Is the linking number via Seifert surfaces well defined?

Question

Let $i(K, F_L)$ be the signed count of intersections of an oriented knot $K$ with a Seifert surface $F_L$. (That is, $F_L$ is an oriented compact surface with boundary $L$ for some knot $L$.) I want to prove that $i(K,F_L)$ is independent of the choice of Seifert surface, to establish that the linking number $lk(K,L)$ is well defined.

Unfortunately my attempts (which were all very geometric) haven't worked, since I can only seem to prove the result for Seifert surfaces obtained via the Seifert algorithm. I've also checked a couple of books (Sporpan - Wild world of four manifolds, Rolfsen - Knots and Links), but neither of these books bother proving it. (I assume it's considered sufficiently obvious for anyone who cares about Seifert surfaces.)

Does every Seifert surface arise as from the Seifert algorithm given some projection? (If so, my proof is done.) I suspect this isn't true - in which case how can we prove that the intersection number is well defined independently of the choice of Seifert surface? This feels like it should be vaguely homological but I haven't been able to figure it out.

Answer 1

Not every Seifert surface arises from Seifert's algorithm. The minimal genus of the Seifert surface from Seifert's algorithm, over every diagram of a knot, is known as the canonical genus. The difference between the genus and the canonical genus can be arbitrarily large.

The usual homological argument is the following. Let $N$ be an open tubular neighborhood of the knot $L$, with $K$ outside of the closure of $N$. The knot exterior $X=S^3-N$ is a compact $3$-manifold with boundary, and in it we have $K$ and $F'_L=F_L\cap X$. Since $F'_L$ is an oriented surface, we may think of it as being a relative homology class $[F'_L]\in H_2(X,\partial X)$. The algebraic intersection form under consideration is $H_1(X)\times H_2(X,\partial X)\to H_0(X)$, where $H_0(X)\cong\mathbb{Z}$. Poincaré dually, this is $H^2(X,\partial X)\times H^1(X)\to H^3(X,\partial X)$ by cup products. If $K$ is a meridian loop for $L$, then it is certainly the case that $i(K,F_L)=\pm 1$, so $[F'_L]$ must not be a divisible element (the cup product is a perfect pairing on the free parts). In particular, Poincaré duality has $H_2(X,\partial X)\cong H^1(X)$, and by Alexander duality $H^1(X)\cong\mathbb{Z}$, so $[F'_L]$ is a generator for $H_2(X,\partial X)$. Which generator it is depends only on the orientation of the surface. The conclusion is that any other Seifert surface gives the same generator, hence the same algebraic intersection numbers.

A more geometric argument comes from understanding how Seifert surfaces are related by "embedded surgeries." Suppose $F'_L$ is a compressible surface, which is to say there is an embedded disk $D\subset X$ such that $D\cap F'_L=\partial D$, $D$ meets $F'_L$ transversely, and $\partial D$ does not bound a disk in $F'_L$. Then we may compress $F'_L$ along $D$ by, roughly speaking, taking two parallel copies of $D$, cutting out the annulus in $F'_L$ that is between them, and then gluing in the two copies of $D$ to form the new surface. As explained in Lickorish's An Introduction to Knot Theory, the equivalence relation on Seifert surfaces generated by compressions (whose inverse operation is called an "embedded arc surgery") and isotopies has only a single equivalence class -- the corresponding effects on Seifert matrices is known as s-equivalence. So: all you need to do to show the intersection number does not depend on the Seifert surface is to show it is unchanged after an isotopy of the surface and after a compression of the surface.