As is standard knowledge - there is a bijective correspondence between compact simply connected simple Lie groups and complex simple Lie algebras. So what is the The Lie group corresponding to the Lie algebra $\frak{so}_n$? Is it the special orthogonal group $SO_n$ or is it the spin group $\mathrm{Spin}_n$? Since the special orthogonal group is not simply-connected, the corresponding Lie group should be $\mathrm{Spin}_n$. But it seems weird to me that the notations do not match . . . I am afraid that I missing something here.
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$\begingroup$ SO(n) is not simply connected; Spin(n) is its simply connected double cover. Spin(n) corresponds to so(n) under the bijection you mentioned. $\endgroup$– Jonny EvansCommented Jan 8, 2022 at 18:27
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4$\begingroup$ It is the spin group. The special orthogonal group has the same Lie algebra, but it is not simply connected. The notation is indeed less than ideal, but given that there are several possible Lie groups for one Lie algebra, a choice has to be made, and the special orthogonal group appears more frequently. $\endgroup$– Gro-TsenCommented Jan 8, 2022 at 18:27
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$\begingroup$ The correspondence (actually, a categorical equivalence) works also with "local lie groups", which always have at least one (really many) extension to global Lie groups. So it works for equivalence classes of (connected) Lie groups with the relation of local isomorphism (they are the quotient of their simple connected covering by a discrete normal subgroup) [Edit: well I see, too many at the same time ...] $\endgroup$– NameNoCommented Jan 8, 2022 at 18:29
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1$\begingroup$ Just to demonstrate that there's nothing especially pernicious going on with $\mathfrak{so}_n$, which equals (= is canonically isomorphic to) $\mathfrak{spin}_n$, we may also want to notice that, say, $\mathfrak{pgl}_n = \operatorname{Lie}(\operatorname{PGL}_n)$ equals $\mathfrak{sl}_n = \operatorname{Lie}(\operatorname{SL}_n)$ (at least, in the sense that the derivative $\mathfrak{sl}_n \to \mathfrak{pgl}_n$ of $\operatorname{SL}_n \to \operatorname{PGL}_n$ is an isomorphism). $\endgroup$– LSpiceCommented Jan 8, 2022 at 20:43
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2$\begingroup$ The reason that we call this Lie algebra $\mathfrak{so}_n$, even though it equals (= is canonically isomorphic to) $\mathfrak{spin}_n$, is probably that the natural $n$-dimensional representation $\mathfrak{spin}_n = \mathfrak{so}_n \to \mathfrak{gl}_n$ is the derivative of an $n$-dimensional representation $\operatorname{SO}_n \to \operatorname{GL}_n$, but not of any representation $\operatorname{Spin}_n \to \operatorname{GL}_n$. $\endgroup$– LSpiceCommented Jan 8, 2022 at 20:44
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