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Let $M \subset \mathbb{R}^d$ be a piecewise smooth $2$-manifold. Let $C$ be a polyhedral complex that covers $\mathbb{R}^d$ and contains faces of dimension $[0,d]$. Since $M$ is a $2$-manifold, we can perturb $M$ so that it only intersects faces of $C$ of dimension $d-2$ or greater. Can we smoothen $M$ into a smooth $2$-manifold $\tilde{M}$ such that $d(M,\tilde{M}) \leq \epsilon$ and $\tilde{M}$ intersects every face of $C$ that is intersected by $M$ and no others?

Blake
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  • Yes, this is a theorem of Whitney's. – Ryan Budney Mar 24 '16 at 03:41
  • Do you have a reference that might be useful? – Blake Mar 24 '16 at 03:43
  • Plenty. http://mathoverflow.net/questions/8789/can-every-manifold-be-given-an-analytic-structure – Ryan Budney Mar 24 '16 at 03:45
  • I believe that M could be smoothed, which seems to be what the result you linked proves, but I don't see how this answer my question about the polyhedral complex. – Blake Mar 24 '16 at 03:49
  • I think generally you will have to create new intersections when you smooth your map. Think of examples where your original map (of $M$) are highly not-transverse on the skeleta. Even for maps into the plane there appear to be problems. – Ryan Budney Mar 25 '16 at 00:33
  • I don't believe that's true. Note that when I say "we can perturb $M$ so that it only intersects faces of $C$ of dimension $d−2$ or greater", I'm also saying that I assume $M$ intersects every face of $C$ transversally. – Blake Mar 25 '16 at 02:11
  • Then just use the word "transversely". Otherwise $M$ could intersect those faces in rather weird ways. – Sebastian Goette Mar 25 '16 at 12:07
  • Ok, so is this true in the case where $M$ is transverse to the faces of $C$? – Blake Mar 25 '16 at 20:56

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