5
$\begingroup$

I am trying to prove that the Lorentz group $SO(3,1)$ is a Lie group. To prove that it is a manifold, I was thinking of proving that it is a closed subgroup of $GL(4,\mathbb{R})$. Firstly, I have not convinced myself that it is in fact closed. If it is, I am not sure how to start that proof. I have also considered proving it is a manifold by means of the constant-rank level set theorem in the way that O(n) is proven to be a regular submanifold of $GL(n,\mathbb{R})$ by devising a constant rank map $f: GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ of which $SO(3,1)$ is the inverse. However, I have yet to find such a map. Does anyone have any hints to get me started on this proof, or a link to an alternate proof?

edit: I am now thinking that the best route to take is to prove that $SO(3,1)$ is the zero set of polynomial equations on $GL(n,\mathbb{R})$. If I can show that, then I can prove that $SO(3,1)$ is closed. However, the definition of $SO(3,1)$ is more complicated than that of $O(n)$ or $SL(n,\mathbb{R})$, so I am still stuck on this proof.

1 Answers 1

8

The Lorentz group $\mathrm{O}(3,1)$ is the zero level set of $$ f : \mathrm{GL}(n)\to\mathrm{GL}(n), \Lambda\mapsto \Lambda \eta \Lambda^T - \eta $$ by definition, where $\eta$ is the usual Minkowski metric of signature $(3,1)$.

  • 2
    This is not enough you also have to prove that $df\neq 0$ everywhere. A much more quick proof is noticing that (a) GL(n) is a Lie group, (b) O(3,1) is evidently closed therein (it evidently includes its limit points). Cartan theorem implies that O(3,1) is a Lie subgroup of GL(n), in particular an analytic (sub)manifold. – 2017-09-21