[math-fun] Mandelbrot''s (frac-7) snowflake fill
Howard Cannon used Julian's fractal Fourier expander to make gosper.org/mandelfou.png illustrating that it fills some areas unboundedly faster than others. (dArea/dt is interesting.) The function has no symmetry, i.e., for no (nontrivial) a, b does Z(t) = a(Z(b(t)), and there is no symmetry in the filled areas of winding number 1 of the Fourier approximations, such as illustrated. (Except for the fundamental circle and ellipse. gosper.org/mandelfou-1-2-3.png) Yet for infinitely many discrete samplings, the filled (polygonal) area does have symmetry, e.g, gosper.org/blusno.gif . --rwg
The plane can be tiled by regular polygons if they are squares, triangles, or hexagons. Any such is a "regular tiling" of the plane. Of course, we can place a countable family of planes with the *square* tiling in 3-space so that every vertex on one plane is a vertex on at least one other plane, using the faces of the cubical tiling of 3-space. Question: --------- Suppose we are given a countably infinite collection of planes all with congruent regular tilings by triangles (or else all by hexagons). Can these tiled planes be placed in 3-space so that each polygon vertex on any plane is also a polygon vertex on at least one more of the planes? To avoid trivial cases, we also require that the planes constitute a discrete set in the space of planes in 3-space.* —Dan ————————————————————————————————————— * This means the existence of a constant K > 0 such that a) any parallel planes are at least a distance K apart, and b) any intersecting planes are at least an angle K apart. ((( The space A_2(3) of affine 2-planes in 3-space can be identified with the space of 3-planes through the point (0, 0, 0, 1) in 4-space {(x,y,z,w)} *except* for the 3-plane w = 1. This shows that A_2(3) may be thought of as the space SO(3)-{I} of nontrivial rotations of 3-space — a 3-dimensional version of the Moebius band. This furnishes A_2(3) with a metric of constant curvature = +1. (Since A_2(3) is not compact it cannot have a metric of constant positive curvature that is complete., i.e., for which all geodesics can be extended arbitrarily far.) Alternatively, A_2(3) can be identified with the result of factoring R^3 out by the action of the group generated by the isometry of R^3 given by g(x,y,z) = (-x,y,z+1) . This furnishes A_2(3) with a complete metric having curvature 0 everywhere. More alternatively, A_2(3) can be identified with the result of factoring H^3 out by the action of its subgroup of isometries** generated by h(z) = 1 + conj(z) . This furnishes A_2(3) with a metric of constant curvature = -1. ————————————————————————————————————— ** H^3 denotes hyperbolic 3-space. Its group of all isometries can be identified with the group generated by all linear fractional transformations of the Riemann sphere S^2 and by conj(). That is because each LFT, as well as conj(), acting on S^2, defines a hyperbolic isometry on the unit 3-ball D^3 having the Poincare metric. )))
For the triangular tiling, take the lattice Z^3 and let the planes be of the form: {(x, y, z) : x + by + cz = d} where b and c are in {+1, -1}, and d is in Z. Then every lattice vertex lies on four planes, and the intersection of every plane with the lattice is a copy of the triangular tiling. -- APG.
Sent: Tuesday, October 25, 2016 at 6:57 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] 3D geometry question
The plane can be tiled by regular polygons if they are squares, triangles, or hexagons. Any such is a "regular tiling" of the plane.
Of course, we can place a countable family of planes with the *square* tiling in 3-space so that every vertex on one plane is a vertex on at least one other plane, using the faces of the cubical tiling of 3-space.
Question: ---------
Suppose we are given a countably infinite collection of planes all with congruent regular tilings by triangles (or else all by hexagons).
Can these tiled planes be placed in 3-space so that each polygon vertex on any plane is also a polygon vertex on at least one more of the planes?
To avoid trivial cases, we also require that the planes constitute a discrete set in the space of planes in 3-space.*
—Dan —————————————————————————————————————
* This means the existence of a constant K > 0 such that
a) any parallel planes are at least a distance K apart,
and
b) any intersecting planes are at least an angle K apart.
((( The space A_2(3) of affine 2-planes in 3-space can be identified with the space of 3-planes through the point (0, 0, 0, 1) in 4-space {(x,y,z,w)} *except* for the 3-plane w = 1.
This shows that A_2(3) may be thought of as the space SO(3)-{I} of nontrivial rotations of 3-space — a 3-dimensional version of the Moebius band. This furnishes A_2(3) with a metric of constant curvature = +1. (Since A_2(3) is not compact it cannot have a metric of constant positive curvature that is complete., i.e., for which all geodesics can be extended arbitrarily far.)
Alternatively, A_2(3) can be identified with the result of factoring R^3 out by the action of the group generated by the isometry of R^3 given by
g(x,y,z) = (-x,y,z+1)
. This furnishes A_2(3) with a complete metric having curvature 0 everywhere.
More alternatively, A_2(3) can be identified with the result of factoring H^3 out by the action of its subgroup of isometries** generated by
h(z) = 1 + conj(z)
. This furnishes A_2(3) with a metric of constant curvature = -1.
————————————————————————————————————— ** H^3 denotes hyperbolic 3-space. Its group of all isometries can be identified with the group generated by all linear fractional transformations of the Riemann sphere S^2 and by conj(). That is because each LFT, as well as conj(), acting on S^2, defines a hyperbolic isometry on the unit 3-ball D^3 having the Poincare metric. ))) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
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Adam P. Goucher -
Bill Gosper -
Dan Asimov