Timeline for trace(xy)=trace(yx) in full generality
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2011 at 16:32 | comment | added | Romanov | This property is called "a tracial property" so I think if a map does not satisfy this property we cannot call it a trace. I am wondering is it possible to classify this category by the property of its trace (analogy to the type of Von Neumann algebras with a "special" trace i.e. finite, semifinite)? | |
| Jul 8, 2011 at 13:33 | history | edited | Ricky |
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| Jul 7, 2011 at 3:30 | comment | added | Marcel Bischoff | It is also true for Ribbon categories, which are in general not symmetric, but just "braided", see eg. Kassel - Quantum Groups. | |
| Jul 6, 2011 at 22:46 | comment | added | David E Speyer | If $X$ is $m \times n$ and $Y$ is $n \times m$, then $\mathrm{Tr}(XY) = \mathrm{Tr}(YX)$. | |
| Jul 6, 2011 at 22:26 | vote | accept | Ricky | ||
| Jul 6, 2011 at 22:12 | comment | added | Ricky | To Qiaochu: I'm completely new to string diagram, so it is possible that my question becomes trivial in that language. I just don't know. To Micheal: what is the trace of a non square matrix? I've always seen the trace of an endomorphism! | |
| Jul 6, 2011 at 22:05 | answer | added | Todd Trimble | timeline score: 17 | |
| Jul 6, 2011 at 21:56 | comment | added | Michael Hardy | Why specify "square" matrices in particular? The occasion for use of this identity with which I am most familiar depends crucially on the matrices not being square. | |
| Jul 6, 2011 at 21:33 | comment | added | Qiaochu Yuan | Have you tried drawing the appropriate string diagrams? | |
| Jul 6, 2011 at 21:31 | history | asked | Ricky | CC BY-SA 3.0 |