How Does SL(2,C) Relate to Its Manifold Structure?

[AlphaNumeric]

According to my notes on SUSY 'as everyone knows, every continuous group defines a manifold', via

$$\Lambda : G \to \mathcal{M}_{G}$$
$$\{ g = e^{i\alpha_{a}T^{a}} \} \to \{ \alpha_{a} \}$$

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

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

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

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

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

I don't see how it's obvious to consider g as the product of two separate 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 $$(x^{0})^{2} - \sum_{i}^{3} |x^{i}|^{2} = 1$$ imply H has $$\mathcal{M}_{H} = \mathbb{R}^{3}$$, it would seem to me the defining equation isn't pointing at $$\mathbb{R}^{3}$$ but something slightly different.

Thanks in advance.

 

[George Jones]

thus giving the full manifold $$\mathcal{M}_{SL(2,\mathbb{C})} = \mathbb{R}^{3} \times S^{3}$$

Care needs to be taken in interpreting what the equals sign means here, as it is being used in a non-standard way. The contextual meaning here is that the two sets are isomorpic as topological spaces.

I don't see how it's obvious to consider g as the product of two separate 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?

The former, I think.

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

Again, the two spaces are isomorphic as topological spaces. To see this, think of

$$(x^{0})^{2} - \sum_{i}^{3} |x^{i}|^{2} = 1$$

(with $x^0 >0$) as a 3-dimensional (as a manifold) hyperboloid that is a subset of $\mathbb{R}^4$.

The mapping

$$\left( x^1 , x^2 , x^3 \right) \mapsto \left( \sqrt{\left(1 - \sum_{i}^{3} |x^{i}|^{2} \right)} , x^1 , x^2 , x^3 \right)$$

is a homeomorphism between $\mathbb{R}^3$ and the hyperboloid.

I might add more of the details sometime in the next few days.

 

[dextercioby]

The decomposition of an SL(2,C) element into a product of an SU(2) and H(2,C) element is known in mathematics as the theorem of polar decomposition for SL(2,C).

Daniel.

 

[AlphaNumeric]

Cheers guys. George, I got the whole 'up to an isomorphism' thing about showing group or manifold relations, it's just I didn't see the particular relation between the surface in M^4 giving R^3 but your description and a discussion I had with a friend cleared it up nicely :)

For someone who disliked 1st year group theory, I'm doing a hell of a lot of it 4 years on! :\