Homogeneous polynomials - Explanation of $f(x,y,z) \in P^k$, then $f(x,y,z)= \sum_{i \geq 0} f_i(x,y) z^{k-i}$

Question

In the document Spectral Geometry in Non-standard Domains at the end of page $37$, they display, without explanations, that if $f(x,y,z) \in P^k$, then $f(x,y,z)= \sum_{i \geq 0} f_i(x,y) z^{k-i}$. I could convince my self, but rigorously, is there anyone could explain to me why is it true in general?

Answer 1

If $f(x,y,z)\in P^k(x,y,z)$, then it must be of the form $$f(x,y,z)=\sum_{k_x+k_y+k_z=k} f_{k_x,k_y,k_z}x^{k_x}y^{k_y}z^{k_z}$$ where the sum runs over all $(k_x,k_y,k_z)\in\mathbb{N}^3$ whose total is $k$. Introducing $i=k-k_z$, we can rewrite this in order to pick out the $z$-dependence:

$$f(x,y,z)=\sum_{k_x+k_y-i=0} f_{k_x,k_y,k-i}x^{k_x}y^{k_y}z^{k-i} =\sum_{i=0}^k z^{k-i}\left(\sum_{k_x+k_y=i} f_{k_x,k_y,k-i}x^{k_x}y^{k_y}\right)$$ where $i=k-k_z$. Identifying each inner sum with $f_i(x,y)$, we see that each is indeed in $P^i(x,y).$

Answer 2

It seems to me that they are considering the polynomial $f$ as a polynomial in $z$ for a moment (looking at $x,y$ and their powers as coefficients) so that this is indeed the expansion of a polynomial of degree $k$ in $z$. Clearly the degree cannot exceed $k$ in $z$ for that would result in $f \notin P^k$, the lesser terms are always accompanied by powers $x^a y^b$ so that $i = a +b$ and each term has degree $k = k-i + a + b$.