# 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?

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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 at 9:10

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 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 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 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.
@James: Maybe it's just the word compact which is confusing. For spaces that are subspaces of $\mathbb{R}^n$, compact = closed and bounded. This is good enough for most physics applications and gives a good intuition. –  Simon Oct 26 '11 at 22:22