User nick salter - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:26:40Z http://mathoverflow.net/feeds/user/960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131080/what-is-an-interpretation-of-the-relation-in-the-cohomology-of-the-pure-braid-gro What is an interpretation of the relation in the cohomology of the pure braid groups? Nick Salter 2013-05-18T18:56:01Z 2013-05-18T20:19:49Z <p>In 1968, Arnol'd proved that the integral cohomology of the pure braid group $P_n$ is isomorphic to the exterior algebra generated by the collection of degree-one classes $\omega_{i,j}\ (1 \le i &lt; j \le n)$, subject to the following relation:</p> <p>$$\omega_{k,l} \omega_{l,m} + \omega_{l,m}\omega_{m,k} + \omega_{m,k}\omega_{k,l} = 0.$$</p> <p>The classes $\omega_{k,l}$ are realizable as differential forms on the pure configuration space of $n$ points in $\mathbb C$ as follows: $$\omega_{k,l} = \frac{1}{2 \pi i} \frac{dz_k - dz_l}{z_k - z_l},$$</p> <p>from which it can be seen that the classes $\omega_{k,l}$ are computing the winding number of the $k^{th}$ point around the $l^{th}$. In his paper (which can be found <a href="http://www.pdmi.ras.ru/~arnsem/Arnold/arnold_MZ69e.pdf" rel="nofollow">here</a>), Arnol'd merely remarks that the above relation can be seen to hold for the forms $\omega_{k,l}$ by direct computation.</p> <p>My question is, does the above relation have an interpretation in the context of winding numbers? How would I know to find this relation if I were trying to compute the cohomology of the pure braid group on a desert island?</p> http://mathoverflow.net/questions/8326/is-there-an-approach-to-understanding-solution-counts-to-quadratic-forms-that-doe Is there an approach to understanding solution counts to quadratic forms that doesn't involve modular forms? Nick Salter 2009-12-09T07:23:18Z 2013-04-23T20:38:06Z <p>Given a quadratic form Q in k variables, there is an associated theta series $$\theta_Q(z) = \sum_{x\in \mathbb{Z}^k}q^{Q(x)}$$ where $q = e^{2\pi i z}$ which is a modular form of weight $k/2$. Thus the coefficient on $q^n$ counts the number of solutions in $\mathbb{Z}^k$ to $Q(x) = n$; denote this quantity $r_Q(n)$. The level of these modular forms is related to the level of their corresponding quadratic forms. Since these modular forms live inside finite-dimensional vector spaces, one question to ask is whether there are any linear relationships among them. At least among the ternary forms for which I've looked at this computationally, there often is a great deal of linear dependence among forms of this type that live in the same space, and so there are a lot of linear relationships among the $r_Q(n)$ for various $Q$.</p> <p>My question is if there are "other reasons" (i.e. not related to modular considerations) to expect that the $r_Q(n)$ for various $Q$ should be related in this fashion. Is there a reason <em>a priori</em> to expect that these theta series should be heavily linearly-dependent on one another? Is there a more combinatorial approach (or an alternative number-theoretic approach) to counting solutions to quadratic forms that could suggest the form that these relationships might take? I'm asking because in the course of my computations with ternary forms I observed that twisting forms by quadratic characters always (at least for all the examples I computed) gave a form that was linearly dependent on untwisted forms that lived in the same space. Is there a reason to suspect why this should always be the case?</p> http://mathoverflow.net/questions/121927/does-there-exist-a-surface-bundle-over-a-surface-of-genus-at-least-2-that-fibers Does there exist a surface bundle over a surface of genus at least 2 that fibers in three distinct ways? Nick Salter 2013-02-15T17:39:53Z 2013-02-27T17:24:15Z <p>Let $$\Sigma_g \to E \to \Sigma_h$$ be a surface bundle over a surface. Unless otherwise stated, I'll assume $g, h \ge 2$. The theory of Thurston norm shows that surface bundles over $S^1$ often fiber in infinitely many ways (e.g. with fibers of infinitely many genera). For surface bundles over surfaces, however, the Euler characteristic (which is the