User sinai robins - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:37:07Z http://mathoverflow.net/feeds/user/12909 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114079/rational-viewing-points-in-a-polygon Rational viewing points in a polygon Sinai Robins 2012-11-21T17:20:06Z 2012-11-21T17:30:23Z <p>We refer to the question posed in <a href="http://mathoverflow.net/questions/112714" rel="nofollow">http://mathoverflow.net/questions/112714</a>, but now ask for constructions or for the existence of rational viewing points. We'll call a point $p$ inside (or on) a polygon $P$ a rational viewing point if all of the angles formed by $p$ together with any two adjacent vertices of $P$ are rational multiples of $\pi$.</p> <p>Problem 1. Suppose we have a convex polygon $P$, and there exists a rational viewing point in (or on) $P$. Must there exist infinitely many rational viewing points in (or on) $P$ ? </p> <p>Problem 2. We observe that whenever we have a vanishing sum of roots of unity, with any real coefficients, then we may consider the individual summands as vertices of a polygon $P$, with the origin as a rational viewing point. In this case, are there always other rational viewing points in (or on) $P$ ?</p> http://mathoverflow.net/questions/112714/seeing-the-vertices-of-a-polygon-with-rational-angles Seeing the vertices of a polygon with rational angles Sinai Robins 2012-11-17T20:00:54Z 2012-11-17T20:33:03Z <p>Given any convex polygon in the plane, is it always possible to find a point $p$ in its interior such that when we draw the line segments from $p$ to each of its vertices, the angles formed at $p$ are all (not necessarily equal) rational multiples of $\pi$? </p> <p>For a triangle $T$, it's easy to construct such a point, namely the Steiner point $p$ will do, enjoying three angles of measure $2\pi/3$ each, between $p$ and any two adjacent vertices of $T$. But is this known in general?</p> http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Zero-free theta functions in the upper half plane Sinai Robins 2012-11-13T16:04:53Z 2012-11-16T01:17:50Z <p>Problem $1$. Which full rank lattices $\Lambda \subset \mathbb R^d$ have their corresponding theta function $\theta_{\Lambda}(\tau):= \sum_{\bf n \in \Lambda } e^{\pi i \tau ||n||^2}$ zero-free in the complex upper half plane $\mathbb H$?</p> <p>For example, for $d=1$, the classical theta function $\theta_{\mathbb Z}(\tau):= \sum_{\bf n \in \mathbb Z } e^{\pi i \tau n^2}$ is the ratio of two Dedekind eta functions, so it is the ratio of two infinite products in q, and hence zero-free inside $\mathbb H$. Similarly, any theta function that can be written as an infinite product in a "nice way" is zero-free.</p> <p>What I would really like is a criterion that, instead of brute-forcing a contour integral to search for zeros, gives us some nice conditions on the lattice $\Lambda$, involving such parameters such as, for example, $vol( \Lambda)$, the successive minima of $\Lambda$ relative to the unit ball, the dual lattice, $Aut(\Lambda)$, etc. </p> <p>EDIT (on Nov 16, 2012). To consider more examples, fix any dimension $d$ and consider any diagonal positive definite quadratic form $a_1 x_1^2 + \dots + a_d x_d^2$, with $a_j$ any fixed positive real numbers. The corresponding lattice $\Lambda$ is therefore the direct sum of $d$ one-dimensional lattices, namely $\sqrt a_1 \mathbb Z \oplus \cdots \oplus \sqrt a_d \mathbb Z$, so that $\theta_\Lambda$ is the product of $d$ one-dimensional theta functions, and hence $\theta_\Lambda$ is nonzero in $\mathbb H$. </p> <p>More generally, if $\Lambda$ is reducible in the sense that it is expressible as the direct sum of some lower dim'l lattices, then we call the corresponding theta function reducible. It is therefore natural to ask the following more particular question.</p> <p>Problem 2. Which irreducible theta functions $\theta_{\Lambda}(\tau)$ are zero-free in $\mathbb H$?</p> http://mathoverflow.net/questions/111747/reciprocity-ehrhart-style-for-real-polytopes Reciprocity (Ehrhart-style) for real polytopes? Sinai Robins 2012-11-07T19:21:24Z 2012-11-12T03:30:23Z <p>Is there some sense in which the well-known Ehrhart reciprocity law for rational, convex, polytopes can be extended to any convex polytope with arbitrary real vertices? </p> <p>In other words, given any real closed (and convex) polytope $P$, is the function that enumerates integer points in dilates of $P$ somehow simply related to the function that enumerates integer points in the dilates of the interior of $P$?