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Feb 7, 2023 at 11:15 history undeleted Martim Pereir
Feb 7, 2023 at 11:14 history deleted Martim Pereir via Vote
Jan 8, 2022 at 20:44 comment added LSpice 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 $\operatorname{Spin}_n \to \operatorname{GL}_n$.
Jan 8, 2022 at 20:43 comment added LSpice 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).
Jan 8, 2022 at 19:11 history closed Gro-Tsen
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Jan 8, 2022 at 18:42 review Close votes
Jan 8, 2022 at 19:17
Jan 8, 2022 at 18:29 comment added NameNo 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 ...]
Jan 8, 2022 at 18:27 comment added Gro-Tsen 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.
Jan 8, 2022 at 18:27 comment added Jonny Evans 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.
Jan 8, 2022 at 18:21 history asked Martim Pereir CC BY-SA 4.0