formula is the identity map on S 1, and ... |
and ( × )∗(ei × ej) = ei α × e j β, hence Φ Ψ d(ei Ψ Φ α × e j β) = d = ( ( = ( since ( )∗(ei × ej) × )∗d(ei × ej) × Ψ )∗(dei × ej + (−1)iei × dej) Ψ Φ ∗(ei)× ∗(ei)× d Φ Ψ ∗(ej) + (−1)i ∗(ej) + (−1)i Ψ × Φ × Ψ Φ ∗(dei)× Ψ Φ ∗(ei)× = d Φ = ∗(dej) ∗(ej ) α× ej = dei Φ Ψ β + (−1)iei α × dej β Φ Ψ )∗ is a chain map by the... |
S 1 by attaching a 2 cell by a map of degree n. Thus X and Y each have CW structures with three cells and so X × Y has nine cells. These are indicated by the dots in the diagram at the right, with X in the horizontal direction and Y in the vertical direction. The arrows denote the nonzero cellular boundary maps. For e... |
next chapter in Theorem 4.8, that every map between CW complexes is homotopic to a cellular map. As we mentioned earlier, a cellular map induces a chain map between cellular chain complexes. It is easy to see from the equivalence between cellular and singular homology that the map on cellular homology induced by a cel... |
this is not always the case, though it is true in Example 3B.3. Our main goal in what follows is to show that the map is always injective, and that its cokernel is. More generally, we consider other coefficients besides iTor Hi(X; Z), Hn−i−1(Y ; Z) Z and show in particular that with field coefficients the map is an isomorp... |
2c ⊗ c′ + (−1)i−1∂c ⊗ ∂c′ + (−1)i∂c ⊗ ∂c′ + c ⊗ ∂2c′ = 0 From the boundary formula ∂(c ⊗ c′) = ∂c ⊗ c′ + (−1)ic ⊗ ∂c′ it follows that the tensor product of cycles is a cycle, and the tensor product of a cycle and a boundary, in either order, is a boundary, just as for the cross product defined earlier. So there is induc... |
� Ci ⊗RHn−i(C ′) = Hi(C) ⊗RHn−i(C ′). Summing over i yields an isomorphism Hn(C ⊗RC ′) ≈, which is the statement of the theorem since there are no Tor terms, Hi(C) = Ci being free. In the general case, let Zi ⊂ Ci and Bi ⊂ Ci denote kernel and image of the boundary homomorphisms for C. These give subchain complexes Z a... |
L Bi ⊗R Hn−i−1(C ′) in−1 ------------→ Zi ⊗R Hn−i−1(C ′) i -→ ··· 0 -→ Coker in -→ Hn(C ⊗R C ′) -→ Ker in−1 -→ 0 The General K¨unneth Formula Section 3.B 275 where Coker in = by Lemma 3A.1. It remains to identify Ker in−1 with Hi(C) ⊗RHn−i(C ′) By the definition of Tor, tensoring the free resolution 0→Bi→Zi→Hi(C)→0 i H... |
groups ilarly we obtain C ′ Hi(C) and Hj(C ′) as chain complexes H(C) and H(C ′) with trivial boundary maps, we thus have chain maps C→H(C) and C ′→H(C ′), whose tensor product is a chain map C ⊗RC ′→H(C) ⊗RH(C ′). The induced map on homology for this last chain map is the desired splitting map since the chain complex... |
formula, since we can homotope arbitrary maps X→X ′ and Y →Y ′ to be cellular by Theorem 4.8, assuring that they induce chain maps of ⊔⊓ cellular chain complexes. With field coefficients the K¨unneth formula simplifies because the Tor terms are always zero over a field: Corollary 3B.7. If F is a field and X and Y are CW com... |
have a reduced K¨unneth formula 0 -→ i L Hi(X; R)⊗R e Hn−i(Y ; R) e -→ Hn(X ∧ Y ; R) -→ iTorR e Hi(X; R), Hn−i−1(Y ; R) -→ 0 If we take Y = S k for example, then X ∧ S k is the k fold reduced suspension of X, and we obtain isomorphisms Hn+k(X ∧ S k; R). Hn(X; R) ≈ e e L The K¨unneth formula and the universal coefficient... |
well as by the algebra of tensor products. For general homology theories this formula can be used as a definition of homology with coefficients. One might wonder about a cohomology version of the K¨unneth formula. Taking coefficients in a field F and using the natural isomorphism Hom(A ⊗ B, C) ≈ Hom A, Hom(B, C), the K¨unne... |
