Frobenius's Theorem/Lemma 1

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


$(1): \quad U = \set {u \in A \setminus \R: u^2 \in \R} \cup \set 0$ is a linear subspace of $A$
$(2): \quad \forall u, v \in U: u v + v u \in \R$
$(3): \quad A = \R \oplus U$
$(4): \quad$ If $A$ is also a division algebra, then every nonzero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.


Proof of First Assertion

$U$ is closed under scalar multiplication.

We have to show that $u, v \in U$ implies $u + v \in U$.

Let $\set {u, v, 1}$ be a linearly dependent set.

Then there exist constants $a, b \in \R$ such that $u = av + b$.

Then $u^2 = a^2 v^2 + 2 a b v + b^2$.

Since $u^2$, $a^2 v^2$, and $b^2$ are all real, it follows that $2 a b v \in \R$, that is, $v \in \R$.

Since $0$ is the only real element of $U$, it follows that $v = 0$.

Reversing $u$ and $v$ in the preceding argument shows that also $u = 0$, so that $u + v = 0 \in U$.

Now let $\set {u, v, 1}$ be a linearly independent set.

We have:

$\paren {u + v}^2 + \paren {u - v}^2 = 2 u^2 + 2 v^2 \in \R$

On the other hand, as $A$ is quadratic there exist $\lambda, \mu \in \R$ such that:

$\paren {u + v}^2 - \lambda \paren {u + v} \in \R$
$\paren {u - v}^2 - \mu \paren {u - v} \in R$


$\lambda \paren {u + v} + \mu \paren {u - v} \in \R$

However, we have that $\set {u, v, 1}$ is linearly independent.

Therefore $\lambda + \mu = \lambda - \mu = 0$, and so $\lambda = \mu = 0$.

This proves that $u \pm v \in U$.

Thus $U$ is indeed a subspace of $A$.


Proof of Second Assertion

Accordingly, from the result of $(1)$:

$\forall u, v \in U: u v + v u = \paren {u + v}^2 - u^2 - v^2 \in \R$.


Proof of Third Assertion

Let $a \in A \setminus \R$.


$\exists \nu \in \R: a^2 - \nu a \in \R$

Therefore, if we set

$u = a - \dfrac \nu 2 \in U$

then $u^2 = a^2 - \nu a + \nu^2/4 \in \R$, so

$a = \dfrac \nu 2 + u \in \R \oplus U$

which proves the assertion.


Proof of Fourth Assertion

This follows directly from the definition of division algebra.

Also, we have that $u^2 \in \R$.

Suppose $u^2 = 0$. Since $u$ is nonzero and since $A$ is a division algebra, there exists an inverse $u^{-1}$ such that $u u^{-1} = 1$.

But then $u = u 1 = u u u^{-1} = 0 \cdot u^{-1} = 0$, which cannot be since $u$ was assumed to be nonzero.

Now suppose $u^2 > 0$.

Then there exists an $\alpha \in \R$ such that $\alpha^2 = u^2$.

But then $\paren {u - \alpha} \paren {u + \alpha} = u^2 - \alpha^2$ would be $0$.

Since $A$ is assumed to be a division algebra, this would imply that either $u = \alpha$ or $u = -\alpha$, which is impossible since $u \in U$.

So $u^2 < 0$ and so $u^2 = -\alpha^2$ with $0 \ne \alpha \in \R$.

Thus, $v = \alpha^{-1} u$ is a desired element.