Is there a ring structure on $S^1=\mathbb{R}/\mathbb{Z}$?
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12$\begingroup$ Yes, but there is not one that interacts nicely with most of the other structure that is already present on the set S^1. $\endgroup$– Tyler LawsonCommented Oct 9, 2022 at 13:21
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3$\begingroup$ I’m voting to close this question because it is one of a sequence where the OP speculates without demonstrating sufficient prior thought. $\endgroup$– Yemon ChoiCommented Oct 9, 2022 at 14:54
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4$\begingroup$ If $X$ is a path-connected topological space carrying a structure of unital ring, then $X$ is contractible (indeed if $a_t$ is a path from $0$ to $1$ then $x\mapsto a_tx$ is a homotopy from a constant to the identity). $\endgroup$– YCorCommented Oct 9, 2022 at 16:34
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$\begingroup$ Dear @YemonChoi. I was not aware that asking questions without demonstrating sufficient prior thought was necessary, since I have found many popular questions on math overflow where no prior thought is demonstrated (for example my question two weeks ago on quadratic forms). I think it was very unfortunate that this question was closed since I would like to see the discussions below unfold way more. The discussion below consists of top mathematicians disagreeing on elementary and important questions, and I am very confused about who is right or not and what to conclude. Best regards, $\endgroup$– Ola SandeCommented Oct 10, 2022 at 8:04
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3$\begingroup$ @OlaSande There is nothing confusing or contradictory here. Your question has been interpreted in two ways: if you are looking for a topological ring with additive structure $S^1$ this is not possible because compact rings are profinite and $S^1$ is not profinite. YCor's reply gives some indication how to prove this in the special case you are considering. If you ignore the natural topology on $S^1$ then there are compatible ring structures and there are many ways to construct them, for example see Goodwillie's comment. Non-unital rings can be called algebras but should not be called rings. $\endgroup$– Peter KrophollerCommented Oct 10, 2022 at 9:14
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2 Answers
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Let's say that a ring is an abelian group with an additional $\mathbf{Z}$-bilinear law.
Listing ring laws on the circle: i.e., describe all continuous $\mathbf{Z}$-bilinear maps $S^1\times S^1\to S^1$. The result is that the only possibility is the zero law.
This is the same as classifying all continuous group homomorphisms $S^1\to \mathrm{Hom}_{\mathrm{TopGrp}}(S^1,S^1)=\mathbf{Z}$. And by compactness there is no nonconstant such continuous homomorphism (well, there is no nonzero homomorphism at all). Hence the only continuous ring (without associativity or unital assumption) law on the circle group is the zero law. In particular there's no unital ring law.
(This works for any ring law $(x,y)\mapsto xy$ assuming only that for each $x$, the map $y\mapsto xy$ is continuous.)
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No. It's compact and compact rings are profinite.
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$\begingroup$ Isn't $\mathbb{R}/\mathbb{Z}$ it a non-unital ring? (I didn't know under what comment I should ask). $\endgroup$ Commented Oct 9, 2022 at 13:23
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3$\begingroup$ I guess any additive (topological) group can be a non-unital ring simply by defining the multiplication to be consistently zero. $\endgroup$ Commented Oct 9, 2022 at 13:25
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4$\begingroup$ As an abelian group it is isomorphic to the product of $\mathbb R$ with something, so it certainly has a unital ring structure compatible with addition. But maybe you wanted it to be compatible with the topology, too. $\endgroup$ Commented Oct 9, 2022 at 13:35
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1$\begingroup$ @OlaSande I am sorry this question got closed because there are still a couple more things that need to be said $\endgroup$ Commented Dec 15, 2022 at 19:02
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1$\begingroup$ @OlaSande First, to a non-mathematician, the circle $S^1$ looks very like a ring insofar as the words \emph{circle} and \emph{ring} might conjure up geometric shapes. So I can't help momentarily imagining that the person who raises this question also notices, with a wry smile, the non-mathematical rather obvious answer 'yes it is'. That raises serious points about the way we use words from everyday language and employ them with strict new definitions. $\endgroup$ Commented Dec 17, 2022 at 21:57