with n→X × Y, and the idea is to subdivide m→X and g : m × m × ∆ ∆ ∆ ∆ appropriate signs. ∆ ∆ m × In the special cases that m or n is 1 we have already seen how to subdivide n into simplices when we constructed prism operators in §2.1. The generalm as ization to ∆ n as w0, w1, ···, wn. Think of the pairs (i, j) with v... |
Rm defined by the inin the following way. We can view ∆ equalities 0 ≤ x1 ≤ ··· ≤ xm ≤ 1, with the vertex vi as the point having coordin ⊂ Rn with coordinates nates m − i zeros followed by i ones. Similarly we have n then consists of (m + n) tuples 0 ≤ y1 ≤ ··· ≤ yn ≤ 1. The product (x1, ···, xm, y1, ···, yn) satisfyin... |
��ni- tion in homology. Given CW complexes X and Y, define a cross product of cellular cochains ϕ ∈ C k(X; R) and ψ ∈ C ℓ(Y ; R) by setting (ϕ× ψ)(ek α × eℓ β) = ϕ(ek α)ψ(eℓ β) and letting ϕ× ψ take the value 0 on (k + ℓ) cells of X × Y which are not the product of a k cell of X with an ℓ cell of Y. Another way of sayin... |
in cellular cohomology. To show this agrees with the earlier definition, we can first reduce to the case that X has trivial (k − 1) skeleton and Y has trivial (ℓ − 1) skeleton via the commutative diagram The left-hand vertical map is surjective, so by commutativity, if the two definitions of cross product agree in the up... |
� 2 (b) = are the identity map of X. p∗ 1 (a) p∗ 2 (b) ∆ Unfortunately, the definition of cellular cross product cannot be combined with ∆ ∆ ∆ to give a definition of cup product at the level of cellular cochains. This is because ∆ is not a cellular map, so it does not induce a map of cellular cochains. It is possible ∆ ... |
ological K¨unneth formula cannot be natural by considering the map f × 11 : M(Zm, n)× M(Zm, n)→S n+1 × M(Zm, n) where f collapses the n skeleton of M(Zm, n) = S n ∪ en+1 to a point. 4. Show that the cross product of fundamental classes for closed R orientable manifolds M and N is a fundamental class for M × N. 5. Show ... |
finition represents something of a middle ground. One could weaken the definition by dropping the condition that the homotopies pre- serve the basepoint e, or one could strengthen it by requiring that e be a strict identity, without any homotopies. An exercise at the end of the section is to show the three possible defini... |
3.D for GLn(R), and the other two cases are similar. Among the simplest H–spaces from a topological viewpoint are the unit spheres S 1 in C, S 3 in the quaternions H, and S 7 in the octonions O. These are H–spaces since the multiplications in these division algebras are continuous, being defined by 282 Chapter 3 Cohomol... |
π ik/q with ρ an arbitrary positive real, k an arbitrary integer, and q a fixed positive integer. The quotient of C∞ − {0} under this identification, an infinite-dimensional lens space L∞ with π1(L∞) ≈ Zq, is therefore also an associative, commutative H–space. This includes RP∞ in particular. The spaces J(X) defined in §3.... |
of particular interest since many of the most important spaces in algebraic topology turn out to be H–spaces. H–Spaces and Hopf Algebras Section 3.C 283 Hopf Algebras Let us look at cohomology first. Choosing a commutative ring R as coefficient ring, we can regard the cohomology ring H ∗(X; R) of a space X as an algebra ... |
⊗ 1. A similar argument shows the component in H 0(X; R) ⊗R H n(X; R) is 1 ⊗ α. We can summarize this situation by saying that H ∗(X; R) is a Hopf algebra, that n≥0 An over a commutative base ring R, satisfying the = 11. This implies that the component of is, a graded algebra A = ∆ ∆ following two conditions: (1) Ther... |
