Timeline for $H^{*}$ algebras as a generalization of $C^{*}$ algebras
Current License: CC BY-SA 3.0
32 events
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| Apr 1, 2023 at 9:32 | vote | accept | Ali Taghavi | ||
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| S Nov 4, 2015 at 8:03 | history | bounty ended | CommunityBot | ||
| S Nov 4, 2015 at 8:03 | history | notice removed | CommunityBot | ||
| Oct 27, 2015 at 13:10 | comment | added | მამუკა ჯიბლაძე | There might be problems with the correct notion of spectrum as there are (noncommutative) polynomials having continuum of roots in quaternions | |
| Oct 27, 2015 at 12:08 | answer | added | Simon Henry | timeline score: 4 | |
| S Oct 27, 2015 at 6:48 | history | bounty started | Ali Taghavi | ||
| S Oct 27, 2015 at 6:48 | history | notice added | Ali Taghavi | Authoritative reference needed | |
| Oct 27, 2015 at 6:47 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 24, 2015 at 20:27 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 22, 2015 at 20:56 | comment | added | Ali Taghavi | @SimonHenry Thanks for your comment.According to the answer to this question let's replace my definition by "Real C* algebra containing a copy of H". then we have automatically a H bimodule, Now what can be said about the spectrum of elements as a subset of H? Is iot non empty and compact? | |
| Oct 21, 2015 at 20:31 | comment | added | eric | @Ali Taghavi -- the definition I'm using is that $A$ is a $B$-algebra if you're given a map $B\to A$. So $H$ is a $C$-algebra. | |
| Oct 21, 2015 at 16:55 | comment | added | Johannes Hahn | Having a left $\mathbb{H}$-module structure is not sufficient imho. When you talk about algebras over noncommutative rings, you generally not want modules, you want a ring-homomorphism $\mathbb{H}\to A$. In the commutative case this is equivalent to the usual formulation in terms of $H$-modules, in the noncommutative case it isn't. Having a ring homomorphism takes care of point 1, point 6 and @SimonHenry s remark about the bimodule structure. | |
| Oct 21, 2015 at 15:55 | comment | added | Christian Remling | @SimonHenry: Yes, you are right, the norm on $\mathbb H$ is the operator norm if the quaternions are realized as matrices. | |
| Oct 21, 2015 at 9:56 | answer | added | André Henriques | timeline score: 6 | |
| Oct 21, 2015 at 8:31 | comment | added | Simon Henry | @AliTaghavi : I'm surprise you don't ask to have an action by $H$ on the right as well. Of course you could define the action on the right by $x \lambda = (\lambda^* x^* )^*$ but then I think you need an axiom to obtain that $\lambda (x \lambda') = (\lambda x) \lambda'$. | |
| Oct 21, 2015 at 5:09 | history | edited | Ali Taghavi |
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| Oct 21, 2015 at 4:49 | comment | added | Ali Taghavi | @ChristianRemling yes, that is the usual norm of $\mathbb{ H}$.. | |
| Oct 20, 2015 at 23:35 | comment | added | Theo Johnson-Freyd | @AndréHenriques May I suggest you upgrade your comment to an answer? | |
| Oct 20, 2015 at 22:43 | comment | added | Ali Taghavi | @AndréHenriques could you please more explain? | |
| Oct 20, 2015 at 21:36 | comment | added | André Henriques | I think that your notion of an $H^*$-algebra is the same thing as a real $C^*$-algebra equipped with a copy of the quaternions in it. | |
| Oct 20, 2015 at 20:41 | comment | added | Ali Taghavi | mathoverflow.net/questions/23478/… | |
| Oct 20, 2015 at 20:40 | comment | added | Ali Taghavi | @eric $H$ is not a complex algebra | |
| Oct 20, 2015 at 20:37 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 20, 2015 at 20:36 | comment | added | eric | As C-algebras even. | |
| Oct 20, 2015 at 20:34 | comment | added | Ali Taghavi | @eric what do you mean by direct sum? (As two vector space over R?) | |
| Oct 20, 2015 at 20:17 | comment | added | eric | I'm not entirely sure how much extra stuff you're getting here. Isn't $H(X)$ just the direct sum of two copies of the continuous maps from $X$ to the complexes? | |
| Oct 20, 2015 at 20:14 | comment | added | Yemon Choi | Warning: the terminology $H^*$-algebra has been used in the past for something completely different, although I guess it is not so commonly studied nowadays | |
| Oct 20, 2015 at 20:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 20, 2015 at 19:55 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 20, 2015 at 19:47 | history | asked | Ali Taghavi | CC BY-SA 3.0 |