User sam nead - MathOverflow

Answer by Sam Nead for Mid point with set square?

2013-04-27T08:09:22Z

I can do it if I am allowed to cheat - in axiom one for the set square (which is the axiom for the straight-edge) I'll allow one of the points to be ideal: that is, on the Gromov boundary of $H^2$. Also, I assume that the set square produces infinite geodesic rays. In particular, this means, in the second axiom, that the given point can be on the given line.

Suppose the segment $[x,y]$ is given. Draw the geodesic ray $P$, based at $x$ and perpendicular to $[x,y]$. Do the same at $y$ to get the ray $Q$, on the same side of $[x,y]$. Connect the ideal endpoint of $P$ to $y$ and the ideal endpoint of $Q$ to $x$. These new rays cross at $z$. Drop a perpendicular from $z$ to $[x,y]$ and we are done.

Answer by Sam Nead for if $S \times \Re$ is diffeomorphic to $T \times \Re$ then are S and T diffeomorphic?

2013-03-29T20:56:43Z

Just to write out Ryan's answer: Let $S$ be the sphere with three closed disks removed. Let $T$ be the torus with one closed disk removed. Note that $T$ is non-planar. Thus $S$ is not homeomorphic to $T$. $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\cross}{\times}$

On the other hand, let $S' = S \cross \RR$ and let $T' = T \cross \RR$. Then $S'$ and $T'$ are both diffeomorphic to the open, genus two handlebody. Thus $S'$ and $T'$ are diffeomorphic to each other.

I agree that this is homework, but it is good homework! What if $S$ and $T$ are required to be compact?

Answer by Sam Nead for If a graph embeds in the projective plane or the torus is there a bound on the number of edge crossings it has in the plane?

2013-03-23T09:31:49Z

There are two ways to read this question. (I'll just ignore the projective plane part.)

Zare's reading: Suppose that a graph $G$ embeds in the torus. Is there an upper bound on the number of edge crossings of $G$ when drawn in the plane?

The answer is "no". You can draw $K_7$ in the torus and then take many parallel edges. All of these graphs embed in the torus, but there is no bound on their crossing number when drawn in the plane.

My reading: What is the largest $k$ so that any graph $G$, with at most $k$ crossings in the plane, embeds in the torus? Is there an upper bound on $k$?

The answer is "yes". Note, if $k < 2$ then $G$ embeds in the torus. However $K_8$ does not embed in the torus, and can be drawn in the plane with only $18$ crossings. So $18$ is an upper bound. I'll guess that this can be improved to $k = 2$.

Answer by Sam Nead for Triangulation of moduli space.

2013-03-17T14:14:39Z

Since $N$ is compact, $S - N$ is not compact. In particular $S - N$ is homeomorphic to a punctured surface: here a punctured surface is a compact surface without boundary minus a finite set of points.

It may help to copy out their definitions word-by-word. Then find an example for each definition.

Answer by Sam Nead for Once punctured torus bundles in snappy/twister

2013-03-06T21:46:46Z

I believe that your question is answered by section (3e) at this page at the Geometry Center. Note that b+ and bo are equivalent as are b- and bn (o and n stand for orientation preserving and reversing, respectively). You can check a few examples using the is_isometric method of Manifold.

Answer by Sam Nead for the carrier graph and Heegaard surface

2013-03-02T13:52:26Z

When the rank of the fundamental group equals the Heegaard genus then there always a carrier graph that embeds in the Heegaard surface $S$. I don't know the answer when the rank is less than the Heegaard genus, but I strongly suspect that the answer is "no". The place to start is the paper of Boileau and Zieschang, titled "Heegaard genus of closed orientable Seifert 3-manifolds". Hyperbolic examples where rank is less than Heegaard genus are given by Tao Li, in his paper "Rank and genus of 3-manifolds".