in R, then the multiplication in R[α] ⊗ R[α] is strictly commutative and αi ⊗ αn−i. In the opposite case that α is odd-dimensional, (α ⊗ 1 + 1 ⊗ α)n = (α2) = (α ⊗ 1 + 1 ⊗ α)2 = α2 ⊗ 1 + 1 ⊗ α2 since (α ⊗ 1)(1 ⊗ α) = α ⊗ α and then (1 ⊗ α)(α ⊗ 1) = −α ⊗ α if α has odd dimension. Thus if we set β = α2, then β βi ⊗ βn−i ... |
algebra is generated as an algebra by primitive elements, then the coproduct ∆ is uniquely determined by the product. This happens in a number of interesting special cases, but certainly not in general, as we shall see. The existence of the coproduct in a Hopf algebra turns out to restrict the multi- ∆ plicative struc... |
by an exercise at the end of this section, the universal cover of an H–space is an H–space, and S 1, S 3, and S 7 are the only spheres that are H–spaces, by the theorem of Adams mentioned earlier. The tensor product A ⊗ B of Hopf algebras A and B is again a Hopf algebra, with ----------------------------→ (A ⊗ A) ⊗ (B... |
xi) involves only xi and terms of smaller dimension. We may assume xn since does not lie in An−1. Since A is associative and commutative, there is a natural F [xn]→An if |xn| is odd. surjection An−1 ⊗ F [xn]→An if |xn| is even, or An−1 ⊗ By induction on n it will suffice to prove these surjections are injective. Thus in ... |
6∈ I implies xn ≠ 0. The relation n = 0 has lower degree than the original relation, and is not the trivial relation since F has characteristic 0, αi ≠ 0 implying iαi ≠ 0 if i > 0. Since we could P assume the original relation had minimum degree, we have reached a contradiction. to a relation α0 + α1xn = 0 gives 0 = α... |
infinite tensor product It is interesting to see what happens to the divided polynomial algebra Z[α] Q[α] is the same as Q[α]. In contrast αi+j, happens to be pi ), as we will show in a moment. However, as Hopf algebras these two objects are different since αpi is primitive in coproduct in [α] when i > 0, since the i≥0 ... |
only in the coefficient of pk, so mod p we have = nk + 1. This conclusion also holds if nk + 1 = p, when the p adic representations of n + pk and n differ also in the coefficient of pk+1. The statement (∗∗) then follows. n+pk n nk+1 nk = Lemma 3C.6. If p is a prime, then k = i nipi and i i kipi with 0 ≤ ni < p and 0 ≤ ki <... |
R) ×--------------→ H∗(X × X; R) µ∗--------------→ H∗(X; R) where the first map is the cross product defined in §3.B. Thus the Pontryagin product consists of bilinear maps Hi(X; R)× Hj(X; R)→Hi+j(X; R). Unlike cup product, the Pontryagin product is not in general associative unless the multiplication µ is associative or... |
the cup product structure, though for some H–spaces it is the other way round. In applications it is often convenient to have the choice of which product structure to use. This calculation immediately generalizes to J(X) where X is any connected CW complex whose cellular boundary maps are all trivial. The cellular bou... |
X) -→ H∗ J(X) J(X) be the homomorphism e H–Spaces and Hopf Algebras Section 3.C 289 where the next-to-last map is induced by the quotient map X n→Jn(X). It is clear that ϕ is a ring homomorphism since the product in J(X) is induced from the natural map Xm × X n→Xm+n. To show that ϕ is an isomorphism, consider the follo... |
R = Z and assume Hn(X; Z) is finitely generated and free for all n. In either case we have H n(X; R) = HomR(Hn(X; R), R), and as a consequence the Pontryagin product H∗(X; R) ⊗ H∗(X; R)→H∗(X; R) and the : H ∗(X; R)→H ∗(X; R) ⊗ H ∗(X; R) are dual to each other, both being incoproduct duced by the H–space product µ : X ×... |
