October 2020 - mathfun@mailman.xmission.com

[math-fun] Zonohedral surfaces
by James Propp 22 Oct '20

22 Oct '20

George Hart’s Celebration of Mind talk (just concluded) makes me wonder: is there a non-self-intersecting torus surface made of parallelograms? Jim Propp

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Dan Asimov 22 Oct '20

22 Oct '20

Tom Karzes wrote: ----- The minimal solution is unique up to multiplication by a unit (1, i, -1, -i). ----- This depends on the point set. If say it's X = {(0,0), (5,0), (10,0), (15,0), (20,0)} Then there are 12 rotations about the origin that carry X into a set of integer points. (By angles N*90º and N*90º ± atan2(3,4), N = 0, 1, 2, 3.) —Dan

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Henry Baker 22 Oct '20

22 Oct '20

I can imagine that 3D is easier, since we only have to solve for *one* rotation. In 4D, there are 2 independent rotations. Perhaps the 4D answer is related to the fact that there are *two* gcd algorithms: a 'left' gcd and a 'right' gcd. So perhaps you have to find both gcd's for the final result? At 10:48 AM 10/22/2020, Fred Lunnon wrote: > Hurwitz quaternions work in 3-space --- see for example my write-up for when >this algorithm last appeared on math-fun, > https://arxiv.org/abs/1202.3198 > >I never needed to consider 4-space, since nobody has managed to discover a >Heronian pentatope! > >WFL > >On 10/22/20, Henry Baker <hbaker1(a)pipeline.com> wrote: >> Hurwitz quaternions might work for 4D. >> >> At 09:32 AM 10/22/2020, Tom Karzes wrote: >>>Good job! That's the solution I found when I first came up >>>with this problem: >>> >>> First translate so that one of the points is at the origin. >>> Then divide all the points by the GCD. >>> >>>That's all there is to it. It doesn't matter which point ends >>>up at the origin, as long as one of them does. The GCD will be >>>the same no matter which one you choose. You can think of the >>>translated set of points as displacement vectors (relative to >>>the point at the origin). >>> >>>Treating the points as Gaussian integers is a key insight. >>>So is translating one of them to the origin, and of course >>>dividing by their GCD. I went out of my way not to mention >>>Gaussian integers in the problem description, just to make >>>the solution a little less obvious. I also made sure my >>>example didn't have any of the points at the origin, again >>>to make it less obvious. >>> >>>The minimal solution is unique up to multiplication by a >>>unit (1, i, -1, -i). >>> >>>Once you have a minimal solution, multiplication by any >>>nonzero Gaussian integer will give you a new scaled/rotated >>>set of integer points with the same shape. If you don't want >>>to rotate, you can just multiply by a positive integer. And >>>of course, you can reflect by taking the complex conjugate. >>> >>>To compute the Gaussian GCD, the standard Euclidean Algorithm >>>works perfectly well. It just needs to operate on Gaussian >>>integers. >>> >>>I actually used this for some raster graphics I was doing, where >>>I had sets of points that I wanted to scale up or down without >>>introducing any distortion. This solved the problem nicely. >>> >>>Any idea how to scale this up to 3 (or more) dimensions? The >>>Gaussian integer approach works great for 2D, but obviously >>>not for higher dimensions. Is there a way to represent higher- >>>dimensional points as, say, orthogonal matrices, or polynomials, >>>or something along those lines, then obtain some sort of GCD and >>>somehow factor it out? I guess part of the difficulty is that >>>rotations don't commute in 3 or more dimensions. >>> >>>Still, the basic problem is perfectly well-defined for higher >>>dimensions, so some solution must be possible. >>> >>>Tom >>> >>>Allan Wechsler writes: >>> > Yes. Then take the GCD of all the remaining points (considered as >>> > Gaussian >>> > integers), and divide out by the GCD to obtain the solution. >>> > >>> > On Thu, Oct 22, 2020 at 1:19 AM Steve Witham <sw(a)tiac.net> wrote: >>> > > Not only does reflection not matter to the problem, >>> > > but translate the problem so that one original point >>> > > is (0, 0) and let the same point in the solution be (0, 0). >>> > > >>> > > Without loss of generality, >>> > > --Steve >>> > > >>> > > > From: Henry Baker <hbaker1(a)pipeline.com> >>> > > > Date: 10/21/20, 10:27 AM >>> > > > To: >>> > > > >>> > > > I'm guessing factorization of Gaussian integers is involved... >>> > > > >>> > > > At 12:51 AM 10/21/2020, Tom Karzes wrote: >>> > > >> Here's a puzzle some of you might like:

