Timeline for tr(ab)=tr(ba), part 2.
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2011 at 20:32 | vote | accept | Bill Johnson | ||
| Oct 23, 2011 at 19:24 | answer | added | Bill Johnson | timeline score: 21 | |
| Oct 6, 2011 at 18:44 | comment | added | Bill Johnson |
That the trace of the zero operator is not well defined on a space that fails the approximation property. That is, if $X$ fails the approximation property, then there are a sequence $f_n$ in $X^*$ and $x_n$ in $X$ s.t. $\sup \|f_n\| <\infty$, $\sum \|x_n\| = 1$, $\sum f_n(x_n) = 1$, yet $\sum f_n(x) x_n = 0$ for all $x$ in $X$.
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| Oct 6, 2011 at 18:15 | comment | added | Mariano Suárez-Álvarez | @Bill, where in Grothendieck's Memoirs were your suspicions of mysteriousness of the trace confirmed? | |
| Oct 6, 2011 at 17:24 | history | edited | Bill Johnson |
Added Tag
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| Sep 29, 2011 at 20:22 | comment | added | Mark Meckes | Yes, motivation is an issue, but I think no more so than with the usual definition. | |
| Sep 29, 2011 at 18:53 | comment | added | Bill Johnson | But introducing a seemingly non linear functional to a linear algebra class is a bit strange. | |
| Sep 29, 2011 at 18:33 | comment | added | Mark Meckes | Fair enough, but I still like it better than the usual elementary definition in terms of matrices. Axler's definition is elementary enough for an undergrad linear algebra class, it is manifestly basis-independent, it's easy to believe from the definition that it's an important and useful quantity, and it generalizes (albeit with work) to Hilbert spaces. I'm not sure I know any other definition meets all those criteria. Of course, if you'd rather drop any of those criteria in favor of getting a definition that generalizes better to Banach spaces, that's understandable. | |
| Sep 29, 2011 at 14:56 | comment | added | Bill Johnson | Hmmm. Sheldon's definition is not a good one because Lidskii is badly wrong in infinite dimensions except for certain spaces that are very close to Hilbert spaces. | |
| Sep 29, 2011 at 14:02 | comment | added | Bill Johnson | Thanks, Matt. Would you please send me that page from your thesis? | |
| Sep 29, 2011 at 13:47 | comment | added | Mark Meckes | @Bill: the flawed proof in the link is of course the elementary one I had in mind for finite dimensions. But I agree with you that the trace is more mysterious than is often recognized. Are you familiar with the textbook "Linear Algebra Done Right" by Axler? He defines the trace of an operator on a finite-dimensional space to be the sum of its eigenvalues, and only after that proves that it can be calculated by summing diagonal matrix entries. | |
| Sep 29, 2011 at 12:41 | comment | added | Matthew Daws | I thought about similar things in my thesis-- the only reference I could find was M. Grosser, "The trace of certain commutators", Rev. Roumaine Math. Pures Appl. 34 (1989), no. 5, 413–418. ams.org/mathscinet-getitem?mr=1021948 This imposes strong conditions on $a$ and $b$ (but the proof is much too complicated, I give an easy, half-page proof in my thesis). So I suspect that the general case might be hard... | |
| Sep 29, 2011 at 2:48 | comment | added | Bill Johnson |
I tried to use approximation properties to reduce to the finite dimensional case, but was not successful. It would be enough to prove that is $ab$ is nuclear and $T_n$ is bounded and tends strongly to the identity on the range space of $b$ and $$T_n$ tends strongly to the identity on the range space of $b^*$, then $tr(aT_n b)$ converges to $tr(ab)$.
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| Sep 29, 2011 at 2:48 | comment | added | Bill Johnson | @Mark Meckes: As a student I found the result as well as the fact that the trace is well defined mysterious. Other students laughed at me, but later, after reading Grothendieck's Memoirs, I realized that I was right. As for a simple proof in finite dimensions, the flawed proof in the link works and is not too bad. | |
| Sep 28, 2011 at 15:26 | comment | added | Mark Meckes | If finite-dimensionality is an allowable approximation property, then I know a very elementary proof... | |
| Sep 28, 2011 at 12:20 | history | edited | Bill Johnson | CC BY-SA 3.0 |
9/28 Fixed punctuation.
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| Sep 28, 2011 at 6:14 | history | asked | Bill Johnson | CC BY-SA 3.0 |