There may be some short-hand / informal statements that are tripping me up, but I am getting confused trying to understand the relationship between Spin(3), SU(2), SO(3), and the unit quaternions.

Trying to find information online, many discussions seem to say SO(3) and SU(2) are isomorphic (for example wikipedia). Mathworld says SU(2) is isomorphic to $O^+_3(2)$ which I'm not quite sure how that relates to SO(3) (I have not seen that notation before). While others state SU(2) is isomorphic to the unit quaternions which are in turn a double cover of SO(3). Which seems to suggest there is a lot of "short-hand" discussion going on and sometimes people say isomorphic ignoring a double cover? Or maybe I just misunderstand, and a double cover doesn't really matter for some reason?

My best effort of trying to figure out what is going on seems to suggest:

$$Spin(3) \cong SU(2) \cong \{q \in \mathbb{H} | q\bar{q}=1 \} \not \cong SO(3)$$

Is that close, or are even more of those actually double covers?
What is the correct relationship between these groups? (and what does $O^+_3(2)$ denote?)

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    $\begingroup$ many discussions seem to say SO(3) and SU(2) are isomorphic (for example wikipedia Within 5 minutes of your post, I word-searched the entire page for "isom" and confirmed that no such claim is made. Of course, I could have overlooked it written in tex. Where do you think it says this? It clearly says elsewhere that the map is 2:1 from SU(2) to SO(3) (not 1:1). $\endgroup$
    – rschwieb
    Apr 4, 2017 at 20:19
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    $\begingroup$ @rschwieb The link I gave is to the section "Isomorphism with su(2)" on the page "Rotation_group_SO(3)", and the first line seeming to state exactly that. Well, I guess it actually says "The Lie algebras so(3) and su(2) are isomorphic" ... so maybe the algebra can be isomorphic when the groups are not? I apologize for my confusion, but if the answer is that simple please just post it as an answer. $\endgroup$
    – PPenguin
    Apr 4, 2017 at 20:25
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    $\begingroup$ yes, it's a basic fact that two groups with isomorphic Lie algebras do not need to be isomorphic. One way to see the groups are nonisomorphic is that SU(2) is simply connected but SO(3) is not. The Lie algebra only depends on the part of the group in the connected component of the group's identity. I think reading this passage about the conenction between the two will probably clear up most of your confusion. $\endgroup$
    – rschwieb
    Apr 4, 2017 at 20:37
  • $\begingroup$ @rschwieb Thank you! I looked through some of the other pages, and in all I've seen so far it was me being ignorant of lie algebras and jumping to incorrect conclusions. Do you happen to know what the notation $O^+_3(2)$ denotes? $\endgroup$
    – PPenguin
    Apr 4, 2017 at 20:43

1 Answer 1


$Spin(3), SU(2)$, and the unit quaternions $Sp(1)$ are all isomorphic; this Lie group is also sometimes referred to simply as its underlying manifold $S^3$. $SO(3)$ is diffeomorphic to $\mathbb{RP}^3$ and so is not diffeomorphic to $S^3$, although its double cover is $Spin(3)$ (and hence also $SU(2)$ and $Sp(1)$).

One possible source of confusion is that all of the corresponding Lie algebras are isomorphic, and some sources (especially from physics) do not closely distinguish between Lie groups and their Lie algebras.

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    $\begingroup$ It is true, though, that $SU(2)/\{1,-1\} \simeq SO(3)$, where $SU(2)/\{1,-1\}$ is the factor group of $SU(2)$ with respect to the group $\{1,-1\}$ (where the group operation is the usual multiplication). Proof can be found in "A. J. Hanson, Visualizing Quaternions (Elsevier, Amsterdam, 2006)." for example. $\endgroup$ Jun 7, 2017 at 12:46
  • $\begingroup$ I'm interested in seeing the proof of this! Do you happen to know which section this is in? $\endgroup$
    – hwhorf
    Dec 1, 2019 at 21:08

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