TITLE: Prove the lorentz group is not compact
QUESTION [5 upvotes]: I cant seem to figure how to prove this. I want to show that the Lorentz group $O(3,1)$ is not compact. I was thinking that the best way to show it was not compact was to show it was unbounded. Any ideas?

REPLY [2 votes]: Here is a more conceptual proof of noncompactness which depends somewhat less on working with coordinates for $O(3,1)$.
There is a theorem in topology that if $f : X \to Y$ is a surjective continuous function and if $X$ is compact then $Y$ is compact. You can turn this into a method for proving noncompactness of $X$: if $Y$ is noncompact then $X$ is noncompact.
Now the Lorentz group $O(3,1)$ acts on the Lorentz space $\mathbb{R}^{(3,1)}$. The orbit of the point $(0,0,0,1)$, meaning the subset
$$Y = \{(0,0,0,1) \cdot M \,\bigm|\, M \in O(3,1)\}
$$
is an unbounded and therefore noncompact subset of $\mathbb{R}^{(3,1)}$, namely one of the sheets of a two-sheeted hyperboloid. And the function $O(3,1) \mapsto Y$ defined by $M \mapsto (0,0,0,1) \cdot M$ is a surjective continuous function. Therefore $O(3,1)$ is noncompact.