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olev inequalities let us do so, since if u ∈ W k,p(U ) for all k, then it must be in C m as well. 0 ([0, 1]). A priori, if u ∈ H 1 To see why we should be expected to be able to do that, consider the space H 1 0 ([0, 1]), then we only know it exists as some measurable function, and there is no canonical representative ... |
αuv for all u, v ∈ H – βu2 ≤ B[u, u] (boundedness) (coercivity) Then if f : H → R is a bounded linear map, then there exists a unique u ∈ H such that B[u, v] = f, v for all v ∈ H. Note that if B is just the inner product, then this is the Riesz representation theorem. Proof. By the Riesz representation theorem, we may ... |
ximation, we can assume u ∈ C∞( ¯Q). For x, y ∈ Q, we write u(x) − u(y) = x1 y1 + d dt x2 u(t, x1, . . . , xn) dt d dt d dt u(y1, t, x3, . . . , xn) dt u(y1, . . . , yn−1, t) dt. y2 + · · · + xn yn Squaring, and using 2ab ≤ a2 + b2, we have u(x)2 + u(y)2 − 2u(x)u(y) ≤ n u(t, x1, . . . , xn) dt 2 d dt x1 y1 + · · · + n ... |
djoint if L = L†. Equivalently, if bi ≡ 0. Definition (Positive operator). We say L is positive if there exists C > 0 such that H 1 u2 0 (U ) ≤ CB[u, u] for all u ∈ H 1 Theorem. Suppose L is a formally self-adjoint, positive, uniformly elliptic operator on U , an open bounded set with C 1 boundary. Then we can represent... |
is locally in H 2 to show that the second-derivative is well-behaved near the boundary. loc(U ). So we only have By a partition of unity and change of coordinates, we may assume we are in the case U = B1(0) ∩ {xn > 0}. Let V = B1/2(0) ∩ {xn > 0}. Choose a ζ ∈ C∞ 0 ≤ ζ ≤ 1. c (B1(0)) with ζ ≡ 1 on V and Most of the pro... |
+ (uN )2 e−λt + λ 2 ( ˙uN )2 + aijuN xi uN xj 59 5 Hyperbolic equations III Analysis of PDEs and B = Uτ dt dx 1 2 ˙aijuN xi uN xj − biuN xi ˙uN + (1 − c)uN ˙uN + f ˙uN e−λt. Integrating in time, and estimating as before, for λ sufficiently large, we get 1 2 Στ ( ˙uN )2 + |DuN |2 dx + Uτ ( ˙uN )2 + |DuN |2 + (uN )2 dx dt ... |
orant, 11 majorize, 11 Maxwell’s equation, 6 method of characteristics, 5 minimal surface equation, 6 mollification, 24 mollifier standard, 23 Morrey’s inequality, 34 multi-index notation, 7 non-characteristic surface, 17 ODE autonomous, 9 Cauchy–Kovalevskaya theorem, 10 order of PDE, 5 hyperbolic equation, 59 partial di... |
Homotopy Equivalence Earlier in this chapter the main tool we used for constructing homotopy equiva- lences was the fact that a mapping cylinder deformation retracts onto its ‘target’ end. By repeated application of this fact one can often produce homotopy equivalences be- tween rather different-looking spaces. However... |
d to a homotopy ft : X→Y of the given f0 . If the pair (X, A) is such that this extension problem can always be solved, one says that (X, A) has the homotopy extension property. Thus (X, A) has the homotopy extension property if every pair of maps X × {0}→Y and A× I→Y that agree on A× {0} can be extended to a map X × I... |
f homotopy equivalences X→Y and Y →Z is a 3. homotopy equivalence X→Z . Deduce that homotopy equivalence is an equivalence relation. (b) Show that the relation of homotopy among maps X→Y is an equivalence relation. (c) Show that a map homotopic to a homotopy equivalence is a homotopy equivalence. 4. A deformation retra... |
B1 + B′ 1 is deformable to B2 , linking A twice. Similarly, B1 + B−1 can be deformed to the loop B0 , unlinked from A . More generally, we see that Bm + Bn can be deformed to Bm+n for arbitrary integers m and n . Note that in forming sums of loops we produce loops that pass through the base- point more than once. This... |
et of all homotopy classes [f ] of loops f : I→X at the basepoint x0 is denoted π1(X, x0) . Proposition 1.3. π1(X, x0) is a group with respect to the product [f ][g] = [f g] . This group is called the fundamental group of X at the basepoint x0 . We will see in Chapter 4 that π1(X, x0) is the first in a sequence of group... |
