Division Algebras and Parallelizable Spheres, Part II Seminartalk by Jerome Wettstein April 5, 2018 1 A quick Recap of Part I We are working on proving the following Theorem: Theorem 1.1. The following statements are true only if n = 1, 2, 4 or 8: R n is a division algebra S n 1 is parallelizable The work we have done towards proving Theorem 1.1 consisted of first introducing the notion of an H-space which is a weaker than the one of a topological group structure on spheres. We say a sphere S n 1 is a H-space if it admits a continuous multiplication map with a two-sided identity element. This notion has enabled us to prove that if either one of the conclusions of Theorem 1.1 holds for some n N, then necessarily S n 1 can be equipped with the structure of a H-space. We have immediately concluded that due to properties related to external products and Bott periodicity that this implies that only spheres of odd dimension can carry H-structures. The next important idea has been to associate to maps g : S n 1 S n 1 S n 1 a new map ĝ : S 2n 1 S n by identifying S 2n 1 = (D n D n ) = D n D n D n D n and defining ĝ on both parts of the boundary by scaling of one of the coordinates. We are especially interested in the case where the map g is defining an H-structures on spheres. The importance of this construction is due to the fact that for n = 2k even, and ĝ : S 4k 1 S 2k as constructed before, we can use ĝ to attach a 4k-cell e 4k to S 2k leading to a space called Cĝ. We used this together with short considerations of the quotient to find a SES: 0 K(S 4k ) q by the usual argument involving quotient- and subspaces. This sequence naturally leads to the Hopf invariant of ĝ by defining: i K(Cĝ) K(S 2k ) 0, (1) α = q ( (H 1) 2k), i (β) = (H 1) k, (2) thus α is the image of our preferred generator of K(S 4k ) under the pullback induced by the quotient map and β maps to our preferred generator in K(S 2k ) under the inclusion map. The Hopf invariant of ĝ is then defined as the integer h, such that: β 2 = hα. (3) We remind ourselves that h is independent of the choice of β and that there is such an integer due to exactness of the sequence. The key observation is now that if g is a H-space structure, then ĝ has Hopf invariant h = ±1. Therefore, Theorem 1.1 will be a consequence of the following Theorem of Adam: Theorem 1.2. There exists a map f : S 4n 1 S 2n with Hopf invariant h = ±1 only if n = 1, 2, 4. Note that this result implies Theorem 1.1 bacause it limits the possible integers leading to division algebra structures to the ones presented above. Trivially, for these integers we are able to describe the division algebra structures directly. Therefore, we already know that H-space structures exist in every one of the remaining possible cases by the well-known division algebras R, C, H and O. The main idea is now to prove the existence of a family of ring homomorphisms, the Adams operations, in order to prove Theorem 1.2.
2 ADAMS OPERATIONS 2 2 Adams Operations The key result for the rest of the proof of Theorem 1.1 is the following: Theorem 2.1. There exists a family of ring homomorphisms ψ k : K(X) K(X), defined for all compact, Hausdorff spaces X and all k N 0, such that the following conditions are satisfied: (i.) ψ k f = f ψ k, k 0 and for all continuous maps f : X Y (ii.) ψ k (L) = L k for all line bundles L over X (iii.) ψ k ψ l = ψ kl, k, l 0 (iv.) ψ p (α) α p mod p, for all p prime We remind ourselves, that ψ p (α) α p mod p means that there exists a β K(X) such that: ψ p (α) α p = pβ. (4) Furthermore, we remind ourselves that as previously defined, we have for all k 0 and all vector bundles E over X: ψ k (E) := s k (λ 1 (E),..., λ k (E)), (5) where s k denotes the k-th Newton polynomial and λ j (E) is the exterior k-power bundle of E. If we know that E is a direct sum of line bundles, i.e. E = L 1 L n, then we have by using the definition of the Newton polynomials: ψ k (E) = L k 1 + + L k