Lorentz group is non-compact

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Hello! I need to show that Lorentz Group is non compact, but has 4 connected components. The way I was thinking to do it is to write the relation between the elements of the 4x4 matrices and based on that, associated it with a known topological space, based on the determinant and the value of the (0,0). However if I am not wrong it would be a 16 dimensional space, so I kinda got scared. I assume there is an easier way. Can someone help me a bit, give send me a link with the proof? Thank you!
 

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jim mcnamara
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If you were to explicitly state the problem - a little more formally please - it might help. I do not know an answer, but fine tuning the question always seems to help. Thanks.
 
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WWGD
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I assume you see a matrix as an element of ## \mathbb R^{n^2} ##? If so, then you just need to show closedness and boundedness. Otherwise, please explain.
 
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George Jones
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Hello! I need to show that Lorentz Group is non compact
I assume you see a matrix as an element of ## \mathbb R^{n^2} ##? If so, then you just need to show closedness and boundedness. Otherwise, please explain.
So (by Heine-Borel) a subset of ## \mathbb R^{16} ## that is not(closed and bounded) needs to be found, e.g., the set of boosts in the x direction.
 
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WWGD
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So (by Heine-Borel) a subset of ## \mathbb R^{16} ## that is not(closed and bounded) needs to be found, e.g., the set of boosts in the x direction.
And notice, since projection operator is continuous, if subset is compact, its projection to any coordinate must also be compact -- closed and bounded.
 
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samalkhaiat
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Hello! I need to show that Lorentz Group is non compact, but has 4 connected components. The way I was thinking to do it is to write the relation between the elements of the 4x4 matrices and based on that, associated it with a known topological space, based on the determinant and the value of the (0,0). However if I am not wrong it would be a 16 dimensional space, so I kinda got scared. I assume there is an easier way. Can someone help me a bit, give send me a link with the proof? Thank you!
A Lie group is called compact if it is compact as a manifold. For example, [itex]SU(2)[/itex] is compact because (topologically) it can be identified with the 3-sphere [itex]S^{3}[/itex] which is compact. The Lorentz group [itex]SO(1,3)[/itex] is not compact because it can “essentially” be written as a product of the non-compact space (of boosts) [itex]\mathbb{R}^{3}[/itex] with the compact space (of rotations) [itex]S^{3}[/itex]: [tex]SO(1,3) \cong SL(2 , \mathbb{C}) / \mathbb{Z}_{2} \cong \mathbb{R}^{3} \times S^{3} / \mathbb{Z}_{2} \ .[/tex]

The proper mathematical proof goes as follow: Recall that a subset [itex]\mathcal{U}[/itex] of [itex]\mathbb{R}^{n}[/itex] is compact if and only if, it is closed and bounded, i.e., if and only if, for every sequence [itex]a_{m} \in \mathcal{U} \subset \mathbb{R}^{n}[/itex], there exists a subsequence which converges to some [itex]a \in \mathcal{U}[/itex]. For simplicity, consider the 2-dimensional Lorentz group [itex]SO(1,1)[/itex]. Define a sequence of elements [itex]\Lambda_{m} \in SO(1,1)[/itex] by [tex]\Lambda_{m} = \begin{pmatrix} \cosh (m) & \sinh (m) \\ \sinh (m) & \cosh (m) \end{pmatrix} \ .[/tex] Now, since the components of [itex]\Lambda_{m}[/itex] are unbounded, it follows that [itex]\Lambda_{m}[/itex] cannot have convergent subsequence. Thus, [itex]SO(1,1)[/itex] is a non-compact Lie group.
 
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WWGD
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A Lie group is called compact if it is compact as a manifold. For example, [itex]SU(2)[/itex] is compact because (topologically) it can be identified with the 3-sphere [itex]S^{3}[/itex] which is compact. The Lorentz group [itex]SO(1,3)[/itex] is not compact because it can “essentially” be written as a product of the non-compact space (of boosts) [itex]\mathbb{R}^{3}[/itex] with the compact space (of rotations) [itex]S^{3}[/itex]: [tex]SO(1,3) \cong SL(2 , \mathbb{C}) / \mathbb{Z}_{2} \cong \mathbb{R}^{3} \times S^{3} / \mathbb{Z}_{2} \ .[/tex]

The proper mathematical proof goes as follow: Recall that a subset [itex]\mathcal{U}[/itex] of [itex]\mathbb{R}^{n}[/itex] is compact if and only if, it is closed and bounded, i.e., if and only if, for every sequence [itex]a_{m} \in \mathcal{U} \subset \mathbb{R}^{n}[/itex], there exists a subsequence which converges to some [itex]a \in \mathcal{U}[/itex]. For simplicity, consider the 2-dimensional Lorentz group [itex]SO(1,1)[/itex]. Define a sequence of elements [itex]\Lambda_{m} \in SO(1,1)[/itex] by [tex]\Lambda_{m} = \begin{pmatrix} \cosh (m) & \sinh (m) \\ \sinh (m) & \cosh (m) \end{pmatrix} \ .[/tex] Now, since the components of [itex]\Lambda_{m}[/itex] are unbounded, it follows that [itex]\Lambda_{m}[/itex] cannot have convergent subsequence. Thus, [itex]SO(1,1)[/itex] is a non-compact Lie group.
Nice. You can also argue that if the product ## \mathbb R^3 \times \mathbb Z/2 ## were compact, then, using the projection map ( given it is continuous), projection onto first component ## \mathbb R^3 ## would also be compact.
 

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