Continous groups and manifolds

In summary, the author discusses the theorem of polar decompositon which states that a SL(2,C) element can be decomposed into a product of an SU(2) and H(2,C) element.
  • #1
AlphaNumeric
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According to my notes on SUSY 'as everyone knows, every continuous group defines a manifold', via

[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 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 [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.
 
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  • #2
AlphaNumeric said:
thus giving the full manifold [tex]\mathcal{M}_{SL(2,\mathbb{C})} = \mathbb{R}^{3} \times S^{3}[/tex]

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 [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.

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

[tex](x^{0})^{2} - \sum_{i}^{3} |x^{i}|^{2} = 1[/tex]

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

The mapping

[tex] \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)[/tex]

is a homeomorphism between [itex]\mathbb{R}^3[/itex] and the hyperboloid.

I might add more of the details sometime in the next few days.
 
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  • #3
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.
 
  • #4
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! :\
 

1. What is a continuous group?

A continuous group is a mathematical concept that describes a set of objects or transformations that can be combined together in a continuous manner. This means that the elements of the group can be smoothly varied and the group operation remains well-defined.

2. What is a manifold?

A manifold is a type of mathematical space that is locally similar to Euclidean space, but may have a more complicated global structure. In other words, a manifold is a space that can be locally described using coordinates, but may have a more complex shape overall.

3. How are continuous groups and manifolds related?

Continuous groups can act on manifolds, meaning that the elements of the group can be used to transform points on the manifold. This allows for the study of symmetries and transformations of manifolds, which has many applications in physics and other fields.

4. What is the significance of continuous groups and manifolds in physics?

In physics, continuous groups and manifolds are used to describe the symmetries and transformations of physical systems. This is important because symmetries often correspond to conserved quantities, and understanding these symmetries can provide insights into the behavior of physical systems.

5. Can you give an example of a continuous group and its corresponding manifold?

One example is the special orthogonal group in three dimensions (SO(3)), which consists of all rotations in three-dimensional space. The corresponding manifold is the three-dimensional sphere (S3), which represents all possible orientations of a rigid body. The group acts on the manifold by transforming the orientation of the body.

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