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.
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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
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$\begingroup$ 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.) $\endgroup$ Commented May 30, 2010 at 13:49
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$\begingroup$ The link to the paper at
springerlink.comis broken, but it can also be found at doi:10.1007/BF01111717. Corresponding zbMATH Open review is at Zbl 0128.25602. $\endgroup$ Commented Jun 18, 2022 at 2:57
$\mathbb{C}$) 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? $\endgroup$