finitely generated free R module in each dimension. Then the product π : A ⊗ A→A and coproduct ∗ : A∗ ⊗ A∗→A∗ that give A∗ the : A→A ⊗ A have duals π ∗ : A∗→A∗ ⊗ A∗ and structure of a Hopf algebra. ∆ ∆ Proof: This will be apparent if we reinterpret the Hopf algebra structure on A formally as a pair of graded R module h... |
j(−1)|a′′ P route gives first j b′ i b′′ and π ⊗ π this becomes ∆ (a) ⊗ j which is (a) ∆ (b). (ab), while the lower, then after applying τ ∆ i a′ j ⊗ b′′ i ⊗ a′′ j b′ j i, P P P Condition (1) for A dualizes to (2) for A∗, and similarly (2) for A dualizes to (1) for A∗. Condition (3) for A dualizes to (3) for A∗. ⊔⊓ ∆ ∆ ... |
J(S n); R) from the geometric calculation H∗(J(S n); R) ≈ R[x] in Example 3C.7. As another application, recall from earlier in this section that RP∞ and CP∞ are H–spaces, so from their H–Spaces and Hopf Algebras Section 3.C 291 cup product structures we can conclude that the Pontryagin rings H∗(RP∞; Z2) and H∗(CP∞; Z) ... |
group with respect to the multiplication induced by the H–space structure, all the path- components must be homotopy equivalent. [Homotopy-associative means associative up to homotopy.] 4. Show that an H–space or topological group structure on a path-connected, locally path-connected space can be lifted to such a stru... |
·· + a1xb1 + a0 of nonzero degree with coefficients in H has a root in H. [See Theorem 1.8.] 11. If T n is the n dimensional torus, the product of n circles, show that the Pontryagin ring H∗(T n; Z) is the exterior algebra Z[x1, ···, xn] with |xi| = 1. 12. Compute the Pontryagin product structure in H∗(L; Zp) where L is ... |
basic objects in algebraic and geometric topology. The orthogonal group O(n) can be defined as the group of isometries of Rn fixing the origin. Equivalently, this is the group of n× n matrices A with entries in R such that AAt = I, where At is the transpose of A. From this viewpoint, O(n) is topologized as a subspace of... |
same planes through decreasing angles. Another reason why SO(n) is connected is that it has a CW structure with a single 0 cell, as we show in Proposition 3D.1. An exercise at the end of the section is to show that a topological group with a finite-dimensional CW structure is an orientable manifold, so SO(n) is a close... |
ified. The map ϕ is clearly injective since the axis of a nontrivial rotation is uniquely determined as its fixed point set, and ϕ is surjective since by easy linear algebra each nonidentity element 294 Chapter 3 Cohomology of SO(3) is a rotation about some axis. It follows that ϕ is a homeomorphism RP3 ≈ SO(3). SO(4) is... |
S 3 × S 3 gives the universal cover S 3→SO(3), so SO(3) is isomorphic to the quotient group of S 3 by the normal subgroup {±1}.. Restricting to identity components, we obtain a homeo- v, (1, v2, v3, v4) It Using octonions one can construct in the same way a homeomorphism SO(8) ≈ S 7 × SO(7). But in all other cases SO(... |
ne a map ρ : P I = P i1 × ··· × P im→SO(n) by letting ρ(v1, ···, vm) be the composition If ϕi : Di→P i is the standard characteristic map for the i cell of ρ(v1) ··· ρ(vm). P i, restricting to the 2 sheeted covering projection ∂Di→P i−1, then the product ϕI : DI→P I of the appropriate ϕij ’s is a characteristic map for... |
P n−1 ⊂ SO(n) fix en, so p(P n−2) = {en}. We claim that p is a homeomorphism from P n−1 − P n−2 onto S n−1 − {en}. This can be seen as follows. Thinking of a point in P n−1 as a vector v, the map p takes this to ρ(v)(en) = r (v)r (e1)(en), which equals r (v)(en) since en is in the hyperplane orthogonal to e1. From the ... |
