May 2020 - mathfun@mailman.xmission.com

Re: [math-fun] The other kind of codimension
by Dan Asimov 24 May '20

24 May '20

Why not tell us which vector spaces the class number is a ratio of. —Dan ----- Here’s an example of a context in which such a ratio is relevant (unless my understanding of Hilbert class fields is wrong, in which case someone please set me straight): The class number h of an algebraic number field K can be understood as a ratio of the dimensions of two vector spaces, though normally we would express this in terms of fields rather than vector spaces, saying that the Hilbert class field of K is a degree-h extension of K. Could we call K a “codegree-h” “intension” of the Hilbert class field of K? Jim Propp On Sun, May 24, 2020 at 3:33 PM James Propp <jamespropp(a)gmail.com> wrote: > If W is a subspace of V, is there a name for dim V / dim W? And more > generally for the ratio of the dimensions of two things, one of which sits > inside the other? > > (This question is inspired by my misrecollection of the nature of knotting > in higher dimensions, which is the same as Cris Moore‘s: we both recalled > that 2 was involved, but we thought it was relevant as an approximate > ratio, rather than as a difference!) > > I could defend the term “codegree”, but I’m hoping someone else already > coined an obscure but uncontroversial word for dimension-ratios. -----

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[math-fun] The other kind of codimension
by James Propp 24 May '20

24 May '20

If W is a subspace of V, is there a name for dim V / dim W? And more generally for the ratio of the dimensions of two things, one of which sits inside the other? (This question is inspired by my misrecollection of the nature of knotting in higher dimensions, which is the same as Cris Moore‘s: we both recalled that 2 was involved, but we thought it was relevant as an approximate ratio, rather than as a difference!) I could defend the term “codegree”, but I’m hoping someone else already coined an obscure but uncontroversial word for dimension-ratios. Jim Propp

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Re: [math-fun] Clarification on Elliptic Knot Periods
by Brad Klee 24 May '20

24 May '20

Hi Dan, The complex period of the knot is determined by its winding numbers multiplied by the real and complex periods of the Riemann surface it is drawn upon, then added together. The Riemann surface I used is a complex extension of an elliptic curve in Harold Edwards’s normal form, see also: https://www.ams.org/journals/bull/2007-44-03/S0273-0979-07-01153-6/ It is a torus and can be drawn parametrically in four dimensions using elliptic functions such as defined by Edwards. To do so, you need to calculate real and complex periods. When z is chosen real between 0 and 1, the periods in Edwards’s model are TR=4*K(z) and TC=i*2*K(1-z). (The factor of 2 comes from a difference of harmonic frequencies around hyperbolic points.) The elliptic functions are defined over the complex plane but they are doubly periodic, with periods TR and TC. Torus knots (n,m) can be drawn by taking non-crossing paths through the complex plane that close modulo the period lattice. The trajectory through the complex plane determines a trajectory through C^2, which we can then map somehow into R^3. (This part is not easy, but doesn’t affect the period integral.) To calculate the period, we simply count n real divisions horizontal and m complex divisions vertical, until we land on the same location in the period rectangle. The period is then T=n*TR+m*TC, and that is the amount of complex time it takes to get once around the curve. In this first example I choose z=1/2 so that we get a period square, and n=3, m=4 to accentuate the dihedral symmetry. To reiterate, the period is: T=(3*4+4*2*i)*K(1/2). On the same torus, other periods would be something like T=(n*4+m*2*i)*K(1/2). On other similar tori, we would have to write T=(n*4*K(z)+m*2*i*K(1-z)). Another example drawing, z=1/4: https://0x0.st/ip1O.png (looks slightly better for low z, maybe) For the trefoil knot it would be more natural to use a cubic elliptic curve with triangular dihedral symmetry. There is a choice of coordinates where TR=K3(z) and TC=(sqrt(3)/3)*K3(1-z), with K3(z)=2*Pi*2F1(1/3,2/3,1,z), so we can calculate the period of a (n,m) knot, on any surface indexed by z: T = 2*Pi*n*2F1(1/3,2/3,1,z) + 2*Pi*(sqrt(3)/3)*2F1(1/3,2/3,1,1-z). It would be awesome to see a symmetry-preserving drawing of the (2,3) knot on an elliptic curve, but not so easy, due to the difficulty of the following problem: Given a form for elliptic curves: z = 2H= q^2 + p^2 - sqrt(4/27)*(3*q*p^2 - q^3) find the parameterizing elliptic function p(t) and q(t). (Also Is z=2H a normal form for elliptic curves?) --Brad On May 24, 2020, at 12:43 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: Brad, I don't yet understand your definition of the period of a knot. Could you please tell me (or math-fun) what that definition is? (I did peek at your nicely illustrated paper on github, but did not readily extract such a definition.) Thanks, Dan

