Well, Papa, can you multiply triplets?

We show that the classical algebra of quaternions is a commutative Z2×Z2×Z2-graded algebra. A similar interpretation of the algebra of octonions is impossible. This note is our “private investigation” of what really happened on the 16th of October, 1843 on the Brougham Bridge when Sir William Rowan Hamilton engraved on a stone his fundamental relations: i = j = k = i · j · k = −1. Since then, the elements i, j and k, together with the unit, 1, denote the canonical basis of the celebrated 4-dimensional associative algebra of quaternions H. Of course, the algebra H is not commutative: the relations above imply that the elements i, j, k anticommute with each other, for instance i · j = −j · i = k. Yes, but... 1. The algebra of quaternions is commutative Our starting point is the following amazing observation. The algebra H is, indeed, commutative. Let us introduce the following “triple degree”: (1) σ(1) = (0, 0, 0), σ(i) = (0, 1, 1), σ(j) = (1, 0, 1), σ(k) = (1, 1, 0), then, quite remarkably, the usual product of quaternions satisfies the graded commutativity condition: (2) p · q = (−1) q · p, where p, q ∈ H are homogeneous (i.e., proportional one of the basic vector) and where 〈 , 〉 is the scalar product of 3-vectors. Indeed, 〈σ(i), σ(j)〉 = 1 and similarly for k, so that i, j and k anticommute with each other, but 〈σ(i), σ(i)〉 = 2. The product i · i of i with itself is commutative and similarly for j and k, without any contradiction. The degree (1) viewed as an element of the abelian group Z2 ×Z2 ×Z2 satisfies the following linearity condition (3) σ(x · y) = σ(x) + σ(y), ha l-0 03 35 50 2, v er si on 1 29 O ct 2 00 8 2 SOPHIE MORIER-GENOUD VALENTIN OVSIENKO for all homogeneous x, y ∈ H. The relations (2) and (3) together mean that H is a Z2 × Z2 × Z2-graded commutative algebra. We did not find the above observation in the literature (see however [1] for a different “abelianization” of H in terms of a twisted Z2 × Z2 group algebra, see also [2, 3, 4]). Its main consequence is a systematic procedure of quaternionization (similar to complexification). Indeed, many classes of algebras allow tensor product with commutative algebras. Let us give an example. Given an arbitrary real Lie algebra g, the tensor product gH := H ⊗R g is a Z2 × Z2 × Z2-graded Lie algebra. If furthermore g is a real form of a simple complex Lie algebra, then gH is again simple. The above observation gives a general idea to study graded commutative algebras over the abelian group Γ = Z2 × · · · × Z2 } {{ } n times . One can show that, in some sense, this is the most general grading, in the gradedcommutative-algebra context, but we will not provide the details here. Let us mention that graded commutative algebras are essentially studied in the case Γ = Z2 (or Z), almost nothing is known in the general case. 2. ...but not the algebra of octonions After the quaternions, the next “natural candidate” for commutativity is of course the algebra of octonions O. However, let us show that The algebra O cannot be realized as a graded commutative algebra. Indeed, recall that O contains 7 mutually anticommuting elements e1, . . . , e7 such that (el) 2 = −1 for l = 1, . . . , 7 that form several copies of H (see [2, 3] for beautiful introduction to the octonions). Assume there is a grading σ : el 7→ Γ with values in an abelian group Γ, satisfying (2) and (3). Then, for three elements el1 , el2 , el3 ∈ O such that el1 · el2 = el3 , one has σ(el3) = σ(el1) + σ(el2). If now el4 anticommutes with el1 and el2 , then el4 has to commute with el3 because of the linearity of the scalar product. This readily leads to a contradiction. 3. Multiplying the triplets Let us now take another look at the grading (1). It turns out that there is a simple way to reconstitute the whole structure of H directly from this formula. First of all, we rewrite the grading as follows: (4) 1 ↔ (0, 0, 0), i ↔ (0, 12, 13), j ↔ (11, 0, 13), k ↔ (11, 12, 0). Second of all, we define the rule for multiplication of triplets. This multiplication is nothing but the usual operation in Z2 ×Z2 ×Z2, i.e., the component-wise addition (modulo 2), for instance, (11, 0, 0) · (11, 0, 0) = (0, 0, 0), (11, 0, 0) · (0, 12, 0) = (11, 12, 0), ha l-0 03 35 50 2, v er si on 1 29 O ct 2 00 8 WELL, PAPA, CAN YOU MULTIPLY TRIPLETS? 3 but with an important additional sign rule. Whenever we have to exchange “leftto-right” two units, 1n and 1m with n > m, we put the “−” sign, for instance (0, 12, 0) · (11, 0, 0) = −(11, 12, 0), since we exchanged 12 and 11. One then has for the triplets in (4): i · j ↔ (0, 12, 13) · (11, 0, 13) = (11, 12, 0) ↔ k, since the total number of exchanges is even (12 and 13 were exchanged with 11) and j · i ↔ (11, 0, 13) · (0, 12, 13) = −(11, 12, 0) ↔ −k, since the total number of exchanges is odd (13 was exchanged with 12). In this way, one immediately recovers the complete multiplication table of H. Remark 3.1. The above realization is of course related to the embedding of H into the associative algebra with 3 generators ε1, ε2, ε3 subject to the relations εn = 1, εnεm = −εmεn, for n 6= m. This embedding is given by i 7→ ε2ε3, j 7→ ε1ε3, k 7→ ε1ε2 and is well-known. Everybody knows the famous story of Hamilton and his son asking his father the same question every morning: “Well, Papa, can you multiply triplets?” and always getting the same answer: “No, I can only add and subtract them”, with a sad shake of the head. This story has now a happy end. As we have just seen, Hamilton did nothing but multiplied the triplets. Or should we rather say added and subtracted them? Acknowledgments. We are grateful to S. Tabachnikov for helpful suggestions.