Timeline for Injectivity of Fermion algebras
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2018 at 16:57 | comment | added | Mateusz Wasilewski | It might be good to look at the original Bożejko and Speicher papers ("An example of a generalized Brownian motion") and also at the paper written by Bożejko, Kümmerer and Speicher ("q-Gaussian processes: non-commutative and classical aspects"). | |
| Mar 26, 2018 at 1:46 | comment | added | Mathbuff | @Wasilewski. Thank you a lot. For now all of my doubts are cleared. Can you suggest some good references (lecture note) for $q$-Fock spaces? THank you once again. | |
| Mar 24, 2018 at 22:35 | comment | added | Mateusz Wasilewski | Well, if the dimension is $n$, then wedging (taking the exterior product) more than $n$ vectors gives zero, so the exterior algebra is equal to a finite sum of exterior powers. | |
| Mar 24, 2018 at 11:40 | comment | added | Mathbuff | @Wasilewski. Yes. Sorry. I actually mean hyperfiniteness of Fermion algebras only. But why if we start from a finite dimensional Hilbert space, then the antisymmetric Fock space (the exterior algebra) is finite dimensional? | |
| Mar 24, 2018 at 11:27 | comment | added | Mateusz Wasilewski | I don't know what you mean by hyperfiniteness of the antisymmetric Fock space but, anyway, the Fermion algebra is certainly hyperfinite. Note that if you start from a finite dimensional Hilbert space, then the antisymmetric Fock space (the exterior algebra) is finite dimensional, hence the algebra has to be finite dimensional. In general case you get an inductive limit of finite dimensional subalgebras. | |
| Mar 24, 2018 at 6:05 | history | asked | Mathbuff | CC BY-SA 3.0 |