# A Crazy Classification of Normed Division Algebras

*Pre-requisities:* Complex analysis, functional analysis (on Banach spaces), knowledge of rings, fields, and algebras.

Let's be honest - functional analysis is the wellspring for a lot crazy math. Even so, what follows is one of the coolest proofs in functional analysis that I know, detouring through Banach algebra-valued complex analysis to prove something about purely real structures.

## Prelude: Finite-Dimensional Real Division Algebras

Recall that a real division algebra $K$ is an algebra over $\mathbb{R}$ such that every non-zero $x \in K$ has an inverse $y \in K$ (meaning that $xy = yx = 1$). We say that $K$ is a *finite-dimensional* algebra if it is finite-dimensional as a vector space over $\mathbb{R}$. The most common real division algebras are $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$, and each of these shows up in basically every corner of math. Due to the usefulness of these three algebras, you might wonder whether there exist any other real division algebras. We will begin with the finite-dimensional case, where this question can be answered without any fancy analytical techniques. This is often called *Frobenius' Theorem*.

**Theorem** (Finite-Dimensional Real Division Algebras): Let $K$ be a finite-dimensional real algebra. Then, $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$.

*Proof:* Pick some $a \in K$. Since $\mathbb{R}[a]$ is a finite-dimensional commutative $\mathbb{R}$-algebra without zero-divisors, $\mathbb{R}[a]$ is a finite field extension of $\mathbb{R}$. Thus, $\mathbb{R}[a] \cong \mathbb{R}$ or $\mathbb{C}$. In particular, $a$ satisfies a quadratic polynomial in $\mathbb{R}$.

Say that $a \in K$ is *imaginary* if $a^2 \in \mathbb{R}$ and $a^2 \leq 0$. Note that $a$ is imaginary if and only if $a=0$ or $a \not\in \mathbb{R}$ and any isomorphism $K[a] \cong \mathbb{C}$ maps $a$ to an imaginary number. Every element of $K$ can be uniquely decomposed into the sum of a real and imaginary element. We will need the fact that the imaginary elements form a subspace of $K$, or equivalently, that the sum of two imaginary elements is imaginary.

Let $a, b \in K$ be imaginary. If $a$, $b$, and 1 are not linearly independent over $\mathbb{R}$, then they all live within a subalgebra of $K$ isomorphic to $\mathbb{C}$. Hence, $a + b$ is also imaginary.

Suppose on the other hand that $a$, $b$, and 1 are linearly independent over $\mathbb{R}$. There exist $\alpha_1, \alpha_2, \beta_1, \beta_2 \in \mathbb{R}$ such that

and

Taking the sum of these equations, and using the fact that $a$, $b$, and 1 are linearly independent over $\mathbb{R}$, you can conclude that $\alpha_1 = \alpha_2 = 0$. Since $(a+b)^2$ is real, and $a+b$ is not real, $a+b$ must be imaginary.

Let $K_0$ denote the imaginary elements of $K$. Define an inner product on $K_0$ by:

If $\dim K_0 = 0$, then $K \cong \mathbb{R}$, and if $\dim K_0 = 1$, then $K \cong \mathbb{C}$.

If $\dim K_0 > 1$, then pick two unit orthogonal elements $e_1, e_2 \in K_0$. Let $e_3 = e_1 e_2$. It can be shown that, if $x, y \in K_0$ are orthogonal, then $x y = - y x$. It follows that $\mathbb{R}[e_1, e_2, e_3] \cong \mathbb{H}$.

If $\dim K_0 > 3$, then let $e_4$ be a fourth orthonormal vector.

This is a contradiction. Thus, the only possibilities are $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. $\square$.

The above proof can be modified into a proof of the following fact, which we will use later.

**Lemma:** Let $K$ be a real division algebra. Suppose that every $a \in K$ has satisfies a degree-2 equation. Then, $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. $\square$.

While finite-dimensional real division algebras are few, infinite-dimensional real division algebras are abundant. Commutative examples show up often, such as $\mathbb{R}(x)$, $\mathbb{C}(x)$, $\mathbb{R}((x))$, and the field of meromorphic functions on $\mathbb{C}$. Non-commutative examples are also plentiful, such as $\mathbb{H}((X))$. Another cool example is $\mathbb{C}((\sigma))$ where $\sigma z = z^{\star} \sigma$ for all $z \in \mathbb{C}$.