X . The restrictions e e and F |{1}× I are paths lifting constant paths, hence they must also be constant by ft lifts ft e F (s, t) is a homotopy of paths, and the uniqueness part of (a). So ft(s) = e e e since → e e To prove (c) we will first construct a lift e Nt × (at, bt) X for N some neighborhood in Y of a given po... |
le is covered by two closed sets, one of them must contain a pair of antipodal points. This is of course false for nonclosed sets since the circle is the union of two disjoint half-open semicircles. 34 Chapter 1 The Fundamental Group The relation between the fundamental group of a product space and the funda- mental gr... |
roduct ht (ϕtf ) ht gives a homotopy of loops at ϕ0(x0) . Restricting this homotopy to t = 0 and t = 1 , we see that ϕ0∗([f ]) = ⊔⊓ βh . ϕ1∗([f ]) Proof of 1.18: Let ψ : Y →X be a homotopy-inverse for ϕ , so that ϕψ ≃ 11 and ψϕ ≃ 11. Consider the maps ϕ∗------------→ π1 ψ∗------------→ π1 π1(X, x0) X, ψϕ(x0) The compos... |
n ∗α Gα is juxtaposition, (g1 ··· gm)(h1 ··· hn) = g1 ··· gmh1 ··· hn . This product may not be reduced, however: If gm and h1 belong to the same Gα , they should be combined into a single letter (gmh1) according to the multiplication in Gα , and if this new letter gmh1 happens to be the identity of Gα , it should be c... |
, possibly unreduced, that is is equivalent to saying that every [f ] ∈ π1(X) mapped to [f ] by . Surjectivity of has a factorization. We will be concerned with the uniqueness of factorizations. Call two factoriza- Φ Φ tions of [f ] equivalent if they are related by a sequence of the following two sorts of moves or the... |
defined deformation retraction of S 3 − K onto X . Another way of describing the situation would be to say that for an open ε neighborhood N of K bounded by a torus T , the complement S 3 − N is the mapping cylinder of a map T→X . To compute π1(X) we apply van Kampen’s theorem to the decomposition of X as the union of X... |
words, the free group on the generators gα modulo the normal subgroup generated by the words rβ in these generators. Corollary 1.27. The surface Mg is not homeomorphic, or even homotopy equivalent, to Mh if g ≠ h . Proof: The abelianization of π1(Mg) is the direct sum of 2g copies of Z . So if Mg ≃ Mh then π1(Mg) ≈ π1... |
edges so that these edges are identified with points of three strips Ri , Rj , and Rk as in the third figure; namely, two opposite edges of Sℓ are identified with short edges of Rj and Rk and the other two opposite edges of Sℓ are identified with two arcs crossing the interior of Ri . The knot K is now a subspace of X , b... |
raph isomorphism that preserves the labeling and orientations of edges. Thus the covering spaces in (3) and (4) are isomorphic, but not by an isomorphism preserving basepoints, so the two subgroups of π1(X, x0) corresponding to these covering spaces are distinct but conjugate. On the other hand, the two covering spaces... |
omeomorphically to U by p . Each loop in U lifts to a loop in U , and the lifted loop is nullhomotopic in X since π1( X) = 0 . So, composing this nullhomotopy with p , the original loop in U is nullhomotopic in X . e e A locally simply-connected space is certainly semilocally simply-connected. For example, CW complexes... |
tion 1.3 67 and U[γ′] are identified in XH then the whole neighborhoods are identified. Hence the natural projection XH→X induced by [γ] ֏ γ(1) is a covering space. If we choose for the basepoint c at x0 , then the image of p∗ : π1(XH , for a loop γ in X based at x0 , its lift to e of this lifted path in XH is a loop iff ... |
is what one would naturally call an isomorphism of sets with π1(X, x0) action. Thus isomorphic covering spaces have isomorphic actions on fibers. The x0)) . Using the less cumbersome notation γ x0)) = h(Lγ( x0) = h(γ e X1 converse is also true, and easy to prove. One just observes that for isomorphic actions ρ1 and ρ2 ,... |