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Re: [math-fun] Fwd: [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 24 May '20

24 May '20

I strongly disagree with this statement to the extent it refers to pure math, and I think I've been around research mathematics long enough to have some idea what I'm talking about. As we all know, math is hard for many people and, to make matters worse, it's not always taught well, particularly in grade school through high school. And it's a huge subject so that many major areas are not even well understood by established researchers in other parts of math. And finally, the expected audience of math research papers such as a Ph.D. dissertation is other researchers. It's hard enough doing the research in the first place no less trying to address a wide audience who will probably never be interested in one's research. I don't know about other subjects, but I don't think this is true in pure math. Or applied math, for that matter. —Dan ----- This is pretty typical of the academic system, which actively encourages researchers to leave most people behind. -----

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Re: [math-fun] Fwd: [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 23 May '20

23 May '20

I thought what I wrote could hardly be clearer. I'm agreeing that trying to read a math research paper is difficult or impossible without the appropriate background. I'm disagreeing that anyone is intentionally encouraged to do that; rather the inscrutability of math research is built into the definition of research mathematics, which by nature abstruse. —Dan ----- I’m confused, because you seem to be giving supporting evidence. Is your opinion that people should be left behind? Or that “research mathematicians” have no other choice? Or? —Brad > On May 23, 2020, at 9:02 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > > I strongly disagree with this statement to the extent it refers to pure math, and I think I've been around research mathematics long enough to have some idea what I'm talking about. > > As we all know, math is hard for many people and, to make matters worse, it's not always taught well, particularly in grade school through high school. And it's a huge subject so that many major areas are not even well understood by established researchers in other parts of math. And finally, the expected audience of math research papers such as a Ph.D. dissertation is other researchers. It's hard enough doing the research in the first place no less trying to address a wide audience who will probably never be interested in one's research. > > I don't know about other subjects, but I don't think this is true in pure math. Or applied math, for that matter. > > —Dan > > ----- > This is pretty typical of the academic system, which actively encourages > researchers to leave most people behind. -----

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 23 May '20

23 May '20

Knotting happens in codimension 2 regardless of the dimension. So a 2-sphere can be knotted in 4-dimensional space. Here's an example: Think of R^4 as R^2 x R^2. Notice that R^4 - (R^2 x {0}) = R^2 x (R^2 - {0}) and that factor R^2 - {0} is the result of rotating a ray (0, oo) around a circle. Now embed a closed arc in R^4 with its endpoints meeting R^2 x {0} perpendicularly ... and with an overhand knot in its middle. Finally, spin that arc around the R^2 x {0} (using the circular symmetry) to sweep out a surface. This surface is homeomorphic to an ordinary 2-sphere S^2. But it is knotted! Similar methods create even higher-dimensional, codimension-2 knots. —Dan Cris Moore wrote: ----- This is great. But I have a question. I had the impression that d-dimensional surfaces can be “knotted” (homeomorphic but not homotopic to a sphere, if I’m remembering the correct uses of those lovely greek-derived words) in 2d+1 dimensions: 1-dimensional curves in 3 dimensions, 2-dimensional surfaces in 5, and so on. But this artlcle says that 2-dimensional surfaces can be knotted in 4 dimensions. I can see how two spheres (2-dimensional surfaces) can be linked in 5 dimensions: put them in 3 dimensions so that the overlap and intersect in a circle, and then “lift” each point on that circle to a pair of antipodal points on a circle in another 2 dimensions. Then the two spheres cannot be drawn apart without having them intersect in all 5 dimensions at some point. But I don’t see a similar construction in 4 dimensions. Can anyone give a simple example of a knot or link of 2-surfaces in 4 dimensions? -----

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[math-fun] Fwd: [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Allan Wechsler 23 May '20