Answer by Sam Nead for Hyperbolic 3-manifolds with no geometrically finite structure

2013-02-07T11:48:12Z

[Edited several times] As the comments say, the answer to the first and hence to the second question is "no". Suppose that $M$ is the compact manifold and $N$ is its interior. Let $\rho$ be the given hyperbolic structure on $N$. If $M$ is without boundary then the volume of $\rho$ is finite and we are done.

Suppose instead that $M$ has boundary. Since $N$ is hyperbolic, via $\rho$, deduce $M$ is atoroidal (which includes aspherical). Thus $M$ is Haken. Place all tori in the boundary of $M$ into the paring locus $P$. By Thurston's hyperbolization theorem, the interior $N$ admits a hyperbolic metric, $\rho_0$, which is geometrically finite. (The convex core has finite volume and contains all torus boundary components.) See Theorem 1.43 in Kapovich's book.

[A brief note - your hypotheses can be weakened. You assumed (a) $N$ is the interior of a compact manifold and (b) N is hyperbolizable. This can be replaced by (a') $\pi_1(N)$ is finitely generated and the same (b). This is called the "tameness theorem", due to Agol and also Calegari-Gabai.]

In the comments below (above?) Igor asks why an atoroidal manifold with torus boundary, and admitting an essential annulus, is Seifert fibered. This can be found as Lemma 1.16 on page 25 of Hatcher's three-manifold notes.

Applications of knot theory

2010-12-03T22:23:17Z

An answer of André Henriques' inspired the following closely related CW question. Parts of the following is extracted from his answer and my comments.

I regularly teach a knot theory class. Every time, students ask about applications. What should I say?

I have two off-the-cuff replies when students ask. The first is that knot theory is a treasure chest of examples for several different branches of topology, geometric group theory, and certain flavours of algebra. The second is a list of engineering and scientific applications: untangling DNA, mixing liquids, and the structure of the Sun's corona. I'm interested hearing about other applications. I am also interested in hearing your take on the pedagogical issues involved. Thank you!

Answer by Sam Nead for The action of torsion of $MCG(S)$ on curve complex

2012-12-14T10:01:51Z

Here is a way to find lots of examples. Suppose that $\Sigma$ is a surface and suppose that $f$ is a periodic mapping class. Let $S$ be the quotient orbifold $\Sigma/f$. Then taking full preimages gives a quasi-isometric embedding of the curve complex of $S$ into the curve complex of $\Sigma$. See

arXiv:1104.3492 and arXiv:math/0701719

for two different proofs. When $S$ has an infinite diameter curve complex we can pick $a$ and $b$, curves in $S$, that are as well separated as you want. Then the preimages $\alpha$ and $\beta$ are also well separated, and fixed as simplices by $f$. A bit more work (which I haven't done) should give examples where $\alpha$ and $\beta$ are single vertices. There are also examples where $S$ is honestly a surface, not just an orbifold.

Answer by Sam Nead for faraway curves in surface

2012-11-20T23:22:57Z

Just to summarize the above: the answer is "no" because the curve complex has countably many vertices while the boundary is uncountable. To give an easier example of this sort of thing, consider the regular four-valent tree, also known as the Cayley graph of the rank two free group. The tree is countable while its Gromov boundary is a Cantor set, and so uncountable.

Answer by Sam Nead for A question on (1,1) bridge Knot

2012-11-15T19:39:44Z

All 2-bridge knots in $S^3$ are $(1,1)$-knots in $S^3$. This is assigned as an exercise here. All two-bridge knots, other than the $(2,2k+1)$-torus knots, are hyperbolic.

Answer by Sam Nead for Teichmuller Theory introduction

2010-01-02T20:31:34Z

The primer on mapping class groups, by Farb and Margalit.

Answer by Sam Nead for Multiple Dehn twists and minimal position

2012-06-02T17:15:30Z

First (and trivially), yes, there are representatives $b$ and $b'$ of the curve and its image that are in minimal position. But you aren't really asking this. Correct me if I am wrong, but you are actually asking "Is there a small motion of the image of $b$ placing it in minimal position to the original position of $b$?" Said original position is assumed to be minimal with respect to the curves $\{a_i\}$.

