Timeline for $H^{*}$ algebras as a generalization of $C^{*}$ algebras
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2015 at 16:20 | comment | added | André Henriques | Indeed, I followed a suggestion by Theo Johnson-Freyd to upgrade my comment to an answer. | |
| Oct 27, 2015 at 16:14 | comment | added | j0equ1nn | This seems more like a comment than an answer. | |
| Oct 22, 2015 at 6:59 | comment | added | Ali Taghavi | @Andre thank you for your answer. What about the spectrum of elements(non empty compact subsets of H?) | |
| Oct 21, 2015 at 16:06 | comment | added | Simon Henry | @ChristianRemling : It is also true for real $C^*$-algebra that a injective morphism of real $C^*$-algebras is isometric, hence if $H$ sit inside $A$ you have that $\Vert h \Vert $ is indeed what it is suppose to be, the fact $\Vert hx \Vert = \Vert h \Vert \Vert x \Vert$ follow from the fact that $h$ has an inverse whose norm is $1/\Vert h \Vert$ and two application of the multiplicativity of the norm. Of course one need $H$ to sit inside $A$ as a sub $*$-algebra, which include an additional axiom on the behavior of $*$ on $H$ in comparison with what the OP is asking. | |
| Oct 21, 2015 at 15:58 | comment | added | Christian Remling | It's still unclear to me how you get property (2), that $\| ax\| = \| a\| \|x\|$ for $a\in\mathbb H$, if you just know that $\mathbb H$ sits inside your $C^*$ algebra. Could you please elaborate. | |
| Oct 21, 2015 at 10:01 | comment | added | Simon Henry | This only work in the unital case, and even for that there is some axioms missing in the Op definition for this to work. In the general case you could say with a copy of the quaternion within the Multiplier algebra but this for example already contains the action of $H$ on the right which is missing from the OP definition. The condition $(1+a^*a)$ is part of the definition of real $C^*$-algebra and it can be replace by the strong $C^*$-inequality ($\Vert x^* x + y^* y \Vert \geqslant \Vert x \Vert ^2$). | |
| Oct 21, 2015 at 9:56 | history | answered | André Henriques | CC BY-SA 3.0 |