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Recall that the topological genus of a surface (or Euler characteristic) is (in essence) the number of its "holes." Thus the genus of a baseball is 0 while the genus of a donut or handled coffee cup is 1.

Is there a way to calculate the topological genus of surfaces defined by a parametric function? For instance, in this case:

ParametricPlot3D[
  {Cos[θ] + .4 Cos[θ] Sin[φ], Sin[θ] + .4 Sin[θ] Sin[φ], Cos[φ]}, 
  {θ, 0, 2 π}, {φ, 0, 2 π}]

the answer is genus = 1, while in this case:

ParametricPlot3D[{Cos[θ] Sin[φ], 2 Sin[θ] Sin[φ], Cos[φ]}, 
     {θ, 0, 2 π}, {φ, 0, π}]

the answer is genus = 0.

I searched for methods based on RegionFunction but found nothing quite right.

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  • $\begingroup$ Yes! Quite astoundingly there is a way to do just that. Have a look at the Gauss-Bonnet Theorem. You can implement this by finding the Gaussian curvature for your surface parametrisation and integrating it over the whole surface via Integrate or NIntegrate. $\endgroup$ – Thies Heidecke Apr 26 '17 at 18:45
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Integrating the Gauss curvature as suggested in the comments is possible, but horrible (it is a rather complicated function of the parameters, and I would not trust Mathematica to do it right). A better solution is to use Morse Theory. Namely, pick "height function" (the $x$ coordinate might work, but a random linear combination of the three coordinates will work better), compute the critical points of this function (as a function of the parameter), and then use Morse's formula for the Euler characteristic:

$$\chi(M) = \sum_\gamma (-1)^\gamma C_\gamma,$$ where $C_\gamma$ is the number of critical points of index $\gamma$ - in this case, $\gamma$ is one of $0, 1, 2,$ and the three kinds of critical points correspond to maxima, minima, and saddles. In particular, if you take the $x$ coordinate for your first surface, the "Morse function" is

$$ \frac{2}{5} \cos (\theta) \sin (\phi)+\cos (\theta). $$ Its critical points are:

$$ \left\{\left\{\theta\to 0,\phi\to \frac{\pi }{2}\right\},\left\{\theta\to 0,\phi\to \frac{3 \pi }{2}\right\},\left\{\theta\to \pi ,\phi\to \frac{\pi }{2}\right\},\left\{\theta\to \pi ,\phi\to \frac{3 \pi }{2}\right\}\right\}, $$ and the values of the function are $$\frac75, \frac35, -\frac35, -\frac75.$$ The outside ones are a maximum and minimum respectively, the middle one are saddles, for $\chi = 0,$ as expected. (Recall that $\chi(M) = 2 - 2 g.)$

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  • $\begingroup$ Very interesting (+1)... Let me play around with this for a few candidate surfaces before I accept the solution... $\endgroup$ – David G. Stork Apr 26 '17 at 21:06
  • $\begingroup$ Excellent... Perfect. Thanks so much (accept). $\endgroup$ – David G. Stork Apr 26 '17 at 22:25

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