Timeline for Characterizing $\mathbf{R}$ as an ordered group
Current License: CC BY-SA 4.0
18 events
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| Jun 11, 2018 at 19:53 | comment | added | André Henriques | Let $R$ be a factor. Then the collection of all isomorphism classes of $R$-modules under the operation of direct sum is isomorphic to either $(\mathbb N\cup\{\infty\},+)$ (the factor is of type I), or $(\mathbb R_{\ge 0}\cup\{\infty\},+)$ (the factor is of type II), or $(\{0\}\cup\{\infty\},+)$ (the factor is of type III). Note that I'm over-simplifying things a tiny bit by not distinguishing between different sizes (cardinalities) of $\infty$. | |
| Jun 10, 2018 at 21:53 | comment | added | Jochen Glueck | @AndréHenriques: This sounds quite interesting (and quite surprising) to me. Could you elaborate a bit on it (or maybe give a reference where this relation is discussed)? | |
| Jun 10, 2018 at 20:16 | comment | added | André Henriques | Corollary 4 is very closely related to the classification of factors (von Neumann algebras with trivial center) into type I, type II, and type III. | |
| Jun 10, 2018 at 10:32 | history | edited | Jochen Glueck | CC BY-SA 4.0 |
deleted 670 characters in body
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| Jun 10, 2018 at 9:07 | comment | added | Jochen Glueck | @YCor: I was indeed a bit lazy and surpressed the order relation in the notation. Admittedly, this is not particularly good style (even more so as I wrote down the group operation explicitly). Thanks also for your comment about bi-ordered groups. I thought that "ordered group" is usually used synonymously with "bi-ordered" group, but it seems that I was wrong. | |
| Jun 10, 2018 at 8:46 | comment | added | Jochen Glueck | @EmilJeřábek: Oops, you're right, of course. I suspected that there might be a simpler argument than what I was able to come up with... ;-). I'm going to incorporate your argument in the post (giving credit to you, of course), and I'll also clean up the edit mess to make the post better readable for future visitors. | |
| Jun 10, 2018 at 7:23 | comment | added | Emil Jeřábek | @YCor Quite right. The results apply to biordered groups. I'm fairly sure they fail for left-ordered groups. | |
| Jun 10, 2018 at 7:21 | comment | added | Emil Jeřábek | The proof of Theorem 6 is unnecessarily complicated. Let $a>1$. If $\{ a^n:n\in\mathbb N\}$ is not cofinal in $G$, it has a least upper bound $s$. Since $a^{-1}s<s$, we have $a^{-1}s\le a^n$ for some $n$. But then $s\le a^{n+1}<a^{n+2}\le s$, a contradiction. | |
| Jun 9, 2018 at 23:44 | comment | added | YCor | By the way if you don't assume $G$ commutative you'd need to clarify whether "ordered group" means the order is left-invariant or both left and right invariant. Both convention exist, so it's better write left-ordered group and bi-ordered group. | |
| Jun 9, 2018 at 23:42 | comment | added | YCor | What do you denote by $(G,.)$? the dot denotes the pair $(+,\le)$? | |
| Jun 9, 2018 at 22:39 | comment | added | Alec Rhea | Very nice answer, it may be worth mentioning that the embedding theorem attributed to Holder at the beginning is a special case of the Hahn embedding theorem (en.wikipedia.org/wiki/Hahn_embedding_theorem), which offers a nice characterization of all totally ordered commutative groups. | |
| Jun 9, 2018 at 22:13 | history | edited | Jochen Glueck | CC BY-SA 4.0 |
Included the implication "complete => Archimedean" also for non-commutative groups.
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| Jun 9, 2018 at 15:20 | history | edited | András Bátkai | CC BY-SA 4.0 |
added zbl citation
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| Jun 9, 2018 at 14:16 | history | edited | Jochen Glueck | CC BY-SA 4.0 |
Added information on the question whether the Archimedean property is redundant.
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| Jun 9, 2018 at 12:26 | vote | accept | coudy | ||
| Jun 9, 2018 at 11:31 | history | edited | LSpice | CC BY-SA 4.0 |
Operation on H is +, not \cdot
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| Jun 9, 2018 at 11:04 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added Google Books link
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| Jun 9, 2018 at 10:55 | history | answered | Jochen Glueck | CC BY-SA 4.0 |