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 $\bigcap_{\mathcal{N}}B_{\|\cdot\|}$?

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    $\begingroup$ 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? $\endgroup$ Mar 23 at 16:25
  • $\begingroup$ 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? $\endgroup$
    – Yemon Choi
    Mar 24 at 0:03
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    $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Mar 24 at 3:07
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    $\begingroup$ @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. $\endgroup$
    – LSpice
    Mar 24 at 13:57
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    $\begingroup$ @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. $\endgroup$
    – Yemon Choi
    Apr 3 at 16:09

1 Answer 1


Here is a non-trivial condition:

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 $0$.


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