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

25 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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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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