## Posts Tagged ‘knot’

### When k looks and smells like the unknot…

February 14, 2011

Valentine’s day special issue~ ^_^

Professor Gabai decided to ‘do some classical topology before getting into the fancy stuff’ in his course on Heegaard structures on 3-manifolds. So we covered the ‘loop theorem’ by Papakyriakopoulos last week. I find it pretty cool~ (So I started applying it to everything regardless of whether a much simpler argument exists >.<)

Let $M$ be a three dimensional manifold with (non-empty) boundary. In what follows everything is assumed to be in the smooth category.

Theorem: (Papakyriakopoulos, ’58)
If $f: \mathbb{D}^2 \rightarrow M$ extends continuously to $\partial \mathbb{D}$ and the image $f(\partial \mathbb{D}) \subseteq \partial M$ is homotopically non-trivial in $\partial M$. Then in any neighborhood $N(f(\mathbb{D}))$ we can find embedded disc $D \subseteq M$ such that $\partial D$ is still homotopically non-trivial in $\partial M$.

i.e. this means that if we have a loop on $\partial M$ that is non-trivial in $\partial M$ but trivial in $M$, then in any neighborhood of it we can find a simple loop that’s still non-trivial in $\partial M$ and bounds an embedded disc in $M$.

We apply this to the following:

Corollary: If a knot $k \subseteq \mathbb{S}^3$ has $\pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$ then $k$ is the unknot.

Proof: Take tubular neighborhood $N_\varepsilon(k)$, consider $M=\mathbb{S}^3 \backslash \overline{N_\varepsilon(k)}$, boundary of $M$ is a torus.

By assumption we have $\pi_1(M) = \pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$.

Let $k' \subseteq \partial M$ be a loop homotopic to $k$ in $N_\varepsilon(k)$.

Since $\pi_1(M) = \mathbb{Z}$ and any loop in $M$ is homotopic to a loop in $\partial M = \mathbb{T}^2$. Hence the inclusion map $i: \pi_1(\mathbb{T}^2) \rightarrow \pi_1(M)$ is surjective.

Let $l \subseteq \partial M$ be the little loop winding around $k$.

It’s easy to see that $i(l)$ generates $\pi_1(M)$. Hence there exists $n$ s.t. $k'-n \cdot l = 0$ in $\pi_1(M)$. In other words, after $n$ Dehn twists around $l$, $k'$ is homotopically trivial in $M$ i.e. bounds a disk in $M$. Denote the resulting curve $k''$.

Since $k''$ is simple, there is small neighborhood of $k''$ s.t. any homotopically non-trivial simple curve in the neighborhood is homotopic to $k''$. The loop theorem now implies $k''$ bounds an embedded disc in $M$.

By taking a union with the embedded collar from $k$ to $k''$ in $N_\varepsilon(k)$:

We conclude that $k$ bounds an embedded disc in $\mathbb{S}^3 \backslash k$ hence $k$ is the unknot.

Establishes the claim.

Happy Valentine’s Day, Everyone! ^_^

### Cutting the Knot

December 13, 2010

Recently I came across a paper by John Pardon – a senior undergrad here at Princeton; in which he answered a question by Gromov regarding “knot distortion”. I found the paper being pretty cool, hence I wish to highlight the ideas here and perhaps give a more pictorial exposition.

This version is a bit different from one in the paper and is the improved version he had after taking some suggestions from professor Gabai. (and the bound was improved to a linear one)

Definition: Given a rectifiable Jordan curve $\gamma: S^1 \rightarrow \mathbb{R}^3$, the distortion of $\gamma$ is defined as

$\displaystyle \mbox{dist}(\gamma) = \sup_{t,s \in S^1} \frac{d_{S^1}(s,t)}{d_{\mathbb{R}^3}(\gamma(s), \gamma(t))}$.

i.e. the maximum ratio between distance on the curve and the distance after embedding. Indeed one should think of this as measuring how much the embedding ‘distort’ the metric.

Given knot $\kappa$, define the distortion of $\kappa$ to be the infimum of distortion over all possible embedding of $\gamma$:

$\mbox{dist}(\kappa) = \inf\{ \mbox{dist}(\gamma) \ | \ \gamma \ \mbox{is an embedding of} \ \kappa \ \mbox{in} \ \mathbb{R}^3 \}$

It was (somewhat surprisingly) an open problem whether there exists knots with arbitrarily large distortion.