The answer is "no". There is no such small motion. To see this, consider such $b$ and $\{a_i\}$ so that there are two components of $b - n(\cup a_i)$ that are parallel as proper arcs in $S - n(\cup a_i)$.

Answer by Sam Nead for Uniqueness of distance realizing geodesic in hyperbolic surface.

2012-05-02T06:40:45Z

For the pants, yes. In general, no. To prove this for the pants, classify all geodesic arcs and just observe the result. There are many ways to find a "no" example in the general case; the first one that came to my mind was taking a double cover.

EDIT - I see that this is a near-duplicate of a closed question. You could improve your question by giving some motivation. Reading the FAQ will be very useful in writing questions that get good answers. In particular please see http://mathoverflow.net/faq#whatnot

Answer by Sam Nead for Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball

2012-04-25T11:29:08Z

$\newcommand{\RR}{\mathbb{R}}$The other answers are completely general, but there is simpler way if we use the (given) hypothesis that all the action is taking place in $\RR^3$. So, suppose that $T$ is a finite triangulation contained in $\RR^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

the space $|T|$ is a manifold and
the boundary $\partial\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

EDIT x2 - Here is a discussion of the "bit of work". Suppose that $C = |T|$ is a manifold and $S = \partial C$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B \subset \RR^3$. We need to show that $B$ is equal to $C$. By the Jordan–Brouwer separation theorem there are two possibilities. Either $S$ separates $B$ from $C$ or it does not.

In the separating case form $M = B \cup C$. Thus $M$ is a compact three-manifold without boundary, embedded in $\RR^3$. This contradicts invariance of domain. See Corollary 2B.4 of Hatcher's Algebraic topology.

Suppose instead that $B$ and $C$ are on the same side of $S$. It follows that $C \subset B$. We must prove the opposite inclusion. Suppose that $p$ is a point of $B$. Let $r$ be any point of $S$ that is as close as possible to $p$. Let $I = [p,r]$ be the line segment from $p$ to $r$. So $I \subset B$. Order the points of $I$, from $p$, to $r$. Note that $r \in S$ so $r \in C$. Let $J = I \cap C$. Let $q = \inf J$. Since $C$ is a closed subset of $\RR^3$ the set $J$ is closed and thus $q$ lies in $C$. Since $C$ is a manifold there is a neighborhood $V \subset C$ so that $q \in V$. Show that $V \cap I$ is a neighborhood of $q$ in $I$. Thus $q = p$ and we are done.

I don't see how to do the second half with "invariance of domain" directly. I'll also remark that the "bit of work" has now been greatly expanded, and perhaps unnecessarily so. One is supposed to do this sort of thing once and then not worry about it ever again.

Answer by Sam Nead for Efficient topological triangulations of non-convex polyhedra

2012-04-18T20:40:07Z

I've been thinking about the main question in the original post on and off for a few days. All of my efforts have been in the direction of finding enough examples to prove a super-linear lower bound, following Misha's suggestion to use hyperbolic volume. This hasn't worked yet - the problem appears to be tricky! In any case, here is the most appealing of the constructions.

Stick braids

Let $x, y, z$ be the usual coordinates on $\mathbb{R}^3$. Let $D$ be the unit disk in the plane $z = 0$ and let $E$ be the unit disk in the plane $z = 1$. Suppose that $\{a_i\}_1^n \subset D$ is a collection of points. Let $b_i$ be the point in $E$ with the same $x$ and $y$ coordinates as $a_i$. Suppose that $\sigma \in \Sigma_n$ is a permutation. Let $B = B(a, \sigma)$ be the collection of line segments where the $i$'th segment has endpoints $a_i$ and $b_{\sigma(i)}$. If the segments are pairwise disjoint then we call $B$ a stick braid.

