Timeline for Behaviour of Direct limit with quotient and double dual
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2018 at 13:33 | comment | added | Nik Weaver | Yes, because the images of the $A_n$ are dense in $A$. | |
| Oct 12, 2018 at 12:55 | comment | added | Math Lover | @NikWeaver: Thanks I can see that $A_n/I_n$ sits isometrically inside $A/I $ but I don’t see why it’s enough to say that limit is actually $ A/I$? Do we need some denseness argument? | |
| Oct 12, 2018 at 9:35 | comment | added | Nik Weaver | The key point is that a $*$-homomorphism of C*-algebra must be isometric if its kernel is ${0}$ (a powerful fact). So, you have a natural map from each $A_n/I_n$ into $A/I$, check that its kernel is ${0}$, etc. It's routine. | |
| Oct 12, 2018 at 7:47 | comment | added | Math Lover | @NikWeaver: Can you please add few lines why the limit of the sequence is $A/I$? May be this is obvious but since I am beginner so I don’t follow it. Thanks | |
| Oct 3, 2018 at 12:09 | comment | added | Nik Weaver | No problem at all. | |
| Oct 3, 2018 at 12:06 | comment | added | Mateusz Wasilewski | I confused the two, I'm sorry. Of course you're right. | |
| Oct 3, 2018 at 11:30 | comment | added | Nik Weaver | Do you mean the $\phi_n$ or the $\tilde{\phi}_n$? The $\tilde{\phi}_n$ are injective. | |
| Oct 3, 2018 at 9:03 | comment | added | Mateusz Wasilewski | Nik, aren't you assuming that the connecting maps are injective? I think that the conclusion holds anyway, because the union of the images of the $A_n$'s is dense in $A$ (and the same for $I_n$'s and $I$). | |
| Oct 3, 2018 at 3:00 | history | answered | Nik Weaver | CC BY-SA 4.0 |