Frobenius theorem (real division algebras)

In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite-dimensional associative division algebras over the real numbers. According to the theorem, every such algebra is isomorphic to one of the following: These algebras have real dimension $R$, and $C$, respectively. Of these three algebras, $H$ and $1, 2$ are commutative, but $4$ is not.
 * $R$ (the real numbers)
 * $C$ (the complex numbers)
 * $H$ (the quaternions).

Proof
The main ingredients for the following proof are the Cayley–Hamilton theorem and the fundamental theorem of algebra.

Introducing some notation

 * Let $D$ be the division algebra in question.
 * Let $n$ be the dimension of $D$.
 * We identify the real multiples of $1$ with $R$.
 * When we write $a ≤ 0$ for an element $a$ of $D$, we imply that $a$ is contained in $R$.
 * We can consider $D$ as a finite-dimensional $R$-vector space. Any element $d$ of $D$ defines an endomorphism of $D$ by left-multiplication, we identify $d$ with that endomorphism. Therefore, we can speak about the trace of $d$, and its characteristic- and minimal polynomials.
 * For any $z$ in $C$ define the following real quadratic polynomial:
 * $$Q(z; x) = x^2 - 2\operatorname{Re}(z)x + |z|^2 = (x-z)(x-\overline{z}) \in \mathbf{R}[x].$$
 * Note that if $z ∈ C&thinsp;∖&thinsp;R$ then $Q(z; x)$ is irreducible over $R$.

The claim
The key to the argument is the following


 * Claim. The set $V$ of all elements $a$ of $D$ such that $a^{2} ≤ 0$ is a vector subspace of $D$ of dimension $n − 1$. Moreover $D = R ⊕ V$ as $R$-vector spaces, which implies that $V$ generates $D$ as an algebra.

Proof of Claim: Pick $a$ in $D$ with characteristic polynomial $p(x)$. By the fundamental theorem of algebra, we can write


 * $$p(x) = (x-t_1)\cdots(x-t_r) (x-z_1)(x - \overline{z_1}) \cdots (x-z_s)(x - \overline{z_s}), \qquad t_i \in \mathbf{R}, \quad z_j \in \mathbf{C} \setminus \mathbf{R}.$$

We can rewrite $p(x)$ in terms of the polynomials $Q(z; x)$:


 * $$p(x) = (x-t_1)\cdots(x-t_r) Q(z_1; x) \cdots Q(z_s; x).$$

Since $z_{j} ∈ C&thinsp;∖&thinsp;R$, the polynomials $Q(z_{j}; x)$ are all irreducible over $R$. By the Cayley–Hamilton theorem, $p(a) = 0$ and because $D$ is a division algebra, it follows that either $a − t_{i} = 0$ for some $i$ or that $Q(z_{j}; a) = 0$ for some $j$. The first case implies that $a$ is real. In the second case, it follows that $Q(z_{j}; x)$ is the minimal polynomial of $a$. Because $p(x)$ has the same complex roots as the minimal polynomial and because it is real it follows that


 * $$p(x) = Q(z_j; x)^k = \left(x^2 - 2\operatorname{Re}(z_j) x + |z_j|^2 \right)^k$$

for some $k$. Since $p(x)$ is the characteristic polynomial of $a$ the coefficient of $x^{&hairsp;2k −&thinsp;1}$ in $p(x)$ is $tr(a)$ up to a sign. Therefore, we read from the above equation we have: $tr(a) = 0$ if and only if $Re(z_{j}) = 0$, in other words $tr(a) = 0$ if and only if $a^{2} = −|z_{j}|^{2} < 0$.

So $V$ is the subset of all $a$ with $tr(a) = 0$. In particular, it is a vector subspace. The rank–nullity theorem then implies that $V$ has dimension $n − 1$ since it is the kernel of $$\operatorname{tr} : D \to \mathbf{R}$$. Since $R$ and $V$ are disjoint (i.e. they satisfy $$\mathbf R \cap V = \{0\}$$), and their dimensions sum to $n$, we have that $D = R ⊕ V$.

The finish
For $a, b$ in $V$ define $B(a, b) = (−ab − ba)/2$. Because of the identity $(a + b)^{2} − a^{2} − b^{2} = ab + ba$, it follows that $B(a, b)$ is real. Furthermore, since $a^{2} ≤ 0$, we have: $B(a, a) > 0$ for $a ≠ 0$. Thus $B$ is a positive-definite symmetric bilinear form, in other words, an inner product on $V$.

Let $W$ be a subspace of $V$ that generates $D$ as an algebra and which is minimal with respect to this property. Let $e_{1}, ..., e_{n}$ be an orthonormal basis of $W$ with respect to $B$. Then orthonormality implies that:


 * $$e_i^2 =-1, \quad e_i e_j = - e_j e_i.$$

The form of $D$ then depends on $k$:

If $k = 0$, then $D$ is isomorphic to $R$.

If $k = 1$, then $D$ is generated by $1$ and $e_{1}$ subject to the relation $e2 1 = −1$. Hence it is isomorphic to $C$.

If $k = 2$, it has been shown above that $D$ is generated by $1, e_{1}, e_{2}$ subject to the relations
 * $$e_1^2 = e_2^2 =-1, \quad e_1 e_2 = - e_2 e_1, \quad (e_1 e_2)(e_1 e_2) =-1.$$

These are precisely the relations for $H$.

If $k > 2$, then $D$ cannot be a division algebra. Assume that $k > 2$. Define $u = e_{1}e_{2}e_{k}$ and consider $u^{2}=(e_{1}e_{2}e_{k})*(e_{1}e_{2}e_{k})$. By rearranging the elements of this expression and applying the orthonormality relations among the basis elements we find that $u^{2} = 1$. If $D$ were a division algebra, $0 = u^{2} − 1 = (u − 1)(u + 1)$ implies $u = ±1$, which in turn means: $e_{k} = ∓e_{1}e_{2}$ and so $e_{1}, ..., e_{k−1}$ generate $D$. This contradicts the minimality of $W$.

Remarks and related results

 * The fact that $D$ is generated by $e_{1}, ..., e_{k}$ subject to the above relations means that $D$ is the Clifford algebra of $R^{n}$. The last step shows that the only real Clifford algebras which are division algebras are $Cℓ^{0}, Cℓ^{1}$ and $Cℓ^{2}$.
 * As a consequence, the only commutative division algebras are $R$ and $C$. Also note that $H$ is not a $C$-algebra. If it were, then the center of $H$ has to contain $C$, but the center of $H$ is $R$.
 * This theorem is closely related to Hurwitz's theorem, which states that the only real normed division algebras are $R, C, H$, and the (non-associative) algebra $O$.
 * Pontryagin variant. If $D$ is a connected, locally compact division ring, then $D = R, C$, or $H$.
 * Pontryagin variant. If ⇭⇭⇭ is a connected, locally compact division ring, then ᙭᙭᙭, or ᙭᙭᙭.