It follows that the braid closure of $B$ is a link with stick number at most $5n$. As a concrete example, the $(p,q)$-torus knot can be obtained by placing the points $a_i$ at the $p$'th roots of unity, taking $\sigma(i) = i + q$, modulo $p$, and taking a braid closure.

Hyperbolic volume

Now we must use the Euclidean geometry of the braid $B$ to draw conclusions about the hyperbolic volume of the braid closure. Consider the unit disk $D_t$ in the plane $z = t$. As $t$ varies from $0$ to $1$, the points of intersection $D_t \cap B = \{a_i^t\}$ move along straight lines at speeds depending on the slope of the $i$'th strand. Here is a lie: when two points $a_i^t$ and $a_j^t$ come much closer to each other than they are to any of the other points, then there is a definite contribution to hyperbolic volume. Making this precise (ie, actually true) and then finding a braid $B$ that arranges superlinearly many such meetings would give the desired lower bound.

One way to do this would be to take $n$ sufficently large, $\epsilon$ correspondingly small, and take the points $a_i$ to be a generic $\epsilon$--net in $D$. Choose $\sigma$ to be a random permutation. Let $B = B(a, \sigma)$. Take the braid closure and plug everything into SnapPy. I've not tried to do this yet, but it would at least give some data...

Edit

Before writing the above, I had the idea of generating a random stick knot using outer billiards -- namely, let $P$ and $Q$ be concentric spheres and build a knot by taking segments tangent to $Q$ with endpoints on $P$. This has the virtue that when $Q$ has smallish radius, the expected crossing number will be quadratic. But it seems easier to estimate volume using the braid construction, and it is volume that really matters to us.

And then... after all this thinking and writing, I started poking randomly around the web and found O'Rourke's question on our very own MO. O'Rourke gives a very simple model for random stick knots: just bounce around in a sphere. The Thurstons suggest that the expected volume grows as $n^{3/2}$.

Answer by Sam Nead for Efficient topological triangulations of non-convex polyhedra

2012-04-13T13:35:13Z

The question in the comment was as follows.

Is there an infinite family of (hyperbolic) stick knots whose crossing numbers are quadratic in the number of edges?

The answer is yes. Suppose that $T$ is the $(p,q)$-torus knot, with $2 \leq p < q \leq 2p$. Wikipedia tells us that $T$ has crossing number $(q−1)p$ and has stick number $2q$. Taking $p$ and $q$ close together gives a non-hyperbolic example. Now change a few crossings randomly - that changes the stick number by a small constant and makes the knot hyperbolic, almost surely. If you'd rather, you can instead use twisted torus knots. See the paper "The simplest hyperbolic knots" or the paper "The next simplest hyperbolic knots".

However, neither of these obviously give $O(n^2)$ lower bounds for your original question -- these knots typically have very small hyperbolic volume and small triangulations; hence the name of the two papers I cited.

EDIT - My previous sentence can be made even more concrete. Fixing $n$ there are torus knots with stick number $O(n)$, with crossing number $O(n^2)$ and whose complements only require $O(\log(n))$ tetrahedra to triangulate. See the last paragraph of Agol's answer to this question: http://mathoverflow.net/questions/46149/lower-bound-on-number-of-tetrahedra-needed-to-triangulate-a-knot-complement

Of course, these minimal triangulation do not have the desired boundary patterns. But it does indicate that the statement "large crossing number implies large triangulation" fails rather badly.

