Let $A$ be a C$^*$algebra with closed twosided ideal $I$. Set $B=A/I$ and let $\pi:A\to B$ be the quotient map. Suppose that $b\in B$ is quasinilpotent. Does there exist quasinilpotent $a\in A$ such that $\pi(a)=b$?
2 Answers
If I is the compact elements of A and B is the corresponding Calkin Algebra, the answer is yes. Have you looked at BARNES, B. A., MURPHY, G. J., SMYTH, M. R. F. and WEST, T. T., "Riesz and Fredholm theory in Banach algebras" (Research Notes in Mathematics 67, Pitman, 1982?

$\begingroup$ Many thanks. I used to have a copy but I got rid of most of my books years ago, and I had forgotten this result. $\endgroup$ Apr 7, 2022 at 19:33

$\begingroup$ It would be interesting to know of any developments beyond the 1982 result, but in the absence of further information the tick goes to @Derek. $\endgroup$ Apr 9, 2022 at 9:08

$\begingroup$ @DouglasSomerset I was also hoping someone who has more current experience in this area could provide supplementary information or comment on extensions. There is also work by Pedersen in lifting nilpotents using an unrelatedand nonequivalent construction, also from the 1980s or late 1970s. I have asked a colleague and student of Gerry Murphy (unfortunately no longer with us) and will comment again (or they can comment) if there is more recent information. I strongly recommend getting the BMSW monograph, as their journal publications are a bit harder to read. $\endgroup$– DerekApr 9, 2022 at 19:13

$\begingroup$ In case Akemann feels slighted, read "Akemann and Pedersen" for "Pedersen" :) $\endgroup$– DerekApr 9, 2022 at 19:27

$\begingroup$ It was shown in C. L. Olsen and G. K. Pedersen, Coronaalgebras and their applications to lifting problems, Math. Scand. 64 (1989), 6386 that every nilpotent can be lifted; and in T. A. Loring, Lifting solutions to perturbing problems in C∗algebras, volume 8 of Fields Institute Monographs. American Mathematical Society, Providence, RI, 1997 that nilpotents can be lifted with arbitrarily small increase in norm. $\endgroup$ Apr 9, 2022 at 19:52
I am rather certain this is unknown. The issue is that procedures we know to lift the relation $a^{n+1}=0$ do not mesh well with the $a^n=0$ case. This means writing the given quasinilpotent as a limit of nilpotents does not seem to help.