# Do non-commutative algebras with dense commutative subalgebras exist?

Let $A$ be a normed unital algebra. Suppose that $C\subseteq A$ is a commutative subalgebra which is dense in $A$.

I ask myself the following question:
Under the above assumptions, is $A$ necessarily commutative?

If the answer is in the negative, then a counterexample would be greatly appreciated. And then there would also be a follow-up question: Would it make a difference if one assumes e.g. that $A$ is a Banach algebra (or C*-algebra)?

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Let $x, y \in A$. Since $C$ is dense, there are sequences $x_n \to x$, $y_n \to y$ with $x_n, y_n\in C$. Now $$xy = \lim x_ny_n = \lim y_n x_n = yx.$$ – martini Oct 14 '11 at 23:15
Thanks! I tried to do epsilon-stuff with $||xy-yx||$ and got lost. Your method was very quick and nice! – Per Oct 14 '11 at 23:21
@martini, please turn that into an answer so that Per can accept it! – Mariano Suárez-Alvarez Jan 9 '12 at 5:26
@MarianoSuárez-Alvarez: Given that martini's last login was November 1 and that this comment is the last thing that shows up under "activity", I thought that seemed unlikely to happen soon. – Jonas Meyer Jan 9 '12 at 5:54

## 1 Answer

Martini's comment answers the question, exploiting (joint) continuity of multiplication $A\times A\to A$ with respect to the norm. This continuity follows easily from the inequality $\|ab-a'b'\|=\|ab-a'b+a'b-a'b'\|\leq\|a-a'\|\|b\|+\|a'\|\|b-b'\|$.

There are interesting topological algebras in which multiplication is only continuous in each variable separately rather than jointly continuous, so it may be of interest that the result still holds in that case. The argument could be adjusted as follows. Given $x$ and $y$ in $A$, let $(x_n)$ be a net in $C$ converging to $x$. Then for all $z$ in $C$, $xz=\lim_n x_n z =\lim_n zx_n=zx$. So $x$ commutes with every element of $C$. Now let $(y_n)$ be a net in $C$ converging to $y$. Then $xy=\lim_n xy_n=\lim_n y_nx=yx$.

(For example, the same result holds if $A$ is a von Neumann algebra and $C$ is ultraweakly dense.)

For the original problem there is also a direct "epsilon-stuff" method as mentioned in your comment. You can use the following inequality which holds for all $x$ and $y$ in $A$ and for all $x'$ and $y'$ in $C$:

\begin{align*} \|xy-yx\|&=\|xy-x'y+x'y-x'y'+y'x'-y'x+y'x-yx\|\\ &\leq \|x-x'\|\|y\|+\|x'\|\|y-y'\|+\|y'\|\|x'-x\|+\|y'-y\|\|x\|. \end{align*}

Given $x\neq0$ and $y\neq0$ in $A$ and $\varepsilon>0$, there exist $x'$ and $y'$ in $C$ such that $\|x-x'\|\lt \min\{\varepsilon/\|y\|,\|x\|\}$, and $\|y-y'\|<\min\{\varepsilon/\|x\|,\|y\|\}$. This ensures that $\|x'\|\lt 2\|x\|$ and $\|y'\|<2\|y\|$, and from the above inequality you get $\|xy-yx\|<6\varepsilon$. This is messier because it doesn't take advantage of continuity of multiplication, which would give us $x'$ and $y'$ in $C$ such that $\|xy-x'y'\|$ and $\|y'x'-yx\|$ are both small, then requiring only $1$ use of the triangle inequality instead of $3$.

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