Timeline for automorphisms of C*-algebras and partial isometries
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2014 at 19:50 | answer | added | Eusebio Gardella | timeline score: 3 | |
| Jul 30, 2013 at 9:35 | comment | added | Ulrich Pennig | @Gabor: You are right. One can probably reformulate this into a question about the map induced by the inclusion $K_0(A^{\alpha}) \to K_0(A)$ - at least in case of Kirchberg algebras. | |
| Jul 30, 2013 at 6:07 | answer | added | Michael | timeline score: 5 | |
| Jul 29, 2013 at 21:40 | comment | added | Gabor Szabo | Because of your comment, your question really is: Which conditions do you have to impose on $\alpha$ or $A$ such that MvN equivalence of any two projections $p,q\in A^\alpha$ inside $A$ is already a MvN equivalence inside $A^\alpha$? | |
| Jul 29, 2013 at 21:17 | comment | added | Julien | Yes, if you can find a fixed path from $p$ to $q$, you just have to exhibit $p=p_0,p_1,\ldots,p_n=q$ all fixed with $\|p_i-p_{i+1}\|<1$. Then at each step $p_i$ and $p_{i+1}$ are unitarily equivalent via $u_i:=(p_i+p_{i+1}-1)|p_i+p_{i+1}-1|^{-1}$ wich is fixed. So $u=u_1\cdots u_n$ is a fixed unitary such that $q=u^*pu$ and you just have to set $v:=pu$ to get your fixed partial isometry. But this strategy requires at least that $p$ and $q$ be homotopic, which is not guaranteed if they are just MvN equivalent in a purely infinite $C^*$ algebra. | |
| Jul 29, 2013 at 19:41 | comment | added | Ulrich Pennig | I would also be interested in the case $p$ arbitrary, but $q=1$. | |
| Jul 29, 2013 at 19:40 | comment | added | Ulrich Pennig | I think, the argument I sketched also works, if I can find a path from $p$ to $q$, which is fixed under pointwise application of $\alpha$. | |
| Jul 29, 2013 at 19:38 | comment | added | Ulrich Pennig | I am wondering more about the case, where $p$ and $q$ are arbitrary. The question is: Is there a partial isometry $v$ that mediates the equivalence and is fixed by the automorphism. In the example I gave, there is such a $v$. | |
| Jul 29, 2013 at 19:33 | comment | added | Michael | Well, in the very special case $p=q=1$, this is just the question "which unitaries are fixed by $\alpha$?" What sort of condition do you have in mind? Or should this be read as "when does there exist a $v$ fixed by $\alpha$"? | |
| Jul 29, 2013 at 14:02 | history | asked | Ulrich Pennig | CC BY-SA 3.0 |