n. (6) Because the formula becomes incredibly easy in the case of direct sums of line bundles, it would make our lives and the proof of Theorem 2.1 incredibly easy if we were able to just work with such vector bundles. Unfortunately, not all vector bundles over a given X are a direct sum of line bundles, but nevertheless we know something almost as good due to the Splitting Principle: Theorem 2.2. Given a vector bundle E X over a compact, Hausdorff space X, there exists another compact, Hausdorff space F (E) and a continuous map p : F (E) X such that the induced map p : K (X) K (F (E)) is injective and p (E) splits as a direct sum of line bundles over F (E). The proof of this theorem will be given in the next seminar talk, for now we just assume its true and use its convenient implications. Note that due to property (i.) in Theorem 2.1, this means that we have: p ( ψ k (E) ) = ψ k( p (E) ), (7) and as p (E) = L 1 L n for some line bundles L j over F (E), we have by (6): p ( ψ k (E) ) = ψ k( p (E) ) = ψ k( L1 L n ) = L k 1 + + L k n, (8) which, due to the injectivity of p, characterises the expression ψ k (E) uniquely. Using these considerations, the proof of Theorem 2.1 becomes quite simple: Proof. The proof now essentially amounts to check properties (i.) to (iv.) for our maps defined by (5). For property (i.), the statement follows from: f (λ j (E)) = λ j (f (E)), (9) for all j 0 and all vector bundles E which is proven by exploiting the uniqueness property of pullbacks. The rest is then simple insertion into the formula presented in (5): ψ k (f (E)) = s k (λ 1 (f (E)),..., λ k (f (E))) = s k (f (λ 1 (E)),..., f (λ k (E))) = f ( s k (λ 1 (E),..., λ k (E)) ) = f (ψ k (E)), (10)
2 ADAMS OPERATIONS 3 where in the third step we used that the s k are given as polynomials and that f defines a ring homomorphism. We can extend it to all of K(X) by considering differences of vector bundles over X and applying ψ k to both. Moreover, we note that additivity of ψ k can be deduced from the Splitting Principle in Theorem 2.2, more precisely let E 1, E 2 be two vector bundles over X, we want to show: ψ k (E 1 E 2 ) = ψ k (E 1 ) + ψ k (E 2 ), k 0. (11) Now we can pull both vector bundles back to F (E 1 ) under p 1 and see that by Theorem 2.2 p 1(E 1 ) is a direct sum of line bundles over F (E 1 ). Note that because the pullback is compatible with direct sums, the pullback of any direct sum of line bundles remains a direct sum of line bundles over a possibly new space. Thus applying Theorem 2.2 now to pull p 1(E 2 ) back to a direct sum of line bundles on F (E 2 ) by some map p 2 : F (E 2 ) F (E 1 ), we see that both n m (p 1 p 2 ) (E 1 ) = L 1,j, (p 1 p 2 ) (E 2 ) = L 2,j, (12) are direct sums of line bundles over F (E 2 ). Note that thus (p 1 p 2 ) (E 1 ) (p 1 p 2 ) (E 2 ) = (p 1 p 2 ) (E 1 E 2 ) is also a direct sum of line bundles and thus: (p 1 p 2 ) (ψ k (E 1 E 2 )) = ψ k ((p 1 p 2 ) (E 1 E 2 )) = ψ k ((p 1 p 2 ) (E 1 ) (p 1 p 2 ) (E 2 ))) n m = ψ k ( L 1,j L 2,l ) = n m L k 1,j + l=1 l=1 L k 2,l = ψ k ((p 1 p 2 ) (E 1 )) + ψ k ((p 1 p 2 ) (E 2 )) = (p 1 p 2 ) (ψ k (E 1 ) + ψ k (E 2 )), (13) where we used the representation of ψ k on line bundles. Note that due to injectivity of both p 1, p 2, we can conclude: ψ k (E 1 E 2 ) = ψ k (E 1 ) + ψ k (E 2 ). (14) Thus, ψ k induces an additive operation on K(X), as any element in K(X) can be represented as the difference of two vector bundles and applying ψ k to both. Another important property we want to check is multiplicativity, i.e. that ψ k induces a ring homomorphism. The proof here runs along the same lines as for additivity, namely checking it for tensor products of vector bundles and extending the result to K(X) by taking differences of vector bundles to check. We just check the property on vector bundles and as before, we can assume that E 1, E 2 pull back to line bundles over some space Y by a map p. Then note that we have due to the compatibility of direct sum and tensor products: p (ψ k (E 1 E 2 )) = ψ k ((p (E 1 E 2 )) n m = ψ k ( L 1,j L 2,l ) = ψ k ( j,l = j,l l=1 L 1,j L 2,l ) (L 1,j L 2,l ) k = j,l ( = L k 1,jL k 2,l )( L k 1,j j l L k 2,l ) = ψ k (p (E 1 ))ψ k (p (E 2 )) = p (ψ k (E 1 )ψ k (E 2 )), (15)
