Timeline for Which $\ast$-algebras are $C^\ast$-algebras?
Current License: CC BY-SA 4.0
9 events
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| Jan 29, 2022 at 13:17 | comment | added | Uri Bader | Got it. You better work in $X=A_h$, the space of self-adjoint elements, which is a real vector space and eventually extend the real state to $A$ which is $A_h\otimes \mathbb{C}$ as a vector space. Forget about the $\epsilon$ bit in my remark, I was wrong... Anyway, I like this short cut, thank you! | |
| Jan 29, 2022 at 12:48 | comment | added | Tim Campion | Now $P = A_+$ is a positive cone in $X$, and let $Y \subseteq X$ be the real scalars. We can apply the above-linked version of Hahn-Banach to the identity function on $Y$. Complexifying, we get a state on $A$. (You may be write about needing $0 < \epsilon \leq a^\ast a$, but I'm not quite sure.) | |
| Jan 29, 2022 at 12:47 | comment | added | Tim Campion | I'm using the version of Hahn-Banach for a cone rather than a norm -- see e.g. Thm 2.1 here. This does require passing back and forth between real and complex statements. But if $-1 \not \in A_+$, then for $a \in A_+$, if $\lambda a \in A_+$ it follows that $\lambda$ is real. So pick a basis of the $\mathbb R$-vector space $X_0$ spanned by $A_+$ whose generators are positive; then $X_0 \otimes_{\mathbb R} \mathbb C \subseteq A$. Extend this to get a real vector space $X_0 \subseteq X$ with $X \otimes_{\mathbb R} \mathbb C = A$. | |
| Jan 29, 2022 at 8:40 | comment | added | Uri Bader | @TimCampion I am glad that you like it. I think this should be better known. In your last argument, I suppose what you mean by $a^*a>0$, is $\exists \epsilon>0,~a^*a\geq \epsilon$, right? How do you use Hahn Banach without pre-assuming the existence of a norm? | |
| Jan 29, 2022 at 1:05 | comment | added | Tim Campion | I'm having a lot of fun with this! Here is an alternate argument, which unfortunately leans into the hypotheses a bit earlier on. Let $A$ be a $\ast$-algebra over $\mathbb C$, and suppose that $-1$ is not in the positive cone, and that for every $a \in A$, $0 < a^\ast a \leq \alpha$ for some $\alpha \in \mathbb R$. Then by the Hahn-Banach theorem, there exists a state $\rho$ on $A$. By the GNS construction, $A$ now embeds into a $C^\ast$-algebra constructed from $\rho$. (Compare Andre's point below that the "correct" definition of a $C^\ast$ algebra is about acting on a Hilbert space.) | |
| Jan 28, 2022 at 21:54 | comment | added | Tim Campion | If I'm not mistaken, one can define the $C^\ast$ uniformity directly in terms of the postive cone without mentioning the norm at all. Then one can simply say: a $\ast$-algebra has a natural (possibly non-separated) uniform structure, and it is a $C^\ast$-algebra if and only if it is complete (and in particular separated) with respect to this uniformity. Of course, one will probably still talk about the associated norm when proving the equivalence. (Also, thanks for including the condensed version of Cimpric's argument in your paper, that was very helpful!) | |
| Jan 28, 2022 at 21:48 | comment | added | Tim Campion | Thanks, this is even better than I hoped! So basically, the natural cone one gets from the $\ast$ structure seems to work better than the spectral radius formula -- one need only assume that $-1$ is not nonnegative (which I think would mean the cone is the whole algebra? and anyway it's a natural "reality" condition) to get a canonically associated $[0,\infty]$-valued, submultiplicative, power-multiplicative seminorm. Then the $C^\ast$ condition just asks for this intrinsic seminorm to be complete. All very tidy! | |
| Jan 28, 2022 at 21:44 | vote | accept | Tim Campion | ||
| Jan 28, 2022 at 16:15 | history | answered | Uri Bader | CC BY-SA 4.0 |