# Does the orthogonal group act irreducibly on totally symmetric tensors?

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 gives that the corresponding representation in $\operatorname{Sym}^k(V)$ is irreducible (see Chapter 9 in Procesi's Lie Groups book). In the case where $V=\mathbb{C}^d$, Theorem 1.9 of this thesis 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, 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 that is not exhibited between $\operatorname{O}(d)$ and $\operatorname{GL}(\mathbb{R}^d)$.

• This is false. If $k=2$, $\sum_{i=1}^d e_i \otimes e_i$ is an invariant vector. – Will Sawin Sep 12 '17 at 12:33
• Will Sawin's comment corresponds to the fact that the identity matrix is a fixed point of the action of $O(d)$ on the space of symmetric $d\times d$ matrices. – Thomas Richard Sep 12 '17 at 12:41

## 1 Answer

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 $n\ge2$.

• "Classical" is nice. A reference is nicer :) – darij grinberg Sep 12 '17 at 13:30
• Well, "classical" means here "easy to find by searching the web". I found e.g. the Wikipedia article on "spherical harmonics" or section 5.6.4. of the book of Goodman-Wallach. – Friedrich Knop Sep 12 '17 at 13:56
• Let me just add that analogous thing is true for $Sp(n)$ and the symplectic form. For details see again Symmetry and representation theory by Goodman and Wallach. – Vít Tuček Sep 13 '17 at 7:01