The real algebras \(\mathbb R\), \(\mathbb C\), \(\mathbb H\), and \(\mathbb O\) have dimensions \(1,2,4,8\). They form the familiar chain

\[\mathbb R\subset\mathbb C\subset\mathbb H\subset\mathbb O,\]

and as we move from left to right, familiar algebraic properties gradually disappear.

Algebra Commutative Associative Alternative Multiplicative norm
\(\mathbb R\) Yes Yes Yes Yes
\(\mathbb C\) Yes Yes Yes Yes
\(\mathbb H\) No Yes Yes Yes
\(\mathbb O\) No No Yes Yes

Three classical rigidity theorems explain much of this pattern.

Frobenius Theorem. Every finite-dimensional associative division algebra over \(\mathbb R\) is isomorphic to \(\mathbb R\), \(\mathbb C\), or \(\mathbb H\).

Wedderburn's Little Theorem. Every finite division ring is commutative, hence is a finite field.

Hurwitz Theorem. Every finite-dimensional normed division algebra over \(\mathbb R\) is isomorphic to \(\mathbb R\), \(\mathbb C\), \(\mathbb H\), or \(\mathbb O\).

Thus Frobenius and Wedderburn concern associative division algebras, while Hurwitz replaces associativity by the existence of a positive-definite multiplicative quadratic form. The generalized form of Hurwitz' theorem will later require only nondegeneracy.

1. Frobenius Theorem

Let \(D\) be a finite-dimensional associative division algebra over \(\mathbb R\). We wish to prove that \(D\) is one of \(\mathbb R\), \(\mathbb C\), or \(\mathbb H\).

1.1 Every element has degree at most two

Since \(D\) is finite-dimensional over \(\mathbb R\), every \(x\in D\) is algebraic over \(\mathbb R\). Indeed, the elements \(1,x,x^2,\ldots\) must eventually become linearly dependent. Let \(m_x(t)\in\mathbb R[t]\) be the minimum polynomial of \(x\). We claim that \(m_x(t)\) is irreducible over \(\mathbb R\). Suppose \(m_x(t)=f(t)g(t)\), where \(f,g\in\mathbb R[t]\) both have positive degree. Since \(\mathbb R\) lies in the center of \(D\), evaluation at \(x\) gives \(0=m_x(x)=f(x)g(x)\). Since \(D\) is a division algebra, it has no zero divisors, so either \(f(x)=0\) or \(g(x)=0\). This contradicts the minimality of \(m_x\). Hence \(m_x\) is irreducible.

The only irreducible polynomials over \(\mathbb R\) have degree \(1\) or \(2\). Therefore every \(x\in D\) has degree at most \(2\) over \(\mathbb R\). If \(x\notin\mathbb R\), its minimum polynomial is an irreducible quadratic. After completing the square, we may write it as \((t-a)^2+b^2\) with \(b\ne0\). Thus \((x-a)^2=-b^2\). Setting \(i=(x-a)/b\), we obtain \(i^2=-1\). Hence either \(D=\mathbb R\), or \(D\) contains a copy of \(\mathbb C=\mathbb R+\mathbb Ri\).

From now on assume \(D\ne\mathbb R\) and fix \(i\in D\) with \(i^2=-1\).

1.2 The centralizer of \(i\)

Consider the centralizer \(C_D(i)=\{x\in D:xi=ix\}\). It is a division subalgebra of \(D\), and \(\mathbb C=\mathbb R+\mathbb Ri\) lies in its center. Thus \(C_D(i)\) is a finite-dimensional division algebra over \(\mathbb C\). But \(\mathbb C\) is algebraically closed. If \(x\in C_D(i)\), then the minimum polynomial of \(x\) over \(\mathbb C\) is irreducible. Hence it must be linear, so \(x\in\mathbb C\). Therefore \(C_D(i)=\mathbb C\).

1.3 Conjugation by \(i\)

Define an \(\mathbb R\)-linear transformation \(\sigma:D\to D\) by \(\sigma(x)=ixi^{-1}\). Since \(i^{-1}=-i\), we have \(\sigma^2=1\). Hence \(D\) decomposes into the two eigenspaces

\[D=D_+\oplus D_-,\]

where \(D_+=\{x:\sigma(x)=x\}\) and \(D_-=\{x:\sigma(x)=-x\}\). Now \(\sigma(x)=x\) is equivalent to \(ix=xi\), so \(D_+=C_D(i)=\mathbb C\). Similarly, \(\sigma(x)=-x\) is equivalent to \(ix=-xi\). Thus

\[D=\mathbb C\oplus D_-.\]

If \(D_-=0\), then \(D=\mathbb C\). It remains to consider the case \(D_-\ne0\).