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Dan Asimov 22 Oct '20

22 Oct '20

You are of course absolutely right. Sorry about that. I glommed on to the 12 rotations of the set I defined and forgot that we were discussing only the minimal solutions. —Dan Tom Karzes wrote: ----- No, it is independent of the point set. As long as there are at least two distinct points, there are exactly four minimal solutions. Look at the subject of the sentence you quoted: The minimal solution is unique up to multiplication by a unit (1, i, -1, -i). ... ... -----

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Steve Witham 22 Oct '20

22 Oct '20

Not only does reflection not matter to the problem, but translate the problem so that one original point is (0, 0) and let the same point in the solution be (0, 0). Without loss of generality, --Steve > From: Henry Baker <hbaker1(a)pipeline.com> > Date: 10/21/20, 10:27 AM > To: > > I'm guessing factorization of Gaussian integers is involved... > > At 12:51 AM 10/21/2020, Tom Karzes wrote: >> Here's a puzzle some of you might like:

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Re: [math-fun] More than you wanted to know about homology and cohomology
by Dan Asimov 22 Oct '20

22 Oct '20

Brad Klee wrote: ----- I think it would help immensely to get into details and see worked examples. ----- Agreed. I didn't want to make that post too long. ----- For a complexified algebraic curve, what is the algorithm to construct a detailed tiling of the uniform time domain? (You can use triangularization if you want) ----- I'm not aware of any such algorithm. But I'm not sure if I understand exactly what is meant by "complexified algebraic curve". ( There is a very nice class of surfaces known as "complex projective curves", which are smooth surfaces defined by the locus of common zeroes of a finite family of n-1 equations P_j(Z_0, Z_1, ..., Z_n) = 0 on complex projective n-space CP^n, where each P_j is a homogenous polynomial with complex coefficients. For example, the spectacularly beautiful "Klein quartic" is defined as K = {(X : Y : Z) ∊ CP^2 | P(X,Y,Z = 0} where P(X,Y,Z) = X^3 Y + Y^3 Z + Z^3 X. K is topologically a compact orientable surface of genus 3. The group of orientation preserving isometries of K is the simple group of order 168, and it admits a tiling by 24 regular hyperbolic heptagons, 3 per vertex, compatible with these isometries. ) Or what is meant by "the uniform time domain". ----- By detailed I mean that singular points and algebraic cycles should be marked. To step up a dimension, just ask what happens when the curve/surface depends on a shape parameter. If you start with a genus g=0,1, or 2 surface and deform it continuously via the shape parameter, how can the resulting 3-manifold be described in terms of the Thurston Geometries? So yes, it’s more than I want to know, but also less than what I want to know. For me (and many others), more constructive methods = less confusing formalism. ----- How is a continuous deformation of a surface defined? If M_g is your initial surface, will the resulting 3-manifold always be topologically M_g x [0,1] ? My previous post was exclusively about topology, and not at all about geometry. —Dan

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Re: [math-fun] More than you wanted to know about homology and cohomology
by Dan Asimov 22 Oct '20

22 Oct '20

Allan Wechsler wrote: ----- I have been trying to wrap my head around this corner of mathematics for my entire adult life, so I will be re-reading this post carefully in hopes of getting some insight. Unless I missed something, you don't actually define the homology and cohomology groups, do you? You are more saying, "These things exist, but I don't want to get into the details," right? (I think I could reconstruct the definition of the j-th homology group of a simplicial complex, but I'm quite sure I never knew the corresponding definition of the cohomology groups.) ----- I didn't want to get into the details *yet* until later. First I wanted to give a general overview. ----- If I remember correctly, two simplicial complexes that are isomorphic as topological spaces will always have identical homology and cohomology, so for a space X that can be represented as a simplicial complex, it makes sense to talk about the homology and cohomology of the space. But the reverse is not true -- two non-isomorphic spaces might have identical homology or cohomology. So the theory sometimes lets us prove that two spaces are different, but it can only suggest when they might be the same. ----- Yes, topologically equivalent spaces have isomorphic homology, and isomorphic cohomology rings, but having isomorphic homology and/or cohomology doesn't necessarily imply topological equivalence. It's easy to come up with examples of that, because two topological spaces that are merely "homotopy equivalent" always have the same homology and cohomology. Spaces X and Y are called homotopy equivalent when there are maps f : X —> Y and g : Y —> X such that the compositions g o f : X —> X and f o g : Y —> Y are each homotopic to the respective identity maps 1_X and 1_Y. For nice spaces like simplicial complexes, this just amounts to a series of continuous deformations willy nilly: X —> X_1 <— X_2 —> X_3 <— X_4 ... Y that ultimately turn X into Y. (For instance, a disk with a smaller disk removed from its interior is homotopy equivalent to a circle with three hairs sticking out from it.) ----- When you speak of complex surfaces, you mean specifically 2-dimensional spaces, right? Because I recall that the taxonomy for higher-dimensional surfaces is more ... fraught. ----- I don't believe I used the term "complex surfaces". But yes, the higher-dimensional generalization of a surface is an n-manifold: Where a small piece of a surface look like a small piece of the plane, a small piece of an n-manifold looks like a small piece of R^n. So a surface is a 2-dimensional manifold, or 2-manifold for short. —Dan

