12
$\begingroup$

Consider a solid region of Euclidean 3-space, or more precisely, a compact, connected 3-dimensional submanifold $U \subset E^3$ bounded by a smooth oriented surface $\Sigma = \partial U$. Very roughly speaking, can one find a representative of each homotopy class of loops in $U$ as a loop on the boundary? More precisely:

Question: For every loop $\gamma$ in the fundamental group $\pi_1(b,U)$ (where $b$ is any point of $U$), does there exist a loop $\tilde{\gamma}$ homotopic to $\gamma$ that is contained entirely in $\Sigma$?

By homotopic here, we of course mean that $\gamma$ and $\tilde{\gamma}$ are related by a homotopy in $U$, not just in $E^3$. I.e., there exists some continuous map $\Gamma: [0,1] \times S^1 \to U$ such that $\Gamma(0,s) = \gamma(s)$ and $\Gamma(1,s) = \tilde{\gamma}(s)$.

$\endgroup$
  • $\begingroup$ Just to be clear, in my situation I do not require all the loops $\tilde{\gamma}$ to be based at the same point. Thanks! $\endgroup$ – JacquesMartin Dec 4 '18 at 21:14
13
+300
$\begingroup$

In case we do wish to fix a basepoint on $\Sigma$:

Consider a torus $\Sigma=T^2$ in $E^3$, and let $U$ be the closure of the bounded region of $E^3-\Sigma$.

One option is that $U$ is a solid torus, in which case the induced map $\pi_1(\Sigma)\to \pi_1(U)$ is surjective.

The other option is that $U$ is a nontrivial knot complement. For example, the following is a trefoil knot complement:

trefoil knot complement

But $\pi_1(\Sigma)\to\pi_1(U)$ can't possibly be surjective because $\pi_1(U)$ is nonabelian (in fact it is the three-strand braid group) yet $\pi_1(\Sigma)\cong\mathbb{Z}^2$.

In case we do not wish to fix a basepoint on $\Sigma$:

Let $U$ be $B^3$ minus a split unlink. This is a compact submanifold of $\mathbb{R}^3$ with three boundary components: a sphere and two tori. A loop that is the composite of the meridians of the two link components is not homotopic to a loop on the boundary. This is true even homologically.

The following paper has something to say about compact oriented manifolds where every loop is freely homotopic to a loop in the boundary:

Brin, Matthew; Johannson, Klaus; Scott, Peter, Totally peripheral 3-manifolds, Pac. J. Math. 118, 37-51 (1985). ZBL0525.57010.

If every loop in $U$ is freely homotopic to one in $\Sigma$ (that is, if $U$ is "totally peripheral"), then their result implies that there is some component $F$ of $\Sigma$ such that $\pi_1(F)\to\pi_1(U)$ is surjective.

Applying this to the case of a compact component of a torus complement in $E^3$, this only happens if $U$ is a solid torus (the complement of a trivial knot), since the fundamental group of a nontrivial knot complement is nonabelian.

In general, a compact orientable irreducible 3-manifold $U$ is a compression body if there is a boundary component $F\subset \Sigma$ with $\pi_1(F)\to\pi_1(U)$ surjective. Marden's Outer Circles example 3-11 (p. 168) explains how this works. A quick definition: a compression body is the boundary connect sum of a handle body with some number of surface-cross-intervals. All of these can be embedded in $E^3$. This should handle the complete classification of compact $U$ since (1) being in $E^3$ means there are no $S^1\times S^2$ connect summands and (2) $U$ has to be prime for this surjectivity condition to hold due to van Kampen's theorem.


I came up with the following before I dug up the above reference. Consider the trefoil complement again. Loops that are homotopic to boundary loops are called peripheral elements, and they are conjugate to an element in the image of $\pi_1(\Sigma)\to\pi_1(U)$, or equivalently if they are in some peripheral subgroup. In the following we'll see that $\pi_1(U)$ has non-peripheral elements. (Though $\pi_1(U)$ is certainly generated by them.)

