Bott Periodicity Theorem

Home , Bott periodicity theorem, Smash product, Suspension (topology)

Davide Matasci 2018 Bott Periodicity Theorem 2018

Remark 1 Just to recall... Ke(X) can be identiﬁed with the ker(K(X) → K(x0)) for x0 ∈ X. All functions are assumed to be continuous and a vector bundle p: E → B is identiﬁed with E.

Deﬁnition 2 For a topological space (X, τX ) and an equivalence relation ∼ on X we deﬁne a new topology (called quotient topology) τY on Y = X/ ∼ by: −1 τY = {U ⊂ Y | q (U) ∈ τX } where q : X → X/ ∼ is the canonical map. The space (Y, τY ) is called quotient space.

Deﬁnition 3 A map f : X → Y is called quotient map if:

• f is surjective;

−1 • U ∈ τY iﬀ f (U) ∈ τX , where τX , τY are the topologies on X resp. Y .

Remark 4 The canonical map q : X → X/ ∼ is always a quotient map, provided that X/ ∼ is equipped with the quotient topology.

Deﬁnition 5 The cone CX of a space X is deﬁned by CX = (X×[0, 1])/(X× {0}).

Deﬁnition 6 The suspension SX of a space X is deﬁned by taking X × [0, 1] and collapsing X × {0} to a point and X × {1} to another point.

SX can be seen as double cone on X.

Deﬁnition 7 The reduced suspension ΣX of a space X is deﬁned by ΣX = SX/({x0} × [0, 1]) for some x0 ∈ X.

Remark 8 Suspension and reduced suspension are homotopically equivalent.

For the proof, see [1].

Deﬁnition 9 Given f : X → Y , we deﬁne Sf : SX → SY by Sf([x, t]) = [f(x), t].

With this deﬁnition, S becomes a functor T op → T op. SnX is the space obtained by applying S to X n times.

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Deﬁnition 10 Let ϕi : Gi → G(i + 1) be an homomorphism between groups (for all 0 ≤ i ≤ n). The sequence

G1 → G2 → ... → Gn is called exact if ker ϕi+1 = Im ϕi for all i.

Deﬁnition 11 A partition of unity subordinate to an open cover {Uα}α of X is a collection {ϕβ : X → [0, 1]}β such that:

•∀ β∃α such that supp(ϕβ) ⊂ Uα (i.e. each ϕβ is supported in one of the Uα’s); P • β ϕβ = 1, where the sum is ﬁnite in a neighbourhood of each point of X.

Deﬁnition 12 A space (X, τ) is called paracompact if it is Hausdorﬀ and for each open cover {Uα}α of X there is a partition of unity {ϕβ}β subordinate to the cover.

Proposition 13 (1.18 in [1]) Every compact Hausdorﬀ space is paracompact.

For the proof, see [1]. Proposition 14 (2.9 in [1]) Let X be compact and Hausdorﬀ, A ⊂ X be closed subspace. Let ı: A,→ X be the inclusion and let q : X → X/A be the quotient map. Then q A −→ı X −→ X/A induces (applying Ke, a contravariant functor) the exact sequence

Ke(q) Ke(ı) Ke(X/A) −−→ Ke(X) −−→ Ke(A)

Proof Let q∗ = Ke(q) and i∗ = Ke(ı). We only have to show that the sequence is exact, i.e. that Im(q∗) = ker(ı∗). Since qi = 0 it holds that Ke(qi) = i∗q∗ = 0 and thus the inclusion Im(q∗) ⊂ ker(ı∗) is proven. To show the other inclusion, we take an element [E] ∈ ker(i∗) ⊂ Ke(X) and want to ﬁnd an element [F ] ∈ Ke(X/A) such that q∗[F ] := [q ∗ F ] = [E]. This is equivalent (by deﬁnition of []) to the following:

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We take a vector bundle E → X with i∗E ≈ ξ0 (i.e. i∗[E] = 0 = [ξ0]) Ke(A) and look for a vector bundle F → X/A such that q∗F ≈ E. i∗E ≈ ξ0 means that E → X is trivial over A. Let h: p−1(A) → A × Cn be the trivialization over A. Let E/h = E/ ∼ with h−1(x, v) ∼ h−1(y, v) for all x, y ∈ A. Since p(h−1(x, v)) = x for all x ∈ A and q is a projection, the composition

p q E → X → X/A satisﬁes qp(h−1(x, v)) = qp(h−1(y, v)) for all x, y ∈ A and thus induces a projection E/h → X/A.

