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Let $G$ be a finite or compact group and $\rho: G \to \mathrm{U}(d)$ a $d$-dimensional unitary representation of $G$. If $\rho$ is irreducible then the following seems to be true: $$ \mathrm{span}_\mathbb{C} \{\rho(g) : g \in G\} = \mathbb{C}^{d \times d} $$ i.e., the complex linear span of $\rho(g)$, taken over all $g \in G$, coincides with the set of all $d \times d$ complex matrices.

Is this claim true and does this result have a name?

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    $\begingroup$ Burnside's theorem $\endgroup$
    – Benjamin Steinberg
    Commented May 11, 2018 at 13:32
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    $\begingroup$ Or simply that the span is a subalgebra that is invariant under the action of $G$. $\endgroup$
    – David Handelman
    Commented May 11, 2018 at 13:40
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    $\begingroup$ I cannot upvote Benjamin Steinbergs comment enough. I spent many weeks trying to prove a very similar result that also follows from Burnside's theorem. Burnside's theorem is absolutely amazing and deserves to be better known! (I also do not agree with David Handelman above that it is fully obvious that if $\mathbb{C}^d$ is a $G$-irrep so is $\mathbb{C}^{d \times d}$ even if it is true.) A good link to a (modern) proof is this: ac.els-cdn.com/S0024379503007225/… $\endgroup$
    – Vincent
    Commented May 11, 2018 at 13:52
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    $\begingroup$ @BenjaminSteinberg, as a representation theorist, I'm deeply embarrassed to have to ask: what is Burnside's theorem? The one that springs to mind is the solubility of groups with only two distinct prime factors, which surely isn't the relevant one here. $\endgroup$
    – LSpice
    Commented May 11, 2018 at 15:56
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    $\begingroup$ @LSpice, in this case it is the theorem that any subalgebras of matrices over an algebraically closed field that acts irreducibly is the whole algebra of matrices. $\endgroup$
    – Benjamin Steinberg
    Commented May 11, 2018 at 16:13

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This is known as Burnside's theorem. Nowadays people formulate it as any algebra of matrices over an algebraically closed field acting irreduciblly is the whole matrix algebra.

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  • $\begingroup$ An earlier reference might also be useful, say to the 1962 text by Curtis and Reiner. $\endgroup$
    – Jim Humphreys
    Commented May 11, 2018 at 22:57
  • $\begingroup$ P.S. Burnside himself was studying mainly finite groups, so a reference for the compact analogue might also be helpful. (However, the same proof works for both types of group, if I recall correctly.) $\endgroup$
    – Jim Humphreys
    Commented May 11, 2018 at 23:04
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    $\begingroup$ At the end of the day or has nothing to do with groups just algebras. So compact or finite makes no difference. $\endgroup$
    – Benjamin Steinberg
    Commented May 12, 2018 at 0:02
  • $\begingroup$ "Burnside's theorem" by itself leads to different results. The correct search term seems to be "Burnside's theorem on matrix algebras". An elementary proof based on rank 1 matrices, which goes back to Halperin and Rosenthal, is in The simplest proof of Burnside's theorem on matrix algebras. $\endgroup$
    – Conifold
    Commented Sep 9, 2023 at 11:55

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