arm’ of M11 by m holes and the 5 fold symmetry by n fold symmetry. This gives a covering space Mmn+1→Mm+1 . An exercise in §2.2 is to show by an Euler characteristic argument that if there is a covering space Mg→Mh then g = mn + 1 and h = m + 1 for some m and n . As a special case of the final statement of the preceding... |
od for viewing e e e gα |||| rβ XG/G = XG in the following way. Let the vertices of groups geometrically as graphs. Recall from Corollary 1.28 how we associated to each group presentation G = a 2 dimensional cell complex XG with π1(XG) ≈ G by taking a wedge-sum of circles, one for each generator gα , and then attaching... |
if a group G acts freely and properly discontinuously on a Hausdorff space X , then the action is a covering space action. (Here ‘properly discontinuously’ means that each x ∈ X has a neighborhood U such that { g ∈ G | U ∩ g(U) ≠ ∅ } is finite.) In particular, a free action of a finite group on a Hausdorff space is a cove... |
ntal group. Another consequence is that a graph is a tree iff it is simply-connected. Proof: The quotient map X→X/T is a homotopy equivalence by Proposition 0.17. The quotient X/T is a graph with only one vertex, hence is a wedge sum of circles, whose fundamental group we showed in Example 1.21 to be free with basis the... |
y of torus knot complements in Examples 1.24 and 1.35. Namely, we showed that for K the torus knot Km,n there is a deformation retraction of S 3 − K onto a certain 2 dimensional complex Xm,n having contractible universal cover. The homotopy lifting property then implies that the universal cover of S 3 − K is homotopy e... |
equivalent to the B is Γ Γ Example 1B.10. Suppose consists of one central vertex with a number of edges Γ radiating out from it, and the group Gv at this central vertex is trivial, hence also all the edge homomorphisms. Then van Kampen’s theorem implies that π1(K ) is the free product of the groups at all the outer ver... |
1) in K , and two vertices are joined by an edge whenever the two universal covers of K(Gv , 1) ’s corresponding to these vertices are connected by a line segment lifting a line segment in the mapping cylinder structure of , has one vertex for each e Γ Γ a mapping cylinder of K construction of T Corresponding to . The ... |
b−1a are to be regarded as equal if we make a commute with b−1 . These two loops ab−1 and b−1a are really the same circle, just with a different choice of starting and ending point: x for ab−1 and y for b−1a . The same thing happens for all loops: Rechoosing the basepoint in a loop just permutes its letters cyclically, ... |
in pairs. In similar fashion a polygon with any number of sides can be cut along diagonals into triangles, so in fact all closed surfaces can be constructed from triangles by identifying edges. Thus we have a single building block, the triangle, from which all surfaces can be constructed. Using only triangles we could ... |
2L and {a, b, a + b − c} is example, ∂1 = 0 so H 1 (T ) ≈ Z⊕ Z with basis the homology classes [a] ∆ a basis for ∆ and [b] . Since there are no 3 simplices, H 2 (T ) is equal to Ker ∂2 , which is infinite cyclic generated by U − L since ∂(pU + qL) = (p + q)(a + b − c) = 0 only if p = −q . ∆ 1(T ) , it follows that H ∆ ∆... |
1 simplex, a map τi : [v0, v1]→X , and then we have ∂τi = σi − σ0 . i niσi is a Hence ∂ ⊔⊓ boundary, which shows that Ker ε ⊂ Im ∂1 . P P i ni = 0 . Thus i niσi since i niσ0 = i niσi − i niτi P P P P = Proposition 2.8. If X is a point, then Hn(X) = 0 for n > 0 and H0(X) ≈ Z . Proof: In this case there is a unique sing... |
n exact sequence 0→A→B→C→0 as in (iv) is called a short exact sequence. Exact sequences provide the right tool to relate the homology groups of a space, a subspace, and the associated quotient space: Theorem 2.13. If X is a space and A is a nonempty closed subspace that is a deformation retract of some neighborhood in ... |
reduced homology groups for the pair e Hi−1(S n−1) are isomorphisms for all i > 0 (Dn, ∂Dn) , the maps Hi(Dn, ∂Dn) Hi(Dn) are zero for all i . Thus we obtain the calculation since the remaining terms ∂-----→ e Hi(Dn, ∂Dn) ≈ e Z for i = n 0 otherwise Example 2.18. Applying the long exact sequence of reduced homology gro... |
