My question is "Can every finite-dimensional associative $\mathbb{C}$-algebra be equipped with a submultiplicative norm?"

If the answer to my question is yes, then it seems that we can prove that the Gelfand-Mazur theorem implies Frobenius's theorem on complex division algebras:

Recall that a Banach algebra $A$ is both a Banach space and an algebra over a field (which we'll assume to be $\mathbb{C}$), such that $\Vert ab \Vert \le \Vert a \Vert \Vert b \Vert$ for every $a, b \in A$. The last property is known as "submultiplicativity".

(a) Gelfand-Mazur theorem:

If $A$ is a Banach algebra over $\mathbb{C}$ such that every element of $A$ has a multiplicative inverse (i.e. $A$ is a division algebra) then $A$ is isometrically isomorphic to $\mathbb{C}$.

(b) Frobenius theorem: (a slightly weaker version than usual).

If $A$ is a finite-dimensional division algebra over $\mathbb{C}$, then $A$ is isomorphic to $\mathbb{C}$.

Now, any finite-dimensional $\mathbb{C}$-vector space is isomorphic (as a vector space) to $\mathbb{C}^n$ for some $n$. So it's automatically a Banach space, once it is equipped with any of the familiar norms. In general, however, one can introduce a variety of different bilinear product operations on $\mathbb{C}^n$, to turn it into various different $\mathbb{C}$-algebras. It's unclear to me whether or not every algebraic structure on $\mathbb{C}^n$ can be equipped with a norm such that it has the submultiplicative property. If the answer is "Yes it can" then every algebra of the kind described in (b) is also an example of the kind described in (a), and it follows that (a) implies (b).

Note that I could have asked instead: "Without assuming Frobenius's theorem, does every finite-dimensional division algebra over $\mathbb{C}$ have a submultiplicative norm?" But this restriction of the scope makes the question less interesting, and doesn't appear to make it any easier.


Let $A$ be a finite-dimensional associative $\mathbb{C}$-algebra. Suppose $B$ is another such algebra and $f : A \to B$ is an injective algebra homomorphism. Now if there exists a submultiplicative norm for $B$, then we get one for $A$ by restriction. Clearly, we can embed $A$ into a finite dimensional associative $\mathbb{C}$-algebra which has a unit and so we may assume that $A$ is unital. In this case, the map $A \mapsto \text{End}_{\mathbb{C}}(A), a \mapsto (x \mapsto ax)$ is an injective algebra homommorphism. So we are left to show that there exists a submultiplicative norm on $\text{End}_{\mathbb{C}}(A) \cong \mathbb{C}^{n \times n}$ where $n$ is the dimension of $A$. The operator norms (induced from norms on $\mathbb{C}^n$) are example of those.

  • $\begingroup$ I'm typing too slow today... $\endgroup$ Jul 29 '16 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.