−2 to SO(n − 1), so we may assume inductively that the maps ρϕI for I ranging 296 Chapter 3 Cohomology over admissible sequences with first term i1 < n − 1 are the characteristic maps for a CW structure on SO(n − 1), with cells the corresponding products eI. The admissible sequences I with i1 = n − 1 then give disjoint ... |
� R2 by statement (2) for n = 2. Therefore α(w) α(v) r r (v)r (w) = α−1r (z)r (e1)α = r α−1(z) r for x = α−1(z) ∈ Ri+1 and y ∈ Ri. α−1(e1) = r (x)r (y) It remains to show that the map ρ : P n−1× P n−2 × ··· × P 1→SO(n) is cellular. This follows from the inclusions P iP i ⊂ P iP i−1 derived above, together with another ... |
P i and hence in P n−1× P n−2 × ··· × P 1 by Proposition 3B.1. Thus H∗(SO(n); Z2) has a Z2 summand for each cell of SO(n). One can rephrase this The Cohomology of SO(n) Section 3.D 297 as saying that there are isomorphisms Hi(SO(n); Z2) ≈ Hi(S n−1 × S n−2 × ··· × S 1; Z2) for all i since this product of spheres also h... |
of P n−1 × ··· × P 1 and 0 to all other i cells, which are products of lower-dimensional cells and hence map to cells in SO(n) disjoint from ei. j αi P P j αi ··· βim First we will show that the monomials βI = βi1 corresponding to admissible sequences I are linearly independent in H ∗(SO(n); Z2), hence are a vector sp... |
j. The quotient Q of the algebra Z2[β1, β2, ···] by the relations β2 2 = j αi j i = β2i and P βj = 0 for j ≥ n then maps onto H ∗(SO(n); Z2). This map Q→H ∗(SO(n); Z2) P is also injective since the relations defining Q allow every element of Q to be represented as a linear combination of admissible monomials βI, and th... |
(v)w)ρ(v ′) for v ∈ Ri+1, w ∈ Rj+1, and v ′ = r (e1)v. The map f : P i × P j→P j × P i, f (v, w) = (ρ(v)w, v ′), is a homeomorphism since it is the composition of homeomorphisms (v, w) ֏ (v, ρ(v)w) ֏ (v ′, ρ(v)w) ֏ (ρ(v)w, v ′). The first of these maps takes ei × ej homeomorphically onto itself since ρ(v)(ej ) = ej if i... |
product of the corresponding truncated divided polynomial algebras, in other words an exterior algebra as just explained. This is in fact the structure of H∗(SO(n); Z2), so since the Pontryagin product in H∗(SO(n); Z2) determines the coproduct in H ∗(SO(n); Z2) uniquely, it follows that the βi ’s must indeed be primit... |
βi is primitive. ∆ Integer Homology and Cohomology ∆ With Z coefficients the homology and cohomology of SO(n) turns out to be a good bit more complicated than with Z2 coefficients. One can see a little of this complexity already for small values of n, where the homeomorphisms SO(3) ≈ RP3 and SO(4) ≈ S 3 × RP3 would allow ... |
in P 2i × P 2i−1, d(e2i × e2i−1) = de2i × e2i−1 + e2i × de2i−1 = 2e2i−1× e2i−1, hence d(e2ie2i−1) = 0 since P 2i−1P 2i−1 ⊂ P 2i−1P 2i−2. The claim is that there are chain complex isomorphisms SO(n) C∗ C∗ SO(2k + 1) SO(2k + 2) ≈ C 2⊗C 4⊗ ··· ⊗C 2k ≈ C 2⊗C 4⊗ ··· ⊗C 2k⊗C 2k+1 where C 2k+1 has basis e0 and e2k+1. Certain... |
The same strategy applies equally well to cohomology, and the uni∗ versal coefficient theorem gives an isomorphism H ∗ (SO(n); Z). (SO(n); Z) f r ee(SO(n); Z) ≈ H f r ee The proof of the proposition shows that the additive structure of H f r ee ∗ ∗ is fairly simple: H f r ee ∗ H f r ee ∗ (SO(2k + 1); Z) ≈ H∗(S 3 × S 7 ×... |
D.2 gives a homeomorphism ei × ej ≈ ej × ei if i < j, and this homeomorphism has local degree (−1)ij+1 since it is the composition (v, w) ֏ (v, ρ(v)w) ֏ (v ′, ρ(v)w) ֏ (ρ(v)w, v ′) of homeomorphisms with local degrees +1, −1, and (−1)ij. Applying this four times to commute EiEj = e2ie2i−1e2je2j−1 to EjEi = e2je2j−1e2ie... |
