Prove the Lorentz group is not compact 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?
 A: 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.
A: You're on the right track.
Hint Fix a sensible bilinear form of signature $(3, 1)$ on $\mathbb{R}^{3, 1}$, say
$g := -dt^2 + dx^2 + dy^2 + dz^2$, so that $O(g) \cong O(3, 1)$. Then, e.g., in the coordinate basis, the subgroup of $O(3, 1)$ that fixes $\partial_y$ and $\partial_z$ and an orientation is
$$
\left\{\pm\begin{pmatrix} \cosh t & \sinh t & 0 & 0 \\ \sinh t & \cosh t & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\right\} \cong SO(1, 1).
$$

This subgroup is an unbounded subset of $M(4, \mathbb{R}) \cong \mathbb{R}^{16}$, hence so is $O(3, 1)$, but the Heine-Borel Theorem gives that a subset of $\mathbb{R}^n$ is compact iff it is closed and bounded.

A: Hint:
People like Gelfand would sometime call $SL(2, \mathbb{C})$ the Lorentz group. Why? It is the connected component of $e$ in $O(3,1)$. You identify $\mathbb{R}^4$ with the space $\mathcal{H}$ of $2\times 2$ hermitian matrices 
$$ \rho = \left( \begin{array}{cc} x_0 + x_3 & x_1 - i x_2\\ x_1 + i x_2 & x_0 - x_3\end{array}\right)$$
Note that $\det \rho = x_0 ^2 - x_1 ^2 - x_2^2 - x_3^2$. The action of $SL(2,\mathbb{C})$ on hermitian matrices is the natural one
$$g \star \rho = g \cdot \rho \cdot g^*$$
Now you only have to check that $SL(2,\mathbb{C})$ is not compact. 