1.4 The quaternionic direction

Choose \(0\ne j\in D_-\). Then \(ij=-ji\). We claim that \(D_-=\mathbb Cj\). Let \(x\in D_-\). Since both \(x\) and \(j^{-1}\) anticommute with \(i\), their product \(xj^{-1}\) commutes with \(i\). Hence \(xj^{-1}\in C_D(i)=\mathbb C\), so \(x\in\mathbb Cj\). Therefore \(D_-=\mathbb Cj\) and \(D=\mathbb C\oplus\mathbb Cj\). In particular, \(\dim_{\mathbb R}D=4\). If \(z=a+bi\in\mathbb C\), then \(jz=j(a+bi)=aj+bji=(a-bi)j=\bar zj\). Thus the action of \(j\) on \(\mathbb C\) is complex conjugation: \(jz=\bar zj\). This is precisely the relation appearing in the quaternion algebra.

1.5 Determining \(j^2\)

Since \(ij=-ji\), we have \(ij^2=j^2i\), so \(j^2\in C_D(i)=\mathbb C\). On the other hand, \(j^2\) commutes with \(j\). Write \(j^2=z\in\mathbb C\). Since \(jz=\bar zj\) and also \(jz=zj\), we obtain \(\bar z=z\). Hence \(z\in\mathbb R\). Thus \(j^2\in\mathbb R\).

Certainly \(j^2\ne0\). Moreover, \(j^2\) cannot be positive. If \(j^2=a>0\), then \((j-\sqrt a)(j+\sqrt a)=0\), contradicting the absence of zero divisors.

Hence \(j^2<0\). After replacing \(j\) by a nonzero real multiple of \(j\), we may assume \(j^2=-1\).

We now have \(i^2=j^2=-1\) and \(ij=-ji\). Setting \(k=ij\), we obtain the usual quaternion relations \(i^2=j^2=k^2=ijk=-1\). Therefore \(D\simeq\mathbb H\).

We have proved

Frobenius Theorem. Every finite-dimensional associative division algebra over \(\mathbb R\) is isomorphic to exactly one of \(\mathbb R\), \(\mathbb C\), and \(\mathbb H\).

2. Wedderburn's Little Theorem

Let \(D\) be a finite division ring and let \(F=Z(D)\) be its center. Since \(F\) is a finite field, write \(|F|=q\). Let \(n=\dim_FD\). Then \(|D|=q^n\) and \(|D^\times|=q^n-1\). We wish to show that \(n=1\). If \(n=1\), then \(D=F\) and there is nothing to prove. Suppose therefore that \(n>1\).

2.1 Conjugacy classes

Let \(D^\times\) act on itself by conjugation. The elements of \(F^\times\) are precisely the central elements, hence each of them forms a conjugacy class containing one element. Choose representatives \(a_1,\ldots,a_r\) for the noncentral conjugacy classes. For each \(a_i\), let \(C_i=C_D(a_i)\) be its centralizer. Since \(a_i\) is not central, \(C_i\) is a proper division subalgebra of \(D\) containing \(F\). Write \(n_i=\dim_FC_i\). Then \(1\le n_i<n\). Moreover, \(D\) is a vector space over the division ring \(C_i\). If \(m_i=\dim_{C_i}D\), then \(n=m_in_i\). Thus \(n_i\mid n\). The centralizer of \(a_i\) inside the multiplicative group \(D^\times\) is \(C_i^\times\), which has \(q^{n_i}-1\) elements. Hence the conjugacy class of \(a_i\) has size

\[\frac{|D^\times|}{|C_i^\times|}=\frac{q^n-1}{q^{n_i}-1}.\]

Consequently the class equation of \(D^\times\) is

\[q^n-1=q-1+\sum_{i=1}^r\frac{q^n-1}{q^{n_i}-1}.\]

Everything now reduces to elementary arithmetic.

2.2 Cyclotomic polynomials

Let \(\Phi_n(X)\) denote the \(n\)-th cyclotomic polynomial. Recall that \(X^n-1=\prod_{d\mid n}\Phi_d(X).\) Since \(n_i\) is a proper divisor of \(n\), the polynomial \(\Phi_n(X)\) occurs as a factor of \(X^n-1\), but not of \(X^{n_i}-1\). Therefore

\[\Phi_n(q)\mid\frac{q^n-1}{q^{n_i}-1}\]

for every \(i\). Returning to the class equation, every summand in the sum is divisible by \(\Phi_n(q)\). Since \(\Phi_n(q)\) also divides \(q^n-1\), it follows that

\[\Phi_n(q)\mid q-1.\]

We shall show that this is impossible when \(n>1\).