ctive definition ∆ of S with the inductive definition of the barycentric subdivision of a simplex. Let us check that the maps S satisfy ∂S = S∂ , and hence give a chain map from the chain complex LC(Y ) to itself. Since S = 11 on LC0(Y ) and LC−1(Y ) , we certainly have ∂S = S∂ on LC0(Y ) . The result for larger n is giv... |
omplexes: Corollary 2.24. If the CW complex X is the union of subcomplexes A and B , then the inclusion (B, A ∩ B)֓ (X, A) induces isomorphisms Hn(B, A ∩ B)→Hn(X, A) for all n . Proof: Since CW pairs are good, Proposition 2.22 allows us to pass to the quotient spaces B/(A ∩ B) and X/A which are homeomorphic, assuming w... |
lie in some X k , so z represents an element of the kernel of H But we know this map is injective, so z is a simplicial boundary in X k , and therefore in X . ∆ It remains to do the case of arbitrary X with A ≠ ∅ , but this follows from the absolute case by applying the five-lemma to the canonical map from the long exac... |
5]. Degree For a map f : S n→S n with n > 0 , the induced map f∗ : Hn(S n)→Hn(S n) is a homomorphism from an infinite cyclic group to itself and so must be of the form f∗(α) = dα for some integer d depending only on f . This integer is called the degree of f , with the notation deg f . Here are some basic properties of ... |
d homomorphisms Hk(X 0) -→ Hk(X 1) -→ ··· -→ Hk(X k−1) -→ Hk(X k) -→ Hk(X k+1) -→ ··· By what we have just shown these are all isomorphisms except that the map to Hk(X k) may not be surjective and the map from Hk(X k) may not be injective. The first part of the sequence then gives statement (b) since Hk(X 0) = 0 when k ... |
detected by Hn(M) , which is Z if M is orientable and 0 otherwise. This is shown in Theorem 3.26. Example 2.38: An Acyclic Space. Let X be obtained from S 1 ∨ S 1 by attaching two 2 cells by the words a5b−3 and b3(ab)−2 . Then d2 : Z2→Z2 has matrix , with the two columns coming from abelianizing a5b−3 and b3(ab)−2 to 5... |
classical theorem of Reidemeister from the 1930s that Lℓ/m is homeomorphic to Lℓ′/m′ iff m′ = m and ℓ′ ≡ ±ℓ±1 mod m . For example, when m = 7 there are only two distinct lens spaces L1/7 and L2/7 . The ‘if’ part of this theorem is easy: Reflecting the lens through a mirror shows that Lℓ/m ≈ L−ℓ/m , and by interchanging ... |
ns. But this contradicts the fact that Hn(Zm) ⊔⊓ is nonzero for infinitely many values of n . Reflecting the richness of group theory, the homology of groups has been studied quite extensively. A good starting place for those wishing to learn more is the text- book [Brown 1982]. At a more advanced level the books [Adem &... |
f the homology theory we have considered so far that behaves in a very similar fashion and sometimes offers technical advantages. i niσi where each σi is The generalization consists of using chains of the form a singular n simplex in X as before, but now the coefficients ni are taken to lie in a fixed abelian group G rathe... |
ism on Hn when n is odd.] 15. Show that if X is a CW complex then Hn(X n) is free by identifying it with the kernel of the cellular boundary map Hn(X n, X n−1)→Hn−1(X n−1, X n−2) . 16. Let [vi0 , ··· , vik ∆ groups i( n = [v0, ··· , vn] have its natural complex structure with k simplices ] for i0 < ··· < ik . Compute t... |
finition, but interesting homology theories with nontrivial groups in negative dimensions do exist. The third axiom may seem less substantial than the first two, and indeed for finite wedge sums it can be deduced from the first two axioms, though not in general for infinite wedge sums, as an example in the Exercises shows. ... |
position of two functors. The first functor assigns to a space X its singular complex S(X) , a complex, and the second functor assigns to ∆ a complex its simplicial chain complex. This is what the two functors do on ∆ objects, and what they do on morphisms can be described in the following way. A map of spaces f : X→Y i... |
