Generating group homomorphisms between Lie groups

[jostpuur]

Suppose $\mathfrak{g}$ and $\mathfrak{h}$ are some Lie algebras, and $G=\exp(\mathfrak{g})$ and $H=\exp(\mathfrak{h})$ are Lie groups. If

$$\phi:\mathfrak{g}\to\mathfrak{h}$$

is a Lie algebra homomorphism, and if $\Phi$ is defined as follows:

$$\Phi:G\to H,\quad \Phi(\exp(A))=\exp(\phi(A))$$

will $\Phi$ be a group homomorphism?

Since $\exp(A)\exp(B)=\exp(A+B)$ is not true in general, I see no obvious way to prove the claim.

 

[Hurkyl]

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)

 

[jostpuur]

I see. In general the equation

$$\Phi(\exp(A)) = \exp(\phi(A))$$

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?

 

[Hurkyl]

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...

$\exp(A)\exp(B)=\exp(A+B)$

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.