Generating group homomorphisms between Lie groups

  • Thread starter jostpuur
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  • #1
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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.
 

Answers and Replies

  • #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)
 
  • #3
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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?
 
  • #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...
[itex]
\exp(A)\exp(B)=\exp(A+B)
[/itex]
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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