Knots that unknot in a manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:43:26Z http://mathoverflow.net/feeds/question/8707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8707/knots-that-unknot-in-a-manifold Knots that unknot in a manifold Mariano Suárez-Alvarez 2009-12-12T19:57:34Z 2009-12-15T07:20:25Z <p>Take a closed $n$-manifold $M$ and fix a nice $n$-ball $B$ in $M$. How much information about $M$ does the set of knotted $(n-2)$-spheres of $B$ which are unknotted <em>in</em> $M$ give?</p> http://mathoverflow.net/questions/8707/knots-that-unknot-in-a-manifold/8786#8786 Answer by Ryan Budney for Knots that unknot in a manifold Ryan Budney 2009-12-13T19:34:01Z 2009-12-15T07:20:25Z <p>Kevin Walker and a member of the forum (by e-mail) let me know about a flaw in the argument in my previous post. I'll erase it after putting up this post. </p> <p>One of my claims was the if $K$ unknots in $M$ then it unknots in $B$, where $K$ is a knot in a ball $B \subset M$. If $K$ is the unknot in $M$ it bounds a disk $D \subset M$, $K = \partial D$. In dimension $3$ we perturb the disc so that $D \cap \partial B$ is a collection of circles, this allows us to cap off $D \cap B$ with disc(s) parallel to $\partial B$ (an embedded surgery construction), constructing a disc in $B$ that bounds $K$. In higher dimensions this argument runs into trouble as it's not clear if you can perform the required embedded surgeries. </p> <p>I also claimed that the only new isotopy relation in general is that knots in $B$ can be isotopic (in $M$) to their mirror inverse if $M$ is non-orientable. I still suspect this is the case and I think I have an idea for an argument but I should think about it some more before posting. Here is an argument that works in dimension 3:</p> <p>Say $M$ is a compact boundaryless $3$-manifold, $K$ and $K'$ contained in a ball $B$ are isotopic in $M$. Perform the connect-sum decomposition on $M \setminus K$ and $M \setminus K'$. Since this is a unique decomposition (modulo issues in the non-orientable case), in both cases you get a list of prime factors that are identical except in the decomposition of $M \setminus K$ you get a summand that's $B \setminus K$ capped off with a disc, and in the other case you get a summand $B \setminus K'$ capped off with a disc. By the Gordon-Luecke theorem $K$ and $K'$ are either isotopic in $B$ or mirror images of each other. </p> <p>I'm still hopeful I have an argument that works in general. I'll edit this post again tomorrow once I'm certain. </p> <p>edit: final exams and deadlines are eating up my time at the moment but I'll get around to another revision... eventually. </p> http://mathoverflow.net/questions/8707/knots-that-unknot-in-a-manifold/8964#8964 Answer by Greg Kuperberg for Knots that unknot in a manifold Greg Kuperberg 2009-12-15T06:35:57Z 2009-12-15T06:35:57Z <p>Ryan gives a very nice answer in dimension 3, leaving the higher-dimensional case of the question open. I can't discuss the question with as much authority as I would like, but I can start to piece together an answer. My impression is that in any dimension $n$, an $(n-2)$-knot $K$ that is unknotted in $M$ is already unknotted in $B$ in the continuous category. In the smooth and PL categories, it should be the same except in $n=4$ dimensions, where the question runs into a lack of understanding of smooth (or equivalently PL) structures on 4-manifolds.</p> <p>First, my impression from skimming some geometric group theory is that a finitely presented group cannot be an infinite free product. Moreover, that it if it is a finite free product, the factors have unique isomorphism types. If this impression is correct, then you can identify the knot group $\pi_1(S^n \setminus K)$ from the embedded fundamental group $\pi_1(M \setminus K)$. In order to be trivial in $M$, the knot group of $K$ would have to be $\mathbb{Z}$.</p> <p>Then, <a href="http://www.springerlink.com/content/8124686447451824/" rel="nofollow">Aspherical manifolds and higher-dimensional knots</a>, by Bruno Eckmann, begins:</p> <blockquote> <p>E. Dyer and R. Vasquez proved that the complement of a higher-dimensional knot $S^{n-2} \subset S^n$, $n \ge 4$, is never aspherical unless the knot group is infinite cyclic (and hence, for $n \ge 5$, the imbedding is unknotted). [Footnote referring to papers of Levine, Stallings, Wall, Shaneson.] In the present note we give a simple proof of this fact based on some remarks concerning compact $\partial$-manifolds.</p> </blockquote> <p>My impression is that the surgery methods in dimension $n \ge 5$ used to prove the unknottedness assertion are valid in the smooth, PL, and continuous categories. Dimension $n=3$ is of course a special case where you do not need to consider algebraic topology, but instead prove things with direct cut-and-paste arguments as Ryan describes. In dimension $n=4$, the surgery theory is (a) much more difficult in the continuous category, and (b) non-existent in the smooth/PL category.</p> <p>The paper <a href="http://www.springerlink.com/content/q334402828160512/" rel="nofollow">The algebraic characterization of the exteriors of certain 2-knots</a>, by Jonathan Hillman, credits Freedman with the result that a 2-knot with knot group $\mathbb{Z}$ is unknotted. If this requires Freedman's work, then I would think that it is open in the smooth/PL case. What would be open is whether the 4-manifold with boundary $B^3 \times S^1$ has more than one smooth structure. If it does, I have seen a principle that smooth structures on a 4-manifold can merge together when you take a connected sum with manifolds such as $\pm \mathbb{C}P^2$ an $S^2 \times S^2$. I do not know if this principle is established for manifolds with a fundamental group. But at the very least, I have heard that the semigroup of smooth structures on $S^4$ is completely unknown, and its action on the smooth structures on another 4-manifold is completely unknown. So if the smooth Poincare conjecture is sufficiently false, presumably you could have a non-trivial 2-knot in $B$ that becomes trivial in $M$.</p>