Answer by Sam Nead for Finding hyperbolic metrics by approximation

2012-04-16T14:21:35Z

To my knowledge this hasn't been done in theory (although see Harriet Moser's thesis http://www.math.columbia.edu/~moser/). But it certainly has been done in practice by Jeff Weeks' program SnapPea. Note that $\mbox{Isom}(\mathbb{H}^3) = \mbox{PSL}(2, \mathbb{C}) = \Gamma$. So your source group $G = \pi_1(M^3)$ already has a very nice matrix group as a target. SnapPea assumes that the three-manifold $M$ is given as a triangulation. (I am sure that it is not easy to go backwards from a presentation of $G$ to a triangulation of $M$.) After tidying the triangulation (and drilling out a curve if necessary - but lets suppose that $M$ has a single torus boundary component) SnapPea gives all of the tetrahedra in the triangulation the same shape, namely that of the regular ideal tetrahedron. "Developing" in $\mathbb{H}^3$ turns shapes of tetrahedra into matrices, one per generator. Naturally this is not yet a representation of $G$ into $\Gamma$. The failure to be a representation is measured by the failure of the shapes to satisfy the Thurston gluing equations. SnapPea uses a multivariate version of Newton's method to find new shapes for the tetrahedra, hopefully converging to the discrete and faithful representation of $G$ into $\Gamma$.

Details, references, and more can be found in Moser's thesis. However, I'll add a final remark - the convergence properties of SnapPea's method certainly do depend sensitively on the initial triangulation. There are certain triangulations where SnapPea will consistently produce wrong answers. This is not a problem in practice -- you randomize the triangulation a few times and SnapPea typically starts to behave much better. But finding an actual algorithm appears to be difficult. So something mysterious going on.

The problem (of finding the discrete and faithful representation) is easy to solve even for very respectfully sized manifolds (say up to 100 tetrahedra). But why? I certainly don't know. You'd have to ask Thurston, Weeks, Hodgson or some other expert for an opinion.

Answer by Sam Nead for Pleated surfaces do not curl up too much

2012-04-15T08:44:54Z

This is an expansion of Misha's comment. Since $g : (S,\rho) \to N$ is a pleating map we have $g$ is $1$-Lipschitz. That is, for any $x, y \in S$ we have $d_N(g(x), g(y)) \leq d_\rho(x, y)$. In fact, if $\alpha$ is a geodesic arc in $(S,\rho)$ connecting $x$ to $y$, and if $\beta$ is a geodesic arc in $N$, connecting $g(x)$ to $g(y)$ in the relative homotopy class of $g(\alpha)$, then $\ell_N(\beta) \leq \ell_\rho(\alpha)$. Similarly, if $\alpha$ is a geodesic loop in $S$ then the geodesic representative $\beta$, of $g(\alpha)$, has $\ell_N(\beta) \leq \ell_\rho(\alpha)$.

It follows that $R = R_\rho$, the injectivity radius of $(S, \rho)$, is greater than or equal to the injectivity radius of $N$. Let $x, y \in S$ be any pair of points. Let $\alpha$ be the shortest geodesic segment in $S$ connecting $x$ to $y$. For any point $z$ of $\alpha$ let $D \subset S$ be the open ball of radius $R$ about $z$. It follows that $\alpha \cap D$ is a single arc, centered at $z$. Thus the $R/2$ neighborhood of $\alpha$ is an embedded strip with area less than the area of $(S, \rho)$. Since the area of $(S, \rho)$ is $-2\pi\chi(S)$, deduce an upper bound $A$ for the length of $\alpha$ and hence for the diameter of $S$.

So -

Reading the third to last sentence of your post, I think that you may be asking a different question from what you actually wrote in the first half. That is, instead of the distances $d_N(g(x), g(y))$ and $d_\rho(x,y)$, you are interested in upper bounds for certain geodesic arcs connecting the points.... Looking at Minsky's paper, I think that the proof with $x \neq y$ is basically the same compactness argument as his version with $x = y$.

Answer by Sam Nead for Why is the fundamental group of a compact Riemann surface not free ?

2012-04-06T22:24:46Z

Here is a Stallings-style argument. Suppose that $S$ is a closed connected surface. Let $G = \pi_1(S)$. Suppose $T$ is a graph and $v \in T$ a vertex. Let $F = \pi_1(T,v)$ and suppose that $\phi \colon G \to F$ is any homomorphism.

Theorem: If $\phi$ is injective then $S$ is the two-sphere.