product of the characteristics of the base and the fiber) provides an arithmetic constraint on the possible genera of the base and the fiber. When the genus of the base and the fiber are both at least two, this shows that there are only finitely many possible fiber genera, and an analysis of the fundamental group shows that there are only finitely many possibilities for the subgroup of $\pi_1E$ corresponding to the fiber (this follows, for instance, by work of F.E.A. Johnson). Moreover, Johnson's work also shows that when the monodromy representation $\rho: \pi_1 \Sigma_h \to \operatorname{Mod}(\Sigma_g)$ is non-injective, there are at most two fiberings, so that any bundle fibering in at least three ways must have injective monodromy.</p> <p>I am aware of the example of Atiyah-Kodaira, which shows that it is possible for bundles with injective monodromy to fiber in at least two ways (of course, product bundles are a trivial example of multiply-fibered total spaces as well). However, I haven't seen any example of a bundle with three distinct fiberings when the base genus is at least two. When the base genus is one, there are trivial constructions one can do: if $M$ is any three-manifold fibering over $S^1$ in infinitely many ways, then $M\times S^1$ will fiber over $S^1\times S^1$ in infinitely many ways. I don't know of an example of a torus bundle over a surface that fibers in infinitely many ways, but I would be interested to see this, too. </p> <blockquote> <p>With all this said, does anyone know of an example of a surface bundle over a surface of genus $h \ge 2$ that fibers in at least three ways (i.e these are pairwise non-fiberwise diffeomorphic)? What about a surface bundle over a closed $2k$-manifold of nonzero Euler characteristic that fibers in at least $k+2$ ways? (Note that a product of $k+1$ surfaces gives an example where there are $k+1$ fiberings).</p> </blockquote> http://mathoverflow.net/questions/104873/regarding-the-thurston-norm-and-the-ways-that-a-three-manifold-can-fiber-over-the Regarding the Thurston norm and the ways that a three-manifold can fiber over the circle Nick Salter 2012-08-16T22:19:51Z 2012-08-17T03:15:15Z <p>I'm learning about the Thurston norm and am trying to understand the implications that the existence of fibered faces has for the ways in which a given three-manifold $M$ can fiber over the circle. In particular, I am interested in the following question:</p> <blockquote> <p>Let $M$ be a compact, oriented three-manifold without boundary, and suppose that $M$ is irreducible and atoroidal, so that the Thurston norm is non-degenerate. Suppose further that $M$ admits a fibering over $S^1$, so that there exists a fibered face for the unit norm ball. Is it then true that $M$ fibers over $S^1$ with fiber $S_g$ the closed surface of genus $g$ for infinitely many $g$?</p> </blockquote> <p>I believe the answer is yes but I haven't seen a discussion of a result of this type in any of the sources I have consulted (Thurston's paper, <a href="http://www.math.harvard.edu/~ctm/papers/home/text/papers/alex/alex.pdf" rel="nofollow">this paper</a> of McMullen, and the books by Kapovich, Calegari, and Candel-Conlon mentioned in <a href="http://mathoverflow.net/questions/85469/expository-accounts-of-the-thurston-norm" rel="nofollow">this question</a>.) The argument I constructed relies on Theorem 6.1 in the aforementioned McMullen paper, which gives conditions under which a class $\phi \in H^1(M, \mathbb{Z})$ is Poincare-dual to a connected norm-minimizing surface $S$. Specifically, for any $\phi$ that is primitive (i.e. maps onto $\mathbb{Z}$) and for which $b_1(M_\phi)$ is finite (where $M_\phi$ is the cyclic covering of $M$ associated to the kernel of $\phi$), a connected, Thurston-norm minimal $S$ dual to $\phi$ exists. In the case where $\phi$ is associated to a fibration $M \to S^1$, it seems to me that the condition on the finiteness of $b_1$ is automatically satisfied: McMullen remarks that in these circumstances we have $b_1(M_\phi) = b_1(F)$ where $F$ is the fiber of the fibration associated to $\phi$.