\cos((d-1)\theta)\, ,\\
 {\left|\nabla P_d(r,\theta)\right|}={\left|\nabla C_d(r,\theta)\right|} = 2d r^{d-1}\, \label{eq_nabla2vars} ,\\
 \frac{\partial^k}{\partial x^k} P_d(r,\theta)|_{(0,0)} = 2 \binom{d}{k} k! r^{d-k}\sin((d-k)\theta)= \binom d k k! P_{d-k}(r,\theta)\, ,\\
  P_d((t,0)+(r,\theta)) = P_d(r,\theta) + \sum_{k=1}^d \binom{d}{k} t^k P_{d-k}(r,\theta)\, .\end{gathered}$$

### Cone Splitting

By simple algebra, it is easy to see that the set of point wrt which $P$ is homogeneous forms a vector subspace of ${\mathbb{R}}^n$. Indeed, let $P$ be homogeneous wrt $0$ and wrt $x\neq 0$, and pick any $y\in {\mathbb{R}}^n$. By Euler’s formula $$\begin{gathered}
 d\cdot P(y) = \nabla P|_y \cdot y = \nabla P|_y \cdot (y-x)\quad \Longrightarrow \quad \nabla P|_y \cdot x =0\, ,\end{gathered}$$ and thus the partial derivative of $P$ in the $x$ direction vanishes at every point, making $P$ invariant wrt the subspace spanned by $x$. In other words, if $P$ is a harmonic polynomial of degree $d\geq 1$, the set $$\begin{gathered}
 V\equiv{\left\{v\in {\mathbb{R}}^n \ \ s.t.  \ \ \nabla P\cdot v =0\right\}}\end{gathered}$$ is a vector subspace of ${\mathbb{R}}^n$, and it is the invariant subspace of $P$ in the sense that $P(x+v)=P(x)$ for all $x\in {\mathbb{R}}^n$ and $v\in V$.

In the next proposition, we prove an extremely important (and simple) relation between the degree of $P$ and the dimension of $V$.

\[prop\_pn-2\] 