orientable surface. The component of K containing i niσi P the boundary circle is a standard closed orientable surface of some genus g with an open disk removed, by the basic structure theorem for compact orientable surfaces. Giving this surface the cell structure indicated in the figure, it then becomes obvious that f ... |
ridian and longitude circles in the torus, which correspond to α1 and α2 . Van Kampen’s theorem now implies that the inclusion Y0 ֓ Y1 induces an injection of π1(Y0) into π1(Y1) as the infinite cyclic subgroup generated by [α1, α2] . In a similar way we can regard Yn+1 as being obtained from Yn by adjoining 2n copies of... |
0 is Hn+1(P n; Z2) , which vanishes since P n is an n dimensional CW complex. The other terms that are zero are Hi(S n) for 0 < i < n . We assume n > 1 , leaving the minor modifications needed for the case n = 1 to the reader. All the terms that are not zero are Z2 , by cellular homology. Alternatively, this exact sequ... |
tion g can be chosen not just homotopic to f but also close to f if we allow subdivisions of L as well as K . The Lefschetz Fixed Point Theorem This very classical application of homology is a considerable generalization of the Brouwer fixed point theorem. It is also related to the Euler characteristic formula. P For a ... |
let en+1 be an (n + 1) cell of X attached by a map ϕ : S n→X n . The map S n→Yn corresponding to ϕ under the homotopy equivalence Yn ≃ X n is homotopic to a simplicial map f : S n→Yn by the simplicial approximation theorem, and it is not hard to see that the spaces X n ∪ϕ en+1 and Yn ∪f en+1 are homotopy equivalent, w... |
ology arises quite naturally. One of these is Poincar´e duality, the topic of the third section of this chapter. Another is obstruction theory, covered in §4.3. Characteristic classes in vector bundle theory (see [Milnor & Stasheff 1974] or [VBKT]) provide a further instance. From the viewpoint of homotopy theory, cohom... |
of understanding the relationship between the homology groups of a chain complex and the homology groups of the dual complex obtained by applying the functor C֏Hom(C, G) . This is the first topic of the chapter. 190 Chapter 3 Cohomology Homology groups Hn(X) are the result of a two-stage process: First one forms a ∂----... |
et us use the temporary notation H n(F ; G) for the homology group Ker f ∗ n+1/ Im f ∗ n of this dual complex. Note that the group Coker i∗ n−1 that we are interested in is H 1(F ; G) where F is the free resolution in (vi). Part (b) of the following lemma therefore shows that Coker i∗ n−1 depends only on Hn−1(C) and G ... |
0(F ; G) = H ∗ = HomR(H, G) by the exactness of definition we have Ext0 0 ← H ∗← 0 . The real reason why unreduced Ext groups are better than re1 ← F ∗ F ∗ duced groups is perhaps to be found in certain exact sequences involving Ext and 1 ← F ∗ 1 ← F ∗ Hom derived in §3.F, which would not work with the Hom terms replace... |
properties (f g)♯ = g♯f ♯ and 11♯ = 11 imply (f g)∗ = g∗f ∗ and 11∗ = 11, so X ֏ H n(X; G) and (X, A) ֏ H n(X, A; G) are contravariant functors, the ‘contra’ indicating that induced maps go in the reverse direction. The algebraic universal coefficient theorem applies also to relative cohomology since the relative chain g... |
in §2.1. ∆ 8. Many basic homology arguments work just as well for cohomology even though maps go in the opposite direction. Verify this in the following cases: (a) Compute H i(S n; G) by induction on n in two ways: using the long exact sequence of a pair, and using the Mayer–Vietoris sequence. (b) Show that if A is a c... |
d 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 + ··· + (m − 1) is 0 if m is odd and k if m = 2k since the terms 1 and m − 1 cancel, 2 and m − 2 cancel, and so on. Thus, writing α2 for... |
mplex and quaternionic cases could be replaced by any commutative ring R , but not for RPn and RP∞ since a polynomial ring R[α] is strictly commutative, so for this to be a commutative ring in the graded sense we must have either |α| even or 2 = 0 in R . Polynomial rings in several variables also have graded ring struc... |
