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NB: By higher-genus surface, I mean a closed orientable surface of genus at least 2.

This question has come up before on math.SE, and even MathOverflow, but most posters suggested using either Blender or Inkscape. However, I would like to draw these higher-genus surfaces in Mathematica, because I am trying to create a Manipulate which takes as input a word in the fundamental group of such a surface, and outputs the corresponding geodesic, drawn on the surface.

So, for example, let's say I am trying to draw a genus 2 surface. What I am doing now is the following:

torus = ParametricPlot3D[{(2 + Cos[s]) Cos[t], (2 + Cos[s]) Sin[t], Sin[s]}, {t, 0, 2 Pi}, {s, 0, 2 Pi}, Mesh -> None, Axes -> False, Boxed -> False, PlotStyle -> Opacity[.3], RegionFunction -> Function[{x, y, z, u, v}, x < 2]]; antitorus = Graphics3D[ Translate[ GeometricTransformation[torus[[1]], ReflectionTransform[{1, 0, 0}, {2, 0, 0}]], {1, 0, 0}], Boxed -> False, Axes -> False]; bound = ParametricPlot3D[{{t, (2 + Cos[s]) Sqrt[ 1 - 4/((2 + Cos[s])^2)], Sin[s]}, {t, -(2 + Cos[s]) Sqrt[1 - 4/((2 + Cos[s])^2)], Sin[s]}}, {s, 0, 2 Pi}, {t, 2, 3}, PlotStyle -> {Opacity[.7]}, Axes -> False, Boxed -> False, Mesh -> None, PlotPoints -> 100]; Show[antitorus,torus,bound,Lighting->"Neutral"]

This gives me this (not-so-bad!) picture:

enter image description here

I am wondering what other methods there are for creating these surfaces, perhaps with a smoother finished product than the one I currently have.

And of course, ideally, I would eventually draw the two "building blocks" of all such surfaces, the once- and twice-punctured tori. Then I could dynamically build these surfaces on the fly...

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  • $\begingroup$ Presumably, your surface must be connected, too--otherwise there are some obvious and simple solutions :-). $\endgroup$ Commented Mar 8, 2013 at 20:56
  • $\begingroup$ Does Stan Wagon's one-liner double torus look like the sort of thing you are looking for? ContourPlot3D[(x^4 - x^2 + y^2)^2 + 9 z^2 == .04, {x, -1.2, 1.2}, {y, -1, 1}, {z, -.4, .4}, PlotPoints -> 30, Boxed -> False, Axes -> False, ContourStyle -> Yellow] $\endgroup$ Commented Mar 8, 2013 at 21:01
  • $\begingroup$ Stan Wagon's 2 double torus's can be found at blog.wolfram.com/2010/12/17/… $\endgroup$ Commented Mar 8, 2013 at 21:07

6 Answers 6

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If you dig through Eric Weisstein notebook you can find this well parametrized version. I changed parameters and styles a bit to get closer to your shape.

With[{R = 1.2, r = 1/2, a = Sqrt[2]}, ContourPlot3D[-a^2 + ((-r^2 + R^2)^2 - 2 (r^2 + R^2) ((-r - R + x)^2 + y^2) + 2 (-r^2 + R^2) z^2 + ((-r - R + x)^2 + y^2 + z^2)^2) ((-r^2 + R^2)^2 - 2 (r^2 + R^2) ((r + R + x)^2 + y^2) + 2 (-r^2 + R^2) z^2 + ((r + R + x)^2 + y^2 + z^2)^2) == 0, {x, -2 (r + R), 2 (r + R)}, {y, -(r + R), (r + R)}, {z, -r - a, r + a}, BoxRatios -> Automatic, PlotPoints -> 35, MeshStyle -> Opacity[.2], ContourStyle -> Directive[Orange, Opacity[0.8], Specularity[White, 30]], Boxed -> False, Axes -> False]]

enter image description here

OK digging through Eric Weisstein another notebook I figured a "tentative" generalization, - at least it works with n=3 or n=4. The rest needs more time (also look here):

