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I want to show that $\mathfrak{so}(4)\cong \mathfrak{so}(3)\oplus \mathfrak{so}(3)$. I know that as lie groups $SO(4)\cong (SU(2)\times SU(2))/\mathbb{Z}_2$ and that as $SU(2)/\mathbb{Z}_2 \cong SO(3)$.

My idea to do this was to show that $SO(4)\cong SU(2)\mathbb{Z}_2\times SU(2)/\mathbb{Z}_2$ and then the result should follow. But the map from $(SU(2)\times SU(2))/\mathbb{Z}_2$ to $SU(2)\mathbb{Z}_2\times SU(2)/\mathbb{Z}_2$ is only surjective not injective. Hence the map from $SO(4)$ to $SU(2)\mathbb{Z}_2\times SU(2)/\mathbb{Z}_2$ is not an isomorphism.

Is this the wrong approach or have I just made a mistake?

Mike Pierce
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Emily
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    I think one should do this directly on the level of Lie algebras, see this duplicate. Note that $\mathfrak{o}(3)=\mathfrak{so}(3)$. – Dietrich Burde Oct 21 '19 at 12:08
  • I agree with Dietrich. Further, not only is the "obvious" map from $SO(4)$ to $SU(2)/\pm I \times SU(2)/\pm I$ not an isomorphism, no map is an isomorphism: $SO(4)$ has no normal subgroups isomorphic to $SO(3)$, but $SO(3)\times SO(3)$ does. – Jason DeVito - on hiatus Oct 21 '19 at 13:30
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    That question looks useful, I'll look into it.

    Is there any condition (weaker than being an isomorphism) for a homomorphism of lie groups which implies the lie algebras are isomorphic?

     – Emily Oct 21 '19 at 15:52

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Lie algebra structure only uniquely determines the connected component of the identity a lie group. If one shows that the homomorphism restricts to an isomorphism on the connected component of the identity then this would do.

But as suggested it is perhaps easier to d this all on the level of lie algebras.

Emily
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  • Both $SO(4)$ and $SU(2)/\pm\times SU(2)/\pm I$ are connected, so by my above comment, this approach won't work. However, to answer your question in the comments: there is a condition weaker than isomorophism which guarantees isomorhpic Lie algebras: you can require the kernel to be finite (as opposed to being just the identity) and this is still enough to conclude that the Lie algebras are isomorphic. And that will allow you to solve your problem. – Jason DeVito - on hiatus Oct 21 '19 at 20:18
  • @JasonDeVito could you elaborate on this finite kernel condition or give some reference for it? Is it saying that for any lie group homomorphism $\varphi:G\rightarrow H$ if $\ker(phi)$ is finite then $\mathfrak{g}\cong \mathfrak{h}$? – Emily Oct 22 '19 at 07:29
  • To be more precise, I'm saying the following. Suppose $\phi:G\rightarrow H$ is a Lie group homomorphism which contains the identity component of $H$ in its image. If $\ker \phi$ is discrete (e.g., finite), then the induced map $\phi_\ast:\mathfrak{g}\rightarrow \mathfrak{h}$ is an isomorphism. For a reference, see these notes from a class taught by my advisor: https://www.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf, focusing on the part about covering groups. – Jason DeVito - on hiatus Oct 22 '19 at 12:41