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Recall the following theorem due to Burnside:Let $G$ be a finite group and let $V$ be its irreducible complex representation of dimension greater than 1, then the character of this representation is $0$ on some element of $G$. Is this statement still correct if $G$ is any compact Lie group? Thanks.

flag	asked May 28 2010 at 19:08 Bedini 261●1●5

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Some context for the question would be helpful. It seems to reduce to looking at the well-studied finite dimensional representations (over $\mathbb{C}$ ) of a connected semisimple compact Lie group. Why would an answer be interesting? There are similarities with the finite group situation, including orthogonality relations, but also some big differences. As far as I can see, the proof for finite groups doesn't readily translate to compact Lie groups. The answer could well be yes (or no), but does it have implications? – Jim Humphreys May 28 2010 at 22:21

Eric Rowell · Accepted Answer · 2010-05-29 21:07:04Z UTC

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The answer is yes--use the Weyl character formula, for example. See: Patrick X. Gallagher, Zeros of group characters. Math. Z. Volume 87 (1965), Number 3.

answered May 29 2010 at 21:07

Eric Rowell
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	Gallagher's proof combines Burnside's theorem with Weyl's character formula, using some Clifford theory. Since finite groups are also compact, it's probably not realistic to expect a general proof to give a new or simpler proof of Burnside's result. (By the way, Gallagher has spent his career at Columbia, serving as the adviser of Dorian Goldfeld among many others.) – Jim Humphreys May 30 2010 at 13:49
	Great! Thanks a lot Eric. – Bedini May 30 2010 at 18:21

Is there a generalization of Burnside’s theorem for compact Lie groups?

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