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Let $X$ be a smooth surface of genus $g$ and $S^nX$ its n-symmetrical product (that is, the quotient of $X \times ... \times X$ by the symmetric group $S_n$). There is a well known, cool formula computing the Euler characteristic of all these n-symmetrical products:

$$\sum_{d \geq 0} \ \chi \left(X^{[d]} \right)q^d \ \ = \ \ (1-q)^{- \chi(X)}$$

It is known that $S^nX \cong X^{[n]}$, the Hilbert scheme of 0-subschemes of length n over $X$. Hence, the previous formula also computes the Euler characteristic of these spaces.

What about for singular surfaces? More precisely, if $X$ is a singular complex algebraic curve, do you know how to compute the Euler characteristic of its n-symmetrical powers $S^nX$? More importantly: what is the Euler characteristic of $X^{[n]}$, the Hilbert scheme of 0-schemes of length n over $X$?

I guess it is too much to hope for a formula as neat as the one given for the smooth case. Examples, formulas for a few cases or general behaviour (e.g. if for large n, $\chi\left(X^{[n]}\right) = 0)$ are all very welcome!

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    $\begingroup$ James, please be cautious about editing your question after answers have started to appear! VA gave a very nice answer using your original notation, whose meaning you've now changed. $\endgroup$
    – Tim Perutz
    Commented Jan 17, 2010 at 4:32

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For singular plane curves, there is a conjectural formula (due to Alexei Oblomkov and myself) in terms of the HOMFLY polynomial of the links of the singularities. For curves whose singularities are torus knots, i.e. like x^a = y^b for a,b relatively prime, and for a few more singularities, the conjecture has been established. See this preprint.

Edit: More recently I have given a different characterization of these numbers in terms of multiplicities of certain strata in the versal deformation of the singular curve.

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    $\begingroup$ This is a beautiful conjecture!! $\endgroup$
    – Geordie Williamson
    Commented Mar 18, 2010 at 8:38
  • $\begingroup$ Thanks for your answer. Indeed, I attended the talk you gave in London on your conjecture and results! I think that Pandharipande's notes also sheds light to the question: math.princeton.edu/~rahulp/hilbert.pdf $\endgroup$
    – James O
    Commented Mar 18, 2010 at 12:46
  • $\begingroup$ May I ask how you arrived at this question? In particular, does it come from studying the Hilbert schemes of points on surfaces on which the curve may sit? It would be very useful to me to know a formula looking roughly like: "the generating function of Euler characteristics of Hilbert schemes of singular curves on a surface is the trace / vacuum expectation value of some generalized Nakajima operator associated to the singular curve"... $\endgroup$
    – Vivek Shende
    Commented Mar 18, 2010 at 15:37
  • $\begingroup$ I was thinking about some questions on stable pairs invariants - one of the ideas lead to this question. But then the problem sounded too hard (indeed, an open problem) and I started exploring some other direction. So, I am sorry but I don't know any interpretation of this question on the Euler char. in terms of Nakajima operators. Actually, I went towards the study of these operators to avoid these Euler characteristic computations. $\endgroup$
    – James O
    Commented Mar 23, 2010 at 10:35
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The same formula holds for $\chi(X^{[n]})$ of any topological space for which $\chi(X)$ is defined and behaves in the expected way for unions, Cartesian products, and quotients by a finite free action. This includes the category of all algebraic varieties over $\mathbb C$.

For example, $$\chi(X^{[2]})= \frac{\chi(X\times X)- \chi(\operatorname{diag} X)}{2} + \chi(X) = \frac{ \chi(X)^2 + \chi(X) }{2},$$ etc. No matter whether $X$ is a smooth curve or a singular 3-fold.

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    $\begingroup$ Thanks a lot VA, it also helped me better phrase the question to the part I am stuck on - The Hilbert scheme part. Thanks again! Also, sorry for my inappropriate change of notation: $X^{[n]}$ here in VA's argument means $S^nX$. My bad. $\endgroup$
    – James O
    Commented Jan 17, 2010 at 4:48
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(Edited to reflect adjusted question.) The topology of symmetric products of a non-singular curve was worked out by Macdonald, Topology 1 (1962), 319-343. The starting point is the observation (which is quite general, hence applies to singular curves) that the rational cohomology of the symmetric product is the invariant part of the cohomology of the Cartesian product, from which one can work out the Euler characteristic.

The Euler characteristic of the Hilbert scheme of $n$ points on a curve is a rather subtler problem, since these spaces feel the nature of the singularities. In the simplest case, a curve with one node, I wrote down how to describe the Hilbert scheme as a variety with normal crossings here (see Section 3.2 and the appendix). I also described its cohomology. From that you can write down the Euler characteristic.

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It is the partitions of n p(n) as is shown in this paper by De Cataldo Link

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  • $\begingroup$ Va P, for James "a surface" is not a two-dimensional complex surface but a one-dimensional complex curve. Moreover it is singular. $\endgroup$
    – Dmitri Panov
    Commented Jan 28, 2010 at 8:12

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