May 2020 - mathfun@mailman.xmission.com

[math-fun] A Conway-ish approach to the dyadic Eisenstein rationals?
by James Propp 27 May '20

27 May '20

I'm trying to see what sorts of algebraic structures can be built up in a Conway-style "genetic" way. I think I may have an approach to constructing the surcomplex numbers that's different from Conway's, but I'm stuck; maybe one of you can help me see what the next step is (or help me see that the enterprise is doomed). Some background may be helpful here. We say that one dyadic rational number is simpler than another if the first has smaller denominator, or, in the case where they have the same denominator, the first has smaller absolute value. Thus 1/4 is simpler than 3/4 but 1/2 is simpler than both of them. This gives a partial ordering on dyadic rationals. The simplest dyadic rational is 0; the next simplest are +1 and -1; the next simplest are +2, -2, +1/2, and -1/2; etc. Conway approached simplicity via birthday ordinals but this is not necessary. Every open interval I in the dyadic rationals contains a unique simplest number I*, and we have the addition rule I*+J* = ((I*+J) \cap (I+J*))* where the sum of a number and an interval is defined in the obvious way (shift the interval by the number, or if you prefer, take the Minkowski sum of the interval with the singleton set associated with the number). This formula is just Conway's definition of addition in disguise. To rip away the disguise, write I* as {x_L | x_R} where x_L and x_R are the left- and right-endpoints of I respectively. If we write I = (x_L, x_R) and I* = x and J = (y_L, y_R) and J* = y, then Conway's definition of addition tells us that x + y is the simplest dyadic rational that is bigger than both x + y_L and x_L + y and smaller than both x + y_R and x_R + y; that is, it is the simplest dyadic rational in the intersection of the interval (x + y_L, x + y_R) = x + (y_L, y_R) = I* + J and the interval (x_L + y, x_R + y) = (x_L, x_R) + y = I + J*. (Has anyone seen this way of reformulating surreal numbers before?) Now on to my real question. Define the dyadic Eisenstein rationals as the complex numbers of the form a + bw where a and b are dyadic rationals and w is a primitive cube root of 1; define its denominator as the maximum of the denominators of a and b. Every dyadic Eisenstein rational z can be written as a + bw, a + bw^2, or aw + bw^2 with a,b non-negative; accordingly define the "triangle norm" of z as a+b (so that the level sets of the triangle norm are triangles centered at 0 in the complex plane; it's not a true norm since the level sets aren't centrally symmetric). We say that one dyadic Eisenstein rational is simpler than another if the first has smaller denominator, or, in the case where they have the same denominator, the first has smaller triangle norm. The role played by intervals in the dyadic rationals will be played by open equilateral triangles that point to the right (like the "play" icons on many devices); hereafter when I write "triangle" I mean that kind of triangle. I can prove that every triangle T in the dyadic Eisenstein rationals contains a unique simplest number T*. What I haven't been able to prove, but believe to be true, is that for all triangles T,U we have (*) T*+U* = ((T*+U) \cap (T+U*))* If this is true, then there's probably a way to "genetically" build up the Eisenstein rationals (perhaps extending beyond the complex numbers into the surcomplex realm) starting from the empty set, much as Conway built up the surreals. (This leaves out the problem of defining multiplication genetically, but hey, you gotta start somewhere.) I've proved the conjecture for triangles that are symmetric about the real axis; in this case (*) follows from facts about surreal numbers. (I should mention that there's an alternative, simpler way to build up the surcomplex numbers genetically, using rectangles instead of triangles. But then all the theorems one needs to prove are trivialities, as all we're doing is looking at ordered pairs of surreal numbers and doing the obvious things with them. This is pretty close to Conway's construction of the surcomplex numbers by adjoining a formal square root of -1. But what interests me is the idea of genetic construction itself. The only things I've seen genetic constructions for are the surreals and the nimbers.) Jim Propp

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[math-fun] Amazing(?) Binomial Identity For A304126.
by Brad Klee 26 May '20

26 May '20

a[n_] := Sum[Binomial[6 n, n + k] Binomial[4 n, k] (-1)^k, {k, 0, 4 n}] zero[n_] := Binomial[3 n, n] Binomial[6 n, 3 n] - Binomial[5 n, n] a[n] zero /@ Range[0, 10] See also: https://oeis.org/A304126 ( nice! ) In terms of OEIS numbering, A304126 neither divides A005809 nor A066802, yet it does divide their product A113424. This can be proven using Zeilberger's algorithm, or via ExpToODE according to the integral definition, a_n = 1/(2*pi) Int_{0..2*Pi} (2^10*cos(x)^6*sin(x)^4)^n *dx . More on this soon... --Brad

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[math-fun] I just "Julianized" the base 2-i 'four-around-one" spacefill.
by Bill Gosper 26 May '20

26 May '20

It's pretty boring <http://gosper.org/2-ifills.png>. No quadruple points. All triple point inverse image denominators seem to be 5^(n+1) or twice that. —rwg You can find Julian's magic piecewiserecursivefractal in OEIS under Dragon triple points. If you are sufficiently hardcore to want "Eisenstein rationals", ask me for Julian's latest version with extra optional simplification. If you're feeling adventuresome, you could try to capture the definitions of bas2mi and im2sab with FindImageText and TextRecognize on the linked big png.

