In this post, we begin a discussion of Hurwitz’s theorem, which states that the real finite-dimensional unital normed algebras are ℝ, ℂ, ℍ (quaternions) and 𝕆 (octonions). We only prove part of it here and we delay a full proof to a future post. about later.

We begin by defining our objects of interest, as well as formalizing some linear algebra in the setting of k-algebras.

Recall that if V is a vector space over k, then a quadratic form is a map q : V → k such that q(a**x) = a2q(x) for all a ∈ k and x ∈ V. A form induces a symmetric bilinear form ⟨⋅, ⋅⟩ defined by ⟨x, y⟩ = q(x + y) − q(x) − q(y). We say a non-zero element u ∈ V is a unit-vector if q(u) = 1. We say a linear map F : V → V is an orthogonal map if q(F(x)) = q(x) for all x ∈ A. As an example, ℝ is a normed space with norm q(x) = x2 for x ∈ ℝ. Similarly, for ℂ, we have a norm q(a + i**b) = a2 + b2. Note that we do not deal with any square roots.

Remark 1. Linear algebra in this setting works exactly as one would expect. In particular, if (V, q) is such a space, then we can assume that we have an orthnormal basis with respect to the inner product ⟨⋅, ⋅⟩q. If F : V → V is an orthogonal map, then ⟨F**x, F**y⟩ = ⟨x, y⟩ for all x, y ∈ V, and FTF = I.

Let k be a field. A vector space A over k, equipped with a k-linear map A⊗k**A → A is called a k-algebra. We say A is unital if it contains a non-zero member e such that e**v = v for each v ∈ A. We define a normed algebra A to be an algebra equipped with a quadratic form q such that q(x**y) = q(x)q(y) for all x, y ∈ A. We let ⟨⋅, ⋅⟩ denote its inner product.

We now have enough to begin our study properly. Let A be a real finite-dimensional unital normed algebra with unit e. For a non-zero element u ∈ A, define the map A**u : A → A by A**u(x) = u**x for all x ∈ A. Note that A**u is a linear map, and that A**u is orthogonal when u is a unit vector. For an element v ∈ A, define its involution by v* = ⟨v, e⟩ − v. We note the following properties for v ∈ A.

Proposition 1.

v*v = q(v)e 1.

v** = v 1.

Au*−1 = AuT = A**u**

Lemma 1. (Braiding equation) Let u, v be ℝ-linearly independent unit-vectors. Then AuTAv + AvTAu = 0.

Proof. Extend {u, v} to a basis, and denote the basis by {e1, …, e**n}. Let 1 ≤ i < j ≤ n. Note that for v ∈ A, q(Aiv) = q(eiv) = v, so each A**i is orthogonal with respect to q, and hence AiTAi = I.

Now we wish to prove the other identity. To this end, let r1, …, r**n be real numbers. Then

$$\begin{align*} \langle \sum r_i A_i v, \sum r_j A_j v &= ||\sum r_i A_i v||^2 \ &= ||\sum r_i e_i v||^2 \ &= ||\sum r_i e_i||^2 ||v||^2 \ &= ( \sum r_i^2 ) ||v||^2 \end{align*}$$

so ∑rir**j⟨v, AitAj⟩ = (∑r**i2)||v||2. It is worth pausing a minute to reflect on this sum. If we consider the cases i = j and i ≠ j separately, we conclude that $$\sum\limits_{i < j} r_i r_j \langle v, (A_i^T A_j + A_j^T A_i) v\rangle = 0.$$ Choosing r**i = 1 for all i yields ⟨v, (AiTAj + AjTAi)v⟩ = 0 for all v. Thus AiTAj + AjTAi = 0 for each i. ◻

Lemma 2. dim A is an even positive integer or dim A = 1.

Proof. As A is unital, we can assume that last basis vector e**n is a unit. Hence A**n = I. Moreover, if i < n, then AiTAn + AnTAi = 0 i.e. −AiT = A**i so A**i2 = −I hence det (A)2 = (−1)n which proves that n must be even. ◻

Lemma 3. Let A be a unital f.d. k-algebra. Let e ∈ A be the unit, and let u be an orthogonal unit vector. Then u2 = −e and u* = −u.

Proof. By the braiding equation we have AuTAe = −AeTAu i.e. AuT = −A**u so u* = −u. Therefore q(u)e = u*u = −u2 so u2 = −e. ◻

Proposition 2. Let B be a unital associative subalgebra of A. Let e ∈ A be the identity, and let u ∈ A be a unit vector orthogonal to e. Then B ⊕ B**u is a subalgebra of A of dimension 2 ⋅ dim B.

Proof. First note that B ⊕ B**u has dimension 2 ⋅ dim B by the definition of the direct sum, and the fact that if {v**i} is an orthonormal basis of B then {viu} is also an orthonormal basis. Thus, it suffices to show that B ⊕ B**u is closed under multiplication. Indeed, let a + b**u and c + d**u be members of B ⊕ B**u, where a, b, c, d ∈ B. Then (a + b**u)(c + d**u) = a**c + a(d**u) + (b**u)d + (b**u)(d**u) = (a**c − b**d) + (a**d − b**d*)u which is a member of B ⊕ B**u. ◻

Corolary 1. If A is a real unital f.d. algebra, then its dimension over ℝ is a power of 2.

Proof. Let B = ℝ and apply the above construction. ◻