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 $$\sum z^n [S^n \Sigma] = exp\left( \sum \frac{z_i P_\Sigma[i]}{(-1)^{i-1}i} \right) \cdot 1 $$$$\alpha \mapsto exp\left( \sum \frac{z_i P_\alpha[-i]}{(-1)^{i-1}i} \right) \cdot 1 $$ that(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 9984), is well defined if we replace. By one of his results, $\Sigma$ by any 2-homology class$[\Sigma] \mapsto \sum z^i [\Sigma^{[n]}]$ $\alpha$(op. cit. page 99). If so, I presume this satisfy the posed question.