According to my notes on SUSY 'as everyone knows, every continous group defines a manifold', via(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Lambda : G \to \mathcal{M}_{G}[/tex]

[tex]\{ g = e^{i\alpha_{a}T^{a}} \} \to \{ \alpha_{a} \}[/tex]

It gives the examples of U(1) having the manifold [tex]\mathcal{M}_{U(1)} = S^{1}[/tex] and SU(2) has [tex]\mathcal{M}_{SU(2)} = S^{3}[/tex].

It then gives the example of SL(2,C) which I don't follow :

[tex]G = SL(2,\mathbb{C})[/tex] and [tex]g=HV[/tex] where [tex]V = SU(2)[/tex] and [tex]H=H^{\dagger}[/tex], positive and detH = 1.

[tex]H = x^{\mu}\sigma_{\mu} = \left( \begin{array}{cc} x^{0}+x^{3} & x^{1}+ix^{2} \\ x^{1}-ix^{2} & x^{0}-x^{3} \end{array} \right)[/tex]

Therefore [tex](x^{0})^{2} - \sum_{i}^{3} |x^{i}|^{2} = 1[/tex] so the manifold is [tex]\mathbb{R}^{3}[/tex], thus giving the full manifold [tex]\mathcal{M}_{SL(2,\mathbb{C})} = \mathbb{R}^{3} \times S^{3}[/tex]

I don't see how it's obvious to consider g as the product of two seperate group elements. Is this just a case of knowing the answer and skipping part of the derivation or is there something obvious about SL(2,C) which tells you it's manifold is the direct product of two simpler manifolds?

Also, though I might be about to look very stupid, why does [tex](x^{0})^{2} - \sum_{i}^{3} |x^{i}|^{2} = 1[/tex] imply H has [tex]\mathcal{M}_{H} = \mathbb{R}^{3}[/tex], it would seem to me the defining equation isn't pointing at [tex]\mathbb{R}^{3}[/tex] but something slightly different.

Thanks in advance.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Continous groups and manifolds

**Physics Forums | Science Articles, Homework Help, Discussion**