</p> <p>Here's one possible direction, given that we no longer have the usual quasi-polynomial behavior for the integer point enumerator. Does there exist some piecewise smooth function $L(t)$, of a real variable $t$, which satisfies the following properties:</p> <p>(1) $L(t)$ agrees with the integer point enumerator #{${\mathbb Z}^d \cap tP$} at all positive integer values of $t$.</p> <p>(2) $L(-t)$ counts the integer points in the interior of $tP$, for all positive integers $t$.</p> <p>(3) The Fourier transform of $L$ is supported at a countable set of points.</p> <p>Intuitively, for all real and positive $t$, such a smooth function $L(t)$ would approximate the piecewise constant function #{${\mathbb Z}^d \cap tP$}). Property (3) above was suggested by Allen Knutson.</p> <p>(ADDED) Let's try to see what happens in dimension one. So we let $P := [0, \alpha]$, a one-dimensional polytope, with $\alpha$ a positive irrational number. We let $L_P(n)$ be the number of integer points in the dilation $nP$, so that here $L_P(n) = [n\alpha]+1$. For the interior $P^o$, we have $L_{P^o}(n) = [n\alpha]$. Using {x} $= \frac{1}{2} + \sum_{k\in \mathbb Z - {0}} \frac{1}{k}e^{2\pi i k x}$, and $L_P(n)= n\alpha -$ {$n\alpha$}+1, we have: $$L_P(n) = [n\alpha]+1 = n\alpha + \frac{1}{2} - \sum_{k \in \mathbb Z - {0} } \frac{1}{k}e^{2\pi i k n \alpha},$$ and let's use this last expression as the extension of $L_P(n)$ to all real values of $n$. We now check for reciprocity, using this new definition of our extended function $L_P(n)$: $$L_P(-n) = -n\alpha + \frac{1}{2} - \sum_{k \in \mathbb Z - {0} } \frac{1}{k}e^{-2\pi i k n \alpha}$$ $$= -n\alpha + \frac{1}{2} + \sum_{k \in \mathbb Z - {0} } \frac{1}{k}e^{2\pi i k n \alpha} \ \ (\text{replacing } k \in \mathbb Z \ by \ -k \in \mathbb Z)$$ $$= - [ n\alpha]= - L_{P^o}(n),$$ which is a reciprocity law. </p> http://mathoverflow.net/questions/111984/gram-matrix-modulo-4 Gram matrix modulo 4 Sinai Robins 2012-11-10T12:36:13Z 2012-11-10T21:02:41Z <p>Suppose we have a full rank, integer sublattice $L$ of the integer lattice $\mathbb Z^d$, where we fix the dimension $d$. Consider the Gram matrix $M$ of $L$, relative to some basis for $L$, and reduce all the entries of $M$ mod $4$. Is there a nice clean description of all the finite types of such mod $4$ reductions of Gram matrices, as we vary over all full rank integer sublattices $L \subset \mathbb Z^d$ while keeping the dimension $d$ fixed? </p> <p>There is a theorem that attempts to describe the Gram matrix of an integer lattice mod powers of $2$, in J.W.S. Cassells' book Rational quadratic forms", Section VIII.4, p. 117 in this book. But in the beginning of that section he writes "...This section is only for the masochistic". I would be very grateful if anyone has found a cleaner description and/or proof, at least in the mod $4$ case. </p> http://mathoverflow.net/questions/111507/partitions-into-parts-from-an-arithmetic-progresion/111738#111738 Answer by Sinai Robins for Partitions into parts from an arithmetic progresion Sinai Robins 2012-11-07T16:12:28Z 2012-11-08T06:25:36Z <p>Problem 1 is solved completely, in the affirmative, in the following paper of Grosswald:</p> <p>Emil Grosswald, Some theorems concerning partitions, Trans. Amer. Math. Soc. 89, 1958, 113–128.</p> <p>Grosswald in fact gives a very accurate estimate for the asymptotics of $q_R(n)$, showing that they grow exponentially fast with $n$, and generalizes things to the case that R consists of any finite union of arithmetic progressions as well. (If you look at his rather intricate paper, look at the function that he calls $H(x)$).</p> http://mathoverflow.net/questions/111141/counting-lattice-points-inside-an-n-dimensional-tetrahedron/111619#111619 Answer by Sinai Robins for Counting lattice points inside an n-dimensional tetrahedron. Sinai Robins 2012-11-06T05:17:44Z 2012-11-07T14:58:59Z <p>This kind of question has come up in the context of "smooth numbers" and their use in factoring large integers. But even in two dimensions, the real right-angled triangle has posed serious difficulties, starting with a sequence of papers by Hardy and Littlewood:</p> <p>G. H. Hardy and J. E. Littlewood, Some Problems of Diophantine Approximation, in ‘’Proc. 5th Int. Congress of Mathematics” (1912), 223–229.