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.


2 Answers 2


I'd say it's the same, up to (an uninteresting) scaling, and possible inclusion of some torsion information.

My preferred interpretation of the Hilbert polynomial is that it encodes the $K$-class of the subscheme, whereas the degree computes the Chow class. In particular these make sense for any other ambient variety $X$, and if $X$ is projective then one can define "the Hilbert scheme of subschemes of $X$ with $K$-class $\alpha$" and be sure that it, too, exists and is projective.

What I can't remember is whether the Chow groups of weighted projective spaces are actually ${\mathbb Z}$, or whether they're ${\mathbb Z} \oplus ({\mathbb Z}/n)$ for some $n$s. If the latter, then the Chow class of a subvariety would have a usual $\mathbb N$-valued degree but also a value in this torsion group.

  • 1
    $\begingroup$ The Chow groups are indeed free, thanks to a computation of Al-Amrani from the 1980's. (The K_0 is also free, also by Al-Amrani.) $\endgroup$ Commented Nov 21, 2012 at 9:24

It is not clear to me how you intend to define the Hilbert polynomial. There are line bundles $\mathcal O(n)$, but $\mathcal O(n) \otimes \mathcal O(m) \neq \mathcal O(n+m)$, so there is no reason for the Hilbert function to be asymptotically polynomial.

A wild guess would be that the right thing to do is to consider the map from the unweighted projective space $(y_1:\dots : y_n)$ to your weighted projected space $(x_1: \dots : x_n)$ that sets $x_i=y_i^{w_i}$. Pull your variety back along this map, then compute its degree there.

This has the advantage that it computes the right degree for hypersurfaces - the degree of a hypersurface is the (weighted) degree of its defining equation.

  • $\begingroup$ Could you remind me of condition we need to have Hilbert polynomial? $\endgroup$
    – user2013
    Commented Nov 21, 2012 at 5:10
  • $\begingroup$ Hilbert polynomial in its most general form comes from a proper scheme $X$, a coherent sheaf $\mathcal F$, a line bundle $\mathcal L$. Then Hilbert polynomial is defined to be $p(n)=\chi(\mathcal F\otimes \mathcal L^{\otimes n})$. If $\mathcal L$ is ample, then this is eventually equal to $\Gamma(X,\mathcal F\otimes \mathcal L^{\otimes n})$ by whatever vanishing theorem. $\endgroup$
    – Will Sawin
    Commented Nov 21, 2012 at 5:33
  • $\begingroup$ I am aware of the definition of Hilbert polynomial but I wonder where you use the condition $\mathcal{O}(n)\times\mathcal{O}(m)=\mathcal{O}(n+m)$. $\endgroup$
    – user2013
    Commented Nov 21, 2012 at 9:57
  • $\begingroup$ What line bundle do you think is appropriate to use? Maybe the point I'm making is that there is no good thing to define to equal $\mathcal L$. $\endgroup$
    – Will Sawin
    Commented Nov 21, 2012 at 16:30

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