2 ADAMS OPERATIONS 4 where we used that the tensor product of two line bundles is again a line bundle and the definition of the product in K(X). Note that multiplicativity now again follows by injectivity of p. Property (ii.) is clear by definition, property (iii.) can be proven by similar means involving pullbacks to line bundles, as for such bundles the statement surely is true by (ii.) and additivity. The general case follows by splitting and then arguing by injectivity of the map used for the pullback. Lastly, we have to show (iv.). Here again, we use pullback to line bundles because if E splits as a direct sum of line bundles, i.e. E = L 1 L n, then for all primes p: ψ p (E) = ψ p (L 1 L n ) = L p 1 + + Lp n, (16) and by direct expansion, we see that this equals (L 1 + + L n ) p modulo p due to p being prime and some combinatorical identities. The general case follows by the splitting principle and the usual extension argument to all of K(X) Next, we note that the naturality in (i.) implies that ψ k restricts to a well-defined operation on K(X). This is due to the inclusion map i : {x 0 } X being continuous for any x 0 X and using the identification of K(X) with the kernel of this map. Next, for the external product K(X) K(Y ) K(X Y ), denoted as usual by, we have for all k 0: This follows by the definition of the external product which states: ψ k (α β) = ψ k (α) ψ k (β). (17) α β = p 1(α)p 2(β), (18) where p 1, p 2 denote the projections of X Y to its factors. Note that the above inclusion is meant in the sense, that α β K(X Y ) is the unique pullback of the right hand side under the quotient map, for more details see the notes on Bott periodicity. The proof of (17) is straight forward by using multiplicativity and naturality: ψ k (α β) = ψ k (p 1(α)p 2(β)) = ψ k (p 1(α))ψ k (p 2(β)) = p 1(ψ k (α))p 2(ψ k (β)) = ψ k (α) ψ k (β). (19) Using (17), we will be able to compute ψ k for spheres S 2n which satisfy by our previous considerations: K(S 2n ) = Z. (20) Note that ψ k will be just multiplication by an integer for any k 0 due to additivity. More precisely, we have the following lemma: Lemma 2.1. ψ k : K(S 2n ) K(S 2n ) is given by multiplication by k n, for all k 0. Proof. The proof is built upon an inductive argument on the dimension of the sphere. Namely, if we consider first the case n = 1, then we just need to check ψ k (α) = kα for some generator α K(S 2 ). Using our preferred generator α = H 1, where H is the canonical line bundle over S 2 = CP 1. Now by using Theorem 2.1, property (ii.), we have: ψ k (α) = ψ k (H) ψ k (1) = H k 1 = (1 + α) k 1, (21) where we used that H is a complex line bundle and that 1 is the trivial line bundle. Note now, due to the multiplication on K(S 2 ), we have: α j = (H 1) j = 0, j 2. (22)
2 ADAMS OPERATIONS 5 Thus by expanding the expression in (21), we see: which is the desired result. ψ k (α) = 1 + kα 1 = kα, (23) Now for general n, we use induction and the external product K(S 2 ) K(S 2n 2 ) K(S 2n ) which is a ring isomorphism. If the Lemma holds for n 1, then by (17), we see: ψ k (α β) = ψ k (α) ψ k (β) = (kα) (k n 1 β) = k n α β, (24) and since a generator of K(S 2n ) can be represented in this form, namely the external product of an appropriate number of H 1, we conclude for general n. We are now able to finish the proof of Theorem 1.2 and thus of Theorem 1.1. The main idea from here is to apply Lemma 2.1 together with property (iii.) of Theorem 2.1 to conclude the proof: Proof of Theorem 1.2. Let α, β K(C f ) as in the definition of the Hopf invariant h for some map f : S 4n 1 S 2n, thus β 2 = hα. By Lemma 2.1, we have for k 0: ψ k (α) = k 2n α, (25) because α is the image of a generator