</p> <p>G. H. Hardy and J. E. Littlewood , The lattice points of a right-angled triangle, Proc. London Math. Soc. (2) 20 (1921) 15–36.</p> <p>G. H. Hardy and J. E. Littlewood , The lattice points of a right-angled triangle (second memoir), Hamburg Math. Abh. 1 (1922) 212–249.</p> <p>To add flesh to my comment above, suppose you wish to factor a large integer N, and use a fixed "factor base" of primes, so you attempt to write N as a product of these primes to some powers, take logarithms of both sides, divide by log N, and you have an equality of the sort that you are asking about.</p> <p>For more references, there is a sequence of papers trying to solve this problem in the context of a conjecture they call the "Durfree Conjecture" about the genus of algebraic curves. They have a nice bibliography, and one of their more recent papers for this line of research that I could find for you is:</p> <p>Stephen T. Yau and Letian Zhang, AN UPPER ESTIMATE OF INTEGRAL POINTS IN REAL SIMPLICES WITH AN APPLICATION TO SINGULARITY THEORY, Math. Res. Lett. 13 (2006), no. 6, 911–921.</p> <p>The Ehrhart theory can bound such integer counts in real tetrahedra from above and from below, which I've also thought about a bit, but these bounds are of course always asymptotic, as Richard Stanley points out.</p> http://mathoverflow.net/questions/54731/sums-of-fractional-parts-of-linear-functions-of-n/55119#55119 Answer by Sinai Robins for sums of fractional parts of linear functions of n Sinai Robins 2011-02-11T12:31:06Z 2011-02-11T12:31:06Z <p>I looked at the literature just now, and I believe that R.R. Hall may have given a fairly complete answer to these questions. See his 1998 Crelle paper:</p> <p><a href="http://www.reference-global.com/doi/abs/10.1515/crll.1998.035" rel="nofollow">http://www.reference-global.com/doi/abs/10.1515/crll.1998.035</a></p> <p>Regards, Sinai</p> http://mathoverflow.net/questions/114079/rational-viewing-points-in-a-polygon Comment by Sinai Robins Sinai Robins 2012-11-26T15:23:22Z 2012-11-26T15:23:22Z Cute.....but I was genuinely curious about any possible constructions that one can make to generate rational viewing points, and you gave one. Though I agree that it sometimes seems as though one is &quot;tweeking&quot; when one asks a second question, but we learn as we go along, and curiosity about various things/constructions keeps building up. http://mathoverflow.net/questions/114079/rational-viewing-points-in-a-polygon Comment by Sinai Robins Sinai Robins 2012-11-26T15:11:52Z 2012-11-26T15:11:52Z @Fedja: Three no's ? ;) But thanks! Yeah, your argument also shows that for a triangle we do in fact have a dense subset of rational viewing points, by building sliding circles through each pair of vertices and considering their countable network of intersections. But I would conjecture that for any n-gon, with n larger than 3, (particularly interesting is the case n=4) we will never again have a dense subset of rational viewing points. http://mathoverflow.net/questions/112753/can-pure-mathematics-harness-citizen-science/113804#113804 Comment by Sinai Robins Sinai Robins 2012-11-20T01:35:52Z 2012-11-20T01:35:52Z I'll delete the comment. apologies. http://mathoverflow.net/questions/112714/seeing-the-vertices-of-a-polygon-with-rational-angles/112716#112716 Comment by Sinai Robins Sinai Robins 2012-11-18T08:27:37Z 2012-11-18T08:27:37Z @Will: I think your dimension count argument is generally valid, though requires more details I think, as far as showing that most polygons don't have a rational viewing point. On the other hand, it leaves out infinitely many interesting polygons that DO have a rational viewing angle, when we restrict attention to algebraic vertices. But perhaps I should post this as a separate MO question, with more examples, as it seems to be separated from your arguments above. http://mathoverflow.net/questions/112714/seeing-the-vertices-of-a-polygon-with-rational-angles/112716#112716 Comment by Sinai Robins Sinai Robins 2012-11-18T08:23:10Z 2012-11-18T08:23:10Z @O'Rourke and @Gerry: Thank you, and I agree with you and Gerry that some explicit examples of quadrilaterals that don't allow any rational viewing points would be very nice. http://mathoverflow.net/questions/112575/combinatorial-meaning-of-the-functional-equation-for-logarithm Comment by Sinai Robins