2.3 The contradiction

Over \(\mathbb C\), the cyclotomic polynomial factors as \(\Phi_n(X)=\prod_\zeta(X-\zeta),\) where \(\zeta\) runs through the primitive \(n\)-th roots of unity. Since \(n>1\), no primitive \(n\)-th root of unity is equal to \(1\). For every such \(\zeta\), the distance from the real number \(q\) to \(\zeta\) is strictly larger than the distance from \(q\) to \(1\). Hence \(|q-\zeta|>q-1.\) Therefore

\[|\Phi_n(q)|=\prod_\zeta|q-\zeta|>q-1.\]

But we already proved that \(\Phi_n(q)\) divides \(q-1\). A nonzero integer dividing \(q-1\) cannot have absolute value greater than \(q-1\). This contradiction shows that \(n>1\) is impossible. Hence \(n=1\) and therefore \(D=F\). Thus \(D\) is commutative and we have

Wedderburn's Little Theorem. Every finite division ring is commutative.

Equivalently, if \(D\) is a finite associative division algebra, then \(D\) is a finite field.

3. Hurwitz Theorem

We now drop the assumption of associativity and study algebras that carry a multiplicative quadratic form.

3.1 Composition algebras

Let \(F\) be a field with \(\operatorname{char}F\neq2\). A quadratic form on a finite-dimensional \(F\)-vector space \(V\) is a map \(N:V\to F\) satisfying:

  • \(N(\lambda v)=\lambda^2N(v)\) for all \(\lambda\in F\) and \(v\in V\);
  • the map \((v,w)\mapsto N(v+w)-N(v)-N(w)\) is \(F\)-bilinear.

Its polar form is

\[(v,w)=\frac12\bigl(N(v+w)-N(v)-N(w)\bigr).\]

Then \(N(v)=(v,v)\). The quadratic form \(N\) is nondegenerate if \((v,w)=0\) for every \(w\in V\) implies \(v=0\).

A composition algebra over \(F\) is a finite-dimensional unital \(F\)-algebra \(A\), not necessarily associative, together with a nondegenerate quadratic form \(N:A\to F\) such that \(N(xy)=N(x)N(y)\) for all \(x,y\in A\). The map \(N\) is called the norm of \(A\). Setting \(x=y=1\) gives \(N(1)=N(1)^2\), and nondegeneracy forces \(N(1)=1\).

Thus a composition algebra is an algebra in which the norm is multiplicative.

3.2 Polarization

Fix a composition algebra \(A\) with norm \(N\) and polar form \((\;,\;)\).

Replace \(y\) by \(y+z\) in the composition law \(N(xy)=N(x)N(y)\). Using bilinearity of the polar form we obtain

\[(xy,xz)=N(x)(y,z). \tag{1}\] The symmetric version is \((yx,zx)=N(x)(y,z)\).

Polarizing (1) again gives the four-linear identity

\[(xy,zw)+(xw,zy)=2(x,z)(y,w). \tag{2}\]

These identities are the algebraic engine of the whole theory. They allow the norm to control multiplication.

3.3 Trace and conjugation

Define the trace of \(x\) by \(t(x)=2(x,1)\) and its conjugate by \(\bar x=t(x)1-x\).

Specializing the polarized identities gives

\[x+\bar x=t(x)1,\qquad x\bar x=\bar xx=N(x)1. \tag{3}\]

Hence every element satisfies the quadratic equation

\[x^2-t(x)x+N(x)1=0.\]

Consequently each element of a composition algebra generates a subalgebra of dimension at most two. Linearizing the quadratic equation gives

\[xy+yx=t(x)y+t(y)x-2(x,y)1. \tag{4}\]

In particular, if \(t(x)=t(y)=0\) and \((x,y)=0\), then \(xy=-yx\). Using the polarized identities one also checks that conjugation reverses multiplication: \(\overline{xy}=\bar y\,\bar x\).

Finally, specializing (2) yields the adjoint identities

\[(xy,z)=(y,\bar xz)=(x,z\bar y), \tag{5}\]

which will be essential in the doubling argument.

3.4 The alternative laws

A composition algebra need not be associative, but the norm forces a weaker property.

Lemma 1

Every composition algebra satisfies the left and right alternative laws

\[x(xy)=x^2y,\qquad(yx)x=yx^2.\]

Equivalently, the associator \([x,y,z]=(xy)z-x(yz)\) is an alternating function of its three arguments.

