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Let $G$ be a compact group and let $V$ be a finite-dimensional complex vector space with a Hermitian inner product. Let $$\alpha : G \to U(V)$$ be an irreducible unitary representation. Then this gives rise to an action $\beta$ of $G$ on $\mathrm{End}_{\mathbb{C}}(V)$ given by $$ \beta_x(T) = \alpha_x T \alpha_x^*. $$

If $P = P^* = P^2$ is a rank-one projection in $\mathrm{End}_{\mathbb{C}}$, then so is $\beta_x(P)$ for each $x \in G$, i.e. the whole orbit consists of projections. I would like to find conditions which guarantee that these projections span the entire algebra of endomorphisms of $V$.

Question:

Is it known when

$$ \mathrm{span} \{ \beta_x (P) \mid x \in G \} = \mathrm{End}_{\mathbb{C}}(V)? $$

Some partial data:

  1. If $G$ is a compact Lie group and $P$ is the projection onto the span of the highest (or lowest) weight vector, then the condition holds.
  2. If $G = SU(2)$ and $V$ is the three-dimensional irreducible representation, this can fail for some $P$. For instance, take $P$ to be the projection onto the (one-dimensional) space of vectors of weight $0$.

I am mainly interested in the situation for finite groups and for compact Lie groups, but if there is anything more general out there I would of course be interested in that as well.

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  • $\begingroup$ If I understand your question, it's as to when $T$ is a "cyclic vector" for $End(V) = V^\ast \otimes V$, not that that helps me so much. $\endgroup$
    – Allen Knutson
    Oct 18, 2011 at 4:16
  • $\begingroup$ @Allen, yes, that is what I am asking. $\endgroup$
    – MTS
    Oct 18, 2011 at 5:01
  • $\begingroup$ There is something I don't understand in your question. The projections always lie in the proper subspace of Hermitian endomorphisms ($P^*=P$), so they can never span $End(V)$. $\endgroup$
    – Claudio Gorodski
    Oct 18, 2011 at 16:08
  • $\begingroup$ @Claudio, that is not true. For instance if $P$ is Hermitian, $iP$ is anti-Hermitian. Any operator can be decomposed into a linear combination of Hermitian operators: $T = \frac{1}{2}(T + T^*) + i \frac{1}{2i}(T-T^*)$. $\endgroup$
    – MTS
    Oct 18, 2011 at 16:26
  • $\begingroup$ i.e. the Hermitian endomorphisms do not form a subspace. $\endgroup$
    – MTS
    Oct 18, 2011 at 16:27

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