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Home , Alternative algebra, Involution (mathematics), Octonion algebra

OCTONIONS, CUBES, EMBEDDINGS MARTIN H. WEISSMAN MARCH 2, 2009

Octonions

Let k be a field, with char(k) = 2. A Hurwitz algebra over k is a Resulting structures on a Hurwitz algebra A 6 include a trace, finite-dimensional, unital, k-algebra A, together with a N : A k, such that the associated bilinear form is nonde- Tr(a)=N(a + 1) N(a) N(1), generate and:! and a¯ = Tr(a) a. Hurwitz N(xy)=N(x)N(y), for all x, y A. algebras are quadratic: 2 a2 Tr(a)+N(a)=0. Every Hurwitz algebra over k has dimension 1, 2, 4, or 8, as a k vector space. An 8-dimensional Hurwitz k-algebra is often called a Cayley or algebra. The isomorphism class of a Cayley k-algebra A is determined by the isomorphism class of the quadratic space (A, N). The quadratic form N of a Cayley k-algebra is always a Constructing all Cayley algebras (up to isomorphism) Pfister form. Every Pfister form of dimension over Q (or a local field) is not difficult. One construction is the 8 is hyperbolic or anisotropic. There are no anisotropic forms over Qp of dimension Cayley-Dickson process: begin with a algebra B over greater than 4. There are two dimension 8 R Q. Define O = B B as a Q-vector space. Define a Q-algebra Pfister forms over . The isomorphism class of a quadratic form over Q is determined structure on O by: by local invariants. There is no local-global obstruction for Cayley algebras. (u, v) (z, w)=(uz wv¯ , wu + vz¯). · If b B, then b¯ = Tr(b) b denotes the main involution.2 Define the main involution on O by (u, v)=(u¯, v). Status update: The Cayley-Dickson construc- tion is implemented in SAGE by J. Hanke and M. Weissman. A common construction of a “nonsplit” 2 Nonsplit here means that the quadratic form over Q arises as follows: Let P (F2) denote the projective space of is anisotropic 3 lines in F2 (a set with seven elements). Since F2 has one nonzero 2 element, there is a natural Elements of P (F2) are often called e1, e2, e3, e4, e5, e6, e7 in the literature. This construction of an octonion algebra is due to P2(F ) F3 0 . 2 $ 2 { } Freudenthal. Define O to be the Q-vector space whose basis is the set 1 2 2 { }[ P (F2). If ~x P (F2), write e~x for the associated basis element of 2 2 2 O. Define a f : P (F2) P (F2) F2 by: Given a long history of mistakes is the ⇥ ! subject, the signs given by f (~x,~y) should be machine-verified: one should verify that one f (~x,~y)= Â xs(1)ys(1) + xs(1)ys(2) + ys(1)xs(2)xs(3). gets an from this choice of s A signs, by considering quantities like (e e )e , 2 3 ~x ~y ~x which should equal e~x (e~y e~x ). Define a Q-algebra structure on O by identifying 1 as the unit, 001 and defining: (e ) (e )=e ( 1) f (~x,~y). ~x · ~y ~x+~y · The main involution on O is determined by 1¯ = 1 and e = e . ~x ~x 101 011

111 1

100 110 010

Figure 1: The Fano . The label abc refers to the line through (a, b, c) P2(F ). 2 2 octonions, cubes, embeddings 2

Bhargava Cubes

1 The results on this page are all due to Manjul Bhargava Define a 1 Manjul Bhargava, Higher composition Z2 Z2 Z2 laws. I. A new view on Gauss composition, Bhargava cube to be an element C Z Z . and quadratic generalizations, in Annals of 3 2 ⌦ ⌦ The SL2(Z) acts on the Z- of Bhargava cubes, Mathematics, 2004. via the tensor cube of the standard action of SL (Z) on Z2. There 2 f is a unique (up to normalization) quartic polynomial invariant for e the resulting geometric action; its (normalized) value is called the a b discriminant, and denoted D(C). From a Bhargava cube C, one may construct three ordered pairs of matrices by slicing:

C : (M1, N1) or (M2, N2) or (M3, N3).