Question: (Gromov ’83) Does there exist a sequence of knots $(\kappa_n)$ where $\lim_{n \rightarrow \infty} \mbox{dist}(\kappa_n) = \infty$?

Now comes the main result in the paper: (In fact he proved a more general version with knots on genus $g$ surfaces, for simplicity of notation I would focus only on torus knots)

Theorem: (Pardon) For the torus knot $T_{p,q}$, we have

$\mbox{dist}(T_{p,q}) \geq \frac{1}{100} \min \{p,q \}$

.

To prove this, let’s make a few observations first:

First, fix a standard embedding of $\mathbb{T}^2$ in $\mathbb{R}^3$ (say the surface obtained by rotating the unit circle centered at $(2, 0, 0)$ around the $z$-axis:

and we shall consider the knot that evenly warps around the standard torus the ‘standard $T_{p,q}$ knot’ (here’s what the ‘standard $T_{5,3}$ knot looks like:

By definition, an ’embedding of the knot’, is a homeomorphism $\varphi:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ that carries the standard $T_{p,q}$ to the ‘distorted knot’. Hence the knot will lie on the image of the torus (perhaps badly distorted):

For the rest of the post, we denote $\varphi(T_{p,q})$ by $\kappa$ and $\varphi(\mathbb{T}^2)$ by $T^2$, w.l.o.g. we also suppose $p.

Definition: A set $S \in T^2$ is inessential if it contains no homotopically non-trivial loop on $T^2$.

Some important facts:

Fact 1: Any homotopically non-trivial loop on $\mathbb{T}^2$ that bounds a disc disjoint from $T^2$ intersects $T_{p,q}$ at least $p$ times. (hence the same holds for the embedded copy $(T^2, \kappa)$).

As an example, here’s what happens to the two generators of $\pi_1(\mathbb{T}^2)$ (they have at least $p$ and $q$ intersections with $T_{p,q}$ respectively:

From there we should expect all loops to have at least that many intersections.

Fact 2: For any curve $\gamma$ and any cylinder set $C = U \times [z_1, z_2]$ where $U$ is in the $(x,y)$-plane, let $U_z = U \times \{z\}$ we have:

$\ell(\gamma \cap C) \geq \int_{z_1}^{z_2} | \gamma \cap U_z | dz$

i.e. The length of a curve in the cylinder set is at least the integral over $z$-axis of the intersection number with the level-discs.

This is merely saying the curve is longer than its ‘vertical variation’:

Similarly, by considering variation in the radial direction, we also have

$\ell(\gamma \cap B(\bar{0}, R) \geq \int_0^{R} | \gamma \cap \partial B(\bar{0}, r) | dr$

Proof of the theorem

Now suppose $\mbox{dist}(T_{p,q})<\frac{1}{100}p$, we find an embedding $(T^2, \kappa)$ with $\mbox{dist}(\kappa)<\frac{1}{100}p$.

For any point $x \in \mathbb{R}^3$, let

$\rho(x) = \inf \{ r \ | \ T^2 \cap (B(x, r))^c$ is inessential $\}$

i.e. one should consider $\rho(x)$ as the smallest radius around $x$ so that the whole ‘genus’ of $T^2$ lies in $B(x,\rho(x))$.

It’s easy to see that $\rho$ is a positive Lipschitz function on $\mathbb{R}^3$ that blows up at infinity. Hence the minimum value is achieved. Pick $x_0 \in \mathbb{R}^3$ where $\rho$ is minimized.

Rescale the whole $(T^2, \kappa)$ so that $x_0$ is at the origin and $\rho(x_0) = 1$.

Since $\mbox{dist}(\kappa) < \frac{1}{100}p$ (and note distortion is invariant under scaling), we have

$\ell(\kappa \cap B(\bar{0}, 1) < \frac{1}{100}p \times 2 = \frac{1}{50}p$

Hence by fact 2, $\int_1^{\frac{11}{10}} | \kappa \cap \partial B( \bar{0}, r)| dr \leq \ell(\kappa \cap B(\bar{0}, 1)) < \frac{1}{40}p$

i.e. There exists $R \in [1, \frac{11}{10}]$ where the intersection number is less or equal to the average. i.e. $| \kappa \cap \partial B(\bar{0}, R) | \leq \frac{1}{4}p$

We will drive a contradiction by showing there exists $x$ with $\rho(x) < 1$.