Here's the proof. Let $e$ be any edge of $T$ and let $p \in e$ be the midpoint. The given generating set ${a_i,b_i}$ for $G$ gives a one-skeleton for $S$, with one vertex; call the vertex $u$. Let $Q$ be the single two-cell remaining in $S$.

We may now define a map $f$ from the one-skeleton of $S$ to $T$ by mapping $u$ to $v$ and by mapping the edges of $Q$ to paths in $T$ as instructed by $\phi$. Note that the relation in $G$ is killed by $\phi$ -- so, thinking of the image as a word $w$, we find $w$ is a completely reducible word. The reduction of $w$ tells us how to extend $f$ to the two-cell $Q$. (Draw $Q$. Subdivide and label the edges of $Q$ by their images. The first reduction of $w$ gives a triangle cutting off a pair of these new, smaller edges. Etc.)

It follows that $f$ induces the homomorphism $\phi$. (It does the correct thing to the generators and to the relator.) Now consider $C = f^{-1}(p)$. By construction, $C$ is a collection of circles in $S$. If any component of $C$ is non-trivial in $G$, then $\phi$ is not injective, a contradiction. From the Jordon curve theorem deduce that all components of $C$ bound disks. Thus we may homotope $f$ so that $p$ is not in the image. Thus we may homotope $f$ so that the interior of $e$ is not in the image. Thus we may reduce the number of edges in $T$. We now induct downwards. In the base case, where $T = v$, we find that $f$ is the constant map, so $G$ is the trivial group, so $S$ is the two-sphere.

Answer by Sam Nead for Pseudoanosov mapping torus and length of curves.

2012-03-19T11:52:05Z

You should look at the papers of Yair Minsky. Perhaps the right place to start is "End invariants and the classification of hyperbolic 3-manifolds".

Answer by Sam Nead for Changing Careers: Becoming a Professional Mathematician

2012-02-11T19:22:11Z

In my day a high score on the GRE subject test was necessary but not sufficient for admission to graduate school. Letters of recommendation are also very important. These do not need to be from researchers in mathematics; letters from employers should be fine. However, the letters do need to discuss your suitability for graduate school; letters that discuss your mathematical ability are particularly useful.

Answer by Sam Nead for Relationship between hyperbolicity in group theory and hyperbolicity in geometry

2012-02-05T23:37:19Z

[See Peter Scott's Bulletin article for more information.] Typically, we say an orbifold $Q$ is hyperbolic if it comes to us as a quotient of hyperbolic space $H^n$ by the action of a discrete group $G$ of isometries. If the action $G$ is cocompact then $G$ will be a Gromov hyperbolic group. This is the "easy direction".

On the other hand, if $Q$ is an orbifold with enough topological hypotheses (for example, dimension three, irreducible, "good" as Agol says, perhaps more...) then, if the orbifold fundamental group of $Q$ is Gromov hyperbolic it follows from the geometrization theorem (Perelman and so on) that $Q$ is orbifold homeomorphic to a quotient as in the first paragraph.

So, roughly, the two notions are equivalent. However one direction is easy -- it follows from basic definitions in the field of coarse geometry -- and the other direction is one of the most famous recent results in mathematics.

Answer by Sam Nead for Homotopy class of a homeomorphism

2011-11-29T11:05:01Z

A homeomorphism (of surfaces) is isotopic to a PL homeomorphism. See Theorem A4 of Epstein's "Curves on 2-manifolds and isotopies". He gives a proof. I haven't read the paper recently, but I recall that it is fairly self-contained.

Answer by Sam Nead for Primitive elements in a free group of rank three

2011-11-27T15:17:15Z

There are numerous algorithms to decide this question. At bottom all of these are based on the "monogon" and "bigon" condition: If $\alpha$ is a closed loop on a surface then we can homotope $\alpha$ to realize its minimal self-intersection number by looking for and then removing mongons and bigons. For an example of such a paper, see Chillingworth's "Winding numbers on surfaces. II". For a related discussion, see section 1.2.4 of the "Primer on mapping class groups" by Farb and Margalit.

