Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbb{C}\langle x,y\rangle$ be the group ring of the complex numbers over the free group in $x,y$. Let $len : \langle x,y \rangle \rightarrow \mathbb{N}$ denote the standard word norm and let $\varphi=exp \circ len: \langle x, y \rangle \rightarrow [1,\infty)$. Define the following norm for $\displaystyle \alpha=\sum_{g \in \langle x,y\rangle} a_{g}g \in \mathbb{C}\langle x,y \rangle$:

$$\|\alpha\|=\sum_{g \in \langle x, y \rangle} |a_{g}|\varphi(g)$$

It's not hard to show that $\|\cdot\|$ is indeed a norm and in fact for all $\alpha, \beta \in \mathbb{C}\langle x, y \rangle$ we have $\|\alpha\beta\| \le \|\alpha\|\cdot\|\beta\|$. The natural involution $\displaystyle ^{*}:\sum_{g \in \langle x, y \rangle}a_{g}g \mapsto \sum_{g \in \langle x, y \rangle} \overline{a_{g}}g^{-1}$ has the properties that would make $(\mathbb{C}\langle x, y \rangle,\|\cdot\|,^{*})$ into a Banach *-algebra, including $\|\alpha\|=\|\alpha^{*}\|$. My question is this:

Is $\mathbb{C}\langle x, y \rangle$ complete with respect to $\|\cdot\|$?

I stumbled upon this while working on an undergrad research project but haven't had any functional analysis, so I'm not sure how to go about proving/disproving this. Any help/references would be greatly appreciated.

share|cite|improve this question
When you write $r_g$ do you mean $a_g$? – Qiaochu Yuan Jun 7 '12 at 20:19
Yup, thanks for catching that and thanks for your answer! – Jackson Jun 7 '12 at 20:30
up vote 5 down vote accepted

There are no countable-dimensional Banach spaces. This is a corollary of the Baire category theorem: if $e_1, e_2, ...$ were a basis for a Banach space $B$, then $B$ would be the countable union of the sets $\text{span}(e_1), \text{span}(e_1, e_2), ...$, all of which are nowhere dense, which contradicts BCT3 (in Wikipedia's terminology).

There is an obvious and more naive argument which doesn't work: if I had $e_1, e_2, ...$ as above, normalized to have norm $1$, then isn't it obvious that, say, $\sum \frac{e_n}{2^n}$ doesn't lie in the span of the $e_i$? This follows in the special case that the $e_i$ lie in a Hilbert space and are taken to be orthonormal, but in general you can't conclude this. Why? The naive continuation of the naive argument is that since $B = \text{span}(e_1, e_2, ...)$ there is a linear functional $$e_j^{\ast} : B \to \mathbb{C}$$

which, given a finite sum $\sum c_i e_i$, takes the value $c_j$, so we ought to have $$e_j^{\ast} \left( \sum \frac{e_n}{2^n} \right) = \frac{1}{2^j}.$$

Indeed there is such a linear functional, but it is not guaranteed to be continuous! So the last step fails. (When $H$ is a Hilbert space the inner product can be used to write down this functional and it is of course continuous in this case.)

share|cite|improve this answer
In this particular case the functionals $e_j^{\ast}$ are continuous so the naive argument works and we can take, say, $\sum \frac{x^n}{n!}$ as an example of an element of the closure that doesn't lie in the space, but this general pitfall is worth looking out for. – Qiaochu Yuan Jun 7 '12 at 20:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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