The goal is now to show that E/h → X/A is a vector bundle. This will be the F we’re looking for. It’s enough to ﬁnd a trivialization of E over a neighbourhood U of A, since this induces a trivialization of E/h over U/A. This is in turn enough to prove that E/h → X/A is a vector bundle since over the other points of X/A a local trivialization is already given (induced by the local trivialization of E over X). By the discussion before Lemma 1.1 in [1], there are sections s1, . . . , sn : A → −1 E such that for all a ∈ A : s1(a), . . . , sn(a) are linearly independent in p (a). Let {Uj}j∈J be an open cover of A such that E is trivial over all of the Uj’s. The existence of these sets follows from the fact that E → X is a vector bundle, and J is ﬁnite since A is compact.

−1 si restricts to A ∩ Uj → p (A). Using the trivialization we can regard si −1 as a map : A ∩ Uj → p (aj) for some aj ∈ A ∩ Uj. More precisely we have:

si −1 hj n πj n A ∩ Uj −→ p (A ∩ Uj) −→ (A ∩ Uj) × C −→ C

−1 n where hj : p (Uj) → Uj ×C is a local trivialization and πj is the projection. 0 −1 0 Taking si(x) := hj (aj, πj ◦ hj ◦ si(x)) for some aj ∈ A ∩ Uj we see that si in 0 −1 fact restricts to a single ﬁber: si(x) ∈ p ({aj}) for all x. We can ”recover” 0 si from si with −1 0 si(x) = hj (x, πj ◦ hj ◦ si(x))

Thus, as claimed, we can w.l.o.g. assume that si restricts to a single ﬁber −1 p (aj).

Since X is a normal topological space, by the Tietze extension theorem we can extend these maps to si,j : Uj → E. Formally, I use the fact that A ∩ Uj is closed in Uj with respect to the relative topology.

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Let {ϕj | j ∈ J} ∪ {ϕ} be a partition of unity subordinate to the (open) cover {Uj | j ∈ J} ∪ {X − A}ofX. This exists because X, being Hausdorff compact, is paracompact (see Proposition 1.8). Using this partition we get P that ti := j∈J ϕjsi,j is a section extending sj to a map X → E. −1 Over a fiber p (a) (for an element a ∈ A) these extended section ti form a basis, i.e det(tj : j ∈ J) 6= 0. Since the determinant is continuous (or alternatively, the linear independence is an open condition) this property holds true even in a small neighbourhood of each fiber. This means that −1 n (tj : j ∈ J) is a linear-space isomorphism between p (a) and a × C for all a ∈ Uj. Only remaining to be shown is that q∗(E/h) ≈ E.

E k E/h

q X X/A

This follows from the Universal property of the pullback (the uniqueness- up-to-isomorphism part) because the quotient map E → E/h, denoted by k in the diagram, takes each fiber over an element x ∈ X isomorphically to the fiber over q(x) ∈ X/A (shown above...). Remark 15 ξn denotes the n-dimensional trivial vector bundle. • V ectn(X) = {”n − dim vector bundles over X”}/ ≈ with E ≈ F iff E and F are isomorphic;

• Ke(X) = {”vector bundles over X”}/ ∼ with E ∼ F iﬀ ∃n, m ≥ 0 : E ⊕ ξn ≈ F ⊕ ξm;

0 0 0 0 • K(X) = {E − F }/ ≈K with E − F ≈K E − F iﬀ E ⊕ F ≈s E ⊕ F , n n where E ≈s F iﬀ ∃n ≥ 0 : E ⊕ ξ ≈ F ⊕ ξ .