</p> <p>Now the argument goes as follows: given any primitive $\phi$ in the cone on the fibered face (such $\phi$ exist with arbitrarily large Thurston norm), take the $S$ associated to $\phi$ by the McMullen theorem. By construction, $S$ is dual to $\phi$, which is in turn dual to the fiber $F$ of the associated fibration, so that $S$ and $F$ represent the same homology class. Per a lemma in Calegari's book, a norm-minimizing surface in an irreducible manifold is incompressible. Then, following the remark of Thurston at the beginning of his Section 3, any incompressible surface homologous to a fiber is in fact isotopic to the fiber; in particular they have the same Euler characteristic. Since we can take $|| \phi ||$ to be arbitrarily large, we find that $\chi(S)$, and hence the genus of the fiber, can be arbitrarily large. </p> <p>My question then is whether the answer to my above question is yes, and if so, is the argument I give correct, and is there a simpler way to show it? Should this be surprising? Is there a simple example of this phenomenon, preferably one where the various fiberings are easy to "see"? I know of examples when $M$ has boundary, namely link complements in the three-torus, but I would like an example with closed fibers.</p> http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces/88896#88896 Answer by Nick Salter for An easy way to to explain the equivalence definitions of tangent spaces? Nick Salter 2012-02-19T05:59:00Z 2012-02-19T05:59:00Z <p>Klaus Jänich's undergraduate-level book "Vector Analysis" includes a section showing the equivalence between the three descriptions of the tangent space that you mention. He gives rigorous proofs of everything, and also provides a fair amount of motivation. </p> http://mathoverflow.net/questions/8415/combinatorial-techniques-for-counting-conjugacy-classes Combinatorial Techniques for Counting Conjugacy Classes Nick Salter 2009-12-10T02:36:21Z 2012-02-12T03:22:59Z <p>The number of conjugacy classes in $S_n$ is given by the number of partitions of $n$. Do other families of finite groups have a highly combinatorial structure to their number of conjugacy classes? For example, how much is known about conjugacy classes in $A_n$?</p> http://mathoverflow.net/questions/49595/why-are-graph-imbeddings-defined-the-way-they-are Why are graph imbeddings defined the way they are? Nick Salter 2010-12-16T02:58:13Z 2010-12-17T12:34:30Z <p>In my <a href="http://mathoverflow.net/questions/49328/double-duality-for-geometrically-defined-graph-imbeddings" rel="nofollow">recent question</a> I asked about a proof for the fact that the dual of a dual graph imbedding is equal to the original graph. Thinking about this a little more leads me to wonder why graph imbeddings are defined using rotation systems at all. </p> <p>It seems that the natural thing to do, if one is asked to precisely define a graph imbedding, is the more geometric approach of imbedding edges as simple curves on a pre-existing surface. Indeed, this is just an abstraction of what one is doing when they draw a graph on a sheet of paper. The other attractive feature of defining imbeddings this way is that it is possible to construct imbeddings where the faces are not all 2-cells, by contrast with the rotation system definition. The downside of this approach is that there are some subtleties in defining the dual graph, as <a href="https://docs.google.com/drawings/edit?id=1rzCTS_WmUEFfpMY2b0z9br7oFG0S0xe4vcUmIvuqpWQ&amp;hl=en" rel="nofollow">this picture</a> shows: the imbedding on the right fails to have $G^{**} = G$. </p> <p>On the other hand, it really is easy to show rigorously that the dual graph construction using rotation systems has the property $G^{**} = G$ for an imbedded graph $G$. Is this why combinatorialists usually define graph imbeddings this way? Or are there deeper reasons to avoid the geometric approach?