14. Compositions over General Fields 301 Recall that λ(r, s) = max{ρ(r),ρ(r + 1),...,ρ(n)} as in (12.25). Moreover ρF (n, n) = ρ(n) because the Hurwitz–Radon Theorem holds true over F . We have just proved that if F has characteristic zero then: r # s ≤ r ∗F s and ρF (n, r) ≤ ρ#(n, r). However this algebraic result was proved using non-trivial topology. Is there a truly algebraic proof of the “Hopf Theorem”: r s ≤ r ∗F s for fields of characteristic zero? Does this result remain true if the field has positive characteristic? One productive idea is to apply Pfister’s results on the multiplicative properties of sums of squares. If F is a field (where 2 = 0), define • DF (n) ={a ∈ F : a is a sum of n squares in F }. • Recall that if q is a quadratic form over F then DF (q) is the set of values in F represented by q. The notation above is an abbreviated version: DF (n) = DF (n1). Evaluating one of our bilinear composition formulas at various field elements estab- lishes the following simple result. 14.3 Lemma. If [r, s, n] is admissible over F then for any field K ⊇ F , DK (r) · DK (s) ⊆ DK (n). Some multiplicative properties of these sets DF (n) were proved earlier. The clas- sical n-square identities show that DF (1), DF (2), DF (4) and DF (8) are closed under −1 −1 −1 2 multiplication. Generally a ∈ DF (n) implies a ∈ DF (n) since a = a · (a ) . Therefore those sets DF (n) are groups if n = 1, 2, 4 or 8. In the 1960s Pfister showed m that every DF (2 ) is a group. This was proved above in Exercise 0.5 and more gen- erally in (5.2). Applying this result to the rational function field F(X,Y) provides some explicit 2m-square identities. Of course any such identity for m>3 cannot be bilinear (it must involve some denominators). Here is another proof that [3, 5, 6] is not admissible. We know [3, 5, 7] is admis- sible over any F and therefore DF (3) · DF (5) ⊆ DF (7). In fact we get equality here. = 2 +···+ 2 ∈ 2 + 2 + 2 = Given any a a1 a7 DF (7) we may assume that a1 a2 a3 0 and factor out that term: a2 + a2 + a2 + a2 a = (a2 + a2 + a2) · 1 + 4 5 6 7 . 1 2 3 2 + 2 + 2 a1 a2 a3 The numerator and denominator of the fraction are in DF (4) (at least if the numerator is non-zero). Since DF (4) is a group the quantity in brackets is a sum of 5 squares, so that a ∈ DF (3) · DF (5). When that numerator is zero the conclusion is even easier. Therefore for any field F , DF (3) · DF (5) = DF (7). 302 14. Compositions over General Fields If [3, 5, 6] is admissible over F then by (14.3): DK (6) = DK (7) for every such K ⊇ F. Of course this equality can happen in some cases. For instance if the form 61 is • isotropic over F then it is isotropic over every K ⊇ F and DK (n) = K for every n ≥ 6. On the other hand Cassels (1964) proved that in the rational function field = R + 2 +···+ 2 K (x1,...,xn) the element 1 x1 xn cannot be expressed as a sum of n squares. Applied to n = 6 this shows that DK (6) = DK (7) and therefore [3, 5, 6] is not admissible over R. Cassels’ Theorem was the breakthrough which inspired Pfister to develop his theory of multiplicative forms. He observed that a quadratic form ϕ over F can be viewed in two ways. On one hand ϕ is a homogeneous quadratic polynomial ϕ(x1,...,xn) ∈ F [X]. On the other hand it is a quadratic mapping ϕ : V → F arising from a symmetric bilinear form bϕ : V × V → F , and we speak of subspaces, isometries, etc. We write ϕ ⊂ q when ϕ is isometric to a subform of q. The general result we need is the Cassels–Pfister Subform Theorem. It was stated in (9.A.1) and we state it again here. 14.4 Subform Theorem. Let ϕ, q be quadratic forms over F such that q is anisotropic. Let X = (x1,...,xs) be a system of indeterminates where s = dim ϕ. Then q ⊗F(X) represents ϕ(X) over F(X)if and only if ϕ ⊂ q. 14.5 Corollary. Suppose s and n are positive integers and n1 is anisotropic over F . The following statements are equivalent. (1) s ≤ n. (2) DK (s) ⊆ DK (n) for every field K ⊇ F . 