# Reference for Kaplansky's proof that in $\mathbb{C}[G]$, $ab=1$ implies $ba=1$

The case [of a group ring $R[G]$ where $G$ is an infinite group] where $R$ is the field of complex numbers is probably the one best studied. In this case, Irving Kaplansky proved that if $a$ and $b$ are elements of $\mathbf{C}[G]$ with $ab = 1$, then $ba = 1$. Whether this is true if $R$ is a field of positive characteristic remains unknown.

Wikipedia doesn't list a reference for this specifically though. Does anyone have a reference for this proof? I would appreciate any current related information about this result too.

• Wikipedia does list references. The last book, D.S. Passman, The algebraic structure of group rings, proves it. Corollary 1.9. – user580373 Aug 6 '18 at 21:35
• @spiralstotheleft aHah! Thank you! Is it a nice enlightening proof? Like is it worth me, or you if you'd like, outlining the proof here? Or if not, you should post your comment as an answer. – Mike Pierce Aug 6 '18 at 21:54
• Yes, the proof it not too long and the ideas in it quite useful in general. – user580373 Aug 6 '18 at 21:56
• Apocrypha: the proof is supposed to be in K's book Fields and Rings; I downloaded a copy and have spent about 15 minutes looking for the proof. It may very well be there . . . – Robert Lewis Aug 6 '18 at 22:42
• @RobertLewis Found it in Fields and Rings! Though no proof, only the statement mentioned at the end of Part 2, Chapter 3, Semi-simple rings. (page 122 in my printing) – Mike Pierce Aug 7 '18 at 1:07

Adaptation of the argument from D.S. Passman, The algebraic structure of group rings for easier reference. I cut down some simple computations, and some parallel interesting results, like $tr(e)$ being algebraic with all its conjugates in $[0,1]$ for $e$ idempotent. Feel free to expand them.

Let $K$ be a field of characteristic $0$, and $G$ a (not necessarily finite) group.

Assuming the definitions, for $a=\sum_{x\in G}a_xx$, and $b=\sum_{x\in G}b_xx$ in $\mathbf{C}[G]$ \begin{align}tr(a)&=a_1\\(a,b)&=\sum_{x\in G}a_x\overline{b}_x\\\|a\|&=(a,a)^{1/2}=\left(\sum_{x\in G}|a_x|^2\right)^{1/2}\\|a|&=\sum_{x\in G}|a_x|\end{align} and their basic properties.

Assume that $ab=1$. Then $e=ba$ satisfies $e^2=b(ab)a=ba=e$.

Also $tr(ba)=tr(ab)=1$. The bulk of the proof is showing that traces of idempotent elements are in $[0,1]$ and that only $\mathbf{1}\in K[G]$ has trace $1$ and $\mathbf{0}\in K[G]$ trace $0$.

Let $F$ be a finitely generated extension of $\mathbb{Q}$ that contains the coefficients of $e$. Embedding $F$ in $\mathbb{C}$ one can reduce to the case $K=\mathbb{C}$.

Assume that $e\neq\mathbf{1}$ (it is also $e\neq\mathbf{0}$ since $tr(e)=1$). To get a contradiction it is enough to prove that $tr(e)>0$, because $\mathbf{1}-e\neq\mathbf{0}$ is also an idempotent, $(\mathbf{1}-e)^2=\mathbf{1}-2e+e^2=\mathbf{1}-e$, and then $1-tr(e)=tr(1-e)>0$ implies that $tr(e)<1$.

Proving that $tr(e)>0$.

Call $I=e\mathbb{C}[G]$, let $d\geq0$ be such that $d^2=\inf_{f\in I}(f-\mathbf{1},f-\mathbf{1})=\inf_{f\in I}\|f-\mathbf{1}\|^2$.

For each $0<n\in\mathbb{N}$ take $f_n\in I$ such that $$\|f_n-\mathbf{1}\|^2<d^2+\frac{1}{n^4}$$

For all $b\in I$ $$|(b,f_n-\mathbf{1})|\leq \frac{\|b\|^2}{n^2}$$ and $$\|f_n-\mathbf{1}\|\leq \left(d^2+\frac{1}{n^4}\right)^{1/2}\leq d+1$$

Therefore $\|f_n\|\leq \|\mathbf{1}\|+\|f_n-\mathbf{1}\|\leq d+2$. Since $f_n\in I$ $$|(f_n,f_n-\mathbf{1})|\leq \frac{\|f_n\|}{n^2}\leq \frac{d+2}{n}$$

It follows that $$|\|f_n\|^2-tr(f_n)|=|(f_n,f_n-\mathbf{1})|\leq \frac{d+2}{n}$$

Now, \begin{align}\|f_ne-e\|^2&=((f_n-\mathbf{1})e,(f_n-\mathbf{1})e)\\&=((f_n-\mathbf{1})e\overline{e},(f_n-1))\\&\leq\frac{\|(f_n-\mathbf{1})e\overline{e}\|}{n^2}\\&=\|f_n-\mathbf{1}\|\frac{|e\overline{e}|}{n^2}\\&\leq (d+1)\frac{|e\overline{e}|}{n^2}\end{align}

Finally, \begin{align}|tr(f_ne)-tr(e)|&=|tr(f_ne-e)|\\&\leq\|f_ne-e\|\\&\leq\sqrt{(d+1)|e\overline{e}|}\frac{1}{n}\end{align}

Since $f_n\in I$ and $e^2=2$ it follows that $ef_n=f_n$. Therefore $tr(f_ne)=tr(ef_n)=tr(f_n)$. This means that the last inequality proves that $$tr(e)=\lim_{n\to\infty}tr(f_n)$$ Two inequalities above shows that $$\lim_{n\to\infty}tr(f_n)=\lim_{n\to\infty}\|f_n\|^2$$

Now, \begin{align}\|e\|&\leq\|e-f_ne\|+\|f_ne\|\\&\leq(d+1)\frac{|e\overline{e}|}{n^2}+\|f_n\||e|\end{align}

Taking limits $$tr(e)\geq\frac{\|e\|^2}{|e|^2}>0$$.

• This is an amazing argument; thanks for sharing it. I'm quite surprised that the engine here seems to be clever observations about $\mathbb{C}[G]$ as a complex inner product space. A quick correction, if I'm not mistaken: I think at the beginning of your post, you want $e^{2} = baba = b(ab)a = ba$. – Alex Wertheim Aug 6 '18 at 23:20

Perhaps the following answer your question, or will at least be of use.

Then Lupini, Lecture 201, p.4-5 provides: