User dave futer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:41:05Z http://mathoverflow.net/feeds/user/8183 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127509/diagrammatic-proof-of-unique-prime-decomposition-of-knots/128644#128644 Answer by Dave Futer for Diagrammatic proof of unique prime decomposition of knots Dave Futer 2013-04-24T18:21:41Z 2013-04-24T18:31:12Z <p>I am fairly certain that there's no <b>known</b> diagrammatic proof of the uniqueness of prime decompositions. That answers the "who is it due to?" question.</p> <p>Of course, a diagrammatic argument may still be out there, waiting to be discovered. But, echoing Ryan's comment, I expect that line of argument to be very difficult, for the following reason.</p> <p>Suppose you're looking at a diagram $D(K)$ of some composite knot. Perhaps the diagram even shows $K$ to be a connected sum in some fashion. The crux of what you need to show is that there is no <i>alternate</i> prime decomposition besides the one you see. In other words, you would need to show that any prime decomposition of $K$ is visible in some (suitably nice) diagram.</p> <p>At present, the Jones polynomial and its relatives are only known to place strong restrictions on diagrams for certain classes of knots and links. These include alternating links (where everything is easiest) and, to a smaller extent, adequate and semi-adequate links. It's <i>conjectured</i> that if a semi-adequate knot is composite, every semi-adequate diagram must also be composite -- and that conjecture is probably within reach. (See <a href="http://www.ams.org/mathscinet-getitem?mr=2900614" rel="nofollow">this paper</a> by Ozawa, as well as Problem 10.6 in <a href="http://www.springer.com/mathematics/geometry/book/978-3-642-33301-9" rel="nofollow">this book</a>.) However, any solution is likely to involve essential surfaces, which you want to avoid. Furthermore, the semi-adequate setting is essentially the limit of where current knowledge about Jones-type invariants can reach diagrammatic information.</p> <p>Without these invariants, you find yourself looking at problems like the additivity of crossing number, which are known to be devilishly hard. In particular, if the additivity conjecture does get solved, the solution would surely involve something beyond purely diagrammatic methods.</p> http://mathoverflow.net/questions/127892/untwisting-heegaard-diagrams/128536#128536 Answer by Dave Futer for Untwisting Heegaard diagrams Dave Futer 2013-04-23T20:51:35Z 2013-04-24T15:59:59Z <p>I'm sorry to say that this condition is quite rare in practice. If the splitting has high enough distance in the curve complex, any pair of curves from the two disk sets will intersect a lot, resulting in many rectangles. Furthermore, high-distance splittings are "generic."</p> <p>The word "generic" can be made precise in two ways. By the work of Joseph Maher, random walks in the mapping class group result in a high-distance splitting with probability approaching 1. In a different direction, the work of Lustig and Moriah implies that high-distance splittings are "generic" in a measure-theoretic sense.</p> <p>Here are the references:</p> <p>Maher: <a href="http://dx.doi.org/10.1112/jtopol/jtq031" rel="nofollow">http://dx.doi.org/10.1112/jtopol/jtq031</a></p> <p>Lustig-Moriah: <a href="http://arxiv.org/abs/1002.4292" rel="nofollow">http://arxiv.org/abs/1002.4292</a></p> <p><b>Update:</b> Actually, I am becoming convinced that only <i>very low</i> distance splittings can have rectangles.</p> <p><b>Lemma:</b> Let $S$ be a Heegaard splitting surface of genus $g \geq 5$, with Hempel distance $\geq 2$. Then any Heegaard diagram for $S$ contains rectangles. </p> <p>In other words, for genus $g \geq 5$, any Heegaard diagram without rectangles must come from a weakly reducible splitting. I strongly suspect this is true in every genus.</p> <p><i>Proof:</i> Let $\alpha_1, \ldots, \alpha_g$ and $\beta_1, \ldots, \beta_g$ be the curves of any Heegaard diagram for this splitting. By hypothesis, every $\alpha_i$ intersects every $\beta_j$. Now, cut $S$ along all the $\alpha_i$. We get a sphere with $2g$ holes. The maximal number of disjoint, non-parallel arcs in this surface is $6g-6$. On the other hand, since every $\alpha_i$ intersects every $\beta_j$, there are at least $g^2$ remnants of the $\beta$ curves in this sphere. Since $g^2 > 6g-6$, when $g \geq 5$, some of these arcs must run in parallel. QED</p> http://mathoverflow.net/questions/114591/generic-words-of-given-weight Generic words of given weight Dave Futer 2012-11-26T22:11:32Z 2012-11-29T13:12:00Z <p>Suppose you have an alphabet with countably many letters. Every letter has a particular weight (for instance, as in the game of Scrabble). There are a total of $n^2$ letters that have weight $n$.</p> <p>Given any word in this alphabet, let the weight of that word be the sum of the weights of its letters (again, as in Scrabble). It follows that there are roughly exponentially many words of weight $W$. </p> <p>I am sampling words of weight $W$, uniformly at random. My somewhat vague question is: what does a generic'' word look like, for large $W$? This can be made precise in a few ways:</p> <ol> <li><p>What is the expected value of the number of letters comprising a word of weight $W$? How many of these letters are expected to be distinct?</p></li> <li><p>Does a generic word of weight $W$ have a letter that appears only once?</p></li> <li><p>What is the expectation for the number of letters that appear only once in the word?</p></li> </ol> <p>This is quite far from my field of expertise, so even simple pointers to references are much appreciated.</p> http://mathoverflow.net/questions/110014/examples-of-3-manifolds-with-rfrs-fundamental-group Examples of 3-manifolds with RFRS fundamental group Dave Futer 2012-10-18T13:58:36Z 2012-10-18T14:32:25Z <p>I'm wondering if anyone knows how to <i>construct</i> hyperbolic 3-manifolds whose fundamental group is RFRS. Clearly the recent work of Agol, Wise, etc. says that such manifolds are abundant, and in particular present in every commensurability class. But how do you construct examples?</p> <p>The only examples of RFRS manifolds that I'm aware of are torus knot complements (thanks to Stefan Friedl for pointing this out), but these are of course non-hyperbolic. </p> http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex Nielsen-Thurston classification via the curve complex? Dave Futer 2012-03-15T19:13:35Z 2012-07-06T13:18:15Z <p>I am curious to see if anyone knows a proof of the Nielsen-Thurston classification of mapping classes that does not depend on results in Teichmuller theory. </p> <p>From a naive point of view, translation distances in the curve complex should serve the same purpose as translation distances in Teichmuller space. For instance, if a mapping class fixes a simplex setwise, then it's reducible (actually, reducibles look much simpler from this point of view). Similarly, a mapping class is pseudo-Anosov iff its stable translation length in the curve complex is strictly positive -- but can one prove this without quoting results of Masur-Minsky that need Teichmuller theory anyhow?</p> http://mathoverflow.net/questions/76885/optimal-pants-decompositions-of-a-hyperbolic-surface Optimal pants decompositions of a hyperbolic surface Dave Futer 2011-09-30T20:59:34Z 2011-10-10T18:18:30Z <p>Let $S$ be a hyperbolic surface, which is not the punctured torus or $4$-holed sphere. I am interested in finding a geometrically optimal'' pants decomposition on $S$. </p> <p>Here is a candidate definition. Given a pants decomposition $P$, order the curves of $P$ <i>from longest to shortest</i> (in the hyperbolic metric). Then, pants decompositions $P$ and $P'$ can be compared by comparing the lengths of their curves lexicographically. That is: if the longest curve $c_1$ of $P$ is shorter than the longest curve $c'_1$ of $P'$, then $P$ is better. Or, if $\ell(c_1) = \ell(c'_1)$ and $\ell(c_2) &lt; \ell(c'_2)$, where $c_2, c'_2$ are the second-longest curves of $P$ and $P'$, then $P$ is better. And so on, lexicographically.</p> <p>With this definition, the induced ordering on pants decompositions becomes a well-ordering. More precisely: given a fixed pants decomposition $P$, there are finitely many curves shorter than the longest curve of $P$, hence finitely many better pants decompositions. In particular, there exists an optimal'' decomposition, whose longest curve is no longer than the <a href="http://arxiv.org/abs/0911.5149" rel="nofollow">Bers constant</a>.</p> <p>It is clear that optimal decompositions are not necessarily unique (otherwise, the optimal decomposition would never change as we move in Teichmuller space). But if $P$ and $P'$ are both optimal on a given surface, what can be said about how far apart they are? For example: are their shortest curves necessarily disjoint?</p> http://mathoverflow.net/questions/68288/laminations-as-a-limit-of-ideal-triangulations Laminations as a limit of ideal triangulations Dave Futer 2011-06-20T14:43:38Z 2011-09-02T19:31:28Z <p>I am wondering about the following:</p> <blockquote> <p>Suppose that $S$ is a non-compact hyperbolic surface of finite area. Suppose that $\lambda \subset S$ is a non-trivial, geodesic, measured lamination. Forget the transverse measure. Is there a (non-compact) geodesic lamination $\lambda'$ containing $\lambda$, and a sequence of ideal geodesic triangulations $T_i$, so that $T_i \to \lambda'$ in the Chabauty topology?</p> </blockquote> <p>A little bit of context:</p> <ul> <li>The Chabauty topology is a generalization of the Hausdorff topology to non-compact sets. It is characterized by the condition that every point of $\lambda'$ is the limit of a sequence of points $x_i \in T_i$, and conversely every convergent sequence $x_{i_n} \in T_{i_n}$ limits to a point of $\lambda'$. See, e.g. <a href="http://books.google.com/books?id=MoZlcJv6ww0C&amp;lpg=PR5&amp;dq=canary%20epstein%20green&amp;pg=PA60#v=onepage&amp;q=chabauty&amp;f=false" rel="nofollow">Notes on Notes of Thurston</a>.</li> <li>I am mainly interested in the answer in the setting where $\lambda$ is the pleating lamination on the boundary of the convex core of a quasifuchsian $3$-manifold. This places some additional hypotheses on $\lambda$: for example, it would have to be compact. But the question seems to be intrinsically $2$--dimensional, and it's not clear to me how to use compactness of $\lambda$ as a hypothesis.</li> <li>If the ideal triangulations $T_i$ are replaced by simple closed curves, the result is well-known. So one approach would be to take a sequence of closed curves $C_i$, limiting to $\lambda' \supset \lambda$, approach each $C_i$ by triangulations (twisting more and more), and then take a diagonal sequence of triangulations. But it's not clear that this diagonal sequence even converges.</li> </ul> <p>Anyway, either a reference or a way to argue would be much appreciated!</p> http://mathoverflow.net/questions/68288/laminations-as-a-limit-of-ideal-triangulations/74391#74391 Answer by Dave Futer for Laminations as a limit of ideal triangulations Dave Futer 2011-09-02T19:14:39Z 2011-09-02T19:31:28Z <p>It turns out that the answer to the question is "yes". Saul Schleimer and I needed this result for a paper that we just finished writing, so we ended up sorting it out. The full argument is written down in <a href="http://front.math.ucdavis.edu/1108.5748" rel="nofollow">Lemma A.6 in the Appendix of this paper</a>, so what follows below is an outline. </p> <p>Take a sequence of closed curves $C_i$, which limits to $\lambda$ in the measure topology. This sequence has a subsequence (which I will still call $C_i$) limiting to $\lambda' \supset \lambda$ in the Chabauty topology. Now, each $C_i$ is contained in the Chabauty limit of a sequence of triangulations $T_{i,j}$. This means that one can take a representative triangulation $T_{i,j(i)}$ that is very close to $C_i$, where "very close" can be quantified (say, closer than distance $1/i$) because the Chabauty topology is metrizable. Now, the sequence $T_{i,j(i)}$ will converge to a lamination $\lambda'' \supset \lambda' \supset \lambda$.</p> http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers/69539#69539 Answer by Dave Futer for What are some examples of colorful language in serious mathematics papers? Dave Futer 2011-07-05T14:23:21Z 2011-07-05T14:23:21Z <p>A gem of R.H. Bing:</p> <blockquote> <p>Dimension 4 is the most difficult dimension. It is too <i>old</i> to spank, the way we might deal with the little dimensions 1, 2, and 3; but it is also too <i>young</i> to reason with, the way we deal with the grown-up dimensions 5 and higher.</p> </blockquote> <p>Source here: <a href="http://www.ams.org/journals/bull/2011-48-03/S0273-0979-2011-01320-9/S0273-0979-2011-01320-9.pdf" rel="nofollow">http://www.ams.org/journals/bull/2011-48-03/S0273-0979-2011-01320-9/S0273-0979-2011-01320-9.pdf</a></p> http://mathoverflow.net/questions/69209/3-orbifolds-with-a-seifert-geometry-that-are-not-actually-seifert-fibered 3-orbifolds with a Seifert geometry that are not actually Seifert fibered Dave Futer 2011-06-30T18:09:22Z 2011-06-30T18:47:57Z <p>It is well-known that Seifert fibered $3$--manifolds are geometric: they admit one of the Thurston geometries $S^2 \times R$, $R^3$, $H^2 \times R$, $S^3$, $Nil$, and $PSL(2,R)$. Furthermore, the converse is also true: every $3$--manifold admitting one of these 6 geometries has a Seifert fibration over a $2$--orbifold.</p> <p>In the realm of $3$--orbifolds, it is still true that every Seifert fibered $3$--orbifold is geometric, with the same 6 geometries. However, the converse is false: there exist geometric $3$--orbifolds with one of the 6 "Seifert" geometries, which do not Seifert fiber. My question is:</p> <blockquote> <p>What are the orientable $3$--orbifolds that have a Seifert geometry, but are not Seifert fibered? Do we know a complete list?</p> </blockquote> <p>One example of this weird phenomenon is the figure-8 knot, labeled 3. This orbifold is Euclidean. But it is not Seifert fibered, because any order-3 singular locus must be vertical in a Seifert fibration. Thus, if there was a Seifert fibering of the orbifold, drilling out the singular locus would produce a Seifert fibration on the knot complement, which is absurd because the knot complement is hyperbolic.</p> <p>More generally, Dunbar has classified the spherical $3$--orbifolds that are not Seifert fibered. There are 21 such examples in total: <a href="http://www.ams.org/mathscinet-getitem?mr=1118824" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=1118824</a></p> <p>What is known about the other 5 Seifert geometries? </p> http://mathoverflow.net/questions/68888/is-a-compact-connected-orientable-3-manifold-with-mathbbzk-fundemental-gr/68926#68926 Answer by Dave Futer for Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? Dave Futer 2011-06-27T13:39:22Z 2011-06-27T13:57:33Z <p>Let me address a more general question:</p> <blockquote> <p>To what extent is a closed, connected $M^3$ determined by its fundamental group?</p> </blockquote> <p>Following the <a href="http://en.wikipedia.org/wiki/Geometrization_conjecture" rel="nofollow">Geometrization Theorem</a>, we have a complete answer. There are only two ways in which a closed $3$--manifold $M$ can fail to be determined by its fundamental group:</p> <ol> <li><p>$M$ is a lens space, or a connected sum of something with a lens space. It is <a href="http://en.wikipedia.org/wiki/Lens_space#Classification_of_3-dimensional_lens_spaces" rel="nofollow">well-known</a> that lens spaces are not determined up to homeomorphism by their fundamental groups. Note that in this case, you would see a ${Z}/p$ free factor in your group $G$, so it can't arise in the context of your question.</p></li> <li><p>$M = N_1 \sharp N_2$, where each $N_i$ is orientable, and each is chiral (fails to have an orientation-reversing symmetry). In this case, reversing the orientation on one factor would produce $M' = N_1 \sharp \overline{N_2}$, which is not homeomorphic to $M$ but has the same fundamental group. This also cannot arise in your context, for exactly the reasons that Igor outlined, and because $S^1 \times S^2$ <em>does</em> have an orientation-reversing symmetry.</p></li> </ol> <p>Finally, let me point out that although Geometrization may seem like an overly big hammer for this question, in fact one needs the Poincare conjecture. For, if there existed a fake $3$--sphere, one could take the connect sum with that manifold without altering the group.</p> http://mathoverflow.net/questions/68063/are-there-any-very-hard-unlinks/68065#68065 Answer by Dave Futer for Are there any very hard unlinks? Dave Futer 2011-06-17T15:27:41Z 2011-06-17T16:30:17Z <p>This is not a direct answer to Daniel's question, but it could be potentially useful.</p> <p>Suppose we replace the unlink in the question by a split link $L$ whose components $K_1, \ldots, K_n$ are prime, non-trivial knots. Then, the $K_i$ are separated by a collection of essential two-spheres $S_j$, but these spheres can of course look extremely complicated in a given diagram of $L$. The analogue of Daniel's question in this context is:</p> <blockquote> <p>Does there exist a fast geometric algorithm that will identify the essential two-spheres in the prime decomposition of $S^3 \setminus L$?</p> </blockquote> <p>Now, there is a paper of Marc Lackenby that provides some insight into this question: </p> <pre><code>http://arxiv.org/abs/0805.4706 </code></pre> <p>Starting from an arbitrary diagram of the connected sum $K_1 \sharp \ldots \sharp K_n$, he cuts it into a distant union (split link) $L = K_1 \cup \ldots \cup K_n$, and proves that the essential two-spheres separating these components have to look (relatively) simple in this handle structure. I am very far from an expert on algorithms, but it seems that knowing the essential two-spheres are not too complicated ought to guide a way to "pull them apart", which is what you're after.</p> http://mathoverflow.net/questions/66484/generalizations-of-dehn-nielsen-baer Generalizations of Dehn-Nielsen-Baer Dave Futer 2011-05-30T19:19:53Z 2011-05-30T19:58:29Z <p>For a manifold $M$, define the mapping class group $Mod(M)$ to be the set of self-diffeomorphisms of $M$, modulo isotopy. In symbols, $Mod(M) = \pi_0 Diff(M)$. Of course, every self-diffeomorphism gives an automorphism of $\pi_1(M)$, well-defined up to conjugacy because of basepoint issues. Thus we have a homomorphism $\sigma: Mod(M) \to Out ( \pi_1 M)$.</p> <p>When $M$ is a surface, the Dehn-Nielsen-Baer theorem says $\sigma: Mod(M) \to Out ( \pi_1 M)$ is an isomorphism. My question is: what can be said in higher dimensions?</p> <p>Assuming $M$ is a $K(\pi, 1)$, one can identify $Out ( \pi_1 M)$ with the set of homotopy classes of homotopy equivalences of $M$. Through this lens, injectivity of $\sigma$ is the question of whether two homotopic diffeomorphisms need to be isotopic. Surjectivity of $\sigma$ is the question of whether every homotopy equivalence is homotopic to a self-diffeomorphism.</p> <p>From a naive point of view, both injectivity and surjectivity seem hard. What is known about them?</p> http://mathoverflow.net/questions/63158/in-knot-theory-benefits-of-working-in-s3-instead-of-mathbbr3/63340#63340 Answer by Dave Futer for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$? Dave Futer 2011-04-28T21:02:53Z 2011-04-28T21:02:53Z <p>In the same spirit as Bruno's answer (which focuses on hyperbolic geometry), let me mention a purely topological result. By the <a href="http://en.wikipedia.org/wiki/Lickorish-Wallace_theorem" rel="nofollow">Lickorish-Wallace theorem</a>, every closed, orientable 3-manifold can be obtained using Dehn surgery on a link in $S^3$. This result is central in the study of 3-manifolds, and simply doesn't hold true if one only considers knots in $R^3$.