June 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Frightening new batflu symptom
by Gabe K 12 Jun '20

12 Jun '20

For 15 to be 1.5 percent, the number of cases would have to be 1000. However, it is clearly not. On Thu, Jun 11, 2020 at 6:10 PM Michael Wang <hangenrestaurant(a)gmail.com> wrote: > Wow. > > On Thu, Jun 11, 2020 at 4:05 AM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Health workers coming into contact with infected victims are losing the >> ability to do math: >> >> Santa Clara County >> Confirmed cases: 3,032 — up by 15 (1.5%) since Tuesday >> Hospitalized: 61 — down by 8 since Tuesday, with 22 in ICU beds >> Deaths: 146 — up by 0 since Tuesday >> Actually, this isn't funny. Our former worry, West Nile virus, can permanently >> damage short term >> memory. I fear Santa Clara's Wednesday count of 15 new cases is a >> sampling artifact. Discouragingly, >> they have been running three times that, despite the sunny weather. But >> the hospitalizations >> are declining, in agreement with widespread evidence that the lethality >> of the virus is weakening. >> Let's hope not only thanks to sunshine. —Bill >> > -- Sincerely, Gabe

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[math-fun] Another horribly confusing Quanta article
by Brad Klee 11 Jun '20

11 Jun '20

https://www.quantamagazine.org/the-useless-perspective-that-transformed-mat… "Representation theory is a way of taking complicated objects and “representing” them with simpler objects." In no way does the article support this thesis; in fact, it supports the opposite and more reasonable: "Representation theory is a way of taking [simple] objects and “representing” them with [more complex] objects." The article does give a graphical definition of the triangular dihedral group--almost as simple as algebraic letters with some prescribed abstract meaning. It does not list out six matrices such as: MatrixForm /@ Join[RotationMatrix[2 Pi/3 #] & /@ Range[3], Dot[IdentityMatrix[2] {1, -1}, RotationMatrix[2 Pi/3 #] ] & /@ Range[3]] Out[]={{{-(1/2), -(Sqrt[3]/2)}, {Sqrt[3]/2, -(1/2)}}, {{-(1/2), Sqrt[3]/ 2}, {-(Sqrt[3]/2), -(1/2)}}, {{1, 0}, {0, 1}}, {{-(1/2), -(Sqrt[3]/2)}, {-(Sqrt[3]/2), 1/2}}, {{-(1/2), Sqrt[ 3]/2}, {Sqrt[3]/2, 1/2}}, {{1, 0}, {0, -1}}} Nor does it explain that matrix representations are coordinate dependent, nor that they have inverses and close under matrix multiplication, nor does it explain how the idempotent projectors and their characters would be derived from a matrix representation. Even worse, the article does not explain that Representation theory came into acceptance through its applications in physics. Why even mention Andrew Wiles proof? If you are trying to learn representation theory, the Quanta article looks to be useless, especially because it only references other Quanta articles (probably with similar problems). For a better course on why Representation Theory isn't a useless perspective, try reading William Harter's book: https://modphys.hosted.uark.edu/markup/PSDS_Info.html Which explains in detail how to calculate character tables, and how to use them for something productive. --Brad PS. Don't forget the beetle's have their own spectrum and a very special technique for controlling the Stokes vector: https://www.inaturalist.org/observations/49263113 https://modphys.hosted.uark.edu/pdfs/PSDS_Pdfs/PSDS_Ch.7_(4.21.10).pdf ( See also page 85 / 636 )

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[math-fun] Never Go Down - with link
by Éric Angelini 11 Jun '20

11 Jun '20

(as usual, more on my personal web page with pictures and colors) http:// cinquantesignes.blogspot.com/2020/06/never-go-down. html (the three parts of the link must be concatenated, else my message might be interpreted as spam) Best, É.

