Decomposition of symmetric powers of the standard representation of $SO(n)$ Let $V$ be the n-dimensional standard-representation of $SO(n)$.
Since $SO(n)$ preserves a bilinear form on $V$ there is a trivial 1-dimensional subrepresentation in $S^2V$. So, in general, $S^k V$ seems not to be irreducible.
Is there any known decomposition of the $k$-th symmetric power $S^kV$ into irreducible $SO(n)$-representations? I am coming from a different area and have little knowledge about general representation theory, but any hint or reference is welcome, thanks.
 A: Assuming we're working over the field $\Bbb F = \Bbb C$ or $\Bbb F = \Bbb R$, the decomposition can be deduced from Fulton & Harris' Representation Theory: A First Course, $\S\S$18–19, which treats the representation theory of the complex Lie algebras $\mathfrak{so}_n \Bbb C$; see, e.g., Exercise 19.21.
Given a nondegenerate bilinear form $Q$ on $\Bbb V \cong \Bbb F^n$ (assumed positive definite if $\Bbb F = \Bbb R$), we can regard $\Bbb V$ as the standard representation of $SO(Q) = SO(n)$, $n \geq 3$, and then define for each symmetric power $\mathsf{S}^d \Bbb V$ of $\Bbb V$, $d \geq 2$, the linear contraction (i.e., trace) operator
$$\Phi_d : \mathsf{S}^d \Bbb V \mapsto \mathsf{S}^{d - 2} \Bbb V$$ characterized by
$$\Phi_d(v_1 \odot \cdots \odot v_d) = Q(v_1, v_2) v_3 \odot \cdots \odot v_d .$$ (By symmetry, the contraction map is independent of which indices we contract into $Q$.) Then, it follows from the description of irreducible representations of $SO(n)$ in terms of highest weights that $$\mathsf{S}^d_\circ \Bbb V := \ker \Phi_d \subset \mathsf{S}^d \Bbb V$$
is irreducible.
Applying this decomposition recursively yields a general formula for the decomposition of symmetric powers $\mathsf{S}^d \Bbb V$ of $\Bbb V$ as $SO(n)$-representations:
$$\mathsf{S}^d \Bbb V \cong \mathsf{S}^d_\circ \Bbb V \oplus \mathsf{S}^{d - 2}_\circ \Bbb V \oplus \cdots \oplus \mathsf{S}^{d - 2p + 2}_\circ  \Bbb V\oplus \mathsf{S}^{d - 2p}_\circ \Bbb V ,$$
where $p$ is the largest nonnegative integer such that $2 p \leq d$ and we denote $$\mathsf{S}^0_\circ \Bbb V := \mathsf{S}^0 \Bbb V \cong \Bbb F \qquad \textrm{and} \qquad \mathsf{S}^1_\circ \Bbb V := \mathsf{S}^1 \Bbb V \cong \Bbb V .$$
For the special case $d = 2$, we can identify the decomposition $\mathsf{S}^2 \Bbb V = \mathsf{S}^2_\circ \Bbb V \oplus \mathsf{S}^0 \Bbb V \cong \mathsf{S}^2_\circ \Bbb V\oplus \Bbb F$ with the decomposition of a general symmetric $n \times n$ matrix (over $\Bbb F$) into the sum of a tracefree matrix and scalar matrix. More generally, $\mathsf{S}^d \Bbb V$ contains a copy of the trivial representation iff $d$ is even, in which case that representation is spanned by $Q^{\odot (d / 2)}$.
Since $\Phi_d$ is surjective, we have for $d \geq 2$ that
\begin{multline*}\dim \mathsf{S}^d_\circ \Bbb V = \dim \mathsf{S}^d \Bbb V - \dim \mathsf{S}^{d - 2} \Bbb V \\ = {{n + d - 1}\choose{d}} - {{n + (d - 2) - 1}\choose{d - 2}} = \frac{n + 2 d - 2}{n - 2} {{n + d - 3}\choose{d}} ,\end{multline*} and checking manually shows that the rightmost expression also yields the correct values for $d = 0, 1$: $\dim \mathsf{S}^0_\circ \Bbb V = \dim \Bbb F = 1$ and $\dim \mathsf{S}^1_\circ \Bbb V = \dim \Bbb V = n$.
