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I've shown that we have the following action of $SL_2(\mathbb C)$ on $\mathbb R^4$: $$ \phi\colon SL_2(\mathbb C)\to O(1,3) $$ given by $$ \phi_X\colon W\mapsto XWX^\dagger. $$ Since $SL_2(\mathbb C)$ is connected, and our homormophism is continuous, it follows that $\phi(SL_2(\mathbb C))$ is contained in the identity component $O(1,3)_e$. However, I would like to argue that it is equal to $O(1,3)_e$. How can I do that?

Sha Vuklia
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  • Is remark #21 here relevant? – Karl Apr 12 '20 at 16:41
  • @Karl I'm not sure which remark you're referring to. – Sha Vuklia Apr 12 '20 at 16:47
  • In the "remarks" section. – Karl Apr 12 '20 at 16:54
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    It depends on how much Lie theory you know, see e.g. https://math.stackexchange.com/questions/301289/spinor-mapping-is-surjective. A direct argument will be that the image of $\phi$ is an open subgroup, hence, closed, hence, is the entire $O(1,3)_e$. See also https://math.stackexchange.com/questions/2240334/for-lie-groups-g-and-h-a-homomorphism-fg-rightarrow-h-is-surjective-if?rq=1 – Moishe Kohan Apr 12 '20 at 17:26
  • Thanks, that was exactly what I needed. – Sha Vuklia Apr 15 '20 at 12:08

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