Let $P$ be a nonconstant homogeneous harmonic polynomial. Then $P$ is of degree $1$ if and only if $V$ has dimension $n-1$.

The direct implication is evident. As for the reverse, if $V$ has dimension $n-1$, then $P$ is a harmonic function of one variable, and thus it is linear.

### Space of invariant polynomials

Given $x\in {\mathbb{R}}^n\setminus \{0\}$, we set ${\mathcal{P}}_d(\cancel x)$ to be the subspace of ${\mathcal{P}}_d$ of polynomials invariant with respect to the coordinate $x$. It is clear that this subspace has a uniquely defined orthogonal complement such that $$\begin{gathered}
  {\mathcal{P}}_d = {\mathcal{P}}_d(\cancel x) \oplus {\mathcal{P}}_d(\cancel x)^\perp\, ,
 \end{gathered}$$ where the direct sum is in the sense of $L^2(\partial B)$.

In the next proposition, we characterize the elements of ${\mathcal{P}}_d(\cancel x_1)^\perp$.

The linear function $K:{\mathcal{P}}_{d-1}\to{\mathcal{P}}_d$ defined by $$\begin{gathered}
  K[p]\equiv x_1p - \frac 1 {2d+n-4} {\left|x\right|}^2 \partial_1 p \equiv x_1p +c_{d,n} {\left|x\right|}^2 \partial_1 p 
 \end{gathered}$$ provides a vector space isomorphism between ${\mathcal{P}}_{d-1}$ and ${\mathcal{P}}_{d}(\cancel x_1)$. Moreover $$\begin{gathered}
\label{eq_K-}
 {\left\|K[p]\right\|}\leq {\left\|x_1 p\right\|}_{L^2(\partial B(0,1)}\leq {\left\|p\right\|}\, .\end{gathered}$$

Note that $K$ is, up to multiplicative constants, the *Kelvin transform* defined in [@HFT].

Let $p\in {\mathcal{P}}_{d-1}$, and let $q\in {\mathcal{P}}_d(\cancel x_1)$. We will prove that ${\left\langleK[p]\middle\vertq\right\rangle}=0$ by induction on $d$.

It is clear 