# Orbit of rank-one projection under compact group action

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.

• 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. – Allen Knutson Oct 18 '11 at 4:16
• @Allen, yes, that is what I am asking. – MTS Oct 18 '11 at 5:01
• 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 $End(V)$. – Claudio Gorodski Oct 18 '11 at 16:08
• @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: $T = \frac{1}{2}(T + T^*) + i \frac{1}{2i}(T-T^*)$. – MTS Oct 18 '11 at 16:26
• i.e. the Hermitian endomorphisms do not form a subspace. – MTS Oct 18 '11 at 16:27