It is easy to see that for any finite $n$, the invertible elements in the algebras algebras $\mathbb{C}^n$ (edowed with the maximum norm, say) and $M_n$ (all square matrices) are dense. Thus, by the Artin-Wedderburn all finite-dimensional C*-algebras have dense invertible groups.

Is there an example of a unital, finite-dimensional Banach algebra where the invertible elements are not dense?


No. The spectrum of a non-invertible element is finite. So you can find sequence of numbers in the resolvent set converging to zero. From this, you can cook up a sequence of invertible elements converging to your non-invertible element.

| cite | improve this answer | |
  • $\begingroup$ What if the algebra is real? $\endgroup$ – Tomasz Kania Apr 17 '17 at 20:51
  • 1
    $\begingroup$ The spectrum of an element of a real algebra equals the spectrum in the complexification. You can choose the sequence to be made of real numbers. $\endgroup$ – vap Apr 17 '17 at 21:16
  • $\begingroup$ If $a \in A$ has finite spectrum (wit $A$ real or complex), then there is a very small $\epsilon > 0$ with $0 \notin \sigma(a + \epsilon)$. $\endgroup$ – user42761 Apr 18 '17 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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