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In my (undergraduate) readings in differential geometry, it seems that they assume that only parametrized surfaces are relevant. But what do we do when we get a surface in implicit form such as:

$$f(x,y,z)=0 \tag{?}$$

In very simple cases such as $x+y+z=0$ we can solve for one of the variables, say $z$ and then we obtain a parametrization:

$$\phi(x,y) = (x,y,-x-y)$$

But we can have a very complicated function where this is not possible. In this cases, how do we compute the plethora of fundamental quantities we see in differential geometry? Such as:

  • First fundamental form.
  • Second fundamental form.
  • Normal curvature.
  • Mean curvature.

The ways to compute these quantities I've seen in books mostly depend on we having a parametrized surface. By having the first and second items, I guess I'd be able to compute the other items. Is there a way to do it or this is actually case is ignored because it's impossible in general?

Red Banana
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2 Answers2

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All the formulas you're looking for should be found in the article:

Goldman, R., Curvature formulas for implicit curves and surfaces, Computer Aided Geometric Design - Special issue: Geometric modelling and differential geometry 22, pp. 632–658, 2005.

Goldman himself has it available on Research Gate.

Ivo Terek
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There different ways to do calculations in differential geometry and in any given situation one way is often easier than the others. Here, it depends somewhat on what exactly $f$ looks like. Unfortunately, books usually focus only on one approach and usually it is using a parameterization which is usually the worst one.

With an implicit definition, $\nabla f$ is normal to the surface. From this you can write at each point a formula for an orthonormal frame $(e_1,e_2,e_3)$ where the first two are tangent and the third is is normal. This a way to express the fist fundamental form and Gauss map. From here the best approach is probably one called moving frames. There are some differential geometry texts that take this approach, the best known one is by Barrett O’Neill.

Deane
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