These yields three integer-valued quadratic forms, all of the same g h discriminant D(C): c d Qi(x, y)= det(Mix Niy), for i = 1, 2, 3. Figure 2: A Bhargava cube. Slicing along the The cube C is called projective if the three associated quadratic light line yields the two matrices: forms Q1, Q2, Q3 are primitive. Projective cubes can be brought, ab M = , Z 3 1 cd by the action of SL2( ) , into a “normal form” as displayed in the ✓ ◆ margin. ef N = . The following theorem is the starting point for Bhargava’s work: 1 gh ✓ ◆ 3 Theorem 1 (Bhargava) Fix an integer D. The orbits of SL2(Z) on 0 g the set of projective Bhargava cubes of discriminant D are in one-to-one 1 0 correspondence with the set of triples ([Q1], [Q2], [Q3]) of SL2(Z)- equivalence classes of primitive binary quadratic forms of discriminant D, such that [Q ] [Q ] [Q ]=[1], in the sense of Gauss composition. 1 2 3 f m 0 e The significance of Bhargava’s theorem is that it provides a completely new definition of Gauss composition. Indeed, in any Figure 3: A Bhargava cube in normal form. ( ) The three resulting quadratic forms, all of group G, knowledge of the set of triples g1, g2, g3 such that discriminant D = m2 + 4efg, are: g1 g2 g3 = 1 suffices to determine the group law. In practice, 2 2 Q (x, y)= ex + mxy + fgy , it implies that, given two primitive quadratic forms Q1, Q2, there 1 exists a projective Bhargava cube C in normal form, from which 2 2 Q2(x, y)= fx + mxy + egy , Q3 can be easily computed. Q (x, y)= gx2 + mxy + efy2. For example, consider the following two quadratic forms of 3 discriminant 20: 2 2 Q (x, y)=x + 5y . 0 2 1 Q (x, y)=3x2 2xy + 2y2. 1 0 2 These fit into the cube at the right. The third quadratic form is:

Q (x, y)=2x2 2xy + 3y2. 3 3 2 Since [Q ]=[Q ] and [Q ]=[1], [Q ] [Q ]=[1]. 3 2 1 2 2 0 1 · Figure 4: A Bhargava cube of discriminant 20. octonions, cubes, embeddings 3

Orders and an Embedding Question

Let O denote the nonsplit octonion algebra over Q, with basis 1 P2(F ). We are interested in writing down a maximal order { }[ 2 in O. The naive guess Wng = Z 2 Ze at a maximal ~x P (F2) ~x O 2 order in is incorrect! The correctL maximal order was identified 2 by Coxeter , after many mistaken attempts (e.g. by Kirmse). There 2 H.S.M. Coxeter, Integral Cayley , Duke Math J., 1946 exists a maximal order W, which contains Wng with index 16; Coxeter describes this order explicitly. The history of mistakes, and the fact that The following statements characterize the order W as a lattice our description of O differs from Coxeter’s, suggests we should recompute the correct containing W : if ~x + ~y = ~z, in P2(F ), then the octonion 1/2 ng 2 · maximal order W, and check our work using ( 1 e e e ) is in W. Furthermore, W contains an element computer assistance. First, ± ± ~x ± ~y ± ~z 1 2 w = /2 (e~x + e~y + e~z + ew~ ), for some ~x,~y,~z, w~ P (F2), such that W W 1/2W . · 2 ng ⇢ ⇢ ng N(w)=1. One should check that the lattice W has the The maximal order W (unique up to conjugation by Aut(O/k)) following properties: the lattice, with the is isometric to the E8 root lattice. The theta function of the E8 root form, is isometric to the E8 root lattice (with 240 elements of norm 1). It should also lattice is just the Eisenstein series E4 (of weight 4, level 1); for this be closed under . reason, there is a formula for the of integral octonions (elements of W) of any given norm: # w W : N(w)=n = 240 Â d3. { 2 } · d n |