## The Main Theorem

Define a *normed algebra* $K$ to be an algebra over $\mathbb{R}$ equipped with a linear norm such that $\|1\| = 1$ and $\|a b\| \leq \|a\| \|b\|$. Note that we are excluding other normed rings such as $\mathbb{Z}$ and $\mathbb{Q}_p$ that aren't algebras over $\mathbb{R}$. The prototypical example of a normed algebra is the space of all continuous maps from a normed vector space $X$ to itself, written $B(X)$. If the normed algebra is complete (with respect to its norm), then it is called a *Banach algebra*. If $A$ is a topological space, then the space of bounded continuous real-valued functions, written $C_b(A)$, is a Banach algebra. Every finite-dimensional real algebra can also be turned into a Banach algebra.

Normed algebras abound. But can they also be division algebras? The common suspects ($\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$) are all normed division algebras. Surprisingly, these are the only examples. Despite the utter variety of exotic normed algebras, none of them are also division algebras.

**Theorem** (Normed Division Algebras): Let $K$ be a (real) normed division algebra. Then, $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$ as real algebras.

Before we can prove this, we will need a key lemma about complex Banach algebras, which in turn requires some machinery from complex analysis.

### The Spectrum of a Banach Space

Complex analysis is often taught as the study of holomorphic $\mathbb{C}$-valued functions on some set $U \subset \mathbb{C}$. In truth, however, requiring holomorphic functions to be $\mathbb{C}$-valued is kind of restrictive. A huge chunk of complex analysis is true verbatim when holomorphic $\mathbb{C}$-valued functions are replaced by holomorphic $X$-valued functions, where $X$ is any complex Banach space. In particular, Liouville's Theorem is true in this setting.

**Lemma** (Liouville's Theorem): Let $X$ be a complex Banach space, and let $f: \mathbb{C} \rightarrow X$ be holomorphic. If $f$ is bounded, then it is constant. $\square$.

This has the following important corollary. This is also a foundational theorem in *spectral theory*, but it is the only idea from spectral theory that we will need.

**Lemma** (Non-Empty Spectrum): Let $X$ be a complex Banach algebra, and let $a \in X$. There exists $z \in \mathbb{C}$ such that $a - z$ does not have an inverse in $X$.

*Proof:* Suppose that $a - z$ is invertible for all $z \in X$. It can be shown that $f: z \mapsto (a-z)^{-1}$ is a holomorphic function on $\mathbb{C}$. Asymptotically, $\|f(z)\| \sim \frac{1}{|z|}$ when $|z| \gg \|a\|$. Thus, $f$ is bounded. By Liouville's Theorem, $f$ is constant, which is impossible. Hence, $a-z$ is not invertible for some $z \in \mathbb{C}$. $\square$.

### Complexification

Now let $K$ be a real normed division algebra. Let $\overline{K}$ denote the completion of $K$, and let $L = \overline{K} \otimes_\mathbb{R} \mathbb{C}$. $L$ can be described as all expressions $a + b i$ where $a, b \in \overline{K}$. $L$ is not necessarily a division algebra, but it is a complex Banach algebra.

For any $a \in K$, there exists $z \in \mathbb{C}$ such that $a - z$ is not invertible in $L$. Thus, $(a - z)(a-z^{\star})$ is not invertible in $L$.

Thus, $(a-z)(a-z^{\star})$ is in $K$, and since it is not invertible in $K$. Thus,

During the prelude, we proved that if $K$ is a real division algebra such that every element satisfies a quadratic equation, then $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. $\square$.

## Conclusion

Isn't that crazy? The introduction of complex analysis comes out of nowhere, but it is the perfect tool for this problem. Moreover, I don't know of any more intuitive proof of this fact (although anyone who does is encouraged to send me a message, and I'll add it to this post).

The requirements on $K$ can actually be weakened slightly: if $K$ is an $\mathbb{R}$-algebra and a normed vector space such that multiplication in $K$ is continuous, then $K \cong \mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$.

Anyone who is interested in more complex analysis-functional analysis shenanigans should learn about the *holomorphic functional calculus*, which uses complex analysis to define holomorphic functions like $\log(z)$ and $\zeta(z)$ on (most of) a Banach algebra.

One last fun fact: the above proof can be modified to remove all dependence on the fact that $\mathbb{C}$ is algebraically closed. Every finite field extension of $\mathbb{C}$ is a real division algebra, and can be made into a normed division algebra. Hence, this technique could be used to prove that $\mathbb{C}$ is algebraically closed (but probably shoudn't).