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I am interested in the following situation: I have two codimension-2 knots $K_1$ and $K_2$ in $S^n$ and they are not isotopic. Furthermore, $K_1$ is not isotopic to the mirror image of $K_2$ and vice versa. Could it be that $K_1$ and $K_2$ become isotopic after connect summing (away from the knots) another $n$-manifold $X$ (so that $K_1$ and $K_2$ are then isotopic inside of $X$)?

(In the case where say $K_2$ is the mirror image of $K_1$, we could connect sum on a nonorientable $X$ and going around a nonorientable loop would do the trick -- I don't know any other tricks.)

I would like to know the answer to this question specifically in low-dimensions (<5). My guess: this can't happen in dimension 2 (here changing the codimension to consider curves on the surface -- say for some geometric reasons), this can't happen in dimension 3 (say by Gordon-Lueke and the fact that the fundamental group of the complement says a lot), this can happen in dimension 4 (since it's a jungle out there).

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  • $\begingroup$ What do you mean when you say “this can’t happen in dimension 2 (say for some geometric reason)”? If I understand your question, this would be about embedded $S^0$’s in $S^2$, which for silly reasons are always isotopic. $\endgroup$ Commented Oct 16, 2022 at 14:54
  • $\begingroup$ @AndyPutman Oh yes -- I'm sorry I was totally incorrect in my phrasing there. I was thinking of curves on a surface (just as fun low low dimensional analogy). I'll edit - thank you. $\endgroup$
    – Sprotte
    Commented Oct 16, 2022 at 16:57

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I believe can happen in dimension $4$, and probably in all higher dimensions. Take two inequivalent knots $K, K'$ with the same exterior (Cappell-Shaneson; Gordon). Then $K'$ is obtained from $K$ by a "Gluck twist", in other words trivialize the normal bundle of $K$ (there's only one way to do this), remove it, and glue back in using the diffeomorphism $f: S^1 \times S^2$ given by $f(\theta,z) = (\theta,$ rotation of $z$ by $\theta$).

There's a general principle that connect summing with $\mathbb{C}P^2$ undoes Gluck twists, and I think that this should prove that $K$ and $K'$ are now isotopic in $S^4 \# \mathbb{C}P^2$. I'd have to think about the proof (or maybe someone could point to a reference).

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