orphism when X is a point since it is just the scalar multiplication map R ⊗R H n(Y ; R)→H n(Y ; R) . The following general fact will then imply the theorem. Proposition 3.17. If a natural transformation between unreduced cohomology theories on the category of CW pairs is an isomorphism when the CW pair is (point, ∅) ,... |
use the following commutative diagram: (ii) If we can show all these maps are isomorphisms, then the same argument will apply with i and j interchanged, and the vertical maps in the left column of (i) will be isomorphisms. The left-hand square in (ii) consists of isomorphisms by cellular cohomology. The right-hand ver... |
all the n cells of (S n)m to form the n cell of Jm(S n) , we see from cellular cohomology that q∗(x1) is the sum α1 +···+αm of the generators of H n dual to the n cells of (S n)m . By the same reasoning we have q∗(xk) = ··· αik Jm(S n); Z Jm(S n); Z dual to the kn cell, represented (S n)m; Z αi1 i1<···<ik . by computin... |
determines the subcomplex up to permutation of factors. How- ever, these cohomology rings are still a whole lot less complicated than the general case, where one takes free algebras modulo ideals generated by arbitrary polynomials having all their terms of the same dimension. Let us conclude this section with an exampl... |
ncar´e duality using the notion of dual cell structures. The germ of this idea can be traced back to the five regular Platonic solids: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. Each of these polyhedra has a dual polyhedron whose vertices are the center points of the faces of the given polyhedron.... |
ntinuous map M→MZ of the form x ֏ αx ∈ Hn(M || x) is called a section of the covering space. An orientation of M is the same thing as a section x ֏ µx such that µx is a generator of Hn(M || x) for each x . f One can generalize the definition of orientation by replacing the coefficient group Z by any commutative ring R wit... |
h a single n cell, which is the case for a large number of manifolds. Note that there can be no cells of higher dimension since a cell of maximal dimension produces nontrivial local homology in that dimension. Consider the cellular boundary map d : Cn(M)→Cn−1(M) with Z coefficients. Since M has a single n cell we have Cn... |
element of verify that it is also injective, so H 1 ∆ Σ c (R; G) ≈ G . ∆ Compactly supported cellular cohomology for a locally compact CW complex could be defined in a similar fashion, using cellular cochains that are nonzero on ∆ Poincar´e Duality Section 3.3 243 only finitely many cells. However, what we really need i... |
ws easily from the definition of direct limits. It remains to consider the commutativity of the preceding diagram involving K and L . In the two squares shown, not involving boundary or coboundary maps, it is a triviality to check commutativity at the level of cycles and cocycles. Less trivial is the third square, which... |
onsingularity of the cup product pairing, ϕ is realized by taking cup product with an element β ∈ H n−k(M; Z) and evaluating on [M] , so having a β with 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 pr... |
restricting to a given orientation at each point of M − ∂M . It will not be difficult to deduce the following generalization of Poincar´e duality to manifolds with boundary from the version we have already proved for noncompact manifolds: 254 Chapter 3 Cohomology Theorem 3.43. Suppose M is a compact R orientable n manifo... |
l in Mi .) (a) Show that if M1 and M2 are closed then there are isomorphisms Hi(M1♯M2; Z) ≈ Hi(M1; Z)⊕ Hi(M2; Z) for 0 < i < n , with one exception: If both M1 and M2 are nonorientable, then Hn−1(M1♯M2; Z) is obtained from Hn−1(M1; Z)⊕ Hn−1(M2; Z) by replacing one of the two Z2 summands by a Z summand. [Euler character... |
