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S Feb 6, 2022 at 23:07 history bounty ended CommunityBot
S Feb 6, 2022 at 23:07 history notice removed CommunityBot
Jan 30, 2022 at 5:50 comment added Noah Schweber Put another way, I'm interested in the (hyper)class $\{\mathbb{G}(\mathfrak{A}): \mathfrak{A}$ is an algebra$\}$, and this class is richer when we do require variables to appear.
Jan 30, 2022 at 5:48 comment added Noah Schweber @TimCampion Yes, $y$ appears in that term (and terms representing equal functions aren't equal), but that's a different algebra $\mathfrak{A}'$ with a different associated collection of groups $\mathbb{G}(\mathfrak{A}')$. To see that requiring variables to appear actually makes a difference, note that if we dropped that requirement we would always have every $\mathbb{G}(\mathfrak{A})$ be closed under direct products with $S_n$s, which isn't the case when we do require variables to appear. Basically, requiring variables to appear gives more possibilities for the $\mathbb{G}(\mathfrak{A})$s.
Jan 30, 2022 at 5:45 comment added Tim Campion If $\mathfrak A$ is an algebra, I can enlarge the language of $\mathfrak A$ by a binary function $f$ with the equation $f(x,y) = x$. If I take my term to be $t(x,y) = f(x,y)$, then is $y$ considered to "appear" in this term? If so, then I think the "variable appearance" condition is not very strong (when we have the flexibility to choose our language to include things like $f$). If not, then I don't think I understand the "variable appearance" condition. Basically -- are terms considered equal if they are represent the same function in $\mathfrak A$?
Jan 30, 2022 at 5:37 history edited Noah Schweber CC BY-SA 4.0
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Jan 30, 2022 at 5:35 comment added Noah Schweber @TimCampion Yes, but "algebra" is shorter and standardly used in this sort of context.
Jan 30, 2022 at 5:34 comment added Tim Campion Ah, I see. So you could equally have said "structure in a functional language" everywhere rather than "algebra".
Jan 30, 2022 at 5:32 comment added Noah Schweber @TimCampion I don't talk about varieties anywhere; everything is about individual algebras, not collections of algebras. As to why I'm looking at algebras as opposed to arbitrary first-order structures, they just seem particularly interesting.
Jan 30, 2022 at 5:30 comment added Tim Campion What's the significance of restricting to varieties, rather than (say) arbitrary first-order theories? Is it meant to make the problem more tractable?
Jan 30, 2022 at 5:20 history edited Noah Schweber CC BY-SA 4.0
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Jan 30, 2022 at 5:12 history edited Noah Schweber CC BY-SA 4.0
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Jan 30, 2022 at 0:53 history edited Noah Schweber CC BY-SA 4.0
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S Jan 29, 2022 at 22:06 history bounty started Noah Schweber
S Jan 29, 2022 at 22:06 history notice added Noah Schweber Draw attention
Jan 29, 2022 at 22:06 history edited Noah Schweber CC BY-SA 4.0
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S Jan 6, 2022 at 23:04 history bounty ended CommunityBot
S Jan 6, 2022 at 23:04 history notice removed CommunityBot
Dec 30, 2021 at 6:09 history edited Noah Schweber CC BY-SA 4.0
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S Dec 29, 2021 at 21:23 history bounty started Noah Schweber
S Dec 29, 2021 at 21:23 history notice added Noah Schweber Draw attention
Dec 27, 2021 at 4:26 history became hot network question
Dec 27, 2021 at 4:15 history edited Noah Schweber CC BY-SA 4.0
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Dec 27, 2021 at 2:24 comment added Noah Schweber @YCor I would also be interested in that version of the question, but right now at least I'm primarily thinking about the groups on their own. (And besides, a coarser invariant is also easier to fully analyze.)
Dec 27, 2021 at 0:54 comment added YCor Why asking about isomorphism type as groups, rather than isomorphism type as permutation groups (which sounds more meaningful — I guess this is what's meant by "isomorphism types of subgroups of symmetric group" in Keith Kearnes' answer?
Dec 26, 2021 at 23:25 history rollback Noah Schweber
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Dec 26, 2021 at 23:23 history edited Noah Schweber CC BY-SA 4.0
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Dec 26, 2021 at 22:54 answer added Keith Kearnes timeline score: 13
Dec 26, 2021 at 20:19 history asked Noah Schweber CC BY-SA 4.0