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Generating group homomorphisms between Lie groups

  1. Aug 10, 2010 #1
    Suppose [itex]\mathfrak{g}[/itex] and [itex]\mathfrak{h}[/itex] are some Lie algebras, and [itex]G=\exp(\mathfrak{g})[/itex] and [itex]H=\exp(\mathfrak{h})[/itex] are Lie groups. If

    [tex]
    \phi:\mathfrak{g}\to\mathfrak{h}
    [/tex]

    is a Lie algebra homomorphism, and if [itex]\Phi[/itex] is defined as follows:

    [tex]
    \Phi:G\to H,\quad \Phi(\exp(A))=\exp(\phi(A))
    [/tex]

    will [itex]\Phi[/itex] be a group homomorphism?

    Since [itex]\exp(A)\exp(B)=\exp(A+B)[/itex] is not true in general, I see no obvious way to prove the claim.
     
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  3. Aug 10, 2010 #2

    Hurkyl

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    I expect G = SO(3) and H = SU(2) along with the identity map so(3)-->su(2) constitute a counter-example.

    It feels like cheating, though, since this surely induces the bundle SU(2) --> SO(3) which could be thought of as a two-valued homomorphism from SO(3) to SU(2). (e.g. compare with the square-root function of the complex plane)
     
  4. Aug 10, 2010 #3
    I see. In general the equation

    [tex]
    \Phi(\exp(A)) = \exp(\phi(A))
    [/tex]

    will not give a well defined mapping, because exponential mapping is not injective.

    But for example, how do you prove that the canonical mapping SU(2) -> SO(3) is group homomorphism? It is easy to verify that the mapping su(2) -> so(3) is a Lie algebra isomorphism, but then what?
     
  5. Aug 10, 2010 #4

    Hurkyl

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    I think what gets induced is a submanifold of GxH whose projection onto G is a local homeomorphism.

    I expect the submanifold to be a subgroup.


    Why? By doing the differential geometry to sew the differential facts into an integral whole.

    Or...
    by using the variation on that identity that is valid.


    But I'm mainly running off of intuition here, and this is far from my field of expertise.
     
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