Sinai Robins 2012-11-16T14:22:24Z 2012-11-16T14:22:24Z Nice question - the Lagrange inversion formula (Chapter 5 of Stanley's book Enumerative Combinatorics, Vol. 2) might help here, giving the cofficients of the Log in terms of certain derivatives of the coefficients of its inverse, the exponential. http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Comment by Sinai Robins Sinai Robins 2012-11-15T16:40:01Z 2012-11-15T16:40:01Z @Francois: Thanks for your interesting thoughts here. I'll edit the problem statement to incorporate the idea that reducible lattices allow the possibility of many non-zero theta functions for each dimension $d$. http://mathoverflow.net/questions/112419/an-identity-which-involves-eulers-totient-function Comment by Sinai Robins Sinai Robins 2012-11-15T01:25:28Z 2012-11-15T01:25:28Z Have you tried playing around with a bivariate generating function such as $\sum_{n \geq 1} \frac{\phi(n) y^n}{1-x^n}$? http://mathoverflow.net/questions/111984/gram-matrix-modulo-4/112022#112022 Comment by Sinai Robins Sinai Robins 2012-11-14T16:23:11Z 2012-11-14T16:23:11Z Thanks a lot, Will. This kind of reference, namely the Burton Jones paper, is what I was looking for. http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Comment by Sinai Robins Sinai Robins 2012-11-14T16:00:56Z 2012-11-14T16:00:56Z @Francois: Sounds interesting-can you expand on this point? I'm not that familiar with the Grothendieck group. But it seems that we need to define a reducible theta function to be one that comes from a reducible lattice, versus the corresponding definition for an irreducible theta function. So for reducible $d$-dim'l theta functions $\theta_\Lambda$ that arise from lattices that are direct sums of lower dimensional lattices whose theta functions are in turn zero-free, we still have that $\theta_\Lambda$ is zero-free. It's now natural to ask when the irreducible theta functions are zero-free. http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Comment by Sinai Robins Sinai Robins 2012-11-14T12:00:35Z 2012-11-14T12:00:35Z Wait - some of the remarks above appear to be contradictory - what about the integer lattice $\mathbb Z^d$? We can decompose its theta function since it has a diagonal quadratic form in the exponent: $\theta_{\mathbb Z^d}(\tau) = \theta^d_{\mathbb Z}(\tau)$, and since $\theta_{\mathbb Z}$ is zero free in $\mathbb H$, so is $\theta^d_{\mathbb Z}$, independent of how large $d$ is. :( http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Comment by Sinai Robins Sinai Robins 2012-11-14T10:10:18Z 2012-11-14T10:10:18Z @Francois: Yes, thank you, I've tried this before too, and I guess your suggestion might work out to give an optimal bound on the dimension, hopefully, though it's not clear to me how to find the order of vanishing at the cusps from the data. The wt=$d/2$ for a d-dim'l lattice, and the level $N \leq vol(\Lambda)$,by a known theorem. So I guess this might give a partial answer for large enough volume of the lattice vol(Λ) and large enough dimension d, together with the assumption that the lattice Λ is an even integral lattice. What about odd dimensions, or lattices which are not even integral? http://mathoverflow.net/questions/112295/zero-free-theta-functions-in-the-upper-half-plane Comment by Sinai Robins Sinai Robins 2012-11-14T04:04:05Z 2012-11-14T04:04:05Z Well, I think it's true that it's a form for even integral lattices, which means the following. If we let the lattice be $\Lambda:= A(\mathbb Z^d)$, for a matrix $A$, then $A^t A$ should be an integer matrix, with all of its diagonal elements even integers. Then the associated theta function is a modular form on some $\Gamma_0(N)$. Do you know how large $d$ has to be to insure that it does indeed have zeros ? How about any general information about their location? http://mathoverflow.net/questions/111984/gram-matrix-modulo-4/112022#112022 Comment by Sinai Robins Sinai Robins 2012-11-11T03:47:06Z 2012-11-11T03:47:06Z Great! Let me look it up and see what they have. http://mathoverflow.net/questions/111141/counting-lattice-points-inside-an-n-dimensional-tetrahedron/111148#111148 Comment by Sinai Robins Sinai Robins 2012-11-10T19:32:29Z 2012-11-10T19:32:29Z I'd like to mention that I was always motivated by chapter 4 of Stanley's Enumerative Combinatorics book, for which the 2'nd edition is now available. It's a great resource.