# Lorentz group is non-compact

• I
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
Mentor
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.

fresh_42
Mentor
Hint: Consider the orbit of some vector.

WWGD
Gold Member
2019 Award
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.

George Jones
Staff Emeritus
Gold Member
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.

WWGD
WWGD
Gold Member
2019 Award
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.

samalkhaiat
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, $SU(2)$ is compact because (topologically) it can be identified with the 3-sphere $S^{3}$ which is compact. The Lorentz group $SO(1,3)$ is not compact because it can “essentially” be written as a product of the non-compact space (of boosts) $\mathbb{R}^{3}$ with the compact space (of rotations) $S^{3}$: $$SO(1,3) \cong SL(2 , \mathbb{C}) / \mathbb{Z}_{2} \cong \mathbb{R}^{3} \times S^{3} / \mathbb{Z}_{2} \ .$$

The proper mathematical proof goes as follow: Recall that a subset $\mathcal{U}$ of $\mathbb{R}^{n}$ is compact if and only if, it is closed and bounded, i.e., if and only if, for every sequence $a_{m} \in \mathcal{U} \subset \mathbb{R}^{n}$, there exists a subsequence which converges to some $a \in \mathcal{U}$. For simplicity, consider the 2-dimensional Lorentz group $SO(1,1)$. Define a sequence of elements $\Lambda_{m} \in SO(1,1)$ by $$\Lambda_{m} = \begin{pmatrix} \cosh (m) & \sinh (m) \\ \sinh (m) & \cosh (m) \end{pmatrix} \ .$$ Now, since the components of $\Lambda_{m}$ are unbounded, it follows that $\Lambda_{m}$ cannot have convergent subsequence. Thus, $SO(1,1)$ is a non-compact Lie group.

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