Timeline for Hattori-Stallings trace
Current License: CC BY-SA 4.0
8 events
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| Nov 15, 2023 at 23:14 | comment | added | Benjamin Steinberg | One can then show that if you take the corresponding matrix trace of an element of $EM_n(R)E$, then it gives the Hattori-Stallings trace of the endomorphism and it is independent of the choice of $E$ (again using that $Tr(AB)=Tr(BA)$ for not necessarily square matrices). | |
| Nov 15, 2023 at 23:13 | comment | added | Benjamin Steinberg | There is also a pedestrian way to do the Hattori-Stallings trace. For a square matrix over $R$, one can define the trace as the image in $R/[R,R]$ of the sum of the diagonal entries and then the usual proof shows $Tr(AB )=Tr(BA)$ for not necessarily square matrices. This implies in particular it doesn't depend on the basis and one can check also that it gives the same answer for free modules as you defined above. If $P$ is a finitely generated projective, one can find a square idempotent matrix $E\in M_n(R)$ with $P$ isomorphic to the image of $E$ and with $EM_n(R)E\cong End_R(P)$. (Ctd) | |
| Nov 15, 2023 at 19:37 | comment | added | Michael Hardy |
\mathrm{Tr} and \operatorname{Tr} don't always have the same effect, and in this case in particular the use of the former was the cause of a lack of proper horizontal spacing.
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| Nov 15, 2023 at 19:35 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 12 characters in body
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| Nov 15, 2023 at 18:57 | history | became hot network question | |||
| Nov 15, 2023 at 14:42 | vote | accept | Qwert Otto | ||
| Nov 15, 2023 at 10:56 | answer | added | Maxime Ramzi | timeline score: 6 | |
| Nov 15, 2023 at 9:12 | history | asked | Qwert Otto | CC BY-SA 4.0 |