torusImplicit[{x_, y_, z_}, R_, r_] = (x^2 + y^2 + z^2)^2 - 2 (R^2 + r^2) (x^2 + y^2) + 2 (R^2 - r^2) z^2 + (R^2 - r^2)^2; build[n_] := Module[{f, cp, polys, cartPolys, cartPolys1},(*implicit polynomial*) f = Product[ torusImplicit[{x - 1.5 Cos[i 2 Pi/n], y - 1.5 Sin[i 2 Pi/n], z}, 1, 1/4], {i, 0, n - 1}] - 10; cp = ContourPlot3D[ Evaluate[f == 0], {x, -3, 3}, {y, -3, 3}, {z, -1/2, 1/2}, BoxRatios -> Automatic, PlotPoints -> 35, MeshStyle -> Opacity[.2], ContourStyle -> Directive[Orange, Opacity[0.8], Specularity[White, 30]], Boxed -> False, Axes -> False]]; build[3]

enter image description here

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  • $\begingroup$ Oh this is pretty! $\endgroup$ Commented Mar 8, 2013 at 21:16
  • $\begingroup$ How hard would it be to generalize to higher genus? $\endgroup$ Commented Mar 8, 2013 at 21:18
  • $\begingroup$ @SteveD thx :P well i added some stuff about higher order $\endgroup$ Commented Mar 8, 2013 at 21:40
  • $\begingroup$ Oh wow, the answers on this thread are incredible! $\endgroup$ Commented Mar 8, 2013 at 21:49
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    $\begingroup$ As an alternative, if you replace f with f = Product[torusImplicit[{x - 2.35 i, y, z}, 1, 1/4], {i, 0, n - 1}] - 10, you get the holes in a line. $\endgroup$ Commented Mar 9, 2013 at 20:05
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Quick and dirty: look at the boundary of a tubular neighborhood of a union of circles.

circle[x_, n_: 32] := {x + Cos[#], Sin[#], 0} & /@ Range[0, 2 \[Pi], 2 \[Pi]/n]; Graphics3D[Tube[circle[#, 72], .5] & /@ Range[-3, 3, 2], Boxed -> False]

Image

Space them approximately two units apart (using x) and keep their radii less than $1/2$.

For smooth surfaces--albeit at a price--we may subvert RegionPlot3D to do our work. It's a similar idea, only now we apply a 3D buffer to a circular skeleton rather than using tubular neighborhoods of fixed radius:

d[{x_, y_, z_}, x0_: 0] := Block[{u, v}, {u, v} = {x0, 0} + Normalize[{x - x0, y}]; Norm[{u, v, 0} - {x, y, z}]^2]; RegionPlot3D[Min[d[{x, y, z}, #] & /@ Range[-2, 2, 2]] <= 1/2, {x, -4,4}, {y, -2,2}, {z, -2,2}, BoxRatios -> {4, 2, 2}, Mesh -> None, PlotPoints -> 50, Boxed -> False, Axes -> False]

Genus 3

The argument x0 to d shifts the skeleton's center to x0 along the x-axis. Taking a contour of the shortest distance to a collection of circular skeletons does the job.

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  • $\begingroup$ Set n to $4$ in circle[#, 4] to get an interesting version. $\endgroup$ Commented Mar 8, 2013 at 21:10
  • $\begingroup$ This is a quick, nice way to do it. But the "joins" aren't very smooth. :) $\endgroup$ Commented Mar 8, 2013 at 21:18
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    $\begingroup$ I'll get you smoother joins, Steve--just a minute. $\endgroup$ Commented Mar 8, 2013 at 21:18
  • $\begingroup$ Oh, I wish I could upvote again for the smooth version! $\endgroup$ Commented Mar 8, 2013 at 21:34
  • $\begingroup$ Actually, if you do some algebraic manipulation of the smooth version, you will obtain a solution almost identical (if not identical) to that of Vitaliy Kaurov. That shows how solutions like his can readily be derived and why they do generalize--and how to generalize them. You can also have some fun here by modifying my function d to draw circles in arbitrary locations with arbitrary orientations: with that you can draw pictures to your heart's content, spacing and sizing the holes as you wish, twisting the figure, and so on. $\endgroup$ Commented Mar 8, 2013 at 21:50
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By using the Erdős lemniscate of order n:

