Decomposition of a homogeneous polynomial into a sum of harmonic and $x^2 q$. Consider the space of polynomials $\mathbb R[x_1, \ldots, x_n]$. And let $\eta$ be an $n \times n$ diagonal matrix s.t. $\eta = \mathrm{diag}(\pm 1, \ldots, \pm 1)$. Denote $x^2 = \sum_{a,b=1}^n \eta_{ab} x_a x_b$, $\Delta = \sum_{a,b=1}^n \eta_{ab} \frac{\partial}{\partial x_a} \frac{\partial}{\partial x_a}$. Whether it's true that any homogeneous polynomial $f$ can be uniquely split as $f = f_0 + x^2 f_1$, with $\Delta f_0 = 0$? I know a nice argument with an inner product on polynomials which allows to prove the statement when $\eta$ is positive- or negative-definite. How can I prove it when $\eta$ is indefinite? $\Delta, \frac14 x^2, \frac{d}{2} + N$, where $N$ counts a degree of a polynomial, form a Lie algebra $\mathfrak{sl}(2)$ under commutations. Maybe this can be useful?

If $\eta=I_n$ this is true over $\mathbb{R}$, as you said ( a classic for the theory of harmonic polynomials). But then it is true over $\mathbb{C}$ (take real and imaginary part). Now you can change the system of coordinates over $\mathbb{C}$ to get the result for $\eta$ over $\mathbb{C}$. ( this involves some $i$'s). But then the result is true over $\mathbb{R}$ too.

Note that the result is an algebraic one. It works under more general conditions, where you can use two separate quadratic forms, one for the multiplicator, and the other for the differential operator.