Timeline for Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Current License: CC BY-SA 3.0
9 events
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| May 6, 2015 at 14:39 | comment | added | Phoenix87 | Ah that's right, I forgot to try and play with this one-to-one correspondence around! Many thanks! | |
| May 6, 2015 at 14:08 | comment | added | Hannes Thiel | Yes, a c.p.c. order-zero map $\varphi\colon A\to B$ between two C*-algebras corresponds to a *-homomorphism $\alpha\colon C_0((0,1])\otimes A \to B$ such that $\varphi(a)=\alpha(t\otimes a)$. Assume $B$ is $\sigma$-unital and $b$ is a strictly positive element in $B$. Given a set $F$ in $A$, consider $G=\{\alpha(t\otimes b)\}\cup \varphi(F)$. Then $\varphi(C^*(F))\subset \alpha(C_0((0,1])\otimes C^*(F)) = C^*(G)$. | |
| May 5, 2015 at 15:15 | vote | accept | Phoenix87 | ||
| May 5, 2015 at 15:15 | comment | added | Phoenix87 | Also, about the second part of your answer, is there something already known in that direction for c.p.c. order zero maps between C*-algebras? For example, if $a\in A^+$, then $\phi(C^*(a))\subset C^*(h_\phi,\pi_\phi(a))\cap B$, but are there elements $g_1,\ldots, g_n\in B$ such that $\phi(C^*(a))\subset C^*(g_1,\ldots,g_n)$? | |
| May 4, 2015 at 21:14 | comment | added | Phoenix87 | Thanks for the clarification, I think I can now see why this is the case (just considering real functions on $X$). However your example shows that there might be a suitable choice of exhaustive families for which the property in the OP holds. Of course, for any c.p.c. order zero map $\phi:A\to B$ between C*-algebras one has that $\phi(A)$ is contained in a sub-C*-algebra of $B$ (which is actually the property I'm trying to get to when $A$ and $B$ are local C*-algebras). | |
| May 4, 2015 at 20:56 | history | edited | Hannes Thiel | CC BY-SA 3.0 |
typo fixed
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| May 4, 2015 at 20:55 | comment | added | Hannes Thiel | Yes, this might seem counter-intuitive at first. But every element in $C(X)$ is contained in a finitely generated sub-C*-algebra of $C(X)$, and therefore $C(X)$ is the union (no closure needed) of all its finitely generated sub-C*-algebras. | |
| May 4, 2015 at 16:52 | comment | added | Phoenix87 | Hi Hannes. Many thanks for your answer. In your example I can see how you can get $A$ from the completion of the union of all the finitely generated C*-subalgebras of $A$, but it seems to me that you are somehow implying that one doesn't need to take the norm-completion, and this is not clear to me at the moment. Perhaps what's misleading me is that I'm thinking of $C(X)$ as the infinite tensor product $C(I)^{\otimes\infty}$ with $I=[0,1]$, seen as an inductive limit with the obvious connecting maps. | |
| May 4, 2015 at 12:27 | history | answered | Hannes Thiel | CC BY-SA 3.0 |