</p> http://mathoverflow.net/questions/49328/double-duality-for-geometrically-defined-graph-imbeddings Double duality for "geometrically defined" graph imbeddings Nick Salter 2010-12-14T00:19:11Z 2010-12-14T00:19:11Z <p>I am studying imbeddings of (connected, undirected, unweighted, multi-)graphs on oriented surfaces of arbitrary genus, and I am searching for a reference for the statement that the dual of the dual graph is the original graph. I have looked at several references, but I have not found a statement that is suitably general for my purposes (and even proofs of the statement are surprisingly rare). In Mohar and Thomassen's <em>Graphs on Surfaces</em>, the authors define graph imbeddings purely combinatorially, using rotation systems, and they do not prove that the dual of the dual of a graph is the original graph, merely stating that this is "clear". In Gross and Tucker's <em>Topological Graph Theory</em>, the authors offer a definition of the dual graph which I do not believe is sufficiently precise to avoid cases where the double dual is not the original graph. Nevertheless they assert that the double dual is the original graph, again without proof.</p> <p>I have been considering graph imbeddings from a geometric perspective, where the edges are imbedded in my surface as simple curves, and where two edge imbeddings intersect only possibly at common endpoints. Thus, my definition differs from the combinatorial approach, but it offers an increased flexibility: the combinatorial definition will only yield faces that are all 2-cells, where I would like to be able to consider more general imbeddings. In any case, it seems like some sort of double-duality statement is true for these general imbeddings.</p> <p>Can you point me to a reference which considers graph imbeddings from this geometric perspective, and that actually proves that the double-dual is the original graph? (As a bonus: can someone explain to me whether "imbedding" or "embedding" is the more widely accepted spelling?)</p> http://mathoverflow.net/questions/40337/ingenuity-in-mathematics/47763#47763 Answer by Nick Salter for Ingenuity in mathematics Nick Salter 2010-11-30T07:14:47Z 2010-11-30T07:14:47Z <p>Alex R's answer reminds me of another sort of clever "strategy" problem. The way I've heard it phrased is as follows (with apologies to the vegetarians in the audience):</p> <p>You are given 1000 cups filled with water, exactly one of which (unknown to you) is laced with a lethal poison that is guaranteed to kill within 24 hours if ingested. You are given ten rats with which to determine the poisoned cup. What is the minimum amount of time needed to be certain?</p> <p>One naive approach would be to break up the cups into groups of 100 to be fed to each rat (i.e. rat 1 gets cups 1 through 100, rat 2 gets 101 through 200 etc). After 24 hours one rat will be dead and you have narrowed the poisoned cup to one in 100. Repeat two more times: you have your poisoned cup identified within 72 hours.</p> <p>There is in fact a much better solution, which identifies the cup in the minimal 24 hours. Since $1000~&lt;~2^{10} = 1024$, assign each cup a number expressed in binary. Then give the water in that cup to those rats for which there is a '1' in the binary representation (so, for example, cup 17 = 10001 is fed to rats 1 and 5). After 24 hours, you can simply read off the number of the poisoned cup by the numbers of the dead rats.</p> <p>You can impress your audience with the remarkable (some might say brutal) efficiency of this procedure: to identify one in a million cups you would need only 20 rats (ignoring, of course, any deaths caused by overhydration...)</p> http://mathoverflow.net/questions/8543/learning-to-think-categorically Learning to Think Categorically Nick Salter 2009-12-11T08:06:57Z 2010-11-02T04:31:08Z <p>Up to this point in my education, I have had very little exposure to the language and machinery of category theory, and I would like to rectify this. My goal is to become conversant with some of the standard categorical ideas; I want to be able to think categorically. What are some suggestions people have for the best way to go about this? Would it be better to take things head on, and read something like MacLane's Categories for the Working Mathematician? Or would it be better to study category theory via its application to a body of ideas that I'm already familiar with, such as representation theory? How have you learned to think categorically?</p> <p>---EDIT---</p> <p>Thank you all for the responses. To clarify, my interest in category theory is more of a means than an end - I want to be fluent in understanding and crafting categorical arguments in other contexts. Although I'd love to be persuaded otherwise at the moment I'm not necessarily interested in studying category theory in its own right; somewhat analogously to how one should be well versed in the language of set theory, even if they do not intend to study sets in and of themselves. Thanks to your responses, I have plenty of material to look at.</p> http://mathoverflow.net/questions/37005/the-geometry-of-recurrent-families-of-polynomials The Geometry of Recurrent Families of Polynomials Nick Salter 2010-08-28T23:06:09Z 2010-08-29T04:41:38Z <p>The Chebyshev T-polynomials have at least two natural definitions, either via the characterizing property $\cos(n\theta) = T_n(\cos(\theta))$, which I will call the geometric definition, or via a recurrence relation $T_{n+1} = 2x\ T_n - T_{n-1}$. My question concerns the relationship between these two definitions, and specifically asks if other families of polynomials defined by similar recurrence relations have a natural geometric interpretation, similarly to how $T_n$ expresses the relation between x-coordinates of particular points on the circle.</p> <p>Starting from the geometric definition of $T_n$, it is straightforward to derive the recurrent definition. Is there a natural way of going the other direction? One thought I have had is as follows. If we treat the $T_n$ as elements of the coordinate ring of the circle, then $T_n$ expresses the relationship between the two parameterizations of the x-coordinates given by $\theta \mapsto \cos(\theta)$ and $\theta \mapsto \cos(n\theta)$. Can we do the same sort of thing with other families of polynomials defined by similar recurrence relations (say, for simplicity, second-order linear polynomial recurrences with coefficients of degree $\leq 1$)?</p> <p>One potential obstacle I have encountered is that $\cosh(n\theta) = T_n(\cosh(\theta))$ as well, so that the hyperbola $X^2 - Y^2 = 1$ is just as natural a choice as the circle for a geometric object associated to $T_n$.</p> <p>Here's my main question: given a particular family of polynomials $P_n$ related by a polynomial recurrence of an appropriately "nice" type, can one associate one or more algebraic curves (or other geometric objects) so that $P_n$ expresses some relationship between various points on the curve?</p> http://mathoverflow.net/questions/32397/vector-spaces-without-natural-bases/32399#32399 Answer by Nick Salter for Vector spaces without natural bases Nick Salter 2010-07-18T20:45:06Z 2010-07-18T20:45:06Z <p>To expand on Anon's answer, I'd like to discuss one way in which the lack of a "natural" basis has some utility. A Hamel basis is a basis for $\mathbb{R}$ over $\mathbb{Q}$. Hamel bases are quite useful, due to their interactions with Cauchy functions (real-valued functions that satisfy an "additive" functional equation $f(x+y) = f(x) + f(y)$. This functional equation is equivalent to being linear over $\mathbb{Q}$. Examples of the utility of Cauchy functions abound. One approach to proving that the cube and the tetrahedron are not equidecomposable (Hilbert's 3rd problem) is to pick the $\mathbb{Q}-$linearly independent set ${1, \pi}$ and, by the magic of AC, this extends to a Hamel basis. Setting up the right Cauchy function then resolves the problem. For more on this, see "Conjecture and Proof" by Miklós Laczkovich.</p> http://mathoverflow.net/questions/32324/what-is-known-about-the-transcendence-of-zeroes-of-riemann-zeta What is known about the transcendence of zeroes of Riemann zeta? Nick Salter 2010-07-17T23:11:29Z 2010-07-18T01:50:14Z <p>I was wondering if there are any well-known results or hunches about whether the non-trivial zeroes of Riemann-zeta (or zeta/L-functions in general) are algebraic or not. </p> http://mathoverflow.net/questions/31073/why-is-the-identity-element-of-the-sandpile-group-self-similar/31269#31269 Answer by Nick Salter for Why is the identity element of the sandpile group self-similar? Nick Salter 2010-07-10T05:08:26Z 2010-07-10T05:16:19Z <p>As per your second question, the following algorithm allows one to compute the identity element.</p> <p>Let $c$ denote the maximal stable configuration; i.e. $c = \sum_{v\in V}(d(v)-1) v$ This is always recurrent. Let $a^{\circ}$ denote the stabilization of a configuration $a$. Then this will give you the identity $e$:</p> <p>$e =(2c - (2c)^\circ)^\circ$</p> <p>If you are interested, check out this applet for doing a lot of this stuff (and it's pretty, too): <a href="http://people.reed.edu/~davidp/sand/program/program.html" rel="nofollow">http://people.reed.edu/~davidp/sand/program/program.html</a></p> http://mathoverflow.net/questions/25004/gaining-intuition-for-how-submodules-behave Gaining intuition for how submodules behave Nick Salter 2010-05-17T12:59:59Z 2010-05-17T20:13:43Z <p>I'm studying elementary commutative algebra this semester, largely following Atiyah-MacDonald. I often find myself in a situation where I'm interested in whether some property of an R-module M is inherited by its submodules (e.g. the property of being finitely generated over R) and I feel like I am lacking the necessary intuition to decide whether something is the case or not. In the case of finite-generation, for example, I think that my mental picture of a finitely-generated R-module is still too close to that of a finite-dimensional vector space to be able to intuit the fact that this simply shouldn't hold true in general. So here's my question - when you run across some property for a module and you want to know whether this is inherited by its submodules, how do you begin to think about the problem? Do you have a standard stock of counterexamples (or a procedure of sorts to concoct a counterexample)? Or do you have a more nuanced way of informally thinking about modules that captures more of their behavior? As I only have a semester of commutative algebra under my belt (think Atiyah-MacDonald), I'd appreciate answers that tend towards the more elementary end of the subject, although if you think that it's not possible to gain a good feel for the way that modules behave without diving deeper, I'd like to hear that too. </p> http://mathoverflow.net/questions/12613/different-conceptions-of-z Different Conceptions of Z Nick Salter 2010-01-22T07:13:56Z 2010-01-22T15:54:01Z <p>To the algebraist, $\mathbb{Z}$ is just the free group with one generator. To the algebraic topologist, $\mathbb{Z}$ is just the fundamental group of the circle. To be glib, what do $\mathbb{Z}$ mean to you?</p> http://mathoverflow.net/questions/8609/favorite-popular-math-book/8641#8641 Answer by Nick Salter for Favorite popular math book Nick Salter 2009-12-12T02:17:28Z 2009-12-15T22:20:36Z <p><strong>Title:</strong> Fearless Symmetry</p> <p><strong>Authors:</strong> Avner Ash and Robert Gross</p> <p><strong>Short Description:</strong> Very much a +++ text, these guys actually got a pop-math account of Galois representation theory published! By far the most technically demanding pop-math book I've ever read, (and one that took me about three years to actually finish) it nonetheless makes for a compelling "non-technical" introduction to a beautiful subject.</p> http://mathoverflow.net/questions/8874/what-are-some-slogans-that-express-mathematical-tricks/8893#8893 Answer by Nick Salter for What are some slogans that express mathematical tricks? Nick Salter 2009-12-14T17:48:08Z 2009-12-14T17:48:08Z <p>You <em>must</em> exchange the order of summation in order to prove any identity involving multiple sums.</p> http://mathoverflow.net/questions/6172/right-way-of-introducing-modular-forms-to-undergraduate-audience "Right" Way of Introducing Modular Forms to Undergraduate Audience? Nick Salter 2009-11-19T21:07:24Z 2009-11-20T17:51:51Z <p>I am giving an <em>extremely</em> short talk (~12 minutes) in a few weeks in which I need to be able to introduce and motivate the idea of a classical modular form to an audience of undergraduates in as short a time as possible. I was intending to introduce modular forms as functions on lattices, following Serre's presentation in A Course in Arithmetic - this seems like the shortest path to motivating the particular symmetries that we require of modular forms. Does anyone else have any good ideas? Particularly, if there is a way of motivating forms of higher levels and of half-integral weight through the same idea of a function on lattices, I'd like to hear about it.