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Path connected subgroups of SO(3), please help

  1. Jul 29, 2014 #1
    Hi, I'm giving a talk tomorrow morning, and I'd like to use the following fact: a path-connected subgroup of SO(3) consists of either a) only the identity, b) all the rotations about a single axis, or c) all of SO(3).

    Unfortunately, I can't for the life of me find where I read it, and I'm not 100% sure I'm remembering it correctly. Can anyone provide a reference, a quick proof, or tell me it's wrong? Thanks!
     
  2. jcsd
  3. Jul 30, 2014 #2

    micromass

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    I probably saw this too late to be helpful for your talk, but the fact is true. The easiest proof is by applying Yamabe's theorem which says that any path-connected subgroup of ##SO(3)## corresponds to a subalgebra of ##\mathfrak{so}(3)##. More precisely, if ##H## is a path-connected subgroup of ##SO(3)##, then there is a Lie-subgroup ##\mathfrak{h}## of ##\mathfrak{so}(3)##, such that ##\mathfrak{h} = \langle \textrm{exp}\mathfrak{h}\rangle##.

    So if ##\mathfrak{h}## has dimension ##0##, then you get the identity. If it has dimension ##1##, then you have the rotations with fixed axis. If it has dimensions ##3## then it is the entire group. It cannot have dimension ##2##.

    For more information, see the excellent book by Hilgert and Neeb: "Structure and Geometry of Lie Groups"
     
  4. Jul 31, 2014 #3

    Matterwave

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    I really thought the proof would be easier than this. This fact seemed very intuitive to me.
     
  5. Jul 31, 2014 #4

    micromass

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    There is likely an elementary, direct proof. But this is the fastest way to prove it.
     
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