In Bigelow's Braid groups are linear, he shows that the braid group $B_n$ acts faithfully on $H_2(\tilde{C})$ for some space $\tilde{C}$. The main tools used are objects Bigelow calls forks and noodles. The short set-up is as follows:
First, we have a $4$-manifold $\tilde{C}$ which is acted upon $\mathbb{Z}^2 = \langle q,t\rangle$. We are given two $2$-dimensional manifolds $\tilde{\Sigma}(N)$ and $\tilde{\Sigma}(F)$ that are properly embedded into $\tilde{C}$. Let $\hat{i}(-, -)$ denote algebraic intersection in $\tilde{C}$. Bigelow then defines an element $\langle N, F\rangle$ of $\mathbb{Z}[q^{\pm},t^{\pm}]$ by
$$\langle N,F\rangle := \sum_{a,b\in\mathbb{Z}} \hat{i}(\underbrace{q^at^b\cdot \tilde{\Sigma}(N)}_{\mathbb{Z}^2 \text{ action}}~,~\tilde{\Sigma}(F))~q^at^b.$$
I have two questions about this.
First, Bigelow states
"The problem is that one cannot necessarily defined an algebraic intersection number between two properly embedded surfaces, since it might be possible to eliminate intersections by pushing them off the infinity. We overcome this problem by proving the existence of an immersed closed surface $\tilde{\Sigma}_2(F)$ which is equal to $(1-q)^2(1+qt)\tilde{\Sigma}(F)$ outside a small neighborhood..."
What does Bigelow mean by algebraic intersection number? If it is the cup product of $\tilde{\Sigma}(N)$ and $\tilde{\Sigma}(F)$ as elements of $H^2(\tilde{C})$, I don't see the problem Bigelow seems concerned with. If it isn't, then what is it, and why is there this embedding issue?
Second, why is $\langle N,F\rangle$ well-defined? I'm assuming that $\hat{i}(q^at^b\cdot\tilde{\Sigma}(N), \tilde{\Sigma}(F))$ is nonzero only finitely often so that we don't have an infinite sum, but I don't see why this ought to be true.
After understanding more of this paper, I’m now in a position to answer this.
First, the algebraic intersection number is the cup product. The issue is that we don’t know a priori that these noncompact subsurfaces can be seen as homology classes in $H_2$ (and thus cohomology classes in $H^2$). Bigelow does some computations to prove that there is a well-defined way of doing this.
Secondly, the sum is finite because the algebraic intersection number can also be seen as the signed sum of the points of intersection between these lifted surfaces (where the orientation is induced by the orientation on the disc). Furthermore, these points correspond to points of intersection between the noodle $N$ and the fork $F$. These objects can only intersect finitely often, and thus the sum has only finitely many terms.