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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?

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Only compact groups have finite dimensional unitary representations, AFAIK. If you allow for non-unitary representations then you can find something finite-dimensional. –  Giuseppe Negro Oct 26 '11 at 18:25
to "only compact groups have finite-dimensional representations" : No, all unitary irreducible representations of abelian groups are one dimensional no matter whether they are compact or not. –  jjcale Oct 1 '13 at 9:10
@GiuseppeNegro, the unit operator on a TVS is trivially unitary, hence non-compact groups also have finite dimensional unitary representations, albeit only 1, the so-called trivial representation which takes any element of the group to the unit operator on the TVS in which you build the representation. –  DanielC Feb 17 at 22:31

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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.

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Yes, for some reason I was blithely assuming the periodic identification wasn't important. –  James Oct 28 '11 at 9:50
Is periodic identification the critical thing? or is it just that the interval must be closed. e.g. $v \in [0,c]$ would be compact, but $v \in [0,c)$ is non-compact? –  innisfree Sep 29 '13 at 18:12
In some sense periodic identification is absolutely crucial. There are precisely two connected $1$-dimensional Lie groups: $S^1$ and $\mathbb{R}$. These two are distinguished precisely by 'periodic identification', and furthermore, $S^1$ is compact while $\mathbb{R}$ is not. Even though $[0,c]$ or $[-c,c]$ might be compact, they are not Lie groups. –  Jonathan Gleason Oct 5 '13 at 17:11
I also think it is mis-leading to only say that Lorentz boosts are parametrized by $v\in (-c,c)$. They are, but you mustn't forget that the group law is not simply addition in $\mathbb{R}$; instead, it is given by the usual velocity addition formula in special relativity. It turns out that this group law makes $(-c,c)$ into a Lie group which is isomorphic (as Lie groups) to $(\mathbb{R},+)$ via $\mathrm{arctanh}$. –  Jonathan Gleason Oct 5 '13 at 17:12

$U(1)$ is compact because rotation by 2$\pi$ and 0 are the same thing. So its not topologically an open interval, but a circle.

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The circle is a compact space. Probably a little time spent on the wikipedia page would convince you of this. Be careful about calling a space closed - every space is closed in it's own topology but may be open or closed as a subset of larger space. Perhaps you should invest a little time in learning the basics of topology? It has a very large payoff to effort ratio. –  BebopButUnsteady Oct 26 '11 at 17:53
I completely agree with BebopButUnsteady. Every physicist should know enough point-set topology to understand at least basic manifold theory. –  Jonathan Gleason Oct 26 '11 at 20:21
It's called U(1) in a boatload of physics papers. For examples, look up Stueckelberg mechanism in "U(1)" gauge theory (necessarily noncompact) –  Ron Maimon Oct 28 '11 at 20:17
@wnoise: It is not incorrect usage. It's very good usage that I use all the time. The "U(1)" Lie Algebra (which is trivial) describes both the additive reals and the multiplicative unit complex numbers. Why separate the additive reals? Are they not U(1)? When you compose them, do they not commute? –  Ron Maimon Nov 27 '11 at 22:11
It's funny, but I find calling two things the same name to hinder clear communication. –  wnoise Nov 28 '11 at 1:19

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.

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