For $G$ a compact Lie group, let $\operatorname{mdfr}(G)$ be the minimum dimension of a faithful complex representation of $G$. Is there a bound on $\operatorname{mdfr}(N(H)/H)$ for $H$ a subgroup of $U(n)$ with normalizer $N(H)$?
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1$\begingroup$ You mean, a bound depending on $H$? or a bound depending on $n$ only (i.e., describe the supremum over $H\le U(n)$ of the values $\mathrm{mdfr}(N(H)/H)$)? $\endgroup$– YCorCommented Jan 3, 2022 at 9:59
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$\begingroup$ I am looking for the existence of a uniform bound depending only on n and independent of H. $\endgroup$– rickCommented Jan 3, 2022 at 21:15
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$\begingroup$ $H$ is a closed subgroup? Closed and connected? $\endgroup$– LSpiceCommented Jan 3, 2022 at 21:44
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$\begingroup$ @LSpice I'm pretty sure $H$ is assumed closed, but not assumed connected. $\endgroup$– YCorCommented Jan 3, 2022 at 21:47
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$\begingroup$ The answer would be "yes" if the following stronger fact would hold: the number of components of $N(H)$ is bounded when $H$ ranges over closed subgroups of $U(n)$. Is this true? I see no obvious obstruction ($H$ finite would be a good starting point). $\endgroup$– YCorCommented Jan 3, 2022 at 21:50
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