The next Lemma shows that for contractible subspaces A in X, we may take X/A instead of X as a base for vector bundles. Lemma 16 (2.10 in [1]) Let X and A as in the proposition above. If A is contractible, q : X → X/A induces a bijection q∗ : V ectn(X/A) → V ectn(X) given by Ke(q) for every n ∈ N. Proof We have to ﬁnd an inverse of the map q∗. Let p: E → X be an n-dimensional vector bundle. The image of p under inverse is going to be E/h, with h as in the proposition above (since A is compact and contractible,

Davide Matasci 5 Bott Periodicity Theorem 2018 p is trivial over A by Corollary 1.8 in [1]). As the fact that E/h is a vector bundle is already been proven, only to show is that E 7→ E/h is well deﬁned, i.e. that E/h does not depend on the choice of the map h.

Suppose we have two trivializations h0 and h1. These are homeomor- −1 n −1 phisms p (A) → A × C . Because h1 = (h1h0 )h0, we see that h1 and h0 n −1 diﬀer by a matrix ga ∈ GLn(C ) over each a ∈ A (i.e. in the ﬁber p 8(a)). Let g : A → GLn(C) be the map a 7→ ga. Composing with a contraction of A we see that g is homotopic to a constant map a 7→ α ∈ GLn(C). We −1 then have h1 = (h1(αh0) )αh0. Without loss of generality we can replace h0 with αh0 (as this doesn’t change E/h0) or, alternatively, assume that α = 1. We then have an homotopy from g to a 7→ 1 ∈ GLn(C). This, using the −1 −1 relation h1 = (h1(αh0) )αh0, gives us an homotopy H : [0, 1] × p (A) → n A × C , (t, a) 7→ ht(a) from h0 to h1. Using the same construction of E/h with the homotopy H yields a collection of vector bundles E/ht → (X/A) or, alternatively, a vector bundle ([0, 1] × E)/H → [0, 1] × (X/A). This restricts to E/h0 resp. h1 on the boundaries of [0, 1]. This shows that E/h0 ≈ E/h1.

Still to check is that V ectn(X) → V ectn(X/A),E 7→ E/h is in fact an inverse of q∗. As in the proof of the Proposition 14, it holds that q∗(E/h) ≈ E. To see that q∗(F )/h ≈ F for a vector bundle F → X/A we can take h to be any trivialization of q∗(F ) over A.

q∗(F )/h q∗(F ) F

X/A X X/A

. Here’s some more useful deﬁnitions: Deﬁnition 17 For spaces X and Y :

• X ∨ Y := (X × {y0}) ∪ ({x0} × Y ) with x0 ∈ X and y0 ∈ Y , the wedge sum;

• X ∧ Y := (X × Y )/(X ∨ Y ), the smash product.

Remark 18 It holds that Σ(X ∨ Y ) = ΣX ∨ ΣY .

Proof Let I = [0, 1]. Then (X ∨Y )×I = (X ×I)t(Y ×I)/(x0 ×I ∼ y0 ×I). Quotienting by (X ∨ Y × 0) t (X ∨ Y × 1) t ((x0, y0) × I yields ΣX ∨ ΣY .

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Before stating the main theorem some more preparation is needed. Let A and X be as above. A may not be contractible, hence Lemma 16 cannot be used. The idea is to use cones (which are contractible) so that the Lemma can be applied. Starting from A,→ X we can get A X X ∪ CA (X ∪ CA) ∪ CX ((X ∪ CA) ∪ CX) ∪ C(X ∪ CA) ...

A X X/A SA SX where each of the spaces on the top row is obtained by attaching a cone on the subspace two steps back in the sequence and adding it to its predecessor. The horizontal maps are inclusions and the vertical ones (except the ﬁrst two, which are identities) are quotient maps obtained by collapsing the last added cone of the space at the top.

The fact that this operation really yields the spaces in the second row can be proven by noting that adding a cone and collapsing it is equivalent to quotienting by the base of the cone (this in turn can be easily proven using the First isomorphism theorem). The drawing above should help understand the idea of the construction (the green lines should represent the cone over X ∪ CA).

