Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:

$$0\to A\to C\to B\to 0$$

Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?

Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?

Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?

  • $\begingroup$ I don't know what you mean by a short exact sequence of unital algebras. If $C \to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A \to C$ won't preserve units. $\endgroup$ – Qiaochu Yuan Nov 18 '18 at 21:45
  • $\begingroup$ @Qiaochu In the question we do not assume that $A\to C$ preserve unit. $\endgroup$ – Ali Taghavi Nov 18 '18 at 21:52
  • 1
    $\begingroup$ Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras. $\endgroup$ – user42761 Nov 20 '18 at 11:43
  • 1
    $\begingroup$ This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections. $\endgroup$ – user42761 Nov 22 '18 at 9:16
  • 1
    $\begingroup$ You would also have a counter example, if you can find a $\sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections. $\endgroup$ – user42761 Nov 22 '18 at 9:22

Your Answer

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

Browse other questions tagged or ask your own question.