Timeline for Infinite matrices and the concept of "determinant"
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2023 at 2:58 | comment | added | Anixx | But det(2I) obviously should be 2. | |
| Feb 6, 2013 at 18:32 | comment | added | Terry Tao | It should perhaps also be pointed out that in the Hilbert space case, this notion of determinant is known as the Fredholm determinant: en.wikipedia.org/wiki/Fredholm_determinant | |
| Feb 6, 2013 at 15:24 | comment | added | Mikael de la Salle | ... (continued) but the problem is that this map is not injective in general. This map is injective if and only if $X$ has the Approximation Property (AP) ( en.wikipedia.org/wiki/Approximation_property ). Hence it is not possible to define a reasonable notion of trace on the nuclear operators on a Banach space without AP. This difficulty is completely overlooked in Simon's book. By the way, the wikipedia page on AP also contains mistakes, I will have to clean it when I have time if nobody does it before. | |
| Feb 6, 2013 at 15:19 | comment | added | Mikael de la Salle | It is not a good idea to try to define trace class operators on arbitrary Banach spaces, it is better to restrict yourself to Hilbert spaces. Indeed, there is an issue in your claim "We then take the closure in the space of all operators of the space of finite-rank operators with respect to this trace norm." What you can do is take the completion, say $P$ for "projective", of the finite rank operators for the (correctly defined) trace norm. On $P$ there is a well-defined notion of trace. There is also a norm $1$ map $P \to B(X)$, and its images are called the nuclear operators... | |
| Oct 27, 2009 at 18:25 | vote | accept | Gabe Cunningham | ||
| Oct 23, 2009 at 7:48 | history | answered | Andrew Stacey | CC BY-SA 2.5 |