, with C(0) having basis the 0 cell of SO(n). The direct sum C(0)⊕ ··· ⊕ C(m) is the cellular chain complex of the subcomplex of SO(n) consisting of cells that are products of m or fewer cells ei. In particular, taking m = 2 we have a subcomplex X ⊂ SO(n) whose homology, mod torsion, consists of the Z in dimension zero... |
7, though the process is somewhat laborious and the answer not very neat. Stiefel Manifolds Consider the Stiefel manifold Vn,k, whose points are the orthonormal k frames in Rn, that is, orthonormal k tuples of vectors. Thus Vn,k is a subset of the product of k copies of S n−1, and it is given the subspace topology. As ... |
i1 ··· eim SO(n − k) for n > i1 > ··· > im ≥ n − k, together with the coset SO(n − k) itself as a 0 cell of Vn,k. These sets of cosets are unions of cells of SO(n) since SO(n−k) consists of the cells eJ = ej1 ··· ejℓ with n−k > j1 > ··· > jℓ. This implies that Vn,k is the disjoint union of its cells, and the boundary o... |
= V3,2× S 3 as we noted at the beginning of this section. Also SO(3) = V3,2 and SO(2) = S 1. Exercises 1. Show that a topological group with a finite-dimensional CW structure is an orientable manifold. [Consider the homeomorphisms x ֏ gx or x ֏ xg for fixed g and varying x in the group.] 2. Using the CW structure on SO(... |
·· -→ H n(X; G) -→ H n(X; H) -→ H n(X; K) -→ H n+1(X; G) -→ ··· whose ‘boundary’ map H n(X; K)→H n+1(X; G) is called a Bockstein homomorphism. We shall be interested primarily in the Bockstein β : H n(X; Zm)→H n+1(X; Zm) asm-----→ Zm2 -→ Zm→0, especially when m is sociated to the coefficient sequence 0→Zm prime, but for ... |
is surjective, so exactness, hence β = 0. e β = 0 by e A useful fact about β is that it satisfies the derivation property (∗) β(a ` b) = β(a) ` b + (−1)|a|a ` β(b) which comes from the corresponding formula for ordinary coboundary. Namely, let ϕ and ψ be Zm cocycles representing a and b, and let ψ be lifts of these to ... |
→CP∞. Looking at the cell structure on L described in Example 2.43, we see that each even-dimensional cell of L projects homeomorphically onto the corresponding cell of CP∞. Namely, the 2n cell of L is the homeomorphic image of the 2n cell in S 2n+1 ⊂ Cn+1 formed by the points i |zi|2 = 1 and 0 < θ ≤ π /2, and the cos ... |
infinite-dimensional lens space, and the homotopy type of an infinite-dimensional lens space is determined by its fundamental group since it is a K(π, 1). It follows that the cup product structure on a lens space S 2n+1/Zm with Zm coefficients is obtained from the preceding calculation by truncating via the relation y n+1... |
as G1, G2, ···. For each Gi choose a generator gi and define a corresponding chain complex M(gi) by the following prescription. If gi has infinite order in Gi ⊂ Hni (C), let M(gi) consist of just a Z in dimension ni, with generator zi. On the other hand, if (C), let M(gi) consist of Z ’s in dimensions ni and ni + 1, gi ... |
Zp), and from the definition of Bockstein homomorphisms it follows i ∈ H ni+1(M; Zp). The that latter element is nonzero iff k is not divisible by p2. ⊔⊓ i ∈ H ni+1(M; Z) and β(x∗ β are zero on y ∗ i ) = (k/p)y ∗ i ) = (k/p)y ∗ i and z∗ β(x∗ e e Corollary 3E.4. In the situation of the preceding proposition, H ∗(X; Z) co... |
) = 0, hence Theorem 3.15. The universal coefficient theorem then implies that H ∗(RP∞ × RP∞; Zp) = 0 by H ∗(RP∞ × RP∞; Z) consists entirely of elements of order a power of 2. From Example 3E.1 we know that Bockstein homomorphisms in H ∗(RP∞; Z2) ≈ Z2[x] are given by β(x2k−1) = x2k and β(x2k) = 0. In H ∗(RP∞ × RP∞; Z2) ≈... |
