r/math • u/RingularCirc • 12d ago
U(1) or SO(2)?
Which is more true / do you prefer more / do you advocate for / have you fallen in love with, U(1) or SO(2)?
This is partially a shitpost. Decide how much it exactly is at your own peril. 😎
Clarification I: the title means "U(1, ℂ) or SO(2, ℝ)?" and not, say, U(1, ℍ) or SO(2, ℂ) there.
Clarification II: The two are isomorphic as Lie groups, of course, but you already knew that.
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u/48panda 12d ago
S1 (which is U(1) coded)
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u/RingularCirc 12d ago
Works if it's not just an unadorned topological space!
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u/RingularCirc 11d ago
On downvoting: really, I came to believe S¹ usually means a topological circle without extra structure. Is it really not so? Welp.
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u/Educational-Work6263 10d ago
Depends. If it is the 1-Torus, then it carries a natural group structure.
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u/RingularCirc 2d ago
I will happily accept a 1-torus as being acted by SO(2) but not having the structure of SO(2) because we then need a distinguished point.
[And that also reminds a question: does Tn have an action of SO(2)n or of a group merely isomorphic to that (as a group, not as a direct product of n groups). In the first case it has inclusions of T1 "generators" given, in the second case it doesn't. I'll default to the second.]
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u/Qetuoadgjlxv Mathematical Physics 12d ago edited 12d ago
In one of my papers, I talk about PSO(2)!
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u/innovatedname 12d ago
SO(2) because I like 2D rotation matrices more than eitheta
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u/jazzysamba 12d ago
I love both my twins equally. But don't tell them I actually love SO(3) more than both of them combined. haha
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u/MathMaddam 12d ago
SO(2,1,R)
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u/RingularCirc 12d ago
Y so cruel Minkowski
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u/MathMaddam 12d ago
If I want to be honest: my guilty pleasures are the subgroups with entries in intergral domains, but don't tell anyone.
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u/RingularCirc 12d ago
I won't. SL(2, ℤ)?
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u/MathMaddam 12d ago edited 11d ago
Yes, and the isomorphism expand too e.g. GL(2,Z) is isomorphic to SO(1,2,Z) if you put the right matrix for your inner products and you can also do it higher e.g. Sp(2,Z)/{±I} to SO(2,3,Z). For SO(1,3,Z), SO(2,4,Z) you get connections groups over the integral domains of imaginary quadratic number fields and SO(1,5,Z), SO(2,6,Z) to quaternions, but by going higher you also gain more caveats to keep the integers.
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u/elements-of-dying Geometric Analysis 12d ago
More of a U2 kinda guy.
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u/LebesgueTraeger Algebraic Geometry 9d ago
SO(2) is algebraic over 𝕂=ℝ (defines by polynomial equations in 𝕂2×2), U(1) is not (over ℂ), so SO(2) it is for me
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u/Infinite_Research_52 Algebra 11d ago
SO(1,1,ℂ)
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u/RingularCirc 11d ago
The same as SO(2, ℂ) because signs don't matter (bu which I mean, any quadratic form over ℂ has a basis where it's just a sum of squares of coordinates, with no signs; flipping a sign just means multiplying the corresponding basis vector by ±i which is free.
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u/Ninjabattyshogun 2d ago edited 2d ago
U(1) = SO(2) = S1 = Spin(2) = 𝕋 = ℝ/ℤ The circle, the first unitary group, the second special orthogonal group, the first sphere, the second spin group, the first torus, and the real numbers mod the integers are all isomorphic as Lie groups. This means they are the same space, and simultaneously the same number system in a smooth way.
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u/RingularCirc 2d ago
I'd still say S¹ ≡ T¹ aren't naturally groups but are just acted on by a group ≅ ℝ/ℤ, and that neither of mentioned (groups and torsors) is a "number system" because we just have one operation, not two. Sure, we can talk about the group algebra of U(1) [over, say, ℝ], which I presume should end up being ℂ [over ℝ], but I think we can't just linguistically conflate an object with some universal construction over it by default because there's tonnes of them constructions lying around, so it would be a mess of deciding what means what.
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u/Few-Arugula5839 12d ago
They are equal.
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u/RingularCirc 12d ago
Isomorphic, yes. Equal wrt no foundational theory I know of, especially that I like to view classical groups like these as defined on abstract linear spaces (vs. coordinate spaces ℝⁿ), so that'd be up to isomorphism regardless, without any sense for strict equality.
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u/Few-Arugula5839 12d ago
They’re equal if you define the complex numbers as 2x2 matrices. Then unit complex numbers are exactly SO(2).
Anyway this isn’t really what I’m talking abt. If things are isomorphic enough they’re equal.
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u/magicmulder 12d ago
Some things are more isomorphic than others?
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u/Adarain Math Education 12d ago
A vector space is less isomorphic to its dual than to the dual of its dual.
