Let $K\subset S^3$ be a knot. We denote by $X=S^3\setminus \nu K$ the knot exterior, i.e. the complement of an open tubular neighborhood of $K$. An immersed Seifert surface for a knot $K$ is an immersion $f\colon S\to X$ such that $f(\partial S)$ is a longitude of $K$. We refer to the genus of $S$ as the genus of the immersed Seifert surface.

Gabai showed that the minimal genus of an immersed Seifert surface is the same as the minimal genus of an embedded Seifert surface. I was wondering whether a stronger statement holds: every immersed Seifert surface of minimal genus is an `embedded Seifert surface in disguise', i.e. homotopic (through immersions) to an embedded Seifert surface.

For higher genus I can think of immersed Seifert surfaces that certainly do not from Seifert surfaces, e.g. there exists an immersed Seifert surface $f\colon S\to X_U$ of genus one for the unknot $U$ such that there are curves on $f(S)$ which are non-trivial in $H_1(X_U)$.

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    $\begingroup$ Maybe I'm missing something but why can't you just take a minimal genus embedded SS and then just re-immerse it by inserting some self-intersections? You could have a self-intersection curve parallel to the knot, which bounds an annulus in the surface, whose only double points are on the boundary. This immersed annulus gives a torus in the knot complement, and the torus bounds a solid torus in the knot complement. Initially I do not think there's a path in the space of immersions to an embedding. $\endgroup$ – Ryan Budney Feb 28 '14 at 1:14

I think this should be false for fibered knots. Consider a fibered knot, which is the mapping torus of a mapping class $\phi: S\to S$. Suppose there is a non-separating simple curve $c\subset S$ such that $\phi(c)\cap c=\emptyset$. Then one can form a "crossjoin" surface, by removing annulus neighborhoods $\mathcal{N}(c)$ and $\mathcal{N}(\phi(c))$, and inserting two crossing annuli which connect one boundary of $\mathcal{N}(c)$ to the opposite boundary of $\mathcal{N}(\phi(c))$. This was used by Cooper-Long-Reid to construct immersed surfaces in fibered manifolds. This immersed crossjoin surface will have a curve on it which is homologically non-trivial, and therefore the surface cannot be homotoped to an embedding.

To get an example, then, one would need to find such a fibered knot. The figure 8 doesn't work, but I think there should be an abundance of them, say among positive braids.

One could probably also get this to work with a knot which is only "partially" fibered, so that its complement has a non-trivial product piece of its relative JSJ decomposition. Again, if there is an annulus in this product region which has disjointly embedded boundary in the Seifert surface, then one may form a crossjoin surface.

Addendum: Actually, I realized that there is a simple way to form counterexamples in torus knots. The monodromy of a torus knot is finite-order, say $k$. Take a non-separating simple closed curve, and remove an annulus neighborhood. Then insert an annulus that winds $k$ times around the mapping torus direction, and connect up with the other boundary component of the annular neighborhood. The cross-cut is not needed, since this produces one component which is an immersed torus which is homologically trivial. So this works even for the trefoil knot.

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    $\begingroup$ thanks Ian! I was thinking about this because of the following paper by Ito front.math.ucdavis.edu/1307.5115 where (as far as I can see) he shows in the proof of Theorem 1, that a genus one knot can not have an immersed Seifert surface of genus one with a homologically non-trivial curve. This seems to fit nicely with your answer, you `need some space' to get such immersed Seifert surfaces. $\endgroup$ – Stefan Friedl Feb 28 '14 at 7:21
  • $\begingroup$ In fact Ito proves the statement I claimed for genus one knots which are not trefoils. So no contradiction with what Ian is saying. $\endgroup$ – Stefan Friedl Mar 1 '14 at 19:09

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