Consider the action of the orthogonal group $\operatorname{O}(d)$ on $k$-way tensors $(\mathbb{R}^d)^{\otimes k}$ defined by


$$Q(x\_1\otimes\cdots\otimes x\_k)=Qx\_1\otimes\cdots \otimes Qx\_k$$


and extending linearly. I would like to locate a proof that the invariant subspace of totally symmetric tensors $\operatorname{Sym}^k(\mathbb{R}^d)$ carries an irreducible representation of $\operatorname{O}(d)$.


If we replace $\operatorname{O}(d)$ with $\operatorname{GL}(V)$ for some vector space $V$, then [Schur-Weyl duality](https://en.wikipedia.org/wiki/Schur%E2%80%93Weyl_duality) gives that the corresponding representation in $\operatorname{Sym}^k(V)$ is irreducible (see Chapter 9 in [Procesi's Lie Groups book](http://www.springer.com/us/book/9780387260402)). In the case where $V=\mathbb{C}^d$, Theorem 1.9 of [this thesis](https://arxiv.org/abs/quant-ph/0604183v1) suggests that the restriction of any irreducible representation of $\operatorname{GL}(\mathbb{C}^d)$ to $\operatorname{U}(d)$ is irreducible. The proof is said to be contained in Chapter 12 of Carter, Segals and MacDonald's [Lectures on Lie Groups and Lie Algebras](https://rads.stackoverflow.com/amzn/click/0521499224), but the closest result I could find (Proposition 12.3) gives that every representation of $\operatorname{U}(d)$ is the restriction of a unique holomorphic representation of $\operatorname{GL}(\mathbb{C}^d)$. It is not clear to me that $\operatorname{U}(d)$ then inherits irreducible representations from $\operatorname{GL}(\mathbb{C}^d)$. Furthermore, this result seems to leverage a relationship between $\operatorname{U}(d)$ and $\operatorname{GL}(\mathbb{C}^d)$ known as [complexification](https://www.encyclopediaofmath.org/index.php/Complexification_of_a_Lie_group) that is not exhibited between $\operatorname{O}(d)$ and $\operatorname{GL}(\mathbb{R}^d)$.



Does the orthogonal group act irreducibly on totally symmetric tensors?
It is classical that, as $O(n)$-representation,
$$
\text{Sym}^k(\mathbf R^n)=H^k\oplus qH^{k-2}\oplus q^2H^{k-4}\oplus\ldots
$$
Here $q=x\_1^2+\ldots+x\_n^2$ is the quadratic form defining $O(n)$ and $H^k\subseteq \text{Sym}^k(\mathbf R^n)$ is the space of harmonic polynomials, i.e., polynomials which are killed by the Laplacian $\Delta=\partial\_{x\_1}^2+\ldots+\partial\_{x\_n}^2$. Moreover all spaces $q^lH^k$ are irreducible provided $n\ge2$.