</p> http://mathoverflow.net/questions/5907/is-there-a-name-for-this-topology Is there a name for this topology? Nick Salter 2009-11-18T02:41:50Z 2009-11-18T03:03:45Z <p>Let $X$ be a set and let $f: X\longrightarrow X$ be a function on $X$. Introduce a topology on $X$ by the following basis of open sets: for any subset $S$ of $X$, let $B_S$ be the set of forward images of $S$ under $f$; i.e. $$B_S = \{f^n(s): s\in S, n\in \textbf{Z}^+\}.$$ My question is, is this topology well-known and well-understood? Is there a theory which relates properties of $f$ to the resulting topology?</p> http://mathoverflow.net/questions/5800/generalization-of-the-two-bucket-puzzle/5801#5801 Answer by Nick Salter for Generalization of the two bucket puzzle Nick Salter 2009-11-17T08:27:41Z 2009-11-17T08:27:41Z <p>Not an answer but rather a good thing to look at in connection with the problem-</p> <p><a href="http://numb3rs.wolfram.com/501/puzzle.html" rel="nofollow">http://numb3rs.wolfram.com/501/puzzle.html</a></p> http://mathoverflow.net/questions/121927/does-there-exist-a-surface-bundle-over-a-surface-of-genus-at-least-2-that-fibers Comment by Nick Salter Nick Salter 2013-02-15T22:55:53Z 2013-02-15T22:55:53Z Initially, I thought that the $M\times S^1$ example of a bundle over a torus fibering in infinitely many ways might be used as a starting point in a construction of a bundle with multiple fiberings over a surface of higher genus, say by pulling back along a branched cover of $T^2$. But this turns out not to work: the monodromy will then factor as $\pi_1 \Sigma_g \to \mathbb Z ^2 \to \operatorname{Mod}(\Sigma_g)$, and as the first map certainly cannot be injective, Johnson's result then prevents this bundle from having more than two fiberings. http://mathoverflow.net/questions/104873/regarding-the-thurston-norm-and-the-ways-that-a-three-manifold-can-fiber-over-the/104888#104888 Comment by Nick Salter Nick Salter 2012-08-17T15:18:04Z 2012-08-17T15:18:04Z Oh, that's rather simple. Thanks! http://mathoverflow.net/questions/104873/regarding-the-thurston-norm-and-the-ways-that-a-three-manifold-can-fiber-over-the/104888#104888 Comment by Nick Salter Nick Salter 2012-08-17T14:13:15Z 2012-08-17T14:13:15Z Thank you for your answer. As I think about this more, I realize that the essential point that I'd like to understand better is why the fiber of a fibration associated to a primitive class is connected. I understand how the theorems in Thurston's paper establish the norm-minimality of fibers, but do they address connectedness? http://mathoverflow.net/questions/104873/regarding-the-thurston-norm-and-the-ways-that-a-three-manifold-can-fiber-over-the/104878#104878 Comment by Nick Salter Nick Salter 2012-08-16T23:47:16Z 2012-08-16T23:47:16Z Here's a link to the paper that he cites in his answer: <a href="http://sci-prew.inf.ua/v140/1/S0305004105008868.pdf" rel="nofollow">sci-prew.inf.ua/v140/1/S0305004105008868.pdf</a> http://mathoverflow.net/questions/72729/product-of-non-disjoint-cycles Comment by Nick Salter Nick Salter 2011-08-11T23:29:23Z 2011-08-11T23:29:23Z I'm not sure this is MO-level; you might try asking this at math.stackexchange. All the same, for non-disjoint cycles, you compute their product by starting with 1 on the right, following it through, and repeating: eg. to compute (1352)(256), you see that 1 is sent to 1 by the rightmost cycle, then is sent to 3 by the left, so that the product begins (13..) Then you insert 3, which is sent to 5. 