ee group Cn . Thus Cn ≈ Zn ⊕ Bn−1 , but the chain complex C is not the direct sum of the chain complexes Z and B since the latter have trivial boundary maps but the boundary maps in C may be nontrivial. Now tensor with G to get a commutative diagram 262 Chapter 3 Cohomology (ii) The rows are exact since the rows in (i)... |
) . The vertical isomorphisms come from the natural commutativity of tensor product. Since the squares commute, there is induced a map Tor(A, B)→Tor(B, A) , which is an isomorphism by the fivelemma. 266 Chapter 3 Cohomology 0 -→ F1 if B is torsionfree. Suppose Now we can prove the statement (3) in the torsionfree case. ... |
ation in the sense of linear algebra, ∆ that is, an equivalence class of ordered bases, two ordered bases being equivalent if they differ by a linear transformation of positive determinant. (An ordered basis can be continuously deformed to an orthonormal basis, by the Gram–Schmidt process, and two orthonormal bases are ... |
ram says that the cross products defined using the different CW structures coincide. Cross product is obviously bilinear, or in other words, distributive. It is not hard to check that it is also associative. What about commutativity? If T : X × Y →Y × X is transposition of the factors, then we can ask whether T∗(a× b) eq... |
ula and the universal coefficient theorem. With a little more algebra the formula can be shown to hold more generally for an arbitrary coefficient group G in place of R ; see [Hilton & Wylie 1967], p. 227. The General K¨unneth Formula Section 3.B 277 Hi Hn(X ∧ Y ; R) ≈ There is an analogous formula Hn−i(Y ; R) . As a speHn... |
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 that slant products Hn(X × Y ; R)× H j(Y ; R) -→ Hn−j(X; R), H n(X × Y ; R)× Hj(Y ; R) -→ H n−j(X; R), (ei × ej , ϕ) ֏ ϕ(ej )ei ... |
e (βn) = (β ⊗ 1 + 1 ⊗ β)n = n i Example 3C.2. The exterior algebra ∆ Hopf algebra, with (α) = α ⊗ 1+1 ⊗ α . To verify that i P R[α] on an odd-dimensional generator α is a is an algebra homomorphism (α)2 , or in other words, since α2 = 0 , we need to see (α)2 = (α ⊗ 1 + 1 ⊗ α)2 = ∆ (α)2 is indeed 0 . Note that if α were... |
n practice satisfy this associativity property. Nor is the Pontryagin product generally commutative, even in the graded sense, unless µ is commutative or homotopy-commutative, which is relatively rare for H–spaces. We will give examples shortly where the Pontryagin product is not commutative. 288 Chapter 3 Cohomology I... |
e, 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 structure on its universal cover. [For the group SO(n) considered in the next... |
led admissible, as will the sequence consisting of a single 0 . Proposition 3D.1. The maps ρϕI : DI→SO(n) , for I ranging over all admissible sequences, are the characteristic maps of a CW structure on SO(n) for which the map ρ : P n−1× P n−2 × ··· × P 1→SO(n) is cellular. In particular, there is a single 0 cell e0 = {... |
truncate the polynomial algebra by a relation x2n Z2 [α] , and with Z2 coefficients the latter is an exterior algebra first glance to be unrelated, but in fact the relationship is fairly direct. As we saw in the previous section, the dual of a polynomial algebra Z2[x] is a divided polynomial alZ2 [α0, α1, ···] gebra where... |
here we embed O(n − k) in O(n) as the orthogonal transformations of the first n − k coordinates of Rn . Thus Vn,k can be viewed as the space O(n)/O(n − k) of such cosets, with the quotient topology from O(n) . This is the same as the previously defined topology on Vn,k since the projection O(n)→Vn,k is a surjection of co... |
hand, if (C) , let M(gi) consist of Z ’s in dimensions ni and ni + 1 , gi has finite order k in Hni generated by xi and yi respectively, with ∂yi = kxi . Let M be the direct sum of the chain complexes M(gi) . Define a chain map σ : M→C by sending zi and xi to cycles ζi and ξi representing the corresponding homology class... |
]/(2x, x4, y 4, z2, xz, x3 − y 2) The next figure shows the nontrivial Bocksteins for H ∗(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 ... |