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Re: [math-fun] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 26 May '20

26 May '20

I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}: (x+iy)^2 + (z+iw)^3 = 0 and x^2 + y^2 + z^2 + w^2 = 1 but which is then stereographically projected into R^3 = {(X,Y,Z)} via X = x/(1-w), Y = y/(1-w), Z = z/(1-w). If my algebra is right, the final (two) equations are 4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0 where W is short for (X^2 + Y^2 + Z^2 - 1). I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher". —Dan

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 25 May '20

25 May '20

What do these equations represent? What are X, Y, and Z ? —Dan ----- For the (3,4) knot I calculated that, 0 = -81 X^2 + 117 X^4 - 40 X^6 + 4 X^8 - 81 Y^2 + 234 X^2 Y^2 - 152 X^4 Y^2 + 16 X^6 Y^2 + 117 Y^4 - 152 X^2 Y^4 + 24 X^4 Y^4 - 40 Y^6 + 16 X^2 Y^6 + 4 Y^8 0 = 9 X^2 - 9 X^4 + 9 Y^2 + 14 X^2 Y^2 - 9 Y^4 - 144 Z^2 + 64 X^2 Z^2 + 64 Y^2 Z^2 {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], Sin[4 t]} . The total number of (X,Y) monomials increases quadratically and total (X,Y,Z) monomials increase cubically, as functions of degree. These equations can be solved to give an upper degree bound. For integral calculations, this is a really nice way to represent the knot. We already have a [0,2*pi] parameterization built in, and trigonometric polynomials are easy to integrate. Nice Idea! -----

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[math-fun] trig and knots
by Cris Moore 25 May '20

25 May '20

BTW this trig identity approach also works nicely for “Lissajous figure” knots like 7_4 : https://en.wikipedia.org/wiki/Knot_theory <https://en.wikipedia.org/wiki/Knot_theory> and https://en.wikipedia.org/wiki/Lissajous_knot <https://en.wikipedia.org/wiki/Lissajous_knot> . We just need to vary z[t] to get the overpasses and underpasses we want. - Cris

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by rcs＠xmission.com 25 May '20

25 May '20

[this is from Dan Asimov -- Rich] From: Dan Asimov <dasimov(a)earthlink.net> Yes. The 3 real equations in 4-space Re((x+iy)^p + (z+iw)^q) = 0 Im((x+iy)^p + (z+iw)^q) = 0 x^2 + y^2 + z^2 + w^2 = epsilon^2 also give the (p,q) torus knot. It would be fun to figure out the equations for the stereographic projection of this to R^3. ?Dan Hi Dan, I think it's obvious, but I will explain a few finer points. Call the polynomial equations P1(X,Y) and P2(X,Y,Z). The (3,4) torus knot can be represented as an algebraic variety: K(3,4)={(X,Y,Z) in R^3: P1(X,Y)=0 & P2(X,Y,Z)=0}, and K(3,4)={ (Sin[t] + 2 Sin[3t], Cos[t] - 2 Cos[3t], Sin[4t] ) in R^3 : t in [0,2*pi] }. The proof is to observe that the map to trigonometric polynomial achieves identity after reducing terms. You must also show that the two polynomials are independent, but that follows from deg(P1) > deg(P2), as P2 includes more variables, so cannot be of the form f(Z)*P1. (Also, P1 does not factor to a lower-degree invariant.) The polynomial P1 has a special interpretation. Curve C={(X,Y): P1(X,Y)=0} is the projection of K onto the pane orthogonal to the Z axis. We can then solve P1(X,Y,Z)=0 for Z(X,Y), the height function of knot K, which has two branches. I am fairly sure this is what Cris Moore originally had in mind; though, I don't know the details of his encoding of the trefoil. It depends slightly on the choice of parametric form (but the degrees will be the same, because they must come from Chebyshev polynomials). Per intuition I could guess the existence of an algorithm from any knot parameterized by trig. polynomials to an algebraic variety, but have not worked out details. --Brad On Mon, May 25, 2020 at 10:52 AM Dan Asimov <dasimov(a)earthlink.net> wrote: What do these equations represent? What are X, Y, and Z ? ?Dan ----- For the (3,4) knot I calculated that, 0 = -81 X^2 + 117 X^4 - 40 X^6 + 4 X^8 - 81 Y^2 + 234 X^2 Y^2 - 152 X^4 Y^2 + 16 X^6 Y^2 + 117 Y^4 - 152 X^2 Y^4 + 24 X^4 Y^4 - 40 Y^6 + 16 X^2 Y^6 + 4 Y^8 0 = 9 X^2 - 9 X^4 + 9 Y^2 + 14 X^2 Y^2 - 9 Y^4 - 144 Z^2 + 64 X^2 Z^2 + 64 Y^2 Z^2 {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], Sin[4 t]} . The total number of (X,Y) monomials increases quadratically and total (X,Y,Z) monomials increase cubically, as functions of degree. These equations can be solved to give an upper degree bound. For integral calculations, this is a really nice way to represent the knot. We already have a [0,2*pi] parameterization built in, and trigonometric polynomials are easy to integrate. Nice Idea! -----