23 May '20

Some of us know Dick Koolish -- he posted the following link on the Bolt, Beranek & Newman alumni list. I wonder if Conway heard that his knot had been cracked -- the article came out in February. Also, it was refreshing to read an article that really captured the basic skeleton of an abstruse proof. I like Piccirilli's style. ---------- Forwarded message --------- From: Richard Koolish <koolish(a)dickkoolish.com> Date: Fri, May 22, 2020 at 6:39 PM Subject: [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine To: <xbbn(a)googlegroups.com> https://www.quantamagazine.org/graduate-student-solves-decades-old-conway-k… -- You received this message because you are subscribed to the Google Groups "xBBN" group. To unsubscribe from this group and stop receiving emails from it, send an email to xbbn+unsubscribe(a)googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/xbbn/6A36C8F2-51CA-4F7D-972B-3CEFEC758F3C… .

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Re: [math-fun] Can an equilateral toroidal polyhedron
by Dan Asimov 23 May '20

23 May '20

The torus T_7 below is the dual to what you get by arranging 6 congruent regular hexagons in the plane around a seventh and then identifying its boundary to itself (by identifying 3 pairs of polygonal arcs via translations; there are only two essentially equivalent ways to do this). To see a way to embed T_7 into the 2-skeleton of the 6-simplex S_6 in R^6:* 1) Label the vertices V of S_7 as V = {p_0, ..., p_6} in cyclic order. 2) For each i = 0, ..., 6 consider the set V_i = V - {p_i}. For each pair of consecutive vertices p_j, p_k in V_j in the inherited cyclic order, draw the triangle whose vertices are p_i, p_j, p_k. 3) This surrounds each vertex with 6 triangles any consecutive pair of which shares an edge. 4) The number of triangles thus drawn is 7*6/3 = 14, because each triangle was drawn thrice, one for each vertex. 5) Thus this is T_7. —Dan _____ * You can think of the 7 vertices of the 6-simplex S_6 as (the endpoints of) the standard basis vectors in R^7, all translated by the vector (-1/7,...,-1/7) so they end up in the subspace R^6 of R^7 defined by sum of coordinates = 0. I wrote: ----- ... Call this torus T_7. Its vertices and edges form the complete graph K_7, and its triangles consist of exactly 2/5 of the possible 35 triangles made of 3 vertices. As a combinatorial torus, it is dual to the 7-hexagon (hexagonal) torus, whose hexagons form a maximally symmetrical 7-coloring of the torus. This shows that T_7 with its intrinsic metric (distances measured along its 14 triangles, not across 6-space) is also a hexagonal torus: One obtained by identifying opposite edges of a regular hexagon. Which suggests the question: Can the 7-hexagon hexagonal torus itself be embedded in R^6 so that each hexagon is a bona fide planar regular hexagon? -----

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[math-fun] Covid Modelers Predict Near Future
by Brad Klee 23 May '20

23 May '20

https://projects.fivethirtyeight.com/covid-forecasts/ (This is not an endorsement.) At what point does an extrapolation become a prediction? Given that we have seen more than a month of plateau behavior in USA infection data, “prediction” that the plateau will continue is none too surprising. One way to distinguish between models above is whether or not they have non-negligible probabilities for rates to increase due to reopening. Can anyone predict if the plateau will end? If yes, in how many weeks or months? If no... —Brad

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Re: [math-fun] Complex modulo function?
by Keith F. Lynch 23 May '20

23 May '20

Veit Elser <ve10(a)cornell.edu> wrote: > Keith F. Lynch <kfl(a)KeithLynch.net> wrote: >> The *second* thing everyone asks when they learn about i is which >> number is i and which is -i. Imagine the horror if they go through >> their whole mathematical career getting those two swapped. > I?m curious, what?s the *first* thing? As I had posted just two hours earlier, in the same thread, "Everyone's first thought when they learn about i is to ask what it *is*. To which I can answer that at least I know its last digit, which is 3. :-)" Students get useful answers when they ask what pi and e are, so why not ask what i is? Instead of being given digits, they're told how to use it. And after they use it for a while, they realize they now know what it is. (Except that half of them are actually using -i. :-) ) > It?s actually very cool that the Galois automorphism i <-> -i is > co-opted by physics in the time-reversal transformation. The real > numbers turned out to be incomplete in a very real sense! Think quaternions, not complex numbers. Three interchangeable and symmetrical dimensions, plus one very different dimension with a direction built in.

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