Timeline for unitization-process of unital- and non-unital $C^*$-algebras
Current License: CC BY-SA 3.0
31 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| May 2, 2016 at 11:20 | answer | added | Simon Henry | timeline score: 5 | |
| May 2, 2016 at 6:55 | answer | added | Hanno | timeline score: 2 | |
| Jul 6, 2015 at 7:39 | vote | accept | Sabrina Gemsa | ||
| Jun 26, 2015 at 14:55 | answer | added | Johannes Hahn | timeline score: 5 | |
| Jun 26, 2015 at 8:13 | comment | added | Sabrina Gemsa | thank you very much! This idea sound really good and does satisfy me. Thanks! | |
| Jun 26, 2015 at 8:02 | comment | added | Duchamp Gérard H. E. | @UwF ah ! of course, thanks. I think that your idea is interesting. | |
| Jun 26, 2015 at 6:55 | comment | added | UwF | @DuchampGérardH.E.: Suppose some $(a,\lambda)\in A_1$ acts trivially. Then you apply it first to $(0,1)$ to see that $\lambda=0$ and then to $(a^*,0)$ to get $a=0$. I am not saying that such a construction is better than the classical proof, for this reason I don't want to turn it into an answer, either. | |
| Jun 26, 2015 at 5:55 | comment | added | Duchamp Gérard H. E. | @UwF How do you prove that your action is faithful without returning to the two (classical) cases ? | |
| Jun 25, 2015 at 16:49 | comment | added | Johannes Hahn | @UwF Oh, that's what you meant. I've just tried it and if I didn't overlook something, a trivial modification of the usual proof that $(a,\lambda)\mapsto\|L_(a,\lambda)\|$ is the $C^\ast$-norm in the non-unital case works with your action too. Thank you. Do you want to turn your comment into a proper answer? | |
| Jun 25, 2015 at 15:37 | comment | added | UwF | I thought of taking for $A\oplus\mathbb{C}$ the $C^*$-algebraic direct sum, i.e. $\|(b,\mu)\|=\max(\|b\|,|\mu|)$, and letting $A_1$ act on $A\oplus\mathbb{C}$ as $(a,\lambda)(b,\mu)=((a+\lambda)b,\lambda\mu)$. | |
| Jun 25, 2015 at 15:13 | comment | added | Johannes Hahn | @UwF: Alright then, let $A_1$ act on $A\oplus\mathbb{C}$. What norm do you use on $A\oplus\mathbb{C}$ without being circular or having the same non-uniformity we're trying to avoid? I've tried a few options and I didn't really work out. Maybe I've just chosen the wrong norm... | |
| Jun 25, 2015 at 12:52 | comment | added | Yemon Choi | @UwF True, but I have some sympathy with the OP's desire for a unified or "canonical" or "natural" construction | |
| Jun 25, 2015 at 10:26 | comment | added | UwF | Isnt't this more a question of cosmetics? We know what the $*$-algebra structure of $A_1$ should be, so there exists at most one norm that makes it a $C^*$-algebra. You could make the construction look more uniform, if you let it act on $A\oplus\mathbb{C}$. | |
| Jun 24, 2015 at 23:00 | comment | added | Yemon Choi | To my eye, the underlying question is this: suppose $A$ is a $B^*$-algebra (i.e. a Banach algebra equipped with an isometric conjugate linear involution that satisfies the $C^*$-identity). How do we equip the algebraic unitization $A\oplus {\bf C}$ with a norm that also satisfies the $C^*$-identity? A more general problem, where one replaces $C^*$-algebras by operator algebras, was treated in some old work of Ralf Meyer arxiv.org/abs/funct-an/9710001 -- I think this may also be somewhere in the book of Blecher and le Merdy, but I don't have my copy to hand right now | |
| Jun 24, 2015 at 18:21 | comment | added | Johannes Hahn | Just FYI: I've just changed the nlab page to reflect the fact that $\|L_{(a,\lambda)}\|_{B(A)}$ does not equal $\|(a,\lambda)\|_{A^+}$ in the unital case. | |
| Jun 24, 2015 at 18:11 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
added 199 characters in body
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| Jun 24, 2015 at 18:09 | comment | added | Sabrina Gemsa | thank you, I have looked in Murphy's book too, there is the distinction between A unital and nonunital too and you get different norms here. And Johannes pointed out what I want to know. I try to precise my question. | |
| Jun 24, 2015 at 17:23 | comment | added | Johannes Hahn | @UwF I've looked up 2.1.6 in Murphy's book. He makes the same unnatural distinction to prove that there is a C*-norm on the unitalisation. | |
| Jun 24, 2015 at 16:27 | comment | added | Johannes Hahn | Yes, of course, $A_1$ should always be bigger than $A$. Exactly because the unital/non-unital distinction on the input is completely unnatural when one wants to define a functor $\{C^\ast-algebras\}\to\{C^\ast-algebras with one\}$. Also the relation $(C_0(X))_1 = C(X^+)$ demands that, because the one-point-compactification adds a point whether or not $X$ is compact. I'll try to have look in Murphys book as soon as possible. | |
| Jun 24, 2015 at 16:14 | comment | added | UwF | @JohannesHahn: The proof in Murphy's book answers your question (and the one by the OP), I believe. There are several procedures available to make C*-algebras unital. Passing to the multiplier algebra is probably more elegant (and more functorial) - this is done in Theorem 2.1.5 in Murphy's book (but if you algebra did not have a unit, then it usually adds a lot more than just a unit). If you want to use $A_1$, then you will never keep your original algebra, even if it was already unital. | |
| Jun 24, 2015 at 16:06 | comment | added | Johannes Hahn | @UwF: If I understand the question correctly, then "treating the case..." is exactly what causes confusion here. Why should there be any case distinction in the definitions or the proofs between non-unital algebras that just happen to have forgotten that they really have an identity and non-unital algebras that genuinely do not have one? If an abstract existence proof of the unitalization functor (e.g. from the adjoint functor theorem) works without this completely unnatural distinction, why can't there be an explicit construction of the unitalisation and its norm that does? | |
| S Jun 24, 2015 at 15:48 | history | suggested | Duchamp Gérard H. E. | CC BY-SA 3.0 |
corrected spelling
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| Jun 24, 2015 at 15:38 | review | Suggested edits | |||
| S Jun 24, 2015 at 15:48 | |||||
| Jun 24, 2015 at 9:49 | comment | added | UwF | See Theorem 2.1.6 in the book by Murphy, he treats also the case where $A$ is already unital. | |
| Jun 24, 2015 at 9:44 | answer | added | Duchamp Gérard H. E. | timeline score: 0 | |
| Jun 24, 2015 at 9:09 | comment | added | Sabrina Gemsa | I added the link. Sorry, I was impatient, next time I wait a longer time. Shall I delate my question here? I use MSE and M.O. for the first time. Sorry | |
| Jun 24, 2015 at 9:07 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
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| Jun 24, 2015 at 9:04 | comment | added | Stefan Kohl♦ | Please include the link to your math.stackexchange question -- and be aware that before cross-posting you should wait a reasonable time for an answer on math.stackexchange (say, a week or two). | |
| Jun 24, 2015 at 8:52 | review | First posts | |||
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| Jun 24, 2015 at 8:49 | history | asked | Sabrina Gemsa | CC BY-SA 3.0 |