# 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:
* R (the real numbers)
* C (the complex numbers)
* H (the quaternions)
These algebras have real dimension 1, 2, and 4, respectively. Of these three algebras, R and C are commutative, but H is not.

## 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 ∖ 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 ∖ 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$^{2k−1}$ 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$_k$ 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 e$^2$ $_1$ = −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.