Timeline for Unital $C^{*}$ algebras whose all elements have path connected spectrum
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Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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Apr 6, 2020 at 0:07 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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Mar 12, 2020 at 21:56 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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Mar 12, 2020 at 12:06 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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Feb 23, 2018 at 7:55 | history | edited | Ali Taghavi |
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S Jan 27, 2017 at 18:39 | history | bounty started | Ali Taghavi | ||
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Nov 13, 2016 at 8:56 | comment | added | Ali Taghavi | @JohannesHahn Thanks for your comment. As you pointed out a minimal dynamical system gives a simple C* algebra. But what about the spectrum of its elements(Path connected spectrum)? | |
Nov 7, 2016 at 19:47 | comment | added | Ali Taghavi | @HannesThiel Thank you very much for your comment and your attention. The motivation for the second part of my question, about the reduced $C^*$ algebra, is obvious. Regarding the first part: the motivation is that the product topology of two path connected space is a path connected space. But according to your comment, one can ask that: Is the tensor product of two idempotent less C* algebras, an idempotent less C* algebra? | |
Nov 7, 2016 at 18:50 | comment | added | Hannes Thiel | @AliTaghavi What is your motivation to consider path connected and not just connected spectra? | |
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S Dec 28, 2015 at 14:21 | history | bounty started | Ali Taghavi | ||
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Nov 7, 2015 at 16:29 | comment | added | jjcale | Example of a connected C∗ algebra : See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ? | |
Nov 7, 2015 at 10:28 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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Nov 7, 2015 at 7:47 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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S Nov 7, 2015 at 7:15 | history | bounty started | Ali Taghavi | ||
S Nov 7, 2015 at 7:15 | history | notice added | Ali Taghavi | Authoritative reference needed | |
Jun 13, 2015 at 21:40 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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Mar 4, 2015 at 15:13 | comment | added | Sam Evington | Hannes' observation can be improved: all elements of a C*-algebra with no non-trivial projections have connected spectrum. Indeed given an element with disconnected spectrum the holomorphic functional calculus gives a non-trivial idempotent. This will be similar to some self-adjoint idempotent (see Blackadar's K-Theory book) which will be non-trivial. | |
Mar 3, 2015 at 19:38 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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Feb 26, 2015 at 16:13 | comment | added | Ali Taghavi | @HannesThiel Thank you for the comment. So my second question leads to the following: What is an example of a projectionless simple $C^{*}$ algebra for which the spectrum of all elements are path connected. | |
Feb 25, 2015 at 18:46 | comment | added | Hannes Thiel | The spectrum of a nontrivial projection (i.e., nonzero and not the unit) is disconnected. Thus, a unital C*-algebra with the property you consider cannot contain any nontrivial projections. Conversely, if a C*-algebra contains a normal element with disconnected spectrum, then it contains a nontrivial projection. Thus, in a unital, simple C*-algebras without nontrivial projections (such as the Jiang-Su algebra), at least the spectrum of every normal element is connected. | |
Feb 7, 2015 at 0:45 | comment | added | Ali Taghavi | @JohannesHahn Thanks for your comment. If I am not mistaken, this algebra is simple iff the action of $G$ on $X$ does not have a nontrivial compact invariant set. Now how you controle the spectrum of elements? | |
Feb 7, 2015 at 0:37 | comment | added | Johannes Hahn | Well $\mathbb{C}$ is simple. But I assume you want something less trivial. A stab in the dark: One could imagine putting some group action on $C(X)$ such that none of its non-trivial ideals are $G$-invariant and hope that $C(X)\rtimes G$ is simple. | |
Feb 7, 2015 at 0:33 | comment | added | Ali Taghavi | @JohannesHahn yes I mean without closed two sided ideal. | |
Feb 7, 2015 at 0:31 | comment | added | Johannes Hahn | Oh you meant simple as in simple not as in easy... | |
Feb 7, 2015 at 0:30 | comment | added | Ali Taghavi | @JohannesHahn But $C(X)$ is not simple. | |
Feb 7, 2015 at 0:28 | comment | added | Johannes Hahn | An easy example: $C(X)$ for some path connected compactum $X$ is path connected in this sense because the spectrum of $f\in C(X)$ coincides with the range of $f$. | |
Feb 7, 2015 at 0:24 | history | asked | Ali Taghavi | CC BY-SA 3.0 |