2 +···+ 2 (3) x1 xs is a sum of n squares in the rational function field F(x1,...,xs). Proof. For (3) ⇒ (1), apply the theorem to the forms q = n1 and ϕ = s1. To generalize the proof above that [3, 5, 6] is not admissible, we need to express the product DF (r) · DF (s) as some DF (k). This was done by Pfister (1965a). It is surprising that Pfister’s function is exactly r s, the function arising from the Hopf– Stiefel condition! 14.6 Proposition. DF (r) · DF (s) = DF (r s), for any field F . In the proof we use the fact that DF (m + n) = DF (m) + DF (n). Certainly if c ∈ DF (m+n), then c = a +b where a is a sum of m squares, b is a sum of n squares. The Transversality Lemma (proved in Exercise 1.15) shows that this can be done with a,b = 0. This observation enables us to avoid separate handling of the cases a = 0 and b = 0. 14. Compositions over General Fields 303 Proof of 14.6. Since the field F is fixed here we drop that subscript. We will use the characterization of r s given in (12.10). The key property is Pfister’s observation, mentioned earlier: D(2m) · D(2m) = D(2m). By symmetry we may assume r ≤ s and proceed by induction on r + s. Choose the smallest m with 2m <s≤ 2m+1. We first prove D(r) · D(s) ⊆ D(r s).Ifr ≥ 2m then r + s>2m+1 and r s = 2m+1. Then D(r) · D(s) ⊆ D(2m+1) = D(r s), as hoped. Otherwise r<2m <sand we express s = s + 2m where s > 0. Then r s = r (s + 2m) = (r s) + 2m by (12.10). Therefore D(r) · D(s) = D(r) · (D(s) + D(2m)) ⊆ D(r) · D(s) + D(r) · D(2m) ⊆ D(r s) + D(2m) = D((r s) + 2m) = D(r s). For the equality we begin with a special case. Claim. D(2m) · D(2m + 1) = D(2m+1). For if c ∈ D(2m+1) then c = a + b where a,b ∈ D(2m). Then c = a(1 + b/a) and 1 + b/a ∈ D(2m + 1) since D(2m) is a group. This proves the claim. If r ≥ 2m then r s = 2m+1 and the claim implies that D(r s) = D(2m+1) ⊆ D(r)·D(s). Otherwise r<2m <sand r s = (r s)+2m as before. If c ∈ D(r s) then c = a + b where a ∈ D(r s) and b ∈ D(2m). The induction hypothesis implies that a = a1a2 where a1 ∈ D(r) and a2 ∈ D(s ). Then c = a1(a2 + b/a1) and m m m a2 + b/a1 ∈ D(s ) + D(2 ) · D(r) = D(s ) + D(2 ) = D(s + 2 ) = D(s). Hence c ∈ D(r) · D(s). 14.7 Proposition. If (r s)1 is anisotropic over the field F then r s ≤ r ∗F s. Proof. Suppose [r, s, n] is admissible over F . By (14.3) and (14.6), DK (r s) ⊆ DK (n) for every field K ⊇ F .Ifn<r s then (14.5) provides a contradiction. This provides an algebraic proof of Hopf’s Theorem over R (for normed bilinear pairings). Unfortunately these ideas do not apply over C or over any field of positive characteristic, because n1 is isotropic for every n ≥ 3 in those cases. Pfister’s methods lead naturally to “rational composition formulas”, that is, for- mulas where denominators are allowed. 14.8 Theorem. For positive integers r, s, n the following two statements are equiva- lent. (1) r s ≤ n. (2) DK (r) · DK (s) ⊆ DK (n) for every field K. Furthermore if F is a field where n1 is anisotropic, then the following statements are also equivalent to (1) and (2). Here X = (x1,...,xr ) and Y = (y1,...,ys) are systems of indeterminates. (3) DK (r) · DK (s) ⊆ DK (n) where K = F(X,Y). 2 +···+ 2 2 +···+ 2 = 2 +···+ 2 (4) There is a formula (x1 xr )(y1 ys ) z1 zn where each zk ∈ F(X,Y). 304 14. Compositions over General Fields (5) There is a multiplication formula as above where each zk is a linear form in Y with coefficients in F(X). Proof. The equivalence of (1) and (2) follow from (14.5) and (14.6). Trivially (2) ⇒ (3), (3) ⇒ (4) and (5) ⇒ (4). ⇒ ∈ = 2 ···+ 2 Proof that (4) (5). Given the formula where zk F(X,Y), let α x1 xr . · 2 +···+ 2 = Then α (y1 ys ) is a sum of n squares in F(X,Y). Setting K F(X) this 2 + ··· + 2 is the same as saying: n 1 represents αy1 αys over K(Y). Since n 1 is anisotropic the Subform Theorem 14.4 implies that sα⊂n1 over K.Now interpret quadratic forms as inner product spaces to restate this condition as: there is a K-linear map f : Ks → Kn carrying the form sα isometrically to a subform of n1.