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[math-fun] Frightening new batflu symptom
by Bill Gosper 11 Jun '20

11 Jun '20

Health workers coming into contact with infected victims are losing the ability to do math: Santa Clara County Confirmed cases: 3,032 — up by 15 (1.5%) since Tuesday Hospitalized: 61 — down by 8 since Tuesday, with 22 in ICU beds Deaths: 146 — up by 0 since Tuesday Actually, this isn't funny. Our former worry, West Nile virus, can permanently damage short term memory. I fear Santa Clara's Wednesday count of 15 new cases is a sampling artifact. Discouragingly, they have been running three times that, despite the sunny weather. But the hospitalizations are declining, in agreement with widespread evidence that the lethality of the virus is weakening. Let's hope not only thanks to sunshine. —Bill

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[math-fun] Never Go Down
by Éric Angelini 11 Jun '20

11 Jun '20

Hello Math-Fun, the "Never Go Down" constraint takes you for a walk on a checkerboard, stepping over integers that have one digit per square. The rule is simple: start on the smallest digit and chose the shorter path to visit all of them, never going down. Example: +---+---+---+---+---+ | 1 | 0 | 2 | 1 | 0 | +---+---+---+---+---+ The shorter path will start on the yellow 0, then lead you to the other 0 (3 steps), then to the blue 1 (1 step), to the next 1 (3 steps) and to the final 2 (1 step). The total length of the path is 8. (Please note that the longest legal path has length 11.) We want now to complete this table: Smallest number Biggest number minimal path minimal path +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 1 | | | | 1 | 0 | | | | | 9 | 9 | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 2 | | | 1 | 0 | 0 | | | | 9 | 9 | 9 | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 3 | | | 1 | 0 | 1 | | | | 9 | 8 | 9 | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 4 | | 1 | 0 | 1 | 1 | | | 9 | 9 | 8 | 9 | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 5 | | 1 | 0 | 0 | 1 | | | 9 | 8 | 8 | 9 | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 6 | | 1 | 3 | 2 | 0 | | | | | | | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 7 | | 1 | 3 | 0 | 2 | | | | | | | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 8 | | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+---+ Path lenght 9 | | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+---+ ... | | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+---+ ... | | | | | | | | | | | | We could ask for two more columns: "Smallest number/maximal path" and "Biggest term/maximal path". But enough for today! à+ É. Catapulté de mon aPhone

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Re: [math-fun] [math>-fun] Negative polyhedra
by Dan Asimov 10 Jun '20

10 Jun '20

Since the Gauss-Bonnet theorem states that for a compact surface M, Integral K dA = 2π X(M) (K = Gaussian curvature, X(M) = Euler characteristic of M), we could in the case of constant negative curvature K = -1 (say), divide both sides by K. Then using the face that X(M) = V - E + F, we get area(M) = 2π (-V + E - F). So for a hyperbolic surface, that is, for a surface whose Euler characteristic is negative (the connected sum of ≥ 2 tori or of ≥ 3 projective planes), we could interpret the V, E, and F as negative (V' = -V, E' = -E, F' = -F) to get area(M) = 2π (V' - E' + F') if we like. —Dan ----- Is it in any way meaningful to think of this thing as a polyhedron with -8 pentagonal faces, four of which meet at each of the -10 vertices, joined by -20 edges? -----

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[math-fun] Correction + surface puzzle.
by Dan Asimov 10 Jun '20

10 Jun '20

I sloppily claimed that to define a polyhedron given a bunch of faces, it's necessary only to say which pairs of faces share an edge. This is not true! It's important to say *which* pairs of edges are identified, and also in which direction* each pair is identified. Thanks to Marc LeBrun for pointing out my error. (E.g., to identify the edges of a square by pretending to fold it along a diagonal gives a sphere instead of the usual torus you get by identifying opposite edges by a translation. And identifying each pair of opposite edges by a reversal results in the projective plane.) Puzzle: What surface do you get if you identify opposite edges of a square of side = 1 by a translation followed by sliding along each edge by an amount equal to 1/2 (mod 1)? In other words, if the square is [0,1]x[0,1] * identify (0,y) with (1, y + 1/2 (mod 1)) for each y in [0,1] and * identify (x,0) with (x + 1/2 (mod 1), 1) for each x in [0,1]. —Dan

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Re: [math-fun] Negative polyhedra
by Dan Asimov 10 Jun '20