Let $C_z = B(\bar{0},R) \cap \{z \in [-\frac{1}{10}, \frac{1}{10}] \}$, since

$\int_{-\frac{1}{10}}^{\frac{1}{10}} | U_t \cap \kappa | dt \leq \ell(\kappa \cap B(\bar{0},1) ) < \frac{1}{50}p$

By fact 2, there exists $z_0 \in [-\frac{1}{10}, \frac{1}{10}]$ s.t. $| \kappa \cap B(\bar{0},1) \times \{z_0\} | < \frac{1}{10}p$. As in the pervious post, we call $B(\bar{0},1) \times \{z_0\}$ a ‘neck’ and the solid upper and lower ‘hemispheres’ separated by the neck are $U_N, U_S$.

Claim: One of $U_N^c \cap T^2, \ U_S^c \cap T^2$ is inessential.

Proof: We now construct a ‘cutting homotopy’ $h_t$ of the sphere $S^2 = \partial B(\bar{0}, R)$:

i.e. for each $t \in [0,1), \ h_t(S^2)$ is a sphere; at $t=1$ it splits to two spheres. (the space between the upper and lower halves is only there for easier visualization)

Note that during the whole process the intersection number $h_t(S^2) \cap \kappa$ is monotonically increasing. Since $| \kappa \cap B(\bar{0},R) \times \{z_0\} | < \frac{1}{10}p$, it increases no more than $\frac{1}{5}p$.

Observe that under such ‘cutting homotopy’, $\mbox{ext}(S^2) \cap T^2$ is inessential then $\mbox{ext}(h_1(S^2)) \cap T^2$ is also inessential. (to ‘cut through the genus’ requires at least $p$ many intersections at some stage of the cutting process, but we have less than $\frac{p}{4}+\frac{p}{5} < \frac{p}{2}$ many interesections)

Since $h_1(S^2)$ is disconnected, the ‘genus’ can only lie in one of the spheres, we have one of $U_N^c \cap T^2, \ U_S^c \cap T^2$ is inessential. Establishes the claim.

We now apply the process again to the ‘essential’ hemisphere to find a neck in the $y$direction, i.e.cutting the hemisphere in half in $(x,z)$ direction, then the $(y,z)$-direction:

The last cutting homotopy has at most $\frac{p}{5} + 3 \times \frac{p}{4} < p$ many intersections, hence has inessential complement.

Hence at the end we have an approximate $\frac{1}{8}$ ball with each side having length at most $\frac{6}{5}$, this shape certainly lies inside some ball of radius $\frac{9}{10}$.

Let the center of the $\frac{9}{10}$-ball be $x$. Since the complement of the $\frac{1}{8}$ ball intersects $T^2$ in an inessential set, we have $B(x, \frac{9}{10})^c \cap T^2$ is inessential. i.e.

$\rho(x) \leq \frac{9}{10} <1$

November 8, 2010

As I was trying to understand the Whitehead manifold and related constructions of non-tame manifolds batter, I guess it makes a cool blog post ^^

The Whitehead manifold is an example of a 3-manifold that’s contractable but not homeomorphic to $\mathbb{R}^3$ i.e. the manifold is homotopically equivalent to a point but not tame. (See the earlier post on tameness for more explanations)

Construction:
Take a solid torus $T_1 \subseteq \mathbb{R}^3$, embed a thinner torus $T_2 \subseteq T_1$ as shown:

Iterate the process: at each step, embed solid torus $T_i$ into $T_{i-1}$ so that $T_i$ “links with itself” inside $T_{i-1}$:

Let the Whitehead continuum be the intersection of the $T_i$s.

i.e. $\displaystyle W=\bigcap_{i=1}^\infty T_i$

As some people know, I have a weird hobby of describing strange continua in terms of Cantor sets…So here comes ‘Conan’s translation’ of the Whitehead continuum:

Take two copies of $C \times [0,1]$ where $C$ is the standard middle-third Cantor set. Bend them into Rainbow-shape with the open ends facing each other:

(I think of this as having a width $1$ ‘brush’ with ink only on the points of the Cantor set, and use the brush to draw two semicircles)

Now we connect the open ends: take a width $1/3$ brush and connect the top pair of ends so that they link with each other, and then a width $1/9$ brush for the highest remaining pair, etc. Take the union of all those connecting sets, union a line segment joining the bottom-most pair of points, we get the Whitehead continuum:

The Whitehead manifold is the complement of $W$ in the three-sphere $\mathbb{S}^3$, equipped with the

What’s the fundamental group of the Whitehead manifold?

