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Is it true that a compact group always has a faithful, finite-dimensional unitary representation? If not, are there any reasonably simple counter-examples?

I've done some research and know that every group has some faithful representation, all irreducible reps of a compact group are finite, and that the irreducible reps separate the points of the group. However, that doesn't quite answer the question!

flag	asked May 5 2011 at 10:50 Sabri 55●3

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The additive group of $p$-adic integers is an example, any finite dimensional continuous representation has some $p^n$ in its kernel. – Torsten Ekedahl May 5 2011 at 14:00

Mark Schwarzmann · Accepted Answer · 2011-05-05 11:44:10Z UTC

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A famous theorem is that this is true if and only if $G$ is a Lie group.

answered May 5 2011 at 11:44

Mark Schwarzmann
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Also, the adjective "unitary" is redundant, as every finite-dimensional representation of a compact group is unitarizable. – Mark Schwarzmann May 5 2011 at 11:47

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Does this theorem have a name? Even better, are there any good references for it? – Sabri May 5 2011 at 14:01

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@Sabri: one direction follows from the fact that a closed subgroup of a Lie group is a Lie group (I don't know if this theorem has a name). The other follows from Peter-Weyl. – Qiaochu Yuan May 5 2011 at 16:15

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Sabri: you can find the proof in Folland's text ("A Course in Abstract Harmonic Analysis", theorem 5.13). I don't know of a name for it. The theorem which Qiaochu mentioned, though, is often called the "closed subgroup theorem" or Cartan's theorem. – Mark Schwarzmann May 5 2011 at 16:33

Bugs Bunny · Answer 2 · 2011-05-05 11:08:54Z UTC

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No, it is false! Take a product of alef-2011 copies of $C_2$. It is a bit too big to fit into $GL_n (C)$...

answered May 5 2011 at 11:08

Bugs Bunny
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And I fear it will be falser next year... – Georges Elencwajg May 5 2011 at 11:30

More specifically, it isn't a second-countable group, while every subspace (let alone subgroup) of $GL_n (\mathbb{C}$ is second-countable. – Mark Schwarzmann May 5 2011 at 11:50

Or use Burnsides's theorem that any linear group of finite exponent is finite. – Benjamin Steinberg Oct 12 2011 at 2:47

Alain Valette · Answer 3 · 2011-05-05 12:25:30Z UTC

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See http://mathoverflow.net/questions/61921/on-closed-totally-disconnected-subgroups-of-connected-real-lie-groups/63030#63030: a non-discrete totally diisconnected group has no faithful representation in $GL_n(\mathbb{C})$ .

edited May 5 2011 at 12:25

answered May 5 2011 at 12:14

Alain Valette
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Dick Palais · Answer 4 · 2011-05-05 17:25:53Z UTC

Another famous and interesting NASC for a compact (or even locally compact) group to have a faithful finite dimensional representation is that it "not have arbitrarily small subgroups", i.e., that there exist a neighborhood of the identity with no non-trivial subgroup. This was the way that von Neumann solved the Hilbert 5th Problem in the compact case, and is explained (starting on page 1243) in:

http://www.ams.org/notices/200910/rtx091001236p.pdf

Finite-dimensional faithful representations of compact groups

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