Now we can apply Ke (to the top row). Since cones are contractible the Lemma (16) gives us the sequence (which is exact by Proposition 14): ... → Ke(SX) → Ke(SA) → Ke(X/A) → Ke(X) → Ke(A)

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This extends the exact sequence of Proposition 14. The sequence above can be used as follows:

Example 19 On A ⊂ A ∨ B to get

Ke(A ∨ B) ≈ Ke(A) ⊕ Ke(B)

Proof Since (A ∨ B)/A = B, the sequence above becomes:

Ke(q) Ke(ı) Ke(SA) → Ke(B) −−→ Ke(A ∨ B) −−→ Ke(A) where q : A∨B → B is the quotient map (quotienting A) and ı: A → A∨B is the inclusion. Let q0 : A∨B → A be the quotient map (w.r.t. B) and ı0 : B → A∨B be the inclusion. The goal is now to show that K(ı0)K(q) = id and e e Ke(B) K(q) K(ı) K(ı)K(q0) = id . Then the sequence K(B) −−→e K(A ∨ B) −−→e K(A) e e Ke(A) e e e splits and we have the desired result.

Let’s only prove that K(ı0)K(q) = id , since the other equality is e e Ke(B) analogous: Let [E] ∈ Ke(B). Then Ke(ı0)Ke(q)[E] = Ke(ı0)[q∗E] = [ı0∗q∗E]. This is a vector bundle over B.

ı0∗q∗E q∗E E ı0∗ q∗ p

0 q B ı A ∨ B B

But since we also have E E id p

0 B id B by the Universal property of the Pullback it holds that ı0∗q∗E is isomorphic 0 to E, i.e. Ke(ı )Ke(q)[E] = [E].

Example 20 On A ∨ B ⊂ A × B to get

Ke(X × Y ) ≈ Ke(X ∧ Y ) ⊕ Ke(X) ⊕ Ke(Y )

Proof This uses the previous result, the fact that suspension is homotopically equivalent to the reduced suspension and Remark 18. The idea is the same as above: show that Ke(X ∧Y ) → Ke(X ×Y ) → Ke(X)⊕Ke(Y ) splits.

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Definition 21 Let pZ : X × Y → Z be the projection onto Z ∈ {X,Y } and ∗ pZ = Ke(pZ ). The goal is to define a product Ke(X) ⊗ Ke(Y ) → Ke(X ∨ Y ). ∗ Let a ∈ Ke(X) and b ∈ Ke(Y ). Using the Remark 1 we get that pX (a) is ∗ zero if restricted over {x0} × Y (i.e. zero in K({x0} × Y )) and pY (b) is ∗ ∗ zero in K(X × y0). Hence the external product a ∗ b = µ(a, b) = pX (a)pY (b) is in Ke(X × Y ) = ker(K(X × Y ) → K({x0} × {y0}). By exactness of the sequence above we get exactly one element in Ke(X ∧ Y ) which maps to a ∗ b ∈ Kl(X × Y ). This element is defined to be the reduced external product of a and b. The reduced external product is a map Ke(X) ⊗ Ke(Y ) → Ke(X ∨ Y ) Remark 22 The reduced external product is just a reduction of the external product, as the following diagram shows:

K(X) ⊗ K(Y ) = (Ke(X) ⊗ Ke(Y )) ⊕ Ke(X) ⊕ Ke(Y ) ⊕ Z

∼ ∼ ∼

K(X × Y ) = Ke(X ∧ Y ) ⊕ Ke(X) ⊕ Ke(Y ) ⊕ Z where the non-trivial vertical maps are the external product and the reduced external product. Remark 23 For any compact Hausdorﬀ space X this holds: Sn ∧X = ΣnX.

1 Proof Let I = [0, 1], x0 ∈ X and s0 ∈ S . If n = 1 we have: S1 ∧ X = (S1 × X)/S1 ∨ X 1 1 = (S × X)/((S × {x0}) ∪ ({s0} × X))

= (I × X)/((I × {x0}) ∪ (∂I × X)) = ΣX

1 since S = I/∂I (we take {s0} to be identiﬁed with ∂I). The last equation holds by deﬁnition of ΣX. Assuming the claim holds for some n > 1 we get (using the case where n = 1): Σn+1X = ΣΣnX = Σ(Sn ∧ X) = ((Sn ∧ X) × I)/((Sn ∧ X) × {0}) ∪ ((Sn ∧ X) × {1}) ∪ ({s} × I) = Sn+1 ∧ X by the same argument as above (i.e. using that S1 = I/∂I), where s ∈ Sn∧X. The claim follows by induction on n.