Consider the presentation $G=\pi_1(U)=\langle x,y\mid x^2=y^3\rangle$ (as seen in Hatcher), where $\mu=y^{-1}x$ is a meridian and $\lambda=xy$ is a corresponding longitude (with $\mu^{-5}xy$ being a zero-framed longitude), which together generate a peripheral subgroup. There is a homomorphism $f:G\to \operatorname{GL}(2,\mathbb{Z}[t^{\pm 1}])$ given by \begin{align} x&\mapsto\begin{bmatrix}0&t\\-t^2&0\end{bmatrix}\\ y&\mapsto\begin{bmatrix}0&t\\-t&t\end{bmatrix} \end{align} called the Burau representation, at least after removing the trivial subrepresentation --- this particular presentation comes from https://arxiv.org/abs/math-ph/0103008 via John Baez. (Interesting fact 1: this representation is faithful. Interesting fact 2: if $a:G\to\mathbb{Z}$ is the abelianization with $a(\mu)=1$, then $\det(f(g))=t^{a(g)}$.)

Since $\mu$ and $\lambda$ generate an abelian subgroup, the subgroup's image under $f$ is simultaneously diagonalizable over $\mathbb{Q}(t)$. With $$P=\begin{bmatrix}\frac{1}{1-t}&1\\1&0\end{bmatrix},$$ then \begin{align} f(\mu)&=P\begin{bmatrix}1&0\\0&t\end{bmatrix}P^{-1}\\ f(\lambda)&=P\begin{bmatrix}-t^3&0\\0&-t^2\end{bmatrix}P^{-1}. \end{align} Every peripheral subgroup has an image generated by some conjugate of these generators. In particular, images of peripheral elements are conjugate to a matrix of the form $$(-1)^m\begin{bmatrix}t^{3m}&0\\0&t^{m+n}\end{bmatrix}$$ for some $(m,n)\in\mathbb{Z}^2$. But, $f(x)$ does not diagonalize over $\mathbb{Q}(t)$ since its characteristic polynomial is $a^2+t^3$. Therefore $x$ is not a peripheral element. Similarly, neither is $y$.

(I think there might be a geometric way to see this by thinking about the universal cover of $U$ as $\mathbb{H}^2\times \mathbb{R}$ with the deck transformations being isometries. Peripheral subgroups are lattices inside flat planes (products of hyperbolic lines and $\mathbb{R}$), and conjugation transforms the plane and lattice. I think that some elements are "too close" to the identity to be on any such lattice, and I would appreciate it if anyone could explain the details of this to me.)

$\endgroup$
  • $\begingroup$ (In case it helps anyone else, there is a picture of a knot complement about 1/4 of the way down the page here: ias.edu/ideas/2016/agol-hyperbolic-link-complements, which helped me better understand Kyle's drawing $\endgroup$ – Jason DeVito Dec 4 '18 at 20:54
  • $\begingroup$ I think this requires some small further argument for unbased homotopy. For instance, the map $\Bbb Z/4 \to Q$ embedding the imaginaries into the quaternions is surjective on conjugacy classes. I don't know why the union of peripheral subgroups doesn't cover the whole group at a glance. $\endgroup$ – user98602 Dec 4 '18 at 20:55
  • $\begingroup$ Terrific---many thanks for the quick response. Just to be clear, in my situation I do not require all the loops $\tilde{\gamma}$ to be based at the same point; only the loops $\gamma$. So the question Mike Miller raises is relevant. $\endgroup$ – JacquesMartin Dec 4 '18 at 21:15
  • $\begingroup$ @JasonDeVito I modified the picture in a way that is hopefully clearer. $\endgroup$ – Kyle Miller Dec 5 '18 at 2:31
  • 1
    $\begingroup$ Impressive update. I wish I could upvote again. I would expect that any nontrivial knot has a non-peripheral element of $\pi_1(S^3 \setminus K)$, but I don't know how I would try to prove that. $\endgroup$ – user98602 Dec 5 '18 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.