10 Jun '20

It doesn't seem obvious at all. But consider eight houses that have collided. (Where each house is a pentagon in the plane.) Suppose they've collided at their apices, in two groups of 4 each. At the center of each quartet there is now one vertex of valence = 4, with the adjacent 8 edges (originally the rooves) identified in pairs. Now identify the 4 houses' bases each with the corresponding base of the other group of 4. At this point we have what is essentially a sphere with 4 square holes that are arranged in a square formation. Now identify opposite square holes with each other. The result is surface of genus 2 made of 8 pentagons, 4 per vertex. (This is *not* a regular polyhedron, because not all mappings of one pentagon to any other in any dihedral manner can extend to a mapping of the entire polyhedron that takes pentagons to pentagons.) I'm not certain but I suspect that any conjectured tiling with the correct Euler characteristic and orientability can be realized as the compact surface with those properties. —Dan ----- Very true. So, let's return to the case in my original question (well, the question wasn't _originally_ mine, but you know what I mean) and consider the possibility of a genus-2 surface with 8 pentagonal faces, 4 of which meet at each of its 10 vertices. Should it be obvious to me that this thing exists as an abstract polyhedron? -----

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Re: [math-fun] Negative polyhedra
by Dan Asimov 10 Jun '20

10 Jun '20

It's worth noting that a polyhedron need not be a subset of 3-space. In fact the existence of a polyhedron with its intrinsic properties should be considered a separate issue from whether it can be embedded in 3-space (or n-space, for that matter). To define such a polyhedron abstractly, it's necessary only to specify the faces, and which pairs of faces share an edge. For example, identifying antipodal points of the dodecahedron gives a tiling of the projective plane P^2 by 6 pentagons. This is particularly nice since each pentagon is surrounded by the other 5 of them, realizing the chromatic number 6 of P^2. For any regular p-gon Q in the Euclidean plane R^2 (p ≥ 3), there is an interesting example of an infinite polyhedron X_Q whose faces are copies of Q. The interior angle of Q is (p-2)π/p. Start by placing copies of Q around a vertex of Q until one of the copies finally shares an edge with Q again. (The number of copies will be the smallest multiple of (p-2)/p that is an even integer.) Now imagine doing the same for each copy of Q that is created in this way, until all edges are shared by 2 copies of Q. Of course in most cases there will be rampant overlapping of the copies. But by interpreting each copy of Q as an abstract p-gon, sharing only its edges with other copies, one gets an abstract polyhedron composed of countably many p-gons. This polyhedron is certainly not embedded in the plane, except when Q is a triangle, square, or hexagon. The surface constructed in this way is an example of a orientable infinite regular polyhedron, since mapping Q to any copy of Q (including Q itself) in any of the 2p dihedral ways can be extended to an isometry of the entire constant-curvature surface to itself. If p = 3, 4, or 6, then clearly the surface will just be the plane. Assuming p = 5 or p ≥ 7, the topology of this infinite polyhedron will be that of the unique orientable surface with a countably infinite number of handles and one "end".* This may be thought of as a kind of limit of an n-holed torus stretching to the right, as n –> oo if more tori are repeatedly added by connected sum to the right end. (This surface has been affectionately nicknamed "the Loch Ness monster", though I prefer "Nessie" since it has fewer syllables.) Still assuming p = 5 or p ≥ 7, the surface can be given constant negative curvature of K = -1. By its construction, it has a highly symmetric map onto the Euclidean plane, and it's easy to check that this map preserves angles away from vertices — where the angle is multiplied by some positive integer ≥ 2. This shows that its map to the plane can be interpreted as a highly symmetric holomorphic map of Riemann surfaces. —Dan _____ * An "end" of a non-compact surface is, roughly speaking, a distinct way that it can "go off to infinity". For any closed subset X of the Cantor set K, including K itself, there is a surface of any genus g (including g = oo) that has X as its space of ends. In fact, the genus and the space of ends together determine the topological type of any orientable surface. See <https://en.wikipedia.org/wiki/End_(topology)>. Veit Elser wrote: ----- I think your negative F, E, V counts come from insisting the surface has Euler characteristic 2. But I completely agree that there are more regular polyhedra than the usual five! Who said they had to be finite? We all know the Greeks had issues with infinity. Your "polyhedral foam based on the cubic lattice” should be the 6th regular polyhedron. It has a dual (analogous to the duality between the cube and octahedron) where all faces are regular hexagons and four meet at every vertex (forming a “saddle”). Think of gluing together truncated octahedra by their square faces so only the hexagon faces remain, forming an infinite foam. If you take the quotient by the translations you get a closed surface with F=8 faces, E=24 edges, and V=12 vertices, so F-E+V = -4 = 2-2(3) and the genus is 3 (the three pairs of square faces of the truncated octahedron that are identified form three handles on the sphere). Just think, if Kepler had known about the two infinite regular polyhedra he probably would not have proposed that silly model of the solar system. -----

