Is this an irreducible representation of the symmetric power?

Question

Let $\mathcal{H}$ be a finite dimensional (or separable?) Hilbert space and let $\pi$ be an irreducible representation of G. Let $v\in\mathcal{H}.$ There is an induced representation on $S^n\mathcal{H}.$ Is the subrepresentation generated by $v^{\otimes n}$ irreducible? I've found that this is true if $G$ is the group of invertible linear operators on $\mathcal{H}.$ Is it true in general, or if the representation is unitary?

Answer 1

Your question is missing details: you don't say what $G$ is (is it a Lie group, a compact group, etc), presumably $H$ is supposed to be a representation of $G$ but you don't say that, etc.

If $G = GL(H)$ then $S^n(H)$ is itself irreducible so the subrepresentation generated by $v^n$ is always all of $S^n(H)$ and irreducible, so this case isn't representative of what happens in general.

As stated, the answer to the question is no: we can take $n = 1$ and then you're asking whether the subrepresentation generated by a vector $v$ is always irreducible. It's easiest to say what happens in the finite case: if $G$ is finite then the subrepresentation of a representation $H$ generated by a vector $v \in H$ can be any quotient of the regular representation $\mathbb{C}[G]$, and in particular can be the regular representation itself (which is never irreducible unless $G$ is trivial), if we take $H = \mathbb{C}[G], v = e$.

So a more interesting version of the question is to also assume that $H$ is irreducible. In this case the answer is still no: in fact for $G$ finite, under some additional hypotheses we can find $n$ such that the subrepresentation generated by $v^{\otimes n}$ is the regular representation (so as large as possible). This will require some lemmas.

Lemma 1: Let $V$ be a finite-dimensional vector space over an infinite field, and let $v_1, \dots v_k$ be distinct nonzero vectors in $V$. Then there exists a codimension-$1$ subspace $W$ of $V$ not containing any of the $v_i$.

Proof. This is equivalent to showing that there exists a linear functional $f \in V^{\ast}$ which does not vanish on any of the $v_i$. Also equivalently, the annihilators $\text{Ann}(v_i) = \{ f \in V^{\ast} : f(v_i) = 0 \}$ of each of the $v_i$ are codimension-$1$ (hence proper) subspaces of $V^{\ast}$, and we want to show that their union is not all of $V^{\ast}$. This is the standard fact that a vector space over an infinite field is not a union of finitely many proper subspaces. $\Box$

Lemma 2: Let $V$ be a finite-dimensional vector space over a field $K$ of characteristic zero, and let $v_1, \dots v_k$ be distinct nonzero vectors in $V$, none of which are scalar multiples of each other. Then there exists some $n$ such that the vectors $v_1^n, \dots v_k^n$ are linearly independent in $S^n(V)$.

Proof. It suffices to take $n = k-1$. Since $K$ has characteristic zero, it's infinite, so we can apply Lemma 1 to find a codimension-$1$ subspace $W$ not containing the $v_i$. Let $w_0 \in V$ be any vector not in $W$, so that we can write $V \cong W \oplus W_0$ where $W_0 = \text{span}(w_0)$, and hence write

$$v_i = w_i + c_i w_0$$

where $w_i \in W$ and each $c_i \neq 0$. By dividing by $c_i$ we can assume WLOG that $c_i = 1$. Consider the expansion

$$v_i^n = (w_i + w_0)^n = \sum_{k=0}^n {n \choose k} w_i^k w_0^{n-k}$$

in $S^n(V) \cong \bigoplus_{k=0}^n S^k(W) \otimes S^{n-k}(W_0)$; note that the monomials $w_i^k w_0^{n-k}$ are linearly independent in $S^n(V)$.

Now consider the determinant of the $(n+1) \times k$ (so $k \times k$) matrix whose entries are $w_i^j \in S^n(W)$ (let's say arranged so that the rows are $1, w_i, \dots w_i^{k-1}$). This is the Vandermonde determinant, so it's equal (up to a sign) to $\prod_{i < j} (w_i - w_j) \in S(V)$. Since the $v_i$ aren't scalar multiples of each other, the $w_i$ are distinct (this is an if-and-only-if), so this determinant is nonzero. It follows that the rows of this matrix are linearly independent (over $S(V)$, so in particular over the underlying field $K$), so the same is true of the $v_i^n$, as desired (this is where we need that $K$ has characteristic zero, to divide by the binomial coefficients). $\Box$

Corollary: Let $V$ be a representation of a finite group $G$ over a field $K$ of characteristic zero such that there exists a vector $v \in V$ whose orbit $\{ gv : g \in G \}$ consists of $|G|$ distinct vectors none of which are scalar multiples of each other. Then $v^n$ generates the regular representation of $G$ in $S^n(V)$ for $n = |G| - 1$.

Proof. The subrepresentation of $S^n(V)$ generated by $v^n$ is $\text{span}((gv)^n, g \in G)$. With the given hypotheses Lemma 2 can be applied and implies that $(gv)^n \in S^n(V)$ are linearly independent for $n = |G| - 1$; hence this subrepresentation is the regular representation. $\Box$

The condition needed on the representation $V$ here is related to but stronger than faithfulness: it suffices for $V$ to be faithful and for no element of $G$ to act as multiplication by a scalar, which is guaranteed if $G$ has trivial center. In particular it is possible for $V$ to be irreducible. For example we can take $V = \mathbb{C}^2, G = S_3 \cong D_3$ acting by rotations and reflections of a triangle, and the generic vector $v \in V$ will satisfy the condition above.