Is the Lorentz group compact (and if not, is U(1)?) A common statement in any quantum field theory text is that only compact groups have finite-dimensional representations, and that the Lorentz group is not compact, since it is parameterised by $0\leq (v/c)<1$. Fair enough, but $U(1)$ is parametrised by $0\leq \theta <2\pi$, so why is $U(1)$ compact? Should they be described as closed instead, or am I getting lost?
 A: $U(1)$ is compact because rotation by 2$\pi$ and 0 are the same thing. So its not topologically an open interval, but a circle.
A: Irreducible representations of compact groups must be finite-dimensional.  Compact groups can have infinite-dimensional representations, but those must necessarily be reducible.  Conversely, noncompact groups cannot have nontrivial irreducible unitary representations (I am not 100% positive about this.  Can somebody confirm that this is true?).  The unitarity here is key.  For example, $\mathbb{C}$ is a noncompact Lie group with a nonunitary one-dimensional irreducible representation (the "identity" representation).  Unitarity is a natural condition, however, because for a represetnation $V$ of a finite (or compact) group, you can choose an inner product on $V$ that makes the representation unitary.  In other words, all representations of finite (and compact) groups are essentially unitary.
This is probably not the best way to think about these groups (in terms of being parameterized by numbers, because in general, you'll need more than one chart to cover the manifold), but using your language, because $2\pi$ is identified with $0$ in $U(1)$, where as $v/c=1$ is not identified with $v/c=0$, you can't apply this same logic to both groups.
In any case, $U(1)=S^1$, and so is obviously compact.  The way you describe is probably the best intuitive way to see that the Lorentz group is not compact, and in fact this argument can be made into a proof.
A: The topological manifold of the Lorentz group can be continuously embedded in the metric space $\mathbb{R}^{16}$ together with (metric) topology inherited from $\mathbb{R}$ (direct product topology). The subset of Lorentzian boosts in 1 spatial direction can be parametrized by $\beta =v/c$ and is hence homeomorphic as a topological space with the open unit interval of $\mathbb{R}$ which is non-compact in the topology generated by the interval metric (by Heine-Borel theorem). Thus the Lorentz topological space is non-compact in the metric topology of $\mathbb{R}^{16}$, because there's no finite subcover for the space of Lorentzian boosts in 1 spatial direction. 
