5
$\begingroup$

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!

$\endgroup$
1
  • 4
    $\begingroup$ The additive group of $p$-adic integers is an example, any finite dimensional continuous representation has some $p^n$ in its kernel. $\endgroup$ May 5, 2011 at 14:00

4 Answers 4

13
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ Also, the adjective "unitary" is redundant, as every finite-dimensional representation of a compact group is unitarizable. $\endgroup$
    – Mark
    May 5, 2011 at 11:47
  • 3
    $\begingroup$ Does this theorem have a name? Even better, are there any good references for it? $\endgroup$
    – Sabri
    May 5, 2011 at 14:01
  • 4
    $\begingroup$ @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. $\endgroup$ May 5, 2011 at 16:15
  • 2
    $\begingroup$ 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. $\endgroup$
    – Mark
    May 5, 2011 at 16:33
13
$\begingroup$

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

$\endgroup$
3
  • 9
    $\begingroup$ And I fear it will be falser next year... $\endgroup$ May 5, 2011 at 11:30
  • 1
    $\begingroup$ More specifically, it isn't a second-countable group, while every subspace (let alone subgroup) of $GL_n (\mathbb{C}$ is second-countable. $\endgroup$
    – Mark
    May 5, 2011 at 11:50
  • 1
    $\begingroup$ Or use Burnsides's theorem that any linear group of finite exponent is finite. $\endgroup$ Oct 12, 2011 at 2:47
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

See On closed totally disconnected subgroups of connected real Lie groups : a non-discrete totally diisconnected group has no faithful representation in $GL_n(\mathbb{C})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.