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[math-fun] Z^2 scale/rotate/translate puzzle
by Tom Karzes 22 Oct '20

22 Oct '20

Here's a puzzle some of you might like: You have a finite set of points in Z^2, so they all have integer coordinates. Taken together, the set of points defines an object. The problem is to find a scale/rotation/translation (with no reflection) that maps those points into Z^2, minimizing the size of the object (i.e. minimizing the distances between pairs of points). Example: Suppose you have the following five points: (-3, 6) (-4, -7) (6, 3) (1, 8) (2, 1) These can be scaled/rotated/translated to: (0, 1) (-4, 0) (0, -2) (1, 0) (-1, -1) This scales the original object by 1/sqrt(10), leaving it in its smallest possible form while preserving its shape. Ignoring translation, any optimal solution may be rotated by multiples of 90 degrees to obtain three additional optimal solutions, for a total of four. This may be a well-known problem, but it was unknown to me. My first thought was to do some sort of exhaustive search, but I ended up finding a much better solution. Any thoughts? Tom

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[math-fun] More than you wanted to know about homology and cohomology
by Dan Asimov 21 Oct '20

21 Oct '20

An "n-simplex" for a nonnegative integer n is a generalized triangle: A 0-simplex is a point, a 1-simplex is a closed interval, a 2-simplex is a (filled) triangle, and in general an n-simplex is the convex hull of the n+1 standard basis vectors of (n+1)-dimensional Euclidean space R^(n+1). These n+1 standard basis vectors are the "vertices" of the n-simplex. A k-dimensional face F of an n-simplex S is the k-simplex defined as the convex hull of any k+1 vertices of S. Using any finite collection of n-simplices for various n, most reasonably nice topological spaces can be constructed. A finite "simplicial complex" is just the finite union any collection of n-simplices for various n, under the condition that any two of them intersect in a common k-dimensional face of both of them. Suppose X is a topological space constructed as a finite simplicial complex. Basic invariants of X are its homology groups, one group H_j(X) for each nonnegative integer j up through the dimension of the highest dimensional simplex of X. Each of these groups will turn out to be a finitely generated abelian group. That is, H_j(X) = Z^n + Z_n_1 + ... + Z_n_r where the nonnegative integers n, n_1, ..., n_r (and r) depend on X and j (and we can assume n_1 | n_2 | ... n_r). These groups H_j(X) document the number and type of "holes" that the topological space X has in dimension h. The jth *cohomology* groups H^j(X) have an added feature: if x ∊ H^j(X) and y ∊ H^k(X), these cohomology classes *multiply together* to give a (j+k)-dimensional cohomology class xy ∊ H^(j+k)(X) and this multiplication distributes over the addition in the individual groups. So the cohomology groups all together form a graded ring. Any two very different simplicial complexes may happen to be topologically equivalent, and when they are, their cohomology rings will be isomorphic as graded rings. Particularly interesting topological spaces are the compact surfaces: The orientable ones are the sphere, the torus, and the surface of genus g for any g ≥ 2 (the surface of a button with g holes in it). The non-orientable ones are the projective plane, the Klein bottle, and the "non-orientable surface of genus g", which is the connected sum of g projective planes. Each surface has its own distinct cohomology ring. Any surface could also have a finite number h of holes punched in it, so it ends up with a boundary consisting of h simple closed curves. These surfaces with boundary will also each have their own cohomology rings. It would be interesting to have a table somewhere of all the cohomology rings of all such surfaces. ... And then one can take cartesian products of various of these surfaces and compute their cohomology rings as well ... —Dan

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Dan Asimov 21 Oct '20

21 Oct '20

Tom Karzes wrote: ----- I really just want a single shrunken version of the object. Adding reflection is trivial (just negate x or y), so there's no need to throw reflected solutions into the mix. Once a reference solution is found, the solution set can be quadrupled by rotating by multiples of 90 degrees, and doubled by adding reflection if desired, so there's no reason to use a reflected solution as the reference solution. ----- The solution set can indeed be multiplied by four via 90º rotations ... or multiplied by infinity via adding an arbitrary integer vector. By the way, this is a very interesting problem (which I haven't tried to solve yet)! Tom's question also suggests the corresponding problem for the triangular lattice Z[w] = {P(w) | P(x) is an integer polynomial}. where w = exp(2πi/3). —Dan

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