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For $n>2,$ the spin group Spin$(n)$ is isomorphic to the universal cover of SO$(n)$. In these cases, the universal covers can be explicitly constructed using the Clifford algebra over $\mathbb R^n$. Spin$(2)$ is defined using the Clifford algebra construction, which yields a double-cover of SO$(2)$ [as it does for all SO$(n)$], but this is not the universal cover of SO$(2)$, namely $\mathbb R$. Is there a good reason to define Spin(2) as this double cover, rather than the universal cover? Or if that seems too subjective, then: Is there any special significance to this double-cover of SO(2), say, in representation theory, Lie theory, or topology?

I ask because the universal cover seems more significant to me, although I am somewhat of a novice in this area. But in physics, we care especially about projective representations of Lie groups, which are related (by several results, e.g., Bargmann's theorem) to the universal cover, not necessarily the double cover.

Edit. A related question and generalization (see here) is to include orthogonal groups of indefinite signature in this discussion.

WillG
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    For $n>2$, the universal cover is the double cover, so taking the double cover means there is a consistent algebraic construction for all $n$. Also, this means we will always be compact, which the universal cover of $SO(2)$ is not. – Aaron Dec 05 '22 at 23:16
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    Maybe this is relevant. There is a canonical linear isomorphism between exterior algebra and any Clifford algebra (in characteristic not 2), so we can use that language when talking about Clifford algebras. Speaking just about Euclidean space $\mathbb R^n$, every element of $\mathrm{Spin}(n)\subset\mathrm{Cl}_n(\mathbb R)$ is of the form $e^B$ where $B$ is a bivector, and conversely every bivector gives an element of $\mathrm{Spin}(n)$. The bivectors represent the Lie algebra of $\mathrm{SO}(n)$ using the Clifford product commutator $[B,B']=BB'-B'B$. This is true for all $n$. – Nicholas Todoroff Dec 06 '22 at 00:22
  • The most important thing about the spin groups is the 2-to-1 map down to $SO(n)$, so the double adjective is more important than the universal one (though as you point out, for $n>2$ there isn't any difference at all). – Randall Nov 09 '23 at 15:39
  • @Randall Important how/why? Naïvely, I would think the universal cover is more important, because we care about projective reps of SO$(n)$, and there are theorems showing that these descend from honest reps of the universal cover (I think). – WillG Nov 09 '23 at 15:55
  • $\text{Spin}(2)$ is an $S^1$... – Bob Dobbs Nov 11 '23 at 19:24

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This is what you need to make sense of Dirac operators; this is what you need to see the relation to theta characteristics. These classical topics are more important than the sense that Spin should be a universal cover. (Take it as a mantra: the point is Dirac operators.)

user1128664
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Dec 08 '22 at 13:43
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    The information as written here is helpful and sufficient; I'm not going to add a "let me Google that for you" link, nor am I going to write a short textbook. – user1128664 Dec 08 '22 at 13:58