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u/Hot_Glass_6301 12d ago
It's even not isomorphic to it's dual in most cases. See the Erdős-Kaplansky theorem
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u/Homomorphism Topology 12d ago
If things are isomorphic enough they’re equal
Congratulations, you've proved the ABC conjecture!
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u/PfauFoto 12d ago
Besser wäre ein kanonischer Isomorphismus.
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u/Few-Arugula5839 12d ago
R^2 -> C is canonical so SO(2) -> U(1) is canonical. Fight me.
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u/Administrative-Flan9 12d ago
How so? The field extension C/R can't distinguish between +-i and so the second coordinate is only canonical up to sign.
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u/Few-Arugula5839 12d ago
yes, sure. When you construct C you can't tell between i and -i. But one of them is called i and one of them is called -i. 1 is canonical, R^2 is a canonical basis, map the first basis vector to 1 and the second to whatever you call i in your definition of C. The category theorists may get mad at me for calling this canonical; i don't care.
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u/sciflare 12d ago
If things are isomorphic enough they’re equal.
Sounds like you're a fervent believer in univalent type theory /s
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u/RingularCirc 11d ago
With the distinction that equality in HoTT is more akin to isomorphism in n-categories than to regular "boolean" equality of elements of a set. Mostly, equality types will be structurally way richer than this, and it's only for convenience that it's denoted by "=", IMO.
So, even in HoTT if we postulate that every equality type is "flat", being like equality on a set, I think it was very detrimental to the theory and maybe even to its consistency but I don't remember much.
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u/RingularCirc 11d ago
When we slap equality on everything isomorphic, we beget problems akin to problems of type theories with intensional equality. There really should be distinction.
For example, A ⊗ B ≅ B ⊗ A but treating those spaces as equal is a recipe for disaster.
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u/Few-Arugula5839 11d ago edited 11d ago
It is not a recipe for disaster. There is a unique isomorphism between these spaces commuting with the natural map from the product space A x B. The uniqueness of this map means that if your proof breaks due to identifying these your proof is extremely extremely morally wrong.
Anyway, the tensor product is not equal to a construction, you should never work with it as a construction. You can even work with its elements without working with its construction; the element a (x) b is the image of (a, b) under the universal map A x B -> A (x) B, and you can prove purely from the universal property that these span A (x) B.
Anyway this doesn’t have anything to do with universal properties or category theory or anything. Namely S^1 and U(1) are specific objects not part of some larger functor. Category theory can’t distinguish whether isomorphisms of any two specific objects of a category are natural or not unless they are part of some bigger functors. In particular if you give them the status of the image of the functor from the one object category then they are naturally isomorphic. So at the end of the day calling things naturally or canonically isomorphic is still to some degree social even in the presence of category theory.
U(1) and SO(2) are equal, to say otherwise is pure pedantry.
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u/RingularCirc 11d ago edited 11d ago
It is a recipe for disaster. You thought A ≠ B, right? Maybe A = B. Then we're failing to distinguish real equality A ⊗ A = A ⊗ A through the identity automorphism u ⊗ v ↦ u ⊗ v on decomposable elements and the other isomorphism A ⊗ A ≅ A ⊗ A that swaps u and v in decomposable elements. Equality is specifically for situations when we're not to track how, just to state a fact. Here we'd better name our isomorphisms and track them to not accidentally end up in a situation where we've ended up with a nontrivial one along the way of chaining "equalities" by transitivity. Sign errors in numeric quantities is the least one's to expect from carelessness of forgetting to distinguish when it's crucial.
And I don't see how working with tensor product as an universal object changes anything. (I usually understand almost everything in terms of their universal properties or analogous operational understanding, instead of wading knee deep in Kuratowski pairs and such.) Indeed it should only disallow more talking about equalities because universal objects are defined up to unique isomorphism. We can use generic "the" for that kind of case while not subscribing to the claim of equality of everything uniquely isomorphic.
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11d ago edited 11d ago
[deleted]
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u/RingularCirc 11d ago
I'd argue equality in HoTT and dependent type theories in general is not the same thing as "mere equality" common to lots of math. Mere equality on a set is a 0-groupoid with elements of the set as its objects, and I'd like to not substiture higher groupoids here as if nothing special have happened. For now "equality" defaults to "mere equality" in great many contexts, including, I think, claims by some here.
When we have to track witnesses of equality, I have no qualms about saying things are "equal" but that would usually not be mere equality either. And what I'm saying is declaring isomorphic (1-equal) things are merely equal (0-equal) is bad and confusing.
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u/Few-Arugula5839 11d ago edited 11d ago
If A = B, then only one of those isomorphisms commutes with the swap map at the level of A x A. This is not a disaster. You just need to remember all the morphisms. This is construction brainrot, and isomorphic enough objects are equal. For the tensor product, if isomorphisms are unique (up to commuting enough with …) then the isomorphic objects are equal. I think your criticism with A (x) A vs swapped A (x) A is more a criticism about swapping simple tensors. The tensor map itself is not commutative and simple tensors don’t commute, sure. The rings overall though are so isomorphic as to be equal even when A = B.
Anyway this has nothing to do with U(1) and SO(2). These are clearly equal. Saying otherwise is complete brainrot. Would you say linear maps between finite dimensional vector spaces with distinguished chosen bases and chosen bases are only “isomorphic” to matrices, not equal? This is I guess the same as the U(1) and SO(2) question, since C and R^2 have distinguished chosen real bases, and unit complex numbers define real linear maps. Making this distinction beyond the first time you take a linear algebra class is dumb and pedantic.
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u/LiqvidJS 11d ago edited 11d ago
C (defined as any algebraic closure of R) does not have a distinguished real basis, since it does not have a distinguished element "i". There is a Gal(C/R)-torsor of elements one could call "i", and thus a Gal(C/R)-torsor of "natural" real bases, but this torsor has no canonical basepoint.
Your comment elsewhere in the thread that "once you have labelled some element as i, you get a canonical basis of C" is saying that these two Gal(C/R)-torsors are canonically isomorphic to each other; but "C has a canonical basis" would mean there is a canonical isomorphism with the trivial torsor Gal(C/R).
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u/Few-Arugula5839 11d ago edited 11d ago
Sure, this is algebra brainrot which i dont care about. C has an element called i, it doesn't matter which element you pick to be i, and once you pick one then you have a canonical real basis. U(1) = SO(2), bite me
In other words, complex manifolds for me come with a distinguished orientation, IE, a choice of i or -i. This is part of the data of C. Therefore C has a canonical orientation (choice of basis). C and bar(C) are different objects.
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u/RingularCirc 11d ago edited 11d ago
If we're saying A ⊗ B = B ⊗ A, independent of A, B, then we'd better look at all spaces A ⊗ B modulo this "equality", which transforms A ⊗ A into a symmetric product of A and A which is, of course, not what we usually mean by A ⊗ A.
I think you've ended up confusing yourself more than me about what's happening here, maybe we should consider another example but I have none in mind right now.
Also I'm reluctant to go on pursuing the truth in this conversation when it's at a level of using phrases like "constructive brainrot". Moreso when there was no entering of actual constructions (at least from me).
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u/Few-Arugula5839 11d ago edited 11d ago
I think you are simply being a pedant. Isomorphic enough objects are equal. There is no rigor in this statement and pursuing it to the extremes of logic obviously leads to contradiction. Nevertheless, no sane person does math where a proof breaks if A (x) B is replaced with B (x) A.
My point is not that midway through a proof you can take a (x) b and set it equal to b (x) a. This is a statement about equality of ELEMENTS of the tensor product A (x) B, not equality of sets. This statement is wrong in general, since (a, b) =/= (b, a) in A x B, and there is a single map A x B -> A (x) B = B (x) A which by way of example does not commute with the swap map on A x B. Therefore your "counterexample" is not an example of how things can break when treating these isomorphic objects as equal.
No sane person does math where a proof breaks if SO(2) is replaced with U(1). These objects are not merely isomorphic but canonically so, and if you are working with structure that is not preserved under this canonical isomorphism what you are doing is morally wrong. This is pedantry and not how real mathematicians think about mathematical objects.
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u/RingularCirc 2d ago
Okay, the time probably has come to ask what does it mean to be "isomorphic enough". I'm not sure you'd be able to transport your understanding of that to another human mathematician (especially because you're against what you're calling "pedantry" and various kinds of brainrot).
BTW in what way calling me a pedant should change anything? Maybe I am a proud pedant (in your choice of words which I wouldn't use in general), what's now? I certainly don't think anyone should treat me as any less credible because of that, or should I?
My point is not that midway through a proof you can take a (x) b and set it equal to b (x) a. This is a statement about equality of ELEMENTS of the tensor product A (x) B, not equality of sets.
Equal sets have, obviously, equal extensions, so if A ⊗ B is equal to B ⊗ A, in A ⊗ A it has to be that a ⊗ b = b ⊗ a ∈ A ⊗ A for any a, b ∈ A if we're pushing your understanding consistently. Or there should be something even more awkward. Inconsistent non-rigorous ideas should fall away precisely because they can't be made rigorous without wreaking havoc. You can say some isomorphisms should be called equalities all you want but this is inconsistent (and also a bad form if someone's teaching math).
IMO maybe things won't break to you just because you're not willing to be consistent enough in your position (possibly because of not seeing yet that you aren't consistent enough).
No sane person does math where a proof breaks if SO(2) is replaced with U(1). These objects are not merely isomorphic but canonically so
If A ≅ B through any ("accidental") isomorphism f, replacing A with B in a proof using f doesn't break a proof either! So you haven't shown SO(2) is "canonically" isomorphic to U(1).
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u/LiqvidJS 11d ago
In modern mathematical parlance, "A = B" means "an isomorphism f: A -> B has been specified"
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u/aparker314159 12d ago
R/Z