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[math-fun] Negative polyhedra
by Gareth McCaughan 10 Jun '20

10 Jun '20

In another place, someone asked (approximately) the following question: [question begins] A regular polyhedron in which each {vertex, face} has {a,b} neighbours has vertex, edge and face counts satisfying av = bf = 2e (count pairs (vertex, neighbouring edge) for av=2e and count pairs (face, neighbouring edge) for bf=2e) and of course v-e+f = 2. So (2/a)e-e+(2/b)e = 2 so e = 2ab/(2a+2b-ab) v = 4b/(2a+2b-ab) f = 4a/(2a+2b-ab) and in addition to the "usual" solutions like a=3,b=5 for the regular dodecahedron there are some illegal ones where v,e,f are _negative_. For instance, a=4, b=5 yields e=-20, f=-8, v=-10. Is it in any way meaningful to think of this thing as a polyhedron with -8 pentagonal faces, four of which meet at each of the -10 vertices, joined by -20 edges? [question ends] ... and I was annoyed to find I didn't have a good answer. Purely numerologically, these "virtual" things of genus 0 / characteristic 2 correspond to not-virtual things of genus 2 / characteristic -2, because (-v)-(-e)-(-f) = -2. So this minus-8-faced thing corresponds to some thing that we might hope is a quotient of the hyperbolic plane by some suitable group, having 8 faces, 10 vertices, and 20 edges. There's a tessellation of the hyperbolic plane by right-angled pentagons, which (like our virtual polyhedron) has four pentagons meeting at each vertex; the quotient is an orbifold whose Euler characteristic is -1/4, so it makes some kind of sense that maybe we can put 8 of these together to get a thing of Euler characteristic -2. It's probably obvious to those skilled in the art how one does this, but I was never very skilled in that particular art and have forgotten almost everything now. I can imagine that there might be some more precise way to think of this sort of hyperbolic-plane thing as a "negative" spherical polyhedron, on the super-handwavy grounds that we can think of the faces, edges and vertices of a polyhedron as places where in the sphere we have "excess angle"; e.g., if you take a regular dodecahedron and project it out onto the sphere, then at each vertex you have three spherical pentagons with pi/3 angles meeting neatly, whereas the dodecahedron has three euclidean pentagons with 3pi/5 angles. Something something Gauss-Bonnet something. And, just as the sphere has uniform positive curvature, the hyperbolic plane has uniform negative curvature. So maaaaybe there's some way to think of faces, edges and vertices in the hyperbolic plane as "negative" compared with the sphere because they have angle _defects_ instead of angle _excesses_. But all the details elude me. I bet they're well known to those who know such things, and perhaps some such people are reading this. - Is it true that there's some sort of hyperbolic thing with 8 pentagonal faces meeting four to a vertex? What do you need to quotient the hyperbolic plane by to get it? What does the fundamental region look like? (And does something similar hold for all the negative solutions to the equations above?) - In the case a=6,b=4 there's a famous _infinite_ thing in euclidean 3-space, a sort of polyhedral foam based on the cubic lattice. The equations above suggest that there "should" be a "negative polyhedron" with a=6, b=4, e=-12, v=-4, f=-6. This seems suspiciously similar to the cube, with e=12, v=8, f=6. Is there some sort of quotient of the "mucube" foam that collapses it down to one cube with ?opposite? pairs of vertices identified? Is either the mucube itself or this hypothetical quotient, again, a quotient of the hyperbolic plane by a reflection group? - Is there anything more than numerology to the idea that we can think of these as spherical things with negative numbers of vertices/edges/faces and Euler characteristic +2, rather than (so I assume) hyperbolic things with positive numbers of vertices/edges/faces and Euler characteristic -2? -- g

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