Timeline for Separability axiom really necessary?
Current License: CC BY-SA 3.0
11 events
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| Jun 15, 2022 at 18:23 | comment | added | Pedro Lauridsen Ribeiro | In due time: Wyss's work actually requires a stronger positivity condition than the one I've imposed in my answer in order to obtain continuity - namely, it requires $\omega$ to take non-negative values on the closed cone generated by Hermitian squares $f^*f$ in $\mathfrak{F}$. The weaker positivity condition $\omega(f^*f)\geq 0$ for all $f\in\mathfrak{F}$ is not enough to carry through to the closure of the cone generated by ( = the set of all sums of) Hermitian squares in $\mathfrak{F}$, only to the latter. Obviously, if $\omega$ is known to be continuous, it does carry through, though. | |
| Dec 15, 2013 at 19:42 | comment | added | Pedro Lauridsen Ribeiro | By the way, multiplication in the Borchers-Uhlmann algebra is separately continuous but jointly continuous only in bounded subsets. Therefore, there is no submultiplicative continuous seminorm in the Borchers-Uhlmann algebra, let alone a continuous C*-seminorm. | |
| Dec 15, 2013 at 19:39 | comment | added | Pedro Lauridsen Ribeiro | Just a small update: I've just found out that positivity does imply continuity for linear functionals if we replace test function spaces $\mathscr{D}$ with compact support by test function spaces $\mathscr{S}$ with rapid decrease in the definition of the Borchers-Uhlmann algebra. This follows from the work of W. Wyss ("The Field algebra and its positive linear functionals". Commun. Math. Phys. 27 (1972) 223-234). I don't know if the result still holds for $\mathscr{D}$ as above, but this shouldn't be hard to check. | |
| Dec 12, 2013 at 19:32 | comment | added | moppio89 | I really didn't expect this, thank you! | |
| Dec 12, 2013 at 16:16 | comment | added | Valter Moretti | Thanks. An interesting open problem maybe. It could be very useful to find a definite answer. Ciao, V. | |
| Dec 12, 2013 at 16:11 | comment | added | Pedro Lauridsen Ribeiro | Ciao Valter! No, in the case of topological $*$-algebras continuity does not follow from positivity, unfortunately. The C$*$-condition on the norm is crucial to obtain continuity from positivity. Maurin assumes continuity as a separate condition. Perhaps a similar result can be obtained if you can find a separating family of C$*$-seminorms on the Borchers-Uhlmann algebra. I don't know if this has ever been done... | |
| Dec 12, 2013 at 16:03 | comment | added | Valter Moretti | Sorry, you assumed it! I did not see. However my question remains. | |
| Dec 12, 2013 at 16:01 | comment | added | Valter Moretti | Pedro (Hi I am Valter!) I did not understand if in your general argument you explicitly assume that $\omega$ is continuous (Wightman does for the vacuum state). For $C^*$-algebras positivity implies continuity, but here you only have a topological $^*$-algebra. Does the result hold anyway? | |
| Dec 12, 2013 at 15:44 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
added reference to David Bar Moshe's answer, typos corrected
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| Dec 12, 2013 at 15:38 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
added 369 characters in body
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| Dec 12, 2013 at 15:23 | history | answered | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |