If $\Sigma$ is a smooth complex curve in a smooth projective surface $X$, then we can consider the homology class represented by $\Sigma^{[n]} \subset X^{[n]}$. $\ \ $ Where, $X^{[n]}, \Sigma^{[n]}$ stand for the Hilbert scheme of $n$-points on $X$ and $\Sigma$, respectively. Is it possible to construct a homomorphism function $\Phi_n: \rm{H}_2(X) \rightarrow H_w(X^{[n]})$, such that $[\Sigma] \mapsto [ \Sigma^{[n]} ]$?

$\ \ \ $ One has the following at ones disposal: we have the obvious quotient map $X^n \rightarrow S^nX$ (where $S^nX$ is the symmetric product of $X$). Now, if $\beta \in H_2(X)$, then we can consider the image of $B := \beta \times \cdots \times \beta$ in $H_{2n}(S^nX)$. If $\beta $ can be represented by an algebraic curve, we can take the proper transform of $B$ under the Chow map $X^{[n]} \rightarrow S^nX$. If $\beta$ is not represented by such a curve, is there anything akin to proper transform that one can apply to $B$ to construct the desired homomorphism function $\Phi_n$?

I am interested in studying the intersection theory between the classes $\Phi_n(\beta)$. Nakajima in his book "Lectures on Hilbert schemes of points on surfaces" states the following nice result. If $\Sigma$ and $\Sigma'$ are two smooth curves in $X$, then (page 102 of Nakajima's book):

$$\sum_n z^n \ [\Sigma^{[n]}] \cdot [\Sigma'^{[n]}] = (1+z)^{[\Sigma] \cdot [\Sigma']}$$

Does anyone know if there are related results for singular curves?

As a side remark. the above formula is obvious if $\Sigma$ and $\Sigma'$ are two curves intersecting transversely. All it says is that of the set of $m = [\Sigma]\cdot [\Sigma']$ points were it intersects, we choose $n$ of them (there are $\binom{m}{n}$ of these guys, which is what the formula is giving). But the general proof of the formula is more intricate - one uses a representation of the Heisenberg group on the space $\oplus_n H_*(X^{[n]})$ to derive it. This fancy shmancy approach is more helpful when computing things like the self intersection of $\Sigma^{[n]}$ when $\Sigma$ is a $(-1)$-curve in $X$. From it we get that $[\Sigma^{[n]}] \cdot[\Sigma^{[n]}] = \binom{-1}{n} = (-1)^n$

EDITED: In view of Nakajima's comment below, please replace function for homomorphism when reading the above question. Notice that, as stated in my comment below, the extension of the map $[\Sigma] \rightarrow [\Sigma^{[n]}]$ should be a "nice" one.

EDITED (I am copying my hidden comments here since their maths don't display well) I can explain my motivation. I am working with some moduli spaces of objects on a surface $X$ and out of them I get a homology class $V_n$ in $X^{[n]}$. In nice cases, one can show that these homology classes are $[\Sigma^{[n]}]$, for some curve $\Sigma \subset X$. Or a sum of such classes. Using this classes $V_n$ I am trying to obtain a map $N : H_2(X) \rightarrow \mathbb{Z}$, defined by $N(\beta) := V_n \cdot \Phi_n(\beta)$. Such that, in the nice case when $V_n = [\Sigma^{[n]}]$ and $\beta = [\Sigma']$, then $$N(\beta) = [\Sigma^{[n]}] \cdot [\Sigma'^{[n]}]$$ Then, my problem became what should be the definition of $\Phi_n(\beta)$, when $\beta$ not represented by a curve. Presumably, we should be able to extend $\Phi_n$ to some 2-classes that are not represented by curves since, by perturbing the complex structure, we could start seeing more curves than before. I don't know what should be $\Phi_n(-2H)$. The best I could imagine is that it should satisfy the equation $$[\Sigma^{[n]}] \cdot \Phi_n(-2H) = \binom{\Sigma \cdot (-2H)}{n} $$ but I really don't know what it should be. Thanks a lot again!

EDIT I am now assume that the formula $$\alpha \mapsto exp\left( \sum \frac{z_i P_\alpha[-i]}{(-1)^{i-1}i} \right) \cdot 1 $$ (the definition of the term $P_\alpha[-i]$ can be found in Prof. Nakajima's book "Lectures on Hilbert schemes of points on surfaces" page 84), is well defined. By one of his results, $[\Sigma] \mapsto \sum z^i [\Sigma^{[n]}]$ (op. cit. page 99). If so, I presume this satisfy the posed question.

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    $\begingroup$ What do you mean by `a homomorphism $\Phi_n$' ? The map $\Sigma\to \Sigma^{[n]}$ is not linear in $\Sigma$. $\endgroup$ Commented Feb 13, 2010 at 5:15
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    $\begingroup$ Let $X$ be $P^2$ and $H$ be the hyperplane class. What is your definition of $\Phi_n$ for $2H$ ? Is it the symmetric product of a quadric or twice of the symmetric product of a line ? $\endgroup$ Commented Feb 14, 2010 at 4:45
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    $\begingroup$ Dear Professor Nakajima, for James's definiton, $\Phi_n$ for $2H$ is the symmetric product of the quadric. $\endgroup$ Commented Feb 14, 2010 at 14:06
  • $\begingroup$ Yes, $2H$ should be sent to the symmetric product of the quadric. Thank you Dmitri. I guess my previouss comments were not well written, so I deleated them. What I was trying to say is that I completely agree with that $\Phi_n$ could not be a homomorphism and that I am just looking for a good extension of the function $[\Sigma] \mapsto [\Sigma^{[n]}]$ to any 2-homology class. Thanks a lot Professor Nakajima and Dmitri. $\endgroup$
    – James O
    Commented Feb 14, 2010 at 19:31
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    $\begingroup$ You should not delete the comments. Just add further comments. I have used your definition given in the deleted comment to conclude $\Phi_n(2H)$ is $2(line)^{[n]}$, and get a confusion. I do not understand why you choose $(quadric)^{[n]}$ yet. How about $\Phi_n(-2H)$ ? Is it $-2(line)^{[n]}$ or $-(quadric)^{[n]}$ ? Not only giving the answer, please explain the reason why you choose. $\endgroup$ Commented Feb 14, 2010 at 22:47

1 Answer 1


I would like to make one naive suggestion, related to the work of Voisin, who constructed Hilbert scheme of every almost complex manifolds $X$ of real dimension $4$. And to a more recent work of Julien Grivaux on this topic


As far as I understand Julien can express the cohomology ring of this scheme in the case when $X$ is a symplectic manifold (Theorem 1.1). Symplectic structure is used in order to construct symplectic surfaces on $X$. These surfaces are quite plentiful on $X$ by a theorem of Donaldson. For any such symplectic surface $C$ one can find an almost complex sturcture on $X$, integrable in a neighborhood of $C$. This is discussed Section 3.2 (in particular Theorem 3.5 and Corollary 3.3).

Now if we have a smooth almost complex curve $C$ inside of $X$, such that the complex structure is integrable in its neighborhood then $C^{[n]}$ seem to be well defined as a submanifold of $X^{[n]}$ (the Voisin Hilbert scheme). Moreover, Lemma 3.8 seem to suggest that the homology class of $[C^{[n]}]$ is given by a formula analgous to the one in the case when $X$ is an algebraic surface (the formula of Nakajima).

I have an extremely superficial understanding of Julien's artice and of the whole topic, so I may make some here some silly mistake. Moreover it is not clear for me if after all the class $[C^{[n]}]$ in homologies of $X^{[n]}$ depends only on the homolgy class $[C]$ in $H_2(X)$, but will not depend on the choice of a sympectic curve $C$ that realises it. And, of course, there are some obvious (and not obvious) restictions on $C$, such as $\int_C \omega>0$.

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    $\begingroup$ Perturbing the complex structure into an almost complex structure is irrelevant. The class $-2H$ is not represented by a pseudo-holomorphic curve, as it has negative degree. For your last question, you should read Chapters 7, 8 in the book. $\endgroup$ Commented Feb 16, 2010 at 2:43
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    $\begingroup$ Thank you very much for this comment! Of course, I completely agree, that $-2H$ is not represented by an almost complex curve, this is what I mean when I say that the construction will not work when $\int_S \omega\le 0$. I will read Chapters 7 and 8 of the book $\endgroup$ Commented Feb 16, 2010 at 2:52
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    $\begingroup$ Thank you very much, in advance, for reading my book. I did not intend to advertise the book....... $\endgroup$ Commented Feb 16, 2010 at 3:23

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