To my knowledge, there are still 3 open conjectures concerning simple theories : elimination of hyperimaginaries, Lascar strong type = strong type, and stable forking conjecture (if a type forks over some parameters, this is due to an instance of a stable formula). I haven't think of it very hard, but intuitively elimination of hyperimaginaries implies Lstrong type = strong type, isn't it ? Are there more implications between those 3 conjectures ? On the other hand, do you think the resolution of one of them would be an important breakthrough in model theory ? Or at the contrary would be rather anecdotal ?
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2$\begingroup$ I think that your question would have a better reception if you attempted to write it grammatically without unnecessary and unexplained abbreviations, and with some background for those who might be interested without being experts. Have a look at other questions on the site to get an idea of community standards. Is “srt” short for “srtong”? I have no idea what you mean by “quite anecdotal”. $\endgroup$– Jeremy RickardCommented Apr 22, 2019 at 13:54
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$\begingroup$ rather anecdotal = not of great importance $\endgroup$– huurdCommented Apr 22, 2019 at 14:07
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2$\begingroup$ Just so you know, this is not what "anecdotal" means in English. $\endgroup$– Alex KruckmanCommented Apr 22, 2019 at 18:26
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1$\begingroup$ Interestingly the main online French-English dictionary seem to propose anecdotal as translation for anecdotique, which indeed means "not of great importance" (worth an anecdote), while "anecdotal" seems to mean "based on hearsay rather than facts" (based on anecdotes). $\endgroup$– YCorCommented Apr 23, 2019 at 7:21
1 Answer
You can find a nice summary of the relationships between (variants of) these three conjectures in this paper by Palacín and Wagner.
In particular:
- A theory with stable forking has weak elimination of hyperimaginaries.
- A one-based simple theory with weak elimination of hyperimaginaries has (strong) stable forking.
- If $T$ is $G$-compact (and every simple theory is $G$-compact), then $T$ has elimination of hyperimaginaries if and only if (a) $T$ has weak elimination of hyperimaginaries and (b) Lstp = stp.
So it seems that for simple theories we have:
- Elimination of hyperimaginaries implies Lstp = stp.
- Elimination of hyperimaginaries and one-based implies stable forking.
- Lstp = stp and stable forking implies elimination of hyperimaginaries.
Caveat: I'm not an expert on these issues, and it's possible I've made a silly mistake in collecting these results.
Do you think the resolution of one of them would be an important breakthrough in model theory?
It would certainly be a breakthrough, since these questions seem very hard and have been open for a long time. In my opinion, how important a breakthrough is really depends on the proof, i.e. whether useful new techniques or conceptual frameworks are developed that significantly increase our understanding. It seems likely to me that progress on these conjectures would require these kinds of new ideas - but you never know.
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$\begingroup$ By the way, what are the known examples of simple non-supersimple theories ? $\endgroup$– huurdCommented Apr 22, 2019 at 18:07
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$\begingroup$ A good reference for examples in model theory is forkinganddividing.com. Note that simple non-supersimple includes stable non-superstable, and there are many such examples. "Natural" strictly simple examples are harder to come by, but you can cook up lots of examples by combining supersimple theories with strictly stable theories in various ways. $\endgroup$ Commented Apr 22, 2019 at 18:15
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$\begingroup$ Oh sorry I meant "simple, non supersimple, non stable" of course... do you have any reference talking about "natural" ones (coming from algebra or other branches of mathematics) ? $\endgroup$– huurdCommented Apr 22, 2019 at 18:19
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$\begingroup$ Off the top of my head, I don't know any "natural" examples other than the ones on the page I linked to in my first comment: $\mathbb{Q}$ACFA and imperfect bounded PAC fields. $\endgroup$ Commented Apr 22, 2019 at 18:25