Frobenius's Theorem

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Theorem

An algebraic associative real division algebra $A$ is isomorphic to $\R, \C$ or $\Bbb H$.


Proof

Recall that an algebra $A$ is said to be quadratic if it is unital and the set $\set {1, x, x^2}$ is linearly dependent for every $x \in A$.


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Thus, for every $x \in A$ there exist $\map t x, \map n x \in \R$ such that:

$x^2 - \map t x x + \map n x = 0$

Obviously, $\map t x$ and $\map n x$ are uniquely determined if $x \notin \R$.


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Suppose $x \notin \R$.

Then $x$ can be expressed as $x = a + b i$, with $a, b \in \R$ and $b \ne 0$.

Then:

$x^2 = a^2 - b^2 + 2 a b i$

and:

$x^2 - \map t x x + \map n x = a^2 - b^2 - a \map t x + \map n x + \paren {2 a b - b \map t x} i = 0$

It follows that both:

$(1): \quad 2 a b - b \map t x = 0$

and:

$(2): \quad a^2 - b^2 - a \map t x + \map n x = 0$

$(1)$ leads to:

$\map t x = 2 a$

and $(2)$ leads to;

$\map n x = a^2 + b^2$

Setting $\map t \lambda = 2 \lambda$ and $\map n \lambda = \lambda^2$ for $\lambda \in \R$, we can then consider $t$ and $n$ as maps from $A$ into $\R$.

(In this way $t$ becomes a linear functional).

We call $\map t x$ and $\map n x$ the trace and the norm of $x$ respectively.


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From $x^2 - \paren {x + x^*} x + x^* x = 0$ we see that all algebras $\Bbb A_n$ are quadratic.

Further, every real division algebra $A$ that is algebraic and power-associative (this means that every subalgebra generated by one element is associative) is automatically quadratic.

Indeed, if $x \in A$ then there exists a nonzero polynomial $\map f X \in \R \sqbrk X$ such that $\map f x = 0$.

Writing $\map f X$ as the product of linear and quadratic polynomials in $\R \sqbrk X$ it follows that $\map p x = 0$ for some $\map p X \in \R \sqbrk X$ of degree $1$ or $2$.

In particular, algebraic alternative (and hence associative) real division algebras are quadratic.

Finally, if $A$ is a real unital algebra, that is, an algebra over $\R$ with unity $1$, then we shall follow a standard convention and identify $\R$ with $\R 1$.

Thus we shall write $\lambda$ for $\lambda 1$, where $\lambda \in \R$.


Lemma 1

Let $\struct {A, \oplus}$ be a quadratic real algebra.


Then:

$\quad U = \set {u \in A \setminus \R: u^2 \in \R} \cup \set 0$

is a linear subspace of $A$.

$\Box$


Lemma 2

Let $\struct {A, \oplus}$ be a quadratic real algebra.


Then:

$\forall u, v \in U: u v + v u \in \R$

$\Box$


Lemma 3

Let $\struct {A, \oplus}$ be a quadratic real algebra.


Then:

$A = \R \oplus U$

$\Box$


Lemma 4

Let $\struct {A, \oplus}$ be a quadratic real algebra.

If $A$ is also a division algebra, then every non-zero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.

$\Box$


Lemma 5

Let $A$ be a quadratic real division algebra.

Let:

$U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$

where $\setminus$ denotes set difference.

Suppose $e_1, \ldots, e_k \in U$ are such that:

$\forall i \le k: e_i^2 = -1$
$\forall i, j \le k, i \ne j: e_i e_j = -e_j e_i$


If $U$ is not equal to the linear span of $e_1, \ldots, e_k$, then there exists $e_{k+1} \in U$ such that:

$e_{k+1}^2 = -1$
$\forall i \le k: e_i e_{k+1} = -e_{k+1} e_i$

$\Box$


We have from above that $A$ is quadratic.

We may assume that $n = \dim A \ge 2$.

By Lemma 4 we can fix $i \in A$ such that $i^2 = -1$.

Thus, $A \cong \C$ if $n = 2$.

Let $n > 2$.

By Lemma 5:

$\exists j \in A: j^2 = -1, i j = -j i$

Set $k = ij$.

It can immediately be checked that:

$k^2 = -1$
$k i = j = -i k$
$j k = i = -k j$
$\set {i, j, k}$ is a linearly independent set.

Therefore $A$ contains a subalgebra isomorphic to $\Bbb H$.


Finally, suppose $n > 4$.

By Lemma 5 there would exist $e \in A, e \ne 0$ such that:

$(a): \quad e i = -i e$
$(b): \quad e j = -j e$
$(c): \quad e k = -k e$

However, from $(a)$ and $(b)$ it follows that $e i j = -i e j = i j e$.

Since $i j = k$, this contradicts $(c)$.

It follows that $n \le 4$, and so $\Bbb H$ is the highest order

$\blacksquare$


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Source of Name

This entry was named for Ferdinand Georg Frobenius.


Historical Note

Frobenius's Theorem was proved by Ferdinand Georg Frobenius in $1878$.


Sources

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