, ν 2 + λ2µ + λµ2) where ρ(λ) = x2, ρ(µ) = y 2, ρ(ν) = x2y + xy 2, and the relation ν 2 + λ2µ + λµ2 = 0 holds since (x2y + xy 2)2 = x4y 2 + x2y 4. This calculation illustrates the general principle that cup product structures with Z coefficients tend to be considerably more complicated than with field coefficients. One can ... |
, and 4, and a Z2 in dimension 3. In both cases all cup products of positive-dimensional classes are zero since for dimension reasons the only possible nontrivial product is the square of the 2 dimensional class, but this is zero as one sees by restricting to the subcomplex S 2 × S 2 or S 2 ∨ S 2 ∨ S 4. For the space Y... |
that with Z coefficients, the cup product H 2 × H 5→H 7 is nontrivial for X1 × Y but trivial for X2 × Y. For in H ∗(Xi × Y ; Z2) we have, using the relation (a× b) ` (c × d) = (a ` c)× (b ` d) which follows immediately from the definition of cross product, (1) e2 1 × f 0 ` e2 1 × f 3 = (e2 1 × f 0 ` (e3 × f 2 + e2 (2) e2... |
(β1)β3 + β1β(β3) = β2β3 + β1β4 = β2 1. This Bockstein data allows us to calculate H i(SO(5); Z) modulo odd torsion, with the results indicated in the remainder of the table, where the vertical arrows denote the map ρ. As 1β3 + β5 j α2i j α2i 1, β2 P P = j j we showed in Proposition 3D.3, there is no odd torsion, so thi... |
∗(SO(7); Z2). Here the numbers across the top indicate dimension, stopping with 21, the dimension of SO(7). The labels on the dots refer to the basis of products of distinct βi ’s. For example, the dot labeled 135 is β1β3β5. The left-right symmetry of the figure displays Poincar´e duality quite graphically. Note that t... |
±ℓ′ for §4.2. (a) Let L = Lm(ℓ1, ···, ℓn) and let Z∗ m be the multiplicative group of invertible elements of Zm. Define t ∈ Z∗ m by the equation xy n−1 = tz where x is a generator of H 1(L; Zm), y = β(x), and z ∈ H 2n−1(L; Zm) is the image of a generator of H 2n−1(L; Z). Show that the image τ(L) of t in the quotient gr... |
smash product of k copies of a Moore space M(Zp, n) with p prime. Compute the Bockstein homomorphisms in H ∗(X; Zp) and use this to describe H ∗(X; Z). 4. Using the cup product structure in H ∗(SO(5); Z), show that SO(5) is not homotopy equivalent to the product of any two CW complexes with nontrivial cohomology. Limi... |
), ···). It is easy to see from this definition that every element of lim--→Gi is represented by an element gi ∈ Gi for some i, and two such representatives gi ∈ Gi and gj ∈ Gj define the same element of lim--→Gi iff they have the same image in some Gk under the appropriate composition of αℓ ’s. If all the αi ’s are injec... |
1 ` scope is the quotient space of the disjoint i (Xi × [i, i + 1]) in which each point union (xi, i + 1) ∈ Xi × [i, i + 1] is identified with (fi(xi), i + 1) ∈ Xi+1 × [i + 1, i + 2]. In the mapping telescope T, let Ti be the union of the first i mapping cylinders. This deformation retracts onto Xi by deformation retract... |
group homomorphisms, the inverse limit lim ←-- Gi is defined i Gi consisting of sequences (gi) with αi(gi) = gi−1 for all i. to be the subgroup of There is a natural map λ : H n(X; G)→ lim ←-- H n(Xi; G) sending an element of H n(X; G) to its sequence of images in H n(Xi; G) under the maps H n(X; G)→H n(Xi; G) induced ... |
or more generally for any cohomology theory. Given a sequence of homomorphisms of abelian groups ··· ------→ G2 α2------------→ G1 α1------------→ G0 i Gi→ define a map δ : lim ←-- Gi is the kernel of δ. Denoting the cokernel of δ by lim ←-sequence i Gi by δ(···, gi, ···) = (···, gi − αi+1(gi+1), ···), so that 1Gi, we ... |
