Let $K_t$ be certain $1$ parametric family of knots in $\mathbb{R}^3$. I am wandering what are the precise obstructions for a parametric Seifert surface to exist; i.e. a $1$parametric family of oriented surfaces $S_t$ such that $\partial(S_t)=K_t$. I am specially interested in the case when the parameter space is $\mathbb{S}^1$. What about higher dimensional parametric families?

1$\begingroup$ The answer will depend on your precise formulation, but if your 1parameter family of knots is literally parametrized by an interval $[0,1]$ then by isotopy extension there is a compatible 1parameter family of Seifurt surfaces. Your question becomes more interesting if your parameter is a circle, or some higherdimensional manifold. For the circle, your Seifert surface would have to be compatible with the symmetry of your knot (i.e. diffeomorphism of the knot exterior). There are symmetries that do not have compatible Seifert surfaces. $\endgroup$ – Ryan Budney Oct 17 '19 at 20:53

1$\begingroup$ This phenomenon is ultimately another way to see the difference in the homotopytype between the embedding space of a circle, and the embedding space of a surface. $\endgroup$ – Ryan Budney Oct 17 '19 at 21:04

$\begingroup$ Yes, I see the nontrivial case is exactly when the 1dimensional parameter space is a circle rather than the interval. That was the exact case I was concerned. Thank you for your answer, Ryan. By the way, I would appreciate if you could provide a more explicit description of the notion of compatibility with the symmetry of the knot. Thanks again! $\endgroup$ – X1921 Oct 24 '19 at 11:37

$\begingroup$ I've been meaning to write a paper on this issue, as it's a bit subtle. The answer depends both on the geometrization of the knot exteriors and your choice of Seifert surface. But at its most basic, when you perform isotopy extension on your 1parameter family, it gives you a diffeomorphism of the knot exterior. Compatibility means that diffeomorphism is isotopic to one that preserves your Seifert surface. For a fibred hyperbolic or torus knot, this can be done. But for more complicated knots (and seifert surfaces) the answer is nuanced. $\endgroup$ – Ryan Budney Oct 24 '19 at 17:58

$\begingroup$ Aha, yes, I see. Thank you for your answer. $\endgroup$ – X1921 Nov 2 '19 at 9:51