</p> http://mathoverflow.net/questions/40046/equidistant-points-in-negatively-curved-metric-spaces Equidistant points in negatively curved metric spaces Dave Futer 2010-09-26T19:55:28Z 2010-09-26T20:10:23Z <p>Suppose that $X$ is a simply connected metric space, with a non-positively curved metric (for example, Euclidean or hyperbolic space). Let $A,B,C$ be disjoint, convex sets in $X$, and suppose that the shortest path from $A$ to $B$ passes through $C$. Under these hypotheses, it should follow that there does not exist a point in $X$ that is equidistant to $A$, $B$, and $C$.</p> <p>In the special case where $A,B,C$ are points, this statement amounts to checking inequalities between the sides of a triangle. That is, for any $D \in X$, one of the triangles $ACD$ or $BCD$ -- say, $ACD$ -- will have an obtuse angle at $C$. Then the side $AD$ is longer than $CD$, hence $D$ is not the equidistant point. But I'm stumped about how to show this for more general convex sets.</p> <p>My hunch is that geometers should have encountered this question before. Does anyone have a reference, an argument, or (gasp) a counterexample?</p> http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/34794#34794 Answer by Dave Futer for Math puzzles for dinner Dave Futer 2010-08-06T19:53:20Z 2010-08-06T21:15:14Z <p>Here's one of my favorites. There are 99 bags, each of which contains some number of apples and some number of oranges. Prove that there's a way to select 50 out of the 99 bags, such that these 50 <i>simultaneously</i> contain at least half the total number of apples and at least half the total number of oranges.</p> <p>One fun aspect of this problem is that there are a number of distinct solutions, inspired by different areas of math. I know of at least three...</p> http://mathoverflow.net/questions/34640/local-vs-infinitesimal-rigidity Local vs. infinitesimal rigidity Dave Futer 2010-08-05T15:13:25Z 2010-08-05T20:43:21Z <p>Can someone please explain the difference between local rigidity and infinitesimal rigidity? Does either version of rigidity imply the other?</p> <p>In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension $\geq 3$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction.</p> http://mathoverflow.net/questions/127509/diagrammatic-proof-of-unique-prime-decomposition-of-knots/128644#128644 Comment by Dave Futer Dave Futer 2013-04-24T19:23:44Z 2013-04-24T19:23:44Z Ryan, that's a good point. In fact, it brings up the grey and amorphous boundary of the &quot;diagrammatic&quot; concept. For instance, take Menasco's proof that prime decompositions of alternating knots must be visible in the diagram. That argument is diagrammatic in flavor, but cut-and-paste topology (of the kind that Daniel seems to want to rule out) is also present. So is this in or out? http://mathoverflow.net/questions/127892/untwisting-heegaard-diagrams/128536#128536 Comment by Dave Futer Dave Futer 2013-04-24T13:00:03Z 2013-04-24T13:00:03Z Scott: I take your point, and mostly agree with it. In fact, I would argue that the most interesting splittings are of distance 2. (Either less than 2 or more than 2 provides a lot of additional information.) But by the above update, splittings with rectangles must be weakly reducible -- at least in high genus, and probably in every genus. http://mathoverflow.net/questions/114591/generic-words-of-given-weight/114627#114627 Comment by Dave Futer Dave Futer 2012-11-28T15:24:56Z 2012-11-28T15:24:56Z Thanks Fedja - if posting here is difficult, or if it's something that would be of limited interest to the MO community, feel free to just contact me directly. My email address is easy to google for. http://mathoverflow.net/questions/114591/generic-words-of-given-weight/114627#114627 