erdos[z_, n_] := Abs[z^n - 1]^2 - 1; f[x_, y_, z_] := erdos[x + I y, 3]^2 + (16*Abs[x + I y]^4 + 1)*(z^2 - 0.12^2); ContourPlot3D[ f[x, y, z] == 0, {x, -1.5, 1.5}, {y, -1.5, 1.5}, {z, -1.5, 1.5}, ContourStyle -> {StippleShading[0.5], White}, Lighting -> "Accent", PerformanceGoal -> "Quality", BoxRatios -> Automatic, Axes -> False, Mesh -> None, Boxed -> False]

enter image description here

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The following pokes n holes in flattish blob:

genus[n_] := Module[{pts, fn}, pts = If[n == 1, {0, 0}, Table[2 {Cos[t], Sin[t]}, {t, 2 \[Pi]/n, 2 \[Pi], 2 \[Pi]/n}]]; fn = 10 z^2 + Total[Join[#/n, (2 + 2/n)/#] &[#.# &[{x, y} - #] & /@ pts]]; ContourPlot3D[fn == 18, {x, -4, 4}, {y, -4, 4}, {z, -2.5, 2.5}, Mesh -> None, ContourStyle -> Yellow, BoxRatios -> Automatic, Boxed -> False, Axes -> False] ]

Array of genus 2,..,7 surfaces

Note: The expression fn is $10\,z^2$ plus the sum over all points pts of $k\,d^2 + l/d^2$, where $d$ is the distance to the point (dropping $z$ coordinates) and $k$, $l$ are coefficients depending on the number of holes $n$. The upshot is that the function goes to infinity at the vertical lines through the points and as $(x,y,z)$ moves away from the points.

With[{n = 1}, 10 z^2 + Total[Join[#/n, (2 + 2/n)/#] &[#.# &[{x, y} - #] & /@ {{a, b}}]]] (* -> (-a + x)^2 + (-b + y)^2 + 4/((-a + x)^2 + (-b + y)^2) + 10 z^2 *)
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Wagon's formulation of the double torus, as featured in David's answer, is based on the Gerono lemniscate. I find that using Booth's lemniscate (of which Bernoulli's is a special case) instead, with its adjustable parameters, to be more flexible. Here are two examples:

With[{a = 5, b = 4, c = 5, f = 3/4}, ContourPlot3D[((x^2 + y^2)^2 - a x^2 + b y^2)^2 + c z^2 == f, {x, -5/2, 5/2}, {y, -1, 1}, {z, -1/2, 1/2}, BoxRatios -> Automatic]]

double torus based on Booth lemniscate

With[{a = 3, b = 3, c = 10, f = 1/2}, ContourPlot3D[((x^2 + y^2)^2 - a x^2 + b y^2)^2 + c z^2 == f, {x, -2, 2}, {y, -1, 1}, {z, -1/2, 1/2}, BoxRatios -> Automatic]]

one more double torus

Of course, one can put all this in a Manipulate[] to explore the effects of changing the parameters, but I'll leave that up to the sufficiently interested reader.

Similar things can be done with e.g. rose curves or sinusoidal spirals if higher genus surfaces are desired.

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Here's a double torus by Stan Wagon from the 2010 one-liner competition

ContourPlot3D[(x^4 - x^2 + y^2)^2 + 9 z^2 == 0.04, {x, -1.2, 1.2}, {y, -1, 1}, {z, -0.4, 0.4}]

Mathematica graphics

With Boxed, Axes, and Mesh set to False and the equation = 0.03:

meshless double torus

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  • $\begingroup$ I find this slightly ugly; it is not the usual, voluptuous double-torus I know and love. :) $\endgroup$ Commented Mar 8, 2013 at 21:17
  • $\begingroup$ I know what you mean. It will look slightly better without the mesh, but the shape is more like a bee-stung 8. $\endgroup$ Commented Mar 8, 2013 at 21:52

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