It is known that for affine and projective varities $V$ that it is possible to construct a polynomial such that for all $s$ sufficiently large, the polynomial is equal to the Hilbert function of $V$. The polynomial is called the Hilbert polynomial and it has the property that the degree of the polynomial is equal to the dimension of the variety, and that $(\dim V)!$ times the leading coefficient of the Hilbert polynomial is equal to the degree of the variety.
I am wondering what the analogous result would be for weighted projective varieties. A weighted projective variety is a variety in a weighted projective space, say with weight vector $(w_1, \cdots, w_n) \in \mathbb{Z}_{>0}^n$. Is it still true that the correct notion of degree for a variety $V$ in such a space be the product of $(\dim V)!$ times the leading coefficient of the Hilbert polynomial, or should the weight vector be factored in some how?
Is the notion of 'degree' well defined in this context? What if we restrict to weight vectors of the form $(1, 1, \cdots, k)$ for some $k > 1$?
I've found the following reference:
Sir, Zybnek, Approximate parametrisation of confidence sets (page 88), Computational Methods for Algebraic Spline Surfaces: ESF Explorator Workshop, Springer, 2005
The relevant quote:
"Using the Hilbert function, we can define the degree also for varieties in weighted projective spaces. It is natural to define that a surface $S$ has almost minimal degree $d$ if the leading coefficient of the Hilbert polynomial is $\frac{d}{2}$, and the value of the Hilbert function at $m=1$ is $d+1$."
This seems to imply that Hilbert polynomials are well-defined for weighted projective varieties, and that the degree of the variety can be defined from that polynomial. However, I don't understand why the leading coefficient of the Hilbert polynomial in this case ought to be $d/2$.
Thanks for any references or insights.