Comment by Dave Futer Dave Futer 2012-11-27T14:43:21Z 2012-11-27T14:43:21Z Thank you -- please do add an explanation assuming the basics. I will try to backfill the basics as needed. http://mathoverflow.net/questions/114591/generic-words-of-given-weight/114627#114627 Comment by Dave Futer Dave Futer 2012-11-27T14:38:09Z 2012-11-27T14:38:09Z Also, one phrase that I cannot parse at all is &quot;the mountain pass in the circle method&quot;. Thanks for answering my stupid questions! http://mathoverflow.net/questions/114591/generic-words-of-given-weight/114627#114627 Comment by Dave Futer Dave Futer 2012-11-27T14:36:19Z 2012-11-27T14:36:19Z Thank you, Fedja! As I'm quite ignorant of this area of math, it's somewhat opaque to me why these particular generating functions arise. Perhaps you can point me to a reference that explains the &quot;cookbook&quot; for this type of counting question? http://mathoverflow.net/questions/114591/generic-words-of-given-weight Comment by Dave Futer Dave Futer 2012-11-27T01:49:58Z 2012-11-27T01:49:58Z Nope - all the letters in the alphabet are distinct. But we are allowed to use each one many times. I'm thinking of them as, eg, generators of a group (except we're not simplifying the word at all). http://mathoverflow.net/questions/110014/examples-of-3-manifolds-with-rfrs-fundamental-group/110016#110016 Comment by Dave Futer Dave Futer 2012-10-18T15:04:22Z 2012-10-18T15:04:22Z Henry, thanks! So, in particular, are all the manifolds in your paper with Chesebro and DeBlois RFRS on the nose, without having to say &quot;virtually&quot;? http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex/91325#91325 Comment by Dave Futer Dave Futer 2012-03-16T00:03:01Z 2012-03-16T00:03:01Z Thanks, Igor! . http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex/91333#91333 Comment by Dave Futer Dave Futer 2012-03-15T23:42:59Z 2012-03-15T23:42:59Z Thanks, Lee. What about the converse implication: positive stable translation length implies pseudo-Anosov? Does that require knowing e.g. Erica Klarreich's work? http://mathoverflow.net/questions/82936/is-there-a-known-method-for-finding-the-minimum-bridge-index-of-a-knot/83894#83894 Comment by Dave Futer Dave Futer 2011-12-19T21:05:59Z 2011-12-19T21:05:59Z I'm pretty sure Alex Coward's preprint is still the state of the art today. http://mathoverflow.net/questions/82936/is-there-a-known-method-for-finding-the-minimum-bridge-index-of-a-knot/82941#82941 Comment by Dave Futer Dave Futer 2011-12-19T21:04:57Z 2011-12-19T21:04:57Z Ryan: I completely agree with everything you said, with the proviso that normal surface algorithms become very tricky in the presence of normal tori. (The basic issue is: you want to enumerate things of low complexity, where negative Euler characteristic is your basic measurement of complexity. When you have a normal torus, other surfaces can &quot;spin&quot; around it, meaning the number of low-complexity surfaces in your manifold is unbounded.) So normal surface algorithms will work well if the knot complement is hyperbolic (this is the case in Coward's preprint) and not as well otherwise. http://mathoverflow.net/questions/83426/plagiarism-software Comment by Dave Futer Dave Futer 2011-12-14T16:24:17Z 2011-12-14T16:24:17Z A very uncharitable way to read this question would be: &quot;I know that part of the thesis is copied from other sources, but I'm wondering if a smidgen of plagiarism can fly under the radar.&quot; That's uncharitable, but consistent with the question... http://mathoverflow.net/questions/76885/optimal-pants-decompositions-of-a-hyperbolic-surface/77597#77597 Comment by Dave Futer Dave Futer 2011-10-10T14:29:39Z 2011-10-10T14:29:39Z Very elegant - thank you! http://mathoverflow.net/questions/76885/optimal-pants-decompositions-of-a-hyperbolic-surface Comment by Dave Futer Dave Futer 2011-10-01T13:52:29Z 2011-10-01T13:52:29Z Thanks! The discussion of systoles seems relevant.