Proof

From the quadratic equation, \(x^2=t(x)x-N(x)1\). Using (5),

\[(x(xy),z)=(xy,\bar xz)=t(x)(xy,z)-(xy,xz).\]

By (1), \((xy,xz)=N(x)(y,z)\), and therefore

\[(x(xy),z)=t(x)(xy,z)-N(x)(y,z)=(x^2y,z).\]

Thus \((x(xy)-x^2y,z)=0\) for every \(z\). Nondegeneracy yields the left alternative law. The right alternative law follows similarly.

Hence \([x,x,y]=[y,x,x]=0\). Linearizing these gives \([x,z,y]+[z,x,y]=0\) and \([y,x,z]+[y,z,x]=0\), so the associator changes sign when any two adjacent arguments are swapped. Because \(\operatorname{char}F\neq2\), this means \([x,y,z]\) is alternating. \(\square\)

An immediate consequence is

Artin theorem. Any subalgebra generated by two elements of an alternative algebra is associative.

3.5 Orthogonal doubling

The heart of the proof is that the classical Cayley–Dickson doubling construction appears naturally inside every composition algebra.

Let \(B\) be a proper composition subalgebra of \(A\) such that the restriction of \(N\) to \(B\) is nondegenerate. Orthogonally decompose \(A=B\oplus B^\perp\). Because \(B\) is proper, \(B^\perp\neq0\). Since \(N\) is nondegenerate on \(A\), its restriction to \(B^\perp\) cannot vanish identically. Pick \(u\in B^\perp\) with \(N(u)\neq0\).

As \(1\in B\), we have \((u,1)=0\), so \(t(u)=0\) and \(\bar u=-u\). The quadratic equation gives \(u^2=-N(u)1=\mu1\), where \(\mu=-N(u)\in F^\times\).

For any \(a\in B\), the linearized commutation formula (4) yields \(ua+au=t(a)u\). Hence

\[ua=\bar au,\qquad au=u\bar a. \tag{6}\]

The direct sum. We first show that \(Bu\) is orthogonal to \(B\). For \(a,b\in B\), the adjoint identity (5) gives

\[(au,b) = (u,\bar a b) = 0,\] because \(\bar a b\in B\) and \(u\in B^\perp\). Thus \(Bu\subseteq B^\perp\). Nondegeneracy on \(B\) forces \(B\cap Bu=0\).

If \(au=0\), then by Artin’s theorem the subalgebra generated by \(a\) and \(u\) is associative, and

\[0 = (au)\bar u = a(u\bar u) = N(u)a,\] so \(a=0\). The map \(a\mapsto au\) is therefore injective, and \(\dim_FBu=\dim_FB\).

We therefore obtain the direct sum \(D=B\oplus Bu\), with \(\dim_FD=2\dim_FB\).

The multiplication. A direct computation using (6), alternativity, and (5) gives

\[ \begin{aligned} a(du)&=(da)u,\\ (bu)c&=(b\bar c)u,\\ (bu)(du)&=\mu\,\bar d b, \end{aligned} \]

for all \(a,b,c,d\in B\). Consequently,

\[(a+bu)(c+du)=(ac+\mu\bar d b)+(da+b\bar c)u. \tag{7}\]

Conjugation and norm on \(D\) are \(\overline{a+bu}=\bar a-bu\) and

\[N(a+bu)=N(a)-\mu N(b). \tag{8}\]

These formulas show that \(D\) is closed under multiplication and that the norm is again multiplicative. Hence \(D\) is a composition subalgebra of \(A\). It is precisely the Cayley--Dickson double of \(B\) with parameter \(\mu\).

Thus any proper nondegenerate composition subalgebra forces the next Cayley–Dickson algebra to appear inside \(A\).

3.6 Successive doubling and the possible dimensions

Start with \(B_0=F1\). Its norm is nondegenerate. If \(A=B_0\) we have a \(1\)-dimensional composition algebra.

First doubling. If \(A\neq B_0\), pick \(u_1\in B_0^\perp\) with \(N(u_1)\neq 0\) and form

\[B_1=B_0\oplus B_0u_1.\] Then \(\dim_FB_1=2\). A two-dimensional composition algebra is commutative and associative; it is a quadratic étale algebra.

Second doubling. If \(B_1=A\), we stop. Otherwise pick \(u_2\in B_1^\perp\) with \(N(u_2)\neq 0\) and form

\[B_2=B_1\oplus B_1u_2.\] Then \(\dim_FB_2=4\). This is a quaternion algebra; it is associative but generally not commutative.