There is a connection between Bhargava’s cubes and Cox- eter’s integral octonions, arising from modular forms on excep- 3 3 tional groups. In my work on D4 modular forms , the following M. Weissman, D4 Modular Forms, Amer. J. quantities arise: Suppose that C is a projective cube, in normal of Math., 2006 form, with D = D(C) < 0, as pictured on the right. 0 g Define: 1 0 (a, b, g) W3 such that Emb(C, W)= 2 . Na = e, Nb = f , Ng = g, Tr(abg)=m ⇢ This counts the number of embeddings of “QT-structures” (triples m of quadratic lattices endowed with a trilinear form), from a QT- f 0 e structure coming from a Bhargava cube, into the QT-structure arising from O (with the norm quadratic form and the trace trilin- Figure 5: A Bhargava cube in normal form. 2 ear form). Suppose that D = m + 4efg < 0.

3 For three octonions a, b, g, it happens that Theorem 2 If C and C0 are in the same SL2(Z) -orbit, then Emb(C, W)= Emb(C0, W). The numbers Emb(C, W) are the Fourier coefficients (in- Tr((ab)g)=Tr(a(bg)). dexed by cubes) of a “theta function” for the group Spin4,4. Call this quantity Tr(abg). The quantities Emb(C, W) should be machine-computable, espe- cially for cubes of small (and perhaps prime) discriminant. These quantities should be identifiable with the Fourier coefficients of an Eisenstein seris, but this has not yet been proven. octonions, cubes, embeddings 4

Computations and questions

Bhargava’s theorem, and the previous result, suggests that one should carry out the following: Choose a negative discriminant D. • Choose a triple [Q ][Q ][Q ]=[1] of classes of binary 0 1 • 1 2 3 quadratic forms of negative discriminant D. 1 0 Find a cube C which “encodes” these classes by slicing. • Compute the quantity Emb([Q ], [Q ], [Q ], W)=Emb(C, W). • 1 2 3 Such quantities fit into a general class of embedding problems 1 0 including the following: 0 5

The classical “representation of quadratic forms by quadratic Figure 6: A Bhargava cube in normal form, • forms” problem can be rephrased as counting embeddings with D = 20. It reflects the identity [1] [1] [1]=[1] of a definite quadratic lattice into another definite quadratic · · 0 2 lattice. 1 0 If K is a quadratic imaginary field, and A is a maximal order • in K, and e is the Dirichlet character associated to K/Q, then Emb(A, W) (the number of unital algebra embeddings) is equal to 252L(e, 2) by a result of Gross-Elkies4. 3 2 In particular, I would hypothesize that Emb(C, W) can be com- 0 1 puted via an Euler product. The specific form of this product should be “guessable” from enough computations. Figure 7: A Bhargava cube in normal form, Observe that Emb([Q ], [Q ], [Q ], W) assigns an integer to each with D = 20. It reflects an identity 1 2 3 [Q] [Q]=[1] “collinear” triple in a class group. There are some immediate · 4 B. Gross and N. Elkies, Embeddings into questions: the integral octonions, Pac. J. of Math., 1997 Do these integers depend the elements in the class group? • Suggested computational project, involving Do the integers only depend upon the genus of the quadratic Emb(C, W) for the two cubes above: Count triples (a, b, g) O3 such that Na = 1, forms, or the spinor genus? Nb = 2, Ng =23, Tr(abg)= 2. Then, count triples (a, b, g) O3 such that Na = 1, Do these integers “look different” (bigger or smaller), when Nb = 1, Ng = 5, and2Tr(abg)=0. One • can definitely use tricks to speed up the brute the structure of the class group is different (e.g., cyclic of force! Do these counts yield the same result? order 4 or the Klein 4-group)? Can one see differences in What are the results? the fine structure of the class group from these integers (a la Cohen-Lenstra)? Can one guess a simple Euler product, or less simple but • well-known L-value for the numbers Emb(C, W)? The de- grees of the Euler factors should be guessable by computing dimensions of Lie algebras over finite fields. The fact that these integers are given by an Euler product is probably not difficult Any such answers could be helpful. Since these integers are the to prove. But, there might be some very bad Euler factors, arising from masses of coefficients of a theta function on Spin4,4, analytic techniques are nonmaximal lattices in a rank 3 Hermitian available to estimate (bound) these integers as the discriminant space. grows.