Existence of Seifert surfaces in Ranicki

Question

For knots in $S^3$, I am used to seeing Seifert's algorithm used to prove the existence of Seifert surfaces. However, In Ranicki's book on knots, High Dimensional Knot Theory, he gives a different, seemingly elegant proof, on page XXI. Assuming the map $p:(cl(M-k(N)\times D^2)), k(N)\times S^1$ exists, this seems like a nice proof. But what exactly is this map $p$? Even the existence of a map like $p$ is not obvious to me.

Answer 1

Let's consider the case of an $S^1$ knot $K$ in a $3$-manifold $M$ such that $K$ is nullhomologous (that is, $[K]=0$ in $H_1(M)$). Then $K$ is the boundary of a $2$-chain $S\in C_2(M)$, which is a relative $2$-cycle in $C_2(M,\operatorname{cl}(\nu(K)))$, where $\nu(K)$ denotes an open tubular neighborhood of $K$. By excision, there is a corresponding relative homology class in $H_2(M-\nu(K),\partial(M-\nu(K)))$, and by Poincaré duality, there is a corresponding cohomology class $\alpha$ in $H^1(M-\nu(K))$. What this class measures is the linking number of an element of $H_1(M-\nu(K))$ with $K$.

By the universal coefficient theorem, $$H^1(M-\nu(K)) \cong \hom(H_1(M-\nu(K)),\mathbb{Z}),$$ and since $H_1$ is the abelianization of the fundamental group, $$\hom(H_1(M-\nu(K)),\mathbb{Z})\cong\hom(\pi_1(M-\nu(K)),\mathbb{Z}).$$ Since $\mathbb{Z}=\pi_1(S^1)$ and $S^1$ is an Eilenberg-MacLane space, $$\hom(\pi_1(M-\nu(K)),\mathbb{Z})\cong [M-\nu(K),S^1],$$ where the square bracket notation means homotopy classes of maps $M-\nu(K)\to S^1$. Hence, starting with the class $\alpha$ and feeding it through all of this, we ultimately get a continuous map $f:M-\nu(K)\to S^1$. Tracing through the definitions, the relationship between $\alpha$ and $f$ is that, given a loop $\gamma:S^1\to M-\nu(K)$, $$\alpha([\gamma]) = \deg(f\circ\gamma).$$ Since every continuous map is homotopic to a smooth map, we may assume $f$ is smooth, and then by using transversality, we may assume $f$ is transverse to $1\in S^1$. The preimage $f^{-1}(1)$ is an embedded surface $\Sigma\subset M-\nu(K)$.

One point about this is that $\Sigma$ might meet $\partial (M-\nu(K))$ in a complicated way. We can modify $f$ so that, in a tubular neighborhood $\nu(\partial(M-\nu(K)))\approx S^1\times S^1\times [0,1]$ with the first $S^1$ being the meridian direction, $f(\theta,\phi,t)=\theta$. This can be done at the stage of constructing the isomorphism to $[M-\nu(K),S^1]$; one construction uses a cellular decomposition of $M-\nu(K)$ and builds up a continuous map cell-by-cell, and since there is no obstruction to ensuring this boundary condition at the beginning of the construction, we may construct it that way.

Hence, $\Sigma$ is an oriented surface whose boundary is isotopic in $\operatorname{cl}(\nu(K))$ to $K$ itself.

One last complication is that a Seifert surface is sometimes supposed to be connected. At least we can throw away all the closed components of $\Sigma$. There might be a way to homotope $f$ to effect this modification (this is at least true for $M=S^3$).

By the way, unlike Seifert's algorithm which produces only some possible Seifert surfaces, this procedure can yield all Seifert surfaces. The minimal genus of a Seifert surface from Seifert's algorithm over all diagrams is called the "canonical genus."