Third doubling. If \(B_2=A\), we stop. Otherwise pick \(u_3\in B_2^\perp\) with \(N(u_3)\neq 0\) and form

\[B_3=B_2\oplus B_2u_3.\] Then \(\dim_FB_3=8\). This is an octonion (Cayley) algebra. By Lemma 1 it is alternative. It is not associative: choose \(a,b\in B_2\) with \(ab\neq ba\). By (7), \(a(bu_3)=(ba)u_3\). If \(B_3\) were associative, then \((ab)u_3=a(bu_3)=(ba)u_3\). Since right multiplication by \(u_3\) is injective, this would imply \(ab=ba\), a contradiction. Hence \([a,b,u_3]\neq0\).

We have obtained the chain

\[F=B_0\subset B_1\subset B_2\subset B_3\]

of composition algebras of dimensions \(1,2,4,8\).

3.7 Why the process stops at eight dimensions

It remains to prove that \(B_3\) cannot be proper, i.e., that no composition algebra of dimension \(16\) exists.

Lemma 2

If the Cayley--Dickson double \(D=B\oplus Bu\) of a composition algebra \(B\) is alternative, then \(B\) is associative.

Proof

Write elements of \(D\) as pairs \((a,b)\) with multiplication as in (7)

\[(a,b)(c,d) = (ac + \mu\,\bar d b,\; da + b\bar c).\] Assume \(D\) satisfies the left alternative law \(x(xy)=x^2y\) for all \(x,y\in D\).

Let \(x=(a,b)\) and \(y=(c,0)\). Compute \(x^2\):

\[x^2 = (a^2 + \mu N(b)1,\; t(a)b).\] Then the second component of \(x^2y\) is \(t(a)b\bar c\).

On the other hand, \(xy=(ac,\;b\bar c)\), and

\[x(xy) = \bigl(a(ac) + \mu\overline{b\bar c}b,\; (b\bar c)a + b(\overline{ac})\bigr).\] The second component is \((b\bar c)a+b(\bar c\bar a)\). Equating the second components of \(x^2y\) and \(x(xy)\) gives

\[t(a)b\bar c = (b\bar c)a + b(\bar c \bar a).\] Since \(t(a)1=a+\bar a\) and \(t(a)\) is scalar, the left-hand side is \((b\bar c)a+(b\bar c)\bar a\). Comparing both sides gives

\[(b\bar c)\bar a = b(\bar c \bar a),\] which is precisely \([b,\bar c,\bar a]=0\). Conjugation is a bijection, and \(a,b,c\) are arbitrary; hence \([x,y,z]=0\) for all \(x,y,z\in B\). Therefore \(B\) is associative. \(\square\)

Now suppose \(B_3\subsetneq A\). Orthogonal doubling would produce a composition subalgebra \(B_4=B_3\oplus B_3u_4\) of dimension \(16\). But every composition algebra is alternative, so \(B_4\) would be alternative. Lemma 2 would then imply that \(B_3\) is associative — a contradiction, because we have already shown that the eight-dimensional octonion algebra is not associative.

Hence \(B_3\) cannot be a proper subalgebra. The doubling process terminates after at most three steps, and we have proved:

Generalized Hurwitz Theorem. Every finite-dimensional composition algebra over a field \(F\) with \(\operatorname{char}F\neq2\) has dimension \(1\), \(2\), \(4\), or \(8\).

3.8 The real case

The original Hurwitz theorem concerns normed division algebras over \(\mathbb R\). A normed division algebra is a finite-dimensional real algebra with a multiplicative norm \(N\) that is positive definite. In particular, it is a composition algebra over \(\mathbb R\) whose quadratic form is anisotropic, \(N(x)=0\Rightarrow x=0\). The generalized theorem forces its dimension to be \(1,2,4\), or \(8\). At each doubling step \(u^2=-N(u)1\) with \(N(u)>0\), so after rescaling we may take \(u^2=-1\). The four possibilities are therefore exactly

\[\mathbb R,\quad \mathbb C,\quad \mathbb H,\quad \mathbb O.\]

They correspond respectively to the real numbers, the complex numbers, the quaternions, and the octonions. Thus:

Hurwitz Theorem. Every finite-dimensional normed division algebra over \(\mathbb R\) is isomorphic to exactly one of \(\mathbb R\), \(\mathbb C\), \(\mathbb H\), and \(\mathbb O\).

This completes the table with which we began the post.

References

  1. Nathan Jacobson, Basic Algebra I, 2nd ed., W. H. Freeman, 1985, §§7.6--7.7.