How could I find an orthogonal basis for $H^k$? - HELP!! Let $P^k=$homogeneous polynomials of degree $k$ in $x$, $y$, $z$, $k=0, 1, 2, \dots, $ i.e. $P^k= \text{span} \{x^{k_x}y^{k_y}z^{k_z} : k_x+k_y+k_z=k\}$ and $H^k= \{f \in P^k : \Delta f = 0\}$, where $\Delta$ is the usual Laplacian operator.
Question : Without using the spherical coordinate, if I give $P^3= \{x^3, y^3, z^3, x^2y, x^2z,xy^2,y^2z, y^2z,xz^2,yz^2, xyz\}$, what could be a basis for $H^3$ and why? I think I can use recursively the fact that $H^0=P^0$ and $H^1=P^1$, but I don't even know how. It seems like as $Δ$ is a linear map, and you have bases of $P^k$ and $P^{k−2}$, it is a linear algebra exercise to find a basis of $\ker⁡Δ$ in $P^k$. I think an excellent article for that should be SPHERICAL HARMONICS AND HOMOGENEOUS HARMONIC POLYNOMIALS page $5$, but I am not quite sure how to do it.
Thanks!
 A: Yes, you can consider it a standard (but quite tedious) linear algebra exercise for any fixed $n,k$ (in your example, $n=k=3$). 
More generally, the article you linked provides the following answer on page 3. Although it's not really clear what you mean by "finding a basis". 
$H^k$ is the orthogonal complement of all polynomials of the form $|x|^2 P(x)$, where $P \in \mathcal{P}^{k-2}$. Here the inner product is 
$$ \langle x^\alpha,x^\beta \rangle = \alpha_1! \ldots \alpha_n! \delta_{\alpha,\beta}. $$
A nice proof is given in E.M. Stein's Singular Integrals and Differentiability Properties of Functions. First, any polynomial $P = \sum_\alpha a_\alpha x^\alpha \in \mathcal{P}^k$ corresponds to a differential operator $P(D) = \sum_\alpha a_\alpha D^\alpha$, where $D^\alpha = \frac{\partial}{\partial x_{\alpha_1}} \ldots \frac{\partial}{\partial x_{\alpha_k}}$. This correspodence is a ring homomorphism and in particular $(|x|^2)(D) = \Delta$. 
Then the inner product in $\mathcal{P}^k$ can be introduced as 
$$ \langle P,Q \rangle = P(D) Q. $$
Check that this coincides with the previous one!
Let $Q \in \mathcal{P}^k$. From the identity 
$$ \langle |x|^2P, Q \rangle = P(D) (\Delta Q)  = \langle P, \Delta Q \rangle $$
we see that $Q \perp |x|^2P$ for every $P \in \mathcal{P}^{k-2}$ if and only if $\Delta Q = 0$. 
