Timeline for Extending $*$-morphisms to the multiplier algebras
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2020 at 16:21 | comment | added | Nik Weaver | You are welcome! | |
| Nov 3, 2020 at 15:25 | vote | accept | CommunityBot | ||
| Nov 3, 2020 at 14:57 | comment | added | user167952 | Thanks for the follow up! It's all clear now. | |
| Nov 3, 2020 at 13:34 | comment | added | Nik Weaver | ... so I automatically get the extension to $M(A)$ and then just have to check that it actually maps into $M(B)$. | |
| Nov 3, 2020 at 13:32 | comment | added | Nik Weaver | @user839372 Ruy answered this, but: I put $B$ in $B(K)$ because $A$ is an ideal of $M(A)$, and it is a general fact that any $*$-homomorphism from an ideal of a C*-algebra into $B(K)$ extends to the whole algebra. | |
| Nov 3, 2020 at 11:33 | comment | added | Ruy | Every representation of an ideal extends to a representation of the ambient algebra, so this is why it is useful to represent $B$. But you may also get away without it: by Cohen-Hewitt every $b$ in $B$ may be written on the nose as $b=φ(a_1)b_1$, so if $m$ is a multiplier you may define the extension by $\tildeφ(m)b=φ(ma_1)b_1$. | |
| Nov 3, 2020 at 9:39 | comment | added | David Roberts♦ | @user839372 $M(A)$ is the completion of $A$ in the strict topology, so the extension you seek exists precisely when the original $*$-homomorphism is strictly continuous. You don't need the concrete representation of $B$ at all, I presume that was just motivation. | |
| Nov 3, 2020 at 9:04 | comment | added | user167952 | @Nik Weaver. Thanks for the answer! Two questions (1) Why is it necessary to regard $B \subseteq B(K)$? (where does the proof use it) (2) Why does $A \to B(K)$ extends toa $*$-morphosm $M(A) \to B(K)$? Is this the universal property or something? | |
| Nov 3, 2020 at 5:59 | comment | added | David Roberts♦ | Can't we get away with slightly less: say $\phi(u_\lambda)$ converges strictly to a projection on $B$, where $(u_\lambda)$ is an approximate identity for $A$? | |
| Nov 3, 2020 at 3:39 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 3, 2020 at 2:41 | history | answered | Nik Weaver | CC BY-SA 4.0 |