There is also a more geometric criterion having to do with "linking at infinity." See section 8.2.4 of the Primer.

Answer by Sam Nead for what is the meaning of a curve $C$ representing Identity in fundamental group?

2011-11-25T17:35:10Z

Suppose that $K$ is a simple closed curve in $M^3$. I'll assume that $M$ is orientable, compact, and without boundary. Let $V$ be a closed regular neighborhood of $K$; so $V \cong S^1 \times D^2$ is a solid torus. Let $X$ be the closure of $M - V$; so $X$ is the exterior of $K$. Let $T = X \cap V$; so $T$ is a two-torus. So $\partial X = \partial V = T$ and $M = X \cup_T V$. Note that $T$ is a two-torus. Let $D \subset V$ be a meridian disk; that is, a disk of the form $\lbrace \mbox{pt} \rbrace \times D^2$.

As Igor indicates, the map $\pi_1(T) \to \pi_1(X)$ induced by inclusion has a kernel if and only if there is a embedded disk in $E \subset X$ with boundary on $T$. If $\partial D$ and $\partial E$ meet once then $K$ bounds a disk in $M$.

To recap: the knot $K$ bounds an embedded disk in $M$ if and only if

the map from $\pi_1(T) \to \pi_1(X)$ has kernel and
the curve that dies ($\partial E$) meets the meridian $(\partial D$) exactly once.

Answer by Sam Nead for How to rigorously prove that simple closed curves on a surface are primitive closed curves ?

2011-11-03T22:24:44Z

Suppose that $c = \gamma^n$ in $\pi_1(X)$. Note that, as $\pi_1(X)$ is torsion free and $c$ is assumed to be non-trivial, the element $\gamma$ generates an infinite cyclic subgroup $\langle \gamma \rangle < \pi_1(X)$. Let $A = X^\gamma$ be the cover of $X$ corresponding to the subgroup $\langle \gamma \rangle$. So $\pi_1(A)$ is also infinite cyclic. Since $X$ is orientable, so is $A$. It follows from the classification of surfaces $A$ is a (non-compact) annulus.

Note that $\gamma$ can be lifted to $A$ and this lift, $\gamma'$, is homotopic to the core curve of $A$. Likewise $c$ lifts to a curve $c'$ and we have $c' = (\gamma')^n$ in $\pi_1(A)$. Since $c$ is simple in $X$ the lift $c'$ is simple in $A$. By the intermediate value theorem (sort of!) the only simple curves in $A$ are isotopic to the trivial curve and to the core curve. Thus $n = \pm 1$ and we are done.

Answer by Sam Nead for What could be some potentially useful mathematical databases?

2011-10-31T12:48:14Z

KnotInfo. The three-manifold and knot censuses included with SnapPy. Also relevant is CompuTop, a list of "topological" software.

Answer by Sam Nead for Hilbert's 3rd problem,number theory, motives, cyclic homology,...

2011-10-28T16:45:02Z

"What is motivic measure?" by Hales. http://www.ams.org/journals/bull/2005-42-02/S0273-0979-05-01053-0/home.html

Do finite groups acting on a ball have a fixed point?

2011-10-18T22:46:04Z

Suppose that $G$ is a finite group, acting via homeomorphisms on $B^n$, the closed $n$-dimensional ball. Does $G$ have a fixed point?

A fixed point for $G$ is a point $p \in B^n$ where for all $g \in G$ we have $g\cdot p = p$. Notice that the answer is "yes" if $G$ is cyclic, by the Brouwer fixed point theorem. Notice that the answer is "not necessarily" if $G$ is infinite. If it helps, in my application I have that the action is piecewise linear.

First I thought this was obvious, then I googled around, then I read about Smith theory, and now I'm posting here.

Comment by Sam Nead

2013-03-06T21:51:25Z

I believe the twister page is not relevant -- the b++ etc is "classic" SnapPea notation.

