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I have seen sources claim that $SO^+(1,3) \cong SU(2) \times SU(2)$, but have seen others claim that only their Lie algebras are isomorphic.

  1. Is it true that $SO^+(1,3) \cong SU(2) \times SU(2)$?
  2. If not, is $SO^+(1,3)$ isomorphic to some quotient of $SU(2) \times SU(2)$?
  3. Is the analogous result true for their Lie alebras, i.e. $\mathfrak{so}^+(1,3) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$, or something similar?
  4. Generally, when can you go from a product group isomorphism to a corresponding result for Lie algebras, or vice versa?

I'm led to believe there is some isomorphism of this form since (I think) $$SO^+(1,3) \cong PSL(2,\mathbb{C}), \quad SL(2,\mathbb{C}) \cong SO(4), \quad SO(4) \cong SU(2) \times SU(2)/\{\pm I\}.$$

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$SO^+(3,1)$ is the so-called restricted Lorentz group, which is the identity component of the Lorentz group $SO(3,1)$. It is a six-dimensional real Lie group, which is not simply connected.

  1. Since $SO^+(1,3)$ is not compact, but $SU(2)\times SU(2)$ is compact, the groups cannot be isomorphic as real Lie groups.

  2. We have $SO^+(3,1)\simeq SL(2,\mathbb{C})/\mathbb{Z}_2\simeq SU(2)_{\mathbb{C}}/\mathbb{Z}_2$, i.e., the complexification of the restricted Loretz group satisfies $$SO^+(3,1)_{\mathbb{C}}\simeq (SU(2)_{\mathbb{C}}\times SU(2)_{\mathbb{C}})/\mathbb{Z}_2.$$

  3. In the same way, $\mathfrak{so}^+(3,1)_{\mathbb{C}}\simeq \mathfrak{su}(2)_{\mathbb{C}}\oplus \mathfrak{su}(2)_{\mathbb{C}}\simeq \mathfrak{sl}_2(\mathbb{C})\oplus \mathfrak{sl}_2(\mathbb{C})$.

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  • $\begingroup$ How would these answers change if I replaced $SO^+(1,3)$ with $SO(1,3)$? $\endgroup$ Jan 19, 2015 at 13:22
  • $\begingroup$ For the Lie algebra it would not matter, because we have $\mathfrak{so}^+(3,1)=\mathfrak{so}(3,1)$. $\endgroup$ Jan 19, 2015 at 15:46
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    $\begingroup$ Is $SO(3,1)_\mathbb{C} \cong SU(2)_\mathbb{C} \times SU(2)_\mathbb{C}$? $\endgroup$ Jan 20, 2015 at 11:07
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    $\begingroup$ Yes, it is. The complexification of $SO(3,1)$ is $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$. $\endgroup$ Jan 21, 2015 at 20:33
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    $\begingroup$ No, it is the direct product. $\endgroup$ Dec 15, 2018 at 12:00

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