�-- Gi or lim ←-- ←-- Gi = lim ←-- 1Gi = 0. Q Example 3F.6. Consider the sequence of natural surjections ··· →Zp3→Zp2→Zp with p a prime. The inverse limit of this sequence is a famous object in number theory, Zp. It is actually a commutative called the p adic integers. Our notation for it will be ring, not just a group... |
The next example shows b b how p adic integers can also give rise to a nonvanishing lim ←-- 1 term. p-----→ Z ∞ Z→ 1 term is the cokernel of the map δ : p-----→ Z for p prime. In this case Example 3F.7. Consider the sequence ··· -→ Z the inverse limit is zero since a nonzero integer can only be divided by p finitely of... |
→ ···. Letting Ti be the union of the first i mapping cylinders in the telescope, the inclusions T1 ֓ T2 ֓ ··· induce on H n(−; Z) the sequence p-----→ Z in Example 3F.7. From the theorem we deduce that H n+1(T ; Z) ≈ ··· -→ Z Zp/Z H k(T ; Z) = 0 for k ≠ n+1. Thus we have the rather strange situation that the CW and com... |
inclusion S n֓T of one end of the telescope, then the long exact sequences of homology and cohomology groups for the pair (T, S n) reduce to the short exact sequences at the right. Zp, the p adic integers, b From these examples and the universal coefficient theorem we obtain isomorZp/Z. These can also be derived phisms ... |
, ···) + ···, so Coker ϕ∗ can be identified with Zp by rewriting a sequence (z1, z2, ···) as the p adic number ··· z2z1. b The calculation Ext(Z[1/p], Z) ≈ b Zp/Z is quite similar. A free resolution of Z[1/p] can be obtained from the free resolution of Zp∞ by omitting the first column of the matrix of ϕ and, for convenie... |
are essentially the same except for a dimension shift in the torsion. But matters are more complicated in the nonfinitely generated case. A useful tool for getting a handle on this complication is the following: Proposition 3F.11. Given an abelian group G and a short exact sequence of abelian groups 0→A→B→C→0, there ar... |
The associated long exact sequence of homology groups has six terms, the first three being the kernels of the three vertical maps to A, B, and C, and the last three being the cokernels of these maps. Since the vertical maps to A and C are surjective, the fourth and sixth of the six homology groups vanish, hence also th... |
b Zp)/Z using the second exact sequence in the proposition. In these examples the groups Ext(A, Z) are rather large, and the next result says p Q b b this is part of a general pattern: Proposition 3F.12. If A is not finitely generated then either Hom(A, Z) or Ext(A, Z) is uncountable. Hence if Hn(X; Z) is not finitely g... |
proof: (c) If A is a direct sum of infinitely many nontrivial finite cyclic groups, then Ext(A, Z) is uncountable, the product of infinitely many nontrivial groups Ext(Zn, Z) ≈ Zn. (d) For p prime, Example 3F.10 gives Ext(Zp∞, Z) ≈ Zp which is uncountable. Consider now the map A→A given by a ֏ pa for a fixed prime p. Deno... |
= C1 ⊂ C2 ⊂ ···. If the indices [Ci : Ci−1] involve infinitely ∞ Zp for these many distinct prime factors p then A/C contains an infinite sum p so Ext(A/C, Z) is uncountable by (a) and (c) and hence also Ext(A, Z) by (b). If only finitely many primes are factors of the indices [Ci : Ci−1] then A/C contains a subgroup Zp∞... |
∞, Zp) ≈ Zp. 7. Show that for a short exact sequence of abelian groups 0→A→B→C→0, a Moore space M(C, n) can be realized as a quotient M(B, n)/M(A, n). Applying the long exact Zp for p prime. b sequence of cohomology for the pair M(B, n), M(A, n) with any coefficient group G, deduce an exact sequence 0→Hom(C, G)→Hom(B, G)... |
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