Comment by Sam Nead

2013-03-02T14:01:52Z

Finding geodesics in the curve complex is algorithmic, due to work of Leasure (his thesis) and independently, Shackleton.

Comment by Sam Nead

2013-03-02T13:59:22Z

What is the definition of a "spanning annulus"? Note that any annulus in a handlebody is either compressible or boundary compressible, so the "right" class of annuli may be a bit difficult to chose. For example, do you want to rule out boundary parallel annuli?

Comment by Sam Nead

2013-02-20T13:55:35Z

math.stackexchange.com is a better place for this precise question. Try looking at other questions that are asking the same thing: math.stackexchange.com/…

Comment by Sam Nead

2013-02-18T16:48:48Z

It would require a little bit more abstraction before your questions become mathematics. Perhaps Zipf's law will be a useful place for you to look. en.wikipedia.org/wiki/Zipf%27s_law

Comment by Sam Nead

2013-02-18T16:43:15Z

I would guess that the shortest and most frequent words would pin down the language pretty precisely. The finer details of the word distribution that you are asking about (length, frequency, sentence length, punctuation patterns) will be more useful in identifying the writer.

Comment by Sam Nead

2013-02-13T16:41:19Z

Try some examples, like $(p,p)$ for any $p$, or $(kp, kq)$ for various values of $k, p, q$. What patterns do you notice?

Comment by Sam Nead

2013-02-07T18:58:03Z

[See page 14, paragraph (3) of Hatcher's notes.] Basically, if a boundary torus $T$ compresses, then there is a disk $D$ with various properties. Take a neighborhood $K$ of $T \cup D$ and consider the frontier of $K$ in the manifold $M$. This will be a sphere and so bound a ball in $M$, by irreducibility. Thus $M$ is a solid torus.

Comment by Sam Nead

2013-02-07T18:53:16Z

@Igor - Not quite! For example, the solid torus $S^1 \times B^2$ is irreducible and atoroidal (and hyperbolizable!) but the boundary torus is compressible. There are also Seifert fibered spaces which satisfy the hypothesis and conclusion, but that are of course not hyperbolic.

Comment by Sam Nead

2013-02-07T17:53:17Z

I will add a reference to my answer.

Comment by Sam Nead

2013-02-07T17:13:03Z

There is a purely geometric proof, as well. Suppose $N$ is hyperbolic, and $B_S$ and $B_T$ are disjoint horo-tori about $S$ and $T$. Suppose that $A$ is a compact annulus connecting $B_S$ to $B_T$. Lift everything to the universal cover. A component of the lift of $A$ is a strip (quasi-isometric to a line!) that fellow-travels lines in two distinct horospheres, a contradiction.

Comment by Sam Nead

2013-02-07T17:08:24Z

There are lots of copies of $Z^2$ in $\pi_1(K)$ and all of them have to be parabolic. This leads to contradictions.

Comment by Sam Nead

2013-02-07T17:04:02Z

@Igor - I suggest you ask this (new, to my eyes) question in a separate post. However, very briefly, a hyperbolic manifold is algebraically atoroidal - that is, any $Z^2$ subgroup is parabolic. [This is an exercise in hyperbolic geometry, using the discreteness of the group.] Let's now do just one of the many cases. Suppose that $S$ and $T$ are boundary tori, and $A$ is an annulus between them. Let $K$ be a small neighborhood of $S \cup A \cup T$. Then $K$ is homeomorphic to $P \times I$ where $P$ is a pair of pants. So $\pi_1(K)$ is a rank two free group, crossed with $Z$.

Comment by Sam Nead

2013-02-07T14:54:47Z

Because Theorem 19.6 is a different path to proving the hyperbolization theorem. See the remarks immediately after the proof of Theorem 15.3, at the top of page 372.

Comment by Sam Nead

2013-02-07T14:53:36Z

(@Bruno - One last case: the new torus is parallel into the boundary.)