in K(S 2n ) under a pullback map and thus by naturality of our ringhomomorphisms ψ k, α satisfies this equation. In the same spirit, we find that: ψ k (β) = k n β + µ k α, (26) which follows by considering the pullback map taking β to our preferred generator and applying naturality. This shows that ψ k (β) k n β lies in the kernel of this map and hence in the image of the quotient pullback of the SES. This justifies the remaining term. Thus we conclude by direct computation that for k, l 0: and similarily: ψ k (ψ l (β)) = k n l n β + (k 2n µ l + l n µ l )α, (27) ψ l (ψ k (β)) = k n l n β + (l 2n µ l + k n µ l )α. (28) Note that due to Theorem 2.1, property (iii.), we know that ψ k ψ l = ψ kl = ψ l ψ k, thus by comparing (27) to (28), we see: k n (k n 1)µ l = l n (l n 1)µ k. (29) Now we specialize to the case k = 2, l = 3. This is motivated by ψ 2 (β) = β 2 modulo 2 which in turn is due to property (iv.). Therefore, by definition of the Hopf invariant in (3), we see that h is equal to µ 2 modulo 2. Therefore, if we assume h = ±1, then µ 2 has to be odd. Using now (29) for our specific choice of k, l, we see: 2 n (2 n 1)µ 3 = 3 n (3 n 1)µ 2, (30) and thus 2 n divides the righthandside of the equation above. Note that since µ 2 and 3 n are odd, this means that 3 n 1 is divisible by 2 n. But this only happens if n = 1, 2 or 4 be the Lemma below. Hence, we are done by using the numbertheoretic lemma below. The final piece of the proof is provided by the following lemma: Lemma 2.2. If 2 n divides 3 n 1, then n = 1, 2 or 4. Proof. The proof of this relies on basic number theory and has nothing to do with algebraic geometry, thus we only give a brief sketch: If n = 2 l m for m odd, we consider the highest power of 2 dividing 3 n 1 for some l 0. This relies on induction, namely for l = 0, the highest power is 2 and for l > 0 it is l + 2. Use the following decomposition to see this: 3 2lm 1 = (3 2l 1m 1)(3 2l 1m + 1), (31) for l 1. Now deduce the claim by considering the highest powers of 2 dividing each of the two factors. We can then derive a contradiction by noting that n > l + 2 in most cases.
3 INTRODUCTION TO THE SPLITTING PRINCIPLE 6 3 Introduction to the Splitting Principle We return now to Theorem 2.2 which we still need to prove in order to finish our discussion of parallelizable spheres. A first step towards a proof of said Theorem is provided by the following proposition: Proposition 3.1. If X is a finite cell-complex with n cells, then K (X) is a finitely generated group with at most n generators. Moreover, if all cells of X have even dimension, then K 1 (X) = 0 and K 0 (X) is free abelian with one basis element for each cell. The most important example for our discussion is CP n which has exactly one cell for each even integer smaller than 2n, hence n cells. Thus the proposition above applies in this case. Before we enter the proof, we need some clarification of the notion of a finite cell-complex. Such a complex is a space built from a discrete set by attaching a finite number of cells to it. Note that we do not put any conditions on the dimensions, thus this object is more general than a CW-complex, as cells can also be attached to higher dimensional ones. But finite cell-complexes are always homotopy equivalent to finite CW-complexes by deforming each successive attaching map to be cellular, i.e. mapping lower order cell-skeletons to each other. Example 3.1. Instead of giving a proof of this statement, consider the example of a sphere with a line handle attached. Note that we can homotope the attaching map of the handle to map to any given point on the sphere. Considering the sphere as a 2-cell attached to a point, this enables us to make the attaching map of the line handle cellular. One advantage of the more general notion of cell complex is that we can prove that a space is a finite cellcomplex if it is a fiber bundle over a finite cell-complexes where the fibers are themselves