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Hello, Consider the hilbert polynomial for a projective scheme. The degree, dimension and arithmetic genus extract information from the lowest term and the highest term in the polynomial. What about all other terms? It would seem they encode some more info about our scheme. I could not find any reference to these coefficients, though. So my question is what else can we learn about our scheme from the hilbert polynomial?


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Just a reminder that, the Hilbert polynomial encodes some info of the embedded geometry of the variety, rather than intrinsic geometry (although some invariants it encodes are intrinsic, e.g. dimension and $p_a$). From the polynomial one knows something about how the variety is embedded, and I don't think I would hope to find a lot about the variety itself. – shenghao Apr 3 '11 at 14:55
a reference for the formula for the coefficients of the hilbert polynomial in terms of the arithmetic genera of the linear sections of the embedded variety, as in the nice answers below, is p. 115 of Mumford's Algebraic Geometry I, Complex projective varieties. In general I recommend this source as an introductory yet deep account of many useful concepts. – roy smith Apr 5 '11 at 22:11
up vote 13 down vote accepted

For a curve, that's all of course, since the polynomial is linear. Now let's say $X$ is a smooth surface with an ample divisor $H$ and canonical divisor $K$, we have the Hilbert polynomial $$\chi(\mathcal{O}_X(nH))= \frac{1}{2}nH(nH-K) + \chi(\mathcal{O}_X)$$ by Riemann-Roch. So the linear coefficient gives you the degree of the canonical divisor. In higher dimensions, the more general form of Riemann-Roch $$\chi(\mathcal{O}_X(nH)) = \int_X ch(\mathcal{O}(nH))td(X)$$ tells you that you're basically getting certain Chern numbers in $X$ and $H$ as coefficients.

What is perhaps simpler, is to use the recurrence formula $$\chi(\mathcal{O}_{H\cap Y}(nH)) =\chi(\mathcal{O}_Y(nH))- \chi(\mathcal{O}_Y((n-1)H))$$ to see that the Hilbert polynomial determines and is determined by the arithmetic genera of $X$ and the complete intersections $H$, $H_1\cap H_2$ etc.

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Hi Donu, to answer your comment to my answer: I think I was probably writing up my answer as you were editing yours. Also, I did not mean to contrast my answer to yours. By "another way" I meant other than the original question. Finally, I like your recurrence formula. It is the kind of equality I referred to as transformation rules. Cheers! – Sándor Kovács Apr 4 '11 at 2:30
This reccurence formula is developed in that excellent question…. – se0808 Jun 9 '14 at 19:21

Another way to look at this is the following: In my opinion, the important invariant is the Hilbert polynomial. It is not a single number, but it is still an invariant.

Actually one should be careful with what one means by the Hilbert polynomial. It is really the Hilbert polynomial with respect to an ample line bundle or what's the same an embedding.

Anyway, my point is that the Hilbert polynomial is an invariant that subschemes of projective space have to share in order to be deformation equivalent. (It is not sufficient for that though!)

This implies that When you are constructing moduli spaces you fix the Hilbert polynomial first. The resulting moduli space is still possibly disconnected, but it will be of finite type. (Insert here a longer discussion of Hilbert schemes.)

The fact, as Donu has already pointed out, that for curves the Hilbert polynomial is equivalent to the dimension, degree, and arithmetic genus is sort of a special case. These are obvious invariants that had been studied independently of Hilbert polynomials. It is a reassurance of their importance that they make up the Hilbert polynomials of curves. One can imagine that instead of defining Hilbert polynomials, one could define the various coefficients along the way Donu explains and then get similar results, but I think it would become pretty clumsy that way as if you consider the coefficients individually their transformation rules become pretty complicated.

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Sándor, I missed your comment during my last edit. I agree that it is more important what it does for your you than what it is. – Donu Arapura Apr 3 '11 at 16:38
Thank you, I liked your explanation as well, gave me good intuition. I thought that if someone stumbels upon this question he might be interested in the formulae in Donu's answer. – Blade Apr 4 '11 at 22:05

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