lize a given element (hi) ∈ If all the αi ’s are isomorphisms then lim←-- Gi ≈ G0 and lim←-In fact, lim i Gi as ←-δ(gi) we can take g0 = 0 and then solve α1(g1) = −h0 , α2(g2) = g1 − h1 , ··· . If all the αi ’s are zero then lim Deleting a finite number of terms from the end of the sequence ··· →G1→G0 1Gi are undoes not... |
e, the two rows are exact, as they are simply direct products of copies of the exact sequence 0→A→B→C→0 , in view of the general fact that i Hom(Gi, H) . Enlarging the diagram by zeros above and below, Hom( it becomes a short exact sequence of chain complexes, and the associated long exact iGi, H) = Q L sequence of hom... |
nd 0→pA→A→Ap→0 can be spliced together to yield the exact sequence across the top of the following diagram Transfer Homomorphisms Section 3.G 321 where the map labeled ‘ p ’ is multiplication by p . Use this to show: (a) Ext(A, Z) is divisible iff A is torsionfree. (b) Ext(A, Z) is torsionfree if A is divisible, and the... |
Zp∞ , the union of the increasing sequence Zp ⊂ Zp2 ⊂ Zp3 ⊂ ··· . The first step is to show that H ∗(K(Zp∞ , 1); Zp) ≈ Zp[β] with H∗(K(Zpi , 1); Z) consists of Zpi ’s in odd dimensions. The in|β| = 2 . We know that clusion Zpi ֓ Zpi+1 induces a map K(Zpi , 1)→K(Zpi+1 , 1) that is unique up to homotopy. We can take this ... |
[π ] , which consists of the finite formal sums Cn( ∆ ∆ i miγi with mi ∈ Z and γi ∈ π , with the natural addition i niγi = e i (mi + ni)γi and multiplication i,j minjγiγj . The boundX) are Z[π ] module homomorphisms since the action of P X)→Cn−1( ary maps ∂ : Cn( P π on these groups comes from an action on X , by sendin... |
eserves or reverses orientation in M→M M , that is, whether γ lifts to a closed loop in the orientable double cover or not. As another example, the action of π1(X) on itself by inner automorphisms corresponds to a bundle of groups p : E→X with fibers p−1(x) = π1(X, x) . This example is rather similar in spirit to the ex... |
X→Y induces f ♯ : C n c (X; G) provided that f is proper: The preimage f −1(K) of each compact set K in Y is compact in X . Thus if ϕ ∈ C n(Y ; G) vanishes on chains in Y − K then f ♯(ϕ) ∈ C n(X; G) vanishes on chains in X − f −1(K) . Further, to guarantee that f ≃ g implies f ∗ = g∗ we should restrict attention to ho... |
o this is not a practical way of classifying homotopy types, but it is useful for various more theoretical purposes. This chapter is arranged so that it begins with purely homotopy-theoretic notions, largely independent of homology and cohomology theory, whose roles gradually in- crease in later sections of the chapter... |
f: For surjectivity of p∗ we apply the lifting criterion in Proposition 1.33, which implies that every map (S n, s0)→(X, x0) lifts to ( x0) provided that n ≥ 2 so that S n is simply-connected. Injectivity of p∗ is immediate from the covering homotopy ⊔⊓ property, just as in Proposition 1.31 which treated the case n = 1... |
, and (1)–(3) are each equivalent to saying that each path-component of X contains points in A since D0 is a point and ∂D0 is empty. The pair (X, A) is called n connected if (1)–(4) hold for all i ≤ n , i > 0 , and (1)–(3) hold for i = 0 . Note that X is n connected iff (X, x0) is n connected for some x0 and hence for ... |
entually stationary in the infinite chain of homotopies. 350 Chapter 4 Homotopy Theory To fill in the missing step in this argument we will need a technical lemma about deforming maps to create some linearity. Define a polyhedron in Rn to be a subspace that is the union of finitely many convex polyhedra, each of which is a... |
0) induces isomorphisms on relative as well as absolute homotopy groups. Here is another application of the technique, giving a more geometric interpreta- tion to the homotopy-theoretic notion of n connectedness: Proposition 4.15. If (X, A) is an n connected CW pair, then there exists a CW pair (Z, A) ≃ (X, A) rel A su... |
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