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Can we approximate (arbitrarily closely) a compact $C^1$ hypersurface in Euclidean space without boundary with a polygonal hypersurface, such as a simplicial complex? To clarify, I want to have the $\mathcal{H}^{n-1}$ area of the approximation converge to that of the hypersurface as the diameter of the polygons goes to 0.

If so, can you please provide a reference for the theorem. Thank you kindly.

Note, I know this to be true for $C^2$ since we have positive reach and can create a tubular neighborhood with the normals.

Edited: I think I made a mistake to say "same size" above (although that would be ideal for my situation) so I removed it. So in the case of allowing different sizes, can it be done? In fact, I've looked more carefully and they wouldn't have to be the same shape either. The critical piece is to have a piecewise linear surface composed of $(n−1)$-dimensional polyhedra, meaning the surface is continuous with no "breaks" or "jumps" at the edges, since I will integrate over each piece. Also, there should be a finite number of pieces since the surface is compact, correct? Thanks!

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    $\begingroup$ Any $C^1$ submanifold can be approximated by $C^\infty$ submanifolds because $C^\infty$ maps are dense in $C^1$ maps, and $C^1$ embeddings form an open subset in $C^1$ maps. Incidentally, tubular neighborhood still exists for $C^1$ manifolds, see e.g., mathoverflow.net/questions/286512/… $\endgroup$
    – Igor Belegradek
    Commented Jan 16, 2018 at 13:23
  • $\begingroup$ Yes, I asked the question you reference, but I am talking in the "geometric sense," which does not hold for the tubular neighborhood. My question for you is I don't think a polygonal surface is $C^\infty$ since the edges will meet at angles that aren't smooth, correct? I really need a "tiling" of a $C^1$ hypersurface for the project I am doing. $\endgroup$
    – L P
    Commented Jan 16, 2018 at 22:54
  • $\begingroup$ Sorry that I did not read the question carefully enough. I gather your hypersurfaces all sit in a Euclidean space, and "polygons" are linear $(n-1)$-dimensional polyhedra, is this correct? Would you explain how you approximate a $C^2$ hypersurface by such hypersurfaces with polyhedra of the same shape and size? $\endgroup$
    – Igor Belegradek
    Commented Jan 17, 2018 at 0:41
  • $\begingroup$ Yes, that is correct. I think I made a mistake to say same size (although that would be ideal), but we certainly can do a triangulation of a $C^2$ surface with simplices, as is done in the finite element method for surface PDE's. But can we do this for $C^1$ hypersurfaces? $\endgroup$
    – L P
    Commented Jan 17, 2018 at 1:58
  • $\begingroup$ I see no difference between $C^1$ and $C^2$. Now that I understood the question I think what you want is done in Munkres's "Elementary differential topology", see maths.ed.ac.uk/~aar/papers/munkresdiff.pdf. $\endgroup$
    – Igor Belegradek
    Commented Jan 17, 2018 at 2:10

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It seems that what you are asking is essentially how one can approximate the area of a surface via triangulations. This is a classical topic stretching back more than a century. The famous example of Schwartz Lantern shows that arbitrary triangulations do not work. Young first described conditions which triangulations should satisfy in order to approximate area:

W.H. Young, On the Triangulation Method of Defining the Area of a Surface, 1921.

These approximations have been studied extensively since then. For a description of Young's condition and other references see the following paper posted recently on the arXiv:

Kobayashi and Tsuchiya, Approximating surface areas by interpolations on triangulations, 2017.

See also

L. V. Toralballa, A geometric theory of surface area, 1970

for more references. I think that these papers answer your question. Some of these references are concerned only with 2-D surfaces, but Young's original paper seems to consider arbitrary dimensions. Further, these methods require very little regularity. I am pretty sure that the answer to your question is yes for surfaces in $R^3$, and I think that things should work in all other dimensions as well.

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  • $\begingroup$ It looks like Young's condition is that the normals to the part of the surface over each simplex need to converge to the normals to the simplex as we subdivide the triangulation further and further. I was told just now that we can do this in $\mathbb{R}^n$ since our hypersurface is $C^1$, which implies in a small section it cannot change too quickly, loosely speaking, and the normals will converge if we choose our triangulation in this way. Thanks for your help and the references. $\endgroup$
    – L P
    Commented Jan 23, 2018 at 7:55

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