# Permanent invertible elements

Let $$A$$ be a unital complex algebra with the unit $$\bf1$$. Let $$\mathcal{N}$$ be the family of all norms on $$A$$ making it a unital normed algebra with the same unit $$\bf1$$. Let us put $$B_{\|\cdot\|}=\{x\in A : \|x-{\bf1}\|<1\}$$ where $$\|\cdot\|\in \mathcal{N}$$.

Clearly, the intersection $$\bigcap_{\mathcal{N}}B_{\|\cdot\|}$$ is contained in $$A^{-1}_{\|\cdot\|}$$ where $$A_{\|\cdot\|}$$ is the completion of $$A$$ with respect to norm $$\|\cdot\|$$ in $$\mathcal{N}$$.

Q. Does there exist any non-scalar element in the intersection $$\bigcap_{\mathcal{N}}B_{\|\cdot\|}$$?

• Why is the intersection contained in the units? The inverse $\sum_k x^k$ need not exist if $A$ is not complete w.r.t. to the norm. And it is not always possible to find a norm which makes $A$ complete, e.g. $A$ could be of countable dimension. So why does $\sum_k x^k$ converge if we cannot assume completeness? Mar 23 at 16:25
• Following Denis Serre's correction of an error that I made, I am now wondering: why are you taking the intersection rather than the union, if you are looking for "permanently invertible" elements? Mar 24 at 0:03
• Can I also request to other users that we stop making pettifogging LaTeX changes? Formatting and style are a lot more subjective than some people seem willing to acknowledge, and I think that superfluous prettification mistakes the point of this site Mar 24 at 3:07
• @YemonChoi, re, the change was not intended to be stylistic but rather TeX-based, in that \lVert\rVert rather than \Vert\Vert and $\mathbf1$ rather than ${\bf1}$ are as it were best practices. I hope it hardly changed the rendering. You mentioned spacing around the dots, which did indeed change, but can be restored by \lVert\,\cdot\,\rVert in a way that doesn't incur the other issues of \Vert. But, @‍AliBagheri, I apologise if it was unwelcome. Mar 24 at 13:57
• @LSpice Have finally found some breathing space to catch up on various backlogs, MO being one of them. Thanks for explaining the reasoning and the TeXnical motivation, which I was not aware of. Apr 3 at 16:09

If $$x$$ belongs to this intersection, then $$x$$ commutes with every nilpotent element.
Proof. For every invertible $$a\in A$$, the function $$N(z):=\lVert a^{-1}za\rVert$$ is a norm of unital algebra. Let $$n$$ be a nilpotent element, of order $$k$$, and choose $$a_t=\mathbf1-tn$$ in the construction above, with $$t\in\mathbb R$$ a parameter. Then $$a_t^{-1}xa_t=(\mathbf 1+tn+\dotsb+t^{k-1}n^{k-1})x(\mathbf1-n)=x+t(nx-xn)+\cdots-t^kn^{k-1}xn$$ is a polynomial function of $$t$$. By assumption, $$t\mapsto\lVert a_t^{-1}xa_t\rVert$$ is a bounded (by $$1$$) function, hence the polynomial above needs to be constant. In particular $$nx-xn=0$$.
As a corollary, the answer for the case of $$A=M_r({\mathbb C})$$ with $$r\ge2$$ is as you expected. Write $$X$$ instead of $$x$$ (it is a matrix). Let $$u\in{\mathbb C}^r$$ be a non-zero vector. Choose $$v$$ such that $$v^Tu=0$$ (it exists). Then $$uv^T$$ is nilpotent, hence $$Xuv^T=uv^TX$$, which implies that $$Xu$$ is parallel to $$u$$. Thus every vector is an eigenvector, which implies that $$X$$ is scalar: $$X=\alpha I_r$$.
Addition. Suppose now that $$(A,\|\cdot\|)$$ is a unital Banach algebra. The spectral radius $$r(u)=\lim\inf\|u^k\|^{1/k}$$ is well-defined. As above, $$\cal N$$ contains all norms $$N_a=\|a^{-1}\cdot a\|$$ for $$a\in A^\times$$. If $$x$$ belongs to the OP's intersection, then the set $$\{a^{-1}xa\;|a\in A^\times\}$$ is bounded. Consider an element $$u\in A$$ for which $$r(u)=0$$ ($$u$$ can be be nilpotent, but this is not necessary if $$A$$ is infinite dimensional). Then $${\bf1}-zu$$ is invertible for every $$z\in\mathbb C$$. Thus $$z\mapsto({\bf1}-zu)^{-1}x({\bf1}-zu)$$ is a bounded entire function, hence a constant function, $$\equiv x$$. In other words $$x({\bf1}-zu)\equiv ({\bf1}-zu)x$$, that is $$xu=ux$$. Thus the property mentionned above extends to:
If $$x$$ belongs to this intersection, then $$x$$ commutes with every element of spectral radius $$0$$.