5 is sent to 6 by the first cycle, and is fixed by the second; at this point we have computed it to be (1356..) Six goes to 2 and then to 1, and 2,4 are clearly fixed, so that (1352)(256)=(1356)(2)(4) http://mathoverflow.net/questions/54986/why-is-the-laplacian-ubiquitous Comment by Nick Salter Nick Salter 2011-02-10T04:45:56Z 2011-02-10T04:45:56Z I don't know nearly enough about this to make this an answer, but doesn't it have something to do with the Casimir element? For a lot of the most readily familiar objects, the Casimir element will be some version of the Laplacian, right? http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation Comment by Nick Salter Nick Salter 2011-02-03T23:35:25Z 2011-02-03T23:35:25Z Out of curiosity, why are you interested in such a question? (Lest it not be obvious, I mean absolutely no disrespect in asking. I'm merely interested in whether the (non)existence of such an equation would have interesting consequences.) http://mathoverflow.net/questions/53202/angles-in-an-integral-lattice Comment by Nick Salter Nick Salter 2011-01-25T09:17:49Z 2011-01-25T09:17:49Z Is the arithmetic geometry tag appropriate here? http://mathoverflow.net/questions/53122/mathematical-urban-legends Comment by Nick Salter Nick Salter 2011-01-24T21:04:16Z 2011-01-24T21:04:16Z Perhaps not an urban legend per se, but when I was learning algebra, my professor, in an attempt to impress upon us the necessity of checking that certain maps are well-defined, told us the story of a classmate of his who got several years into his Ph.D. thesis before realizing that the maps he was investigating weren't well defined. Horrified, we asked him if this was true. &quot;No&quot; he said, &quot;but that's one lie you'll never forget!&quot; http://mathoverflow.net/questions/52902/proving-theorems-by-using-functions-with-fixed-points Comment by Nick Salter Nick Salter 2011-01-23T04:11:31Z 2011-01-23T04:11:31Z This may be relevant: <a href="http://mathoverflow.net/questions/19272/applications-of-brouwers-fixed-point-theorem" rel="nofollow" title="applications of brouwers fixed point theorem">mathoverflow.net/questions/19272/&hellip;</a> http://mathoverflow.net/questions/49595/why-are-graph-imbeddings-defined-the-way-they-are/49631#49631 Comment by Nick Salter Nick Salter 2010-12-17T20:53:35Z 2010-12-17T20:53:35Z Thanks for your answer. The last part speaks to the major problems I was having in trying to work with embeddings defined geometrically - it seems like even giving a construction of the dual graph which has the double-duality property is a big mess. On the other hand, I still find it somewhat disappointing that in moving to the combinatorial approach, we lose the ability to do things like embed trees on tori, which seems like something we might want to do from time to time. http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/48808#48808 Comment by Nick Salter Nick Salter 2010-12-09T21:43:05Z 2010-12-09T21:43:05Z @ Max Muller: In the spirit of accuracy (but without wanting to move towards opening up any cans of worms) I believe it is the case that Furstenburg's is not a repackaging of <i>analytic</i> ideas, but rather Euclid's original argument. http://mathoverflow.net/questions/40337/ingenuity-in-mathematics/40400#40400 Comment by Nick Salter Nick Salter 2010-12-01T20:47:53Z 2010-12-01T20:47:53Z While I like this result, I'm not sure if this is the sort of thing that will really &quot;wow&quot; a room full of artists, for whom (I imagine) notions of rationality and irrationality aren't familiar enough for this to be anything more than a remote curiosity. http://mathoverflow.net/questions/47214/how-to-present-mathematics-to-non-mathematicians/47235#47235 Comment by Nick Salter Nick Salter 2010-11-24T23:35:30Z 2010-11-24T23:35:30Z +1 for &quot;marriage-theoretic terminology&quot; :) http://mathoverflow.net/questions/37005/the-geometry-of-recurrent-families-of-polynomials/37008#37008 Comment by Nick Salter Nick Salter 2010-08-28T23:51:50Z 2010-08-28T23:51:50Z Thank you. I wonder if this can be extended to address the second question. If you take $y = \sqrt{x^2 - 1}$ then we have the relation $x^2 - y^2 = 1$, from which the hyperbola arises naturally as the geometric object underlying the Chebyshev polynomials. It seems like a similar technique can be used to introduce a $y$ element that generates a finite extension of $\mathbb{C}[x]$ - an algebraic curve!