Definition by degree and intersection number are equivalent (linking number). [repost]

Question

I will here restate a question I asked earlier. It did not have much succes (probably by an incomplete introduction of the problem on my part).

I am reading a paper by Ricca ( http://www.maths.ed.ac.uk/~aar/papers/ricca.pdf p.1338 on the bottom) on the linking number. I would like to see that its definition as the degree of the map $\Psi$ is equal to that by intersection number.

A link $L$ has two component knots $\gamma_1$ and $\gamma_2$, these denote embeddings of $S^1$ in $\mathbb{R}^3$. Consider $$\Psi: \mathbb{T} \,\,(= C_1\times C_2)\to S^2: (s,t)\mapsto \textbf{n}(s,t) = \frac{\gamma_1(s) - \gamma_2(t)}{\bigl|\gamma_1(s) - \gamma_2(t)\bigr|} $$ Then $\text{deg} \Psi$ is one definition of the linking number. Taking a Seifert Surface $M$ of $\gamma_1$ (i.e. an orientable surface with boundary $\gamma_1$) then its intersection number with $\gamma_2$ will also give the linking number.

Now I would like to prove that these are equivalent. Therefore consider the orientable manifold
$$N = (M\times\gamma_2) \,\, \bigm\backslash \,\bigcup_{m\in M\cap \gamma_2} \mathcal{B}(m,\epsilon) $$ with boundary $$\partial N = (\gamma_1\times\gamma_2) \cup \bigl(\bigcup_{m\in M\cap\gamma_2}\mathcal{S}(m,\epsilon)\bigr) $$ Also consider $\overline{\Psi}:N\to S^2: (x,y)\mapsto \frac{y-x}{|y-x|}$. Its restriction to $\gamma_1\times\gamma_2$ is now $\Psi$.

The degree of $\overline{\Psi}$ restricted to $\partial N$ is now zero. So we get $$\text{deg}\biggl(\overline{\Psi}\biggm| {\gamma_1\times \gamma_2}\biggr) + \text{deg}\biggl(\overline{\Psi}\biggm|{\bigcup_{m\in M\cap \gamma_2} \mathcal{S}(m,\epsilon)} \biggr) =0 \text{.} $$ The first term gives the linking number defined by degree. How do I see that the second term gives the intersection number?

I know this has to do with the orientations, but I don't have enough feeling for this yet, to write this down rigorously. I really hope that someone could explain how this works or at least give me hint. Thanks in advance.

Answer 1

Here's the idea in a more general setting. If you have two oriented submanifolds $X$ and $Y$ intersecting transversally at a point in $\mathbb{R}^n$ (or in an arbitrary oriented manifold but viewed locally as living in $\mathbb{R}^n$ so that you can use vector operations), then the map $\psi: (X \times Y )\setminus (X \cap Y) \to S^{n-1} \subset \mathbb{R}^n$ given by $$(x,y) \mapsto \frac{y-x}{\|y- x\|}$$ encodes the sign of intersections between $X$ and $Y$. It suffices to consider a local model with oriented copies of $\mathbb{R}^k$ and $\mathbb{R}^\ell$ inside $\mathbb{R}^{k+\ell}$. The above map is defined on $(\mathbb{R}^k \times \mathbb{R}^\ell)\setminus \{(0,0)\}$. For convenience, let $\bar S^{n-1}$ and $S^{n-1}$ denote the unit spheres in $\mathbb{R}^k \times \mathbb{R}^\ell$ and $\mathbb{R}^{k+\ell}$, respectively, each oriented as the boundary of the unit ball. Note that the obvious map $\mathbb{R}^k \times \mathbb{R}^\ell \to \mathbb{R}^{k+\ell}$ is orientation-preserving if and only if the intersection $\mathbb{R}^k\cap \mathbb{R}^\ell$ is positive. It follows that the restriction $\bar S^{n-1} \to S^{n-1}$ of this map has degree positive (resp. negative) one when $\mathbb{R}^k$ and $\mathbb{R}^\ell$ intersect positively (resp. negatively). The restriction of $\psi$ to $\bar S^{n-1}$ is simply $(x,y)\mapsto y-x$, and we note that it factors as a composition $\bar S^{n-1} \to \bar S^{n-1} \to S^{n-1}$, where the first map is $(x,y)\mapsto (-x,y)$ and the second map is $(x,y)\mapsto x+y$. The first map has degree $(-1)^k$ and the second map has degree positive (resp. negative) one when the intersection is positive (resp. negative).

Now in our setting we have nearly the same setup, but we'll have to make one minor change that ultimately flips the degree. For each intersection point $(m_\alpha,t_\alpha) \in M \cap \gamma_2$ we remove a neighboring 3-ball from $M \times S^1$, obtaining a new manifold $N$. Note that each spherical boundary component $S^2_\alpha$ is oriented as the boundary of $N$, not the boundary of the (removed) 3-ball, so the orientation of $S^2_\alpha$ is the reverse of the sphere $\bar S^{n-1}$ in the above discussion. Thus, at a positive intersection point, the degree of $\overline{\Psi}$ is $-(1)(-1)^{k=2}=-1$. Simiarly, its degree is $-(-1)(-1)^{k=2}=1$ at negative intersection points.