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Corollary 24 Ke(SnX) ≈ Ke(Sn ∧ X) Proof Using the previous remark and the fact that ΣnX is obtained from SnX by collapsing a contractible subspace (namely an n-disk) we get the result by Lemma 16.

Let H be the canonical line bundle over S2 ≈ CP 1. Theorem 25 (2.11 in [1], Bott Periodicity Theorem) Suppose X is a compact Hausdorﬀ space. Then the homomorphism β : Ke(X) → Ke(S2X) given by a 7→ (H − 1) ∗ a is an isomorphism, where ∗ denotes the reduced external product.

Proof β is given by the composition of the maps

Ke(X) −→Ke(S2) ⊗ Ke(X) −→Ke(S2X) a 7−→ (H − 1) ⊗ a a ⊗ b 7−→ a ∗ b

The ﬁrst map is an isomorphism since Ke(S2) is inﬁnite cyclic (generated by (H − 1)... see Corollary 2.3 in [1]) (the same argument used to show that A = A ⊗ Z for any abelian group A applies here). The second map is also an isomorphism by the Product Theorem, which applies to the reduced case as shown below. From the Product Theorem it follows that

K(X) ⊗ K(S2) → K(X × S2) is an isomorphism. Because K(A) = Ke(A) ⊕ Z and Ke(A) ⊗ Z = Ke(A) for A ∈ {X,S2} we get:

2 2 K(X) ⊗ K(S ) = (Ke(X) ⊕ Z) ⊗ (Ke(S ) ⊕ Z) 2 2 = (Ke(X) ⊗ Ke(S )) ⊕ (Ke(X) ⊗ Z) ⊕ (Z ⊗ Ke(S )) ⊕ (Z ⊗ Z) 2 2 = (Ke(X) ⊗ Ke(S )) ⊕ Ke(X) ⊕ Ke(S ) ⊕ Z and

2 2 K(X × S ) = Ke(X × S ) ⊕ Z

Moreover, using that Ke(X ×S2) ≈ Ke(X ∧S2)⊕Ke(X)⊕Ke(S2) (see Example 20):

2 2 2 K(X × S ) = Ke(X ∧ S ) ⊕ Ke(X) ⊕ Ke(S ) ⊕ Z

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Thus we have an isomorphism (given by the external product) from:

2 2 (Ke(X) ⊗ Ke(S )) ⊕ Ke(X) ⊕ Ke(S ) ⊕ Z to 2 2 Ke(X ∧ S ) ⊕ Ke(X) ⊕ Ke(S ) ⊕ Z By remark 22, we see that the reduced external product Ke(X)⊗Ke(S2) → Ke(X ∧ S2) is also an isomorphism. And from Corollary 24 we know that 2 2 Ke(X ∧ S ) = Ke(S X), which concludes the proof.

Corollary 26 (2.12 in [1]) For all n ≥ 0: ( Z, if n is even Ke(Sn) ≈ 0, if n is odd Moreover, Ke(S2k) is generated by the k-fold reduced external product (H− 1) ∗ ... ∗ (H − 1).

Proof Since SSn = Sn+1 (one can think of the two vertices of the suspension as the two poles of the resulting sphere) it’s enough to show that Ke(S0) = Z and Ke(S1) = 0. 0 0 S = {x0, x1}, so every vector bundle over S consists of a bundle over {x0} and one over {x1}. Let x0 be chosen as the basepoint for the restriction 0 0 map K(S ) → K({x0}). Then the kernel of this map, identiﬁed with Ke(S ), n n n is of the form {C − C | n ∈ N}, where C is a vector bundle over {x1}. It follows that Ke(S0) ≈ Z. The last part, i.e. that Ke(S1) = 0, follows from the fact that all vector 1 1 1 bundles over S are trivial (thus K(S ) = Z, and ker(K(S ) → K(x0)) = 0 1 for any x0 ∈ S ).

References

[1] Allen Hatcher. Vector Bundles and K-Theory. 2017.

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