finite cell-complexes. But this result will be covered in the next seminar talk thus we do not go any further. Proof. The argument uses induction on the number of cells attached. If X is obtained from a subcomplex A by attaching a k-cell, then we consider the exact sequence corresponding to the pair (X, A). Because X/A = S k, using stereographic projection to visualize this, we have K (X/A) = K (S k ) = Z, and thus by the exact sequence: K (X/A) K (X) K (A), (32) we see that K (X) can at most have one generator more than K (A). To finish the proof, we need to check the second part of the statement. For this note that K 1 (X/A) is vanishing if all cells have even dimension and hence K 1 (X) = 0 by induction and exactness. This is due to the structure of the K-groups for spheres of odd dimension. Moreover, we see that then there is a SES: 0 K 0 (X/A) K 0 (X) K 0 (A) 0, (33) and we know K 0 (X/A) = Z. Thus by induction we conclude that K 0 (X) is free with the required number of generators by splitting of the sequence due to K 0 (A) being free. Lastly for this seminar talk, we will now describe the structure of CP n. Note that by Proposition 3.1, we already know about the additive structure of said ring, thus we just need to observe its multiplication. For this we remind ourselves that we can identify CP n with a CW-complex with exactly one cell in each even dimension 2n. This means by Proposition 3.1: K 0 (CP n ) = Z[X]/X n+1, K 1 (CP n ) = 0, (34) as additive groups. Additionally, according to the following proposition, the ring K(CP n ) is as simple as we could hope for: Proposition 3.2. K(CP n ) is isomorphic as a ring to the quotient ring Z[L]/(L 1) n+1, where L is identified with the canonical line bundle over CP n. Proof. The proof is based on induction using the pair (CP n, CP n 1 ) as well as the corresponding SES (by Prop. 3.1): 0 K(CP n, CP n 1 ) K(CP n ) ρ K(CP n 1 ) 0. (35) Note that we use the CW-complex description of the complex space in the identification of the pair above. If we assume that the proposition holds for n 1, i.e. K(CP n 1 ) has the desired structure, then if (L 1) n generates
REFERENCES 7 the kernel of ρ, for any n N, we can conclude by induction and the splitting. Thus we need to check whether (L 1) n generates the kernel of ρ. The idea here is to relate the exterior products of K(S 2 ) and K(CP n ) which is inspired by the analogous generating property for spheres and their external products. We use the usual identification of the complex linear space as quotient of S 2n+1 under the action of S 1. Note that we can equally well work with the boundary of a product of n + 1 closed unit discs D 0 D n in C which is clearly homeomorphic to S 2n+1. By the usual formula for the boundary: (D 0 D n ) = n i=0d 0 D i D n. (36) The S 1 -action respects the decomposition of the boundary and we denote the orbit space under the action of D 0 D i D n by C i. This yields a decomposition of CP n = i C i such that C i C j = C i C j, if i j. Using this decomposition, we are now able to find a monstrous commutative diagram as depicted in [1, p.67] using the inclusions of pairs (D i, D i ) (C 0, i C 0 ) (CP n, C i ) where we denote by i C 0 the i-th boundary part of C 0, i.e.: i C 0 := D 1 D i D n. (37) The commutative diagram involves the sequence of maps between the external products induced by the pairs above as well as the quotient maps from CP n to CP n /C i. Ther result now follows by mapping L 1 upwards in each component of the external product and observing that we get a generator in the upper left corner. Following further along the maps, we eventually found that thus the product (L 1) n gets mapped to a generator of the image of the left map in the SES in (35). Thus, by exactness, we see that it generates the kernel of ρ. This concludes the argument. References [1] A.Hatcher, Vector Bundles and K-Theory, Link: https://www.math.cornell.edu/~hatcher/vbkt/ VBpage.html. (March 2018)