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 25 May '20

25 May '20

I believe Veit's idea will always work for "classical" knots" — knotted topological circles S^1 in 3-space (or more commonly in knot theory, in the 3-sphere S^3). Classical knots are assumed "tame", i.e., equivalent to a knot that is a finite closed polygon. That is because a continuous function can be approximated by a smooth one, and a smooth one by a real analytic one, and a real analytic one by a polynomial. And for a tame knot, any small enough approximation will produce an equivalent knot. So Yes, all classical knots (up to knot equivalence) can be described as the intersection of the zero-loci of two real polynomials P(x,y,z) and Q(x,y,z) defined on R^3. Therefore it makes sense to ask what is the minimum (say) sum of degrees of all such polynomial pairs that define an equivalent knot. There's an invariant called the "genus" of a knot K in R^3 defined as follows: A "Seifert surface" of a knot is a compact orientable surface in R^3 whose boundary is the knot. The genus of a compact orientable surface with boundary = C_1 u ... u C_n (disjoint circles) is the genus of the surface *without* boundary obtained by glueing the boundary of a disk D_j to the circle C_j for each j: the result is an n-holed torus for some n, and the genus is defined as n. A knot has many Seifert surfaces, but among them there is a minimum genus. That's defined as the genus of the knot. I would guess there's a simple relationship between the genus of a knot and the polynomial invariant suggested by Veit. —Dan Adam Goucher wrote: ----- For the trefoil, one way to construct the knot is to take the 2-variable *complex* parameterisation: a^2 = b^3, |a|^2 + |b|^2 = 1 and then 'compile' that into a 4-variable real parameterisation (where the second equation is just the equation for the 3-sphere), and then stereographically project that into a 3-variable real parameterisation. More generally, this idea will work for every torus knot. Veit Elser wrote: ----- Here’s a graduate-student-level question about knots — the deepest I’m capable of: Is it possible to define a “real-algebraic degree” for knots, and if so, has it been tabulated? My idea is to first construct knots as non-singular algebraic varieties, i.e. as the solution set of a pair of polynomial equations in x,y,z, and I’m using non-singular to mean that locally, near every point of the solution set, the linear approximation of the variety is two planes intersecting as a line. For example, x^2+y^2=1 z=0 defines the unknot. The variety defined by the pair of equations will in general have several disconnected components. We only care if *one* of these components is our target knot, and if so, we take some suitable “minimal” combination of the degrees of the two polynomials as the “real-algebraic degree” of our knot, e.g. min(d1+d2), min(max(d1,d2)). By the second definition, the unknot would have degree 2. Two specific questions: 1) Can all knots be constructed in this way? 2) What’s the degree of the trefoil? ----- -----

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 25 May '20

25 May '20

So-called "algebraic" knots are the ones obtainable by intersecting a small 3-sphere S^3(eps) about 0 in complex 2-space C^2 with the locus of a complex polynomial equation P(z,w) = 0 having an isolated singularity (where the locus is not a smooth surface) at 0. The (p,q) torus knot (assuming gcd(p,q) = 1) is obtained from the polynomial P(z,w) = z^p + w^q. that way. (I even made a computer graphic picture of that (2,3) trefoil, using the same polynomial Adam cited, stereographically projected into 3-space, a few months ago.) It's known that algebraic knots form a proper subset of all "iterated torus knots", which are defined via: Start with a torus knot and imagine another torus knot, possibly different, drawn on the boundary of a thin solid-torus neighborhood of that; repeat finitely many times. There are some knots that aren't algebraic. For example, a (p,q) torus knot with p < 0 and q > 1. —Dan

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Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
by Dan Asimov 25 May '20

25 May '20

Yup. A p-sphere and a q-sphere can link when they're embedded disjoint from each other in (p+q+1)-dimensional space. (Even when p = q = 0.) —Dan Cris Moore wrote: ----- Thanks. I wonder where I got this 2d+1 thing… maybe I’m misremembering some other embedding theorem. I do like the 5-dimensional construction of two linked 2-spheres, though, assuming I got that right. -----

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