First discussion of Hurwitz's theorem

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(ax) = 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 + ib) = 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 ⟨Fx, Fy⟩ = ⟨x, y⟩ for all x, y ∈ V, and F^TF = I.

Let k be a field. A vector space A over k, equipped with a k-linear map A⊗_kA → A is called a k-algebra. We say A is unital if it contains a non-zero member e such that ev = v for each v ∈ A. We define a normed algebra A to be an algebra equipped with a quadratic form q such that q(xy) = 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) = ux 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
v^*^* = v
A_u⁻¹ = A_u^T = A_u^*

Lemma 1. (Braiding equation) Let u, v be ℝ-linearly independent unit-vectors. Then A_u^TA_v + A_v^TA_u = 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(A_iv) = q(e_iv) = v, so each A_i is orthogonal with respect to q, and hence A_i^TA_i = 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 ∑r_ir_j⟨v, A_i^tA_j⟩ = (∑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, (A_i^TA_j + A_j^TA_i)v⟩ = 0 for all v. Thus A_i^TA_j + A_j^TA_i = 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 A_i^TA_n + A_n^TA_i = 0 i.e. −A_i^T = 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 A_u^TA_e = −A_e^TA_u i.e. A_u^T = −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 ⊕ Bu is a subalgebra of A of dimension 2 ⋅ dim B.

Proof. First note that B ⊕ Bu 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 {v_iu} is also an orthonormal basis. Thus, it suffices to show that B ⊕ Bu is closed under multiplication. Indeed, let a + bu and c + du be members of B ⊕ Bu, where a, b, c, d ∈ B. Then (a + bu)(c + du) = ac + a(du) + (bu)d + (bu)(du) = (ac − bd) + (ad − bd^*)u which is a member of B ⊕ Bu. ◻

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. ◻