Claim: $T_i^c$ is null-homotopic in $T_{i+1}^c$.

pf: $T_1^c$ is contractible to a loop in $T_{i+1}^c$, hence it suffice to homotope the red loop to a point without touching the black loop:

Note that for homotopy, the loop is allowed to pass through itself: (in contrast to isotopy)

The loop can now be easily contracted:

Hence we deduce the Whitehead manifold is null-homotopic. (by collapsing each $T_1$ at some finite time)

In particular, it has trivial fundamental group! (This might seem hard to believe especially when looking at my Cantor-set picture) Infact for this, we can directly see from the picture that all loop can be homotoped to constant:

Since loop is compact, there is a ‘finest gap’ in the Cantor set which the loop passes through, say it’s a gap with width $1/3^i$. Now by performing the operation above, we can homotope all parts of the loop that goes through the $1/3^i$ to segments that goes through $1/3^{i-1}$-gaps, by having the segments crossing themselves once. Now we pass to the $1/3^{i-2}$-gaps, etc. until the loop lie completely outside the thickened disc which the continua lies in. Once it’s outside the disc, the loop can be contracted.

The manifold is not homeomorphic to $\mathbb{R}^3$ as we can easily see that, unlike in $\mathbb{R}^3$, the red loop in picture cannot be isotoped to a trivial loop.

As an alternative point of view, we should note that in fact $T_i^c$ and $T_{i+1}$ form a thickened Whitehead link:

Since the Whitehead link is symmetric, this gives an simpler (but less direct, in my opinion) way of knowing that red loop is homotopically trivial in the complement of the black loop. (As the black loop is obviously homotopically trivial in the complement of the red loop.)

In light of this, one may construct many different non-tame manifolds with finitely generated fundamental group by embedding a handlebody inside another copy of itself and take the complement of the infinite intersection.

Here is an example of embedding a genus $3$ handlebody. The resulting manifold (after taking the complement of the intersection) is a homotopy genus $2$ handlebody. (As in the whitehead case, $T_i^c$ can be homotoped to a genus $2$ handlebody inside $T_{i+1}^c$. The ‘third loop’ can be unknotted by crossing itself once.) But it’s of coruse not homeomorphic to the genus $2$ handlebody.

April 21, 2010

I’ve been going through Thurston’s book ‘The Geometry and Topology of Three-Manifolds‘ in a reading course with Amie Wilkinson. In Chapter 3, p32, when he’s constructing a hyperbolic structure on the Whitehead link complement, there is a picture on how to glue the 2-cells to the knot, to quite Thurston, ‘the attaching map for the two-cells are indicated by the dotted lines.’ However, for me it’s impossible to see where are the dotted lines going. So I reconstruct it here with some more clear pictures. The construction itself was a cool reading that I wish to share.

First, we have the Whitehead link, looking like the first figure below:

We attach three 1-cells (line segments) as in the second figure, note that the ‘x’ in the middle represents a line segment orthogonal to the screen, connecting the top and bottom line in the figure ‘8’ loop.

Now we will start to attach four 2-cells to the 1-complex above: First, we attach a 2-cell spanning the top part of the figure ‘8’ loop, spanning one side of the middle segment and two sides of the top segment (denote this by cell A):

Do the same with the bottom half (cell B). Note that each cell is attached to three edges, hence they are triangles without vertices in the knot complement with three one-cells attached.

For the other two cells, we attach as follows (cells C and D):

Combining the four 2-cells, we get something like the figure showed below. Note that at the top, cell A is under cell C in the left, intersecting the surface spanned by cells C and D at the edge, and comes above cell D to the right of the edge.

It’s easy to see that the complement of the above 2-complex does not separate $\mathbb{R}^3$, hence it’s a 3-cell with eight faces (i.e. it has to go through both sides of each 2-cell in order to fill the 3-space) each of its face has three edges. Hence we may glue an octahedron to the 2-complex after the gluing, pairs of faces of the octahedron will be identified groups of four edges will be identified to single edges. Hence to put a hyperbolic structure on the link complement, it suffice to put an hyperbolic structure to the octahedron with vertices deleted.

Since each edge is glued up by four edges of the octahedron, it suffice to find an octahedron (without vertices) in the hyperbolic 3-space that has all adjacent faces intersect in dihedral angle $2 \pi / 4$ i.e. all adjecent faces are orthogonal in the hyperbolic space. But this is achieved if we inscribe the regular octahedron into the Klein model (also called projective model of hyperbolic 3-space.

The gluing map for the faces are merely rotations and reflections of the ball which are certainly hyperbolic isometries. Hence this gives a hyperbolic structure to the link complement.