10
$\begingroup$

Assume that $C$ is a convex set in $\mathbb{R}^{n}$ with non empty interior. Then consider its closure, is it a smooth manifold with corners?

Edit: 1) The closure of $C$ should be a smooth manifold with corners equipped with the induced smooth structure from $\mathbb{R}^{n}$.

2) With non empty interior I mean that $C$ contains a non empty open set.

$\endgroup$
  • 4
    $\begingroup$ I understand a manifold with corners to be locally diffeomorphic to $I^k$. So the number of edges at corner points would have to come out right, no? For example, a square pyramid has four edges meeting at its apex, but you'd need three if I understand it correctly. $\endgroup$ – Todd Trimble Sep 28 '15 at 13:40
  • 2
    $\begingroup$ @user71040 I think OP is working in the smooth context, judging by the tags. $\endgroup$ – Todd Trimble Sep 28 '15 at 13:47
  • 2
    $\begingroup$ @user71040 The concept of manifold with corners doesn't really make sense (as far as I know) if you only work with topological spaces and homeomorphisms -- $[0,1)^2$ is homeomorphic to $[0,1) \times (0,1)$. You need something like a stratification at least. $\endgroup$ – Najib Idrissi Sep 28 '15 at 13:56
  • 5
    $\begingroup$ Maybe what you should do, Cepu, is define what you mean exactly by "manifold with corners". Either by incorporating the definition in your question, or by linking to a definition. There seems to be a lot of confusion (for example, I don't know why my first comment doesn't immediately answer the question, except for the possibility you are using some other definition). $\endgroup$ – Todd Trimble Sep 28 '15 at 17:58
  • 2
    $\begingroup$ I believe the answer is no. If you are talking about manifolds with cubical corners, there's an "easy" no answer: just find an example where the stratifications of the boundary are not of cubical type. But even if you allow for more general smooth "manifold with corners" types, you can construct examples where the strata are not smooth-enough. I'll let Cepu write down a definition of "manifold with corners" before considering writing up a proper answer. $\endgroup$ – Ryan Budney Sep 29 '15 at 0:47
19
$\begingroup$

Consider the following curve (very informally described):

  • start from the origin in $\mathbb{R}^2$, then move from one unit up.
  • Turn of an angle $\pi/4$ on the left and move from half of unit.
  • Turn of an angle $\pi/8$ on the left and move from a quarter of a unit.
  • ...
  • Turn of an angle $\pi/2^{n+2}$ on the left and move from $1/2^{n+1}$ unit.
  • ...

If you parametrised this using an interval of length one for each step you get a function from $\mathbb{R}$ to $\mathbb{R}^2$ whose derivative is always in the upper left quarter of the plane and which converge to a point $p=(x,y)$ in the upper left quarter of the plane. Add the segments $(x,t)$ with $t$ between $0$ and $y$ and $(v,0)$ with $v$ between $x$ and $0$.

curve and interior

The inside (in the sense of Jordan's theorem) of this curve is a closed convex subset of $R^2$ of non-empty interior which cannot be endowed with a structure of differentiable manifold with corner which makes the maps to $R^2$ differentiable: Indeed such a structure would have to treat all the angles in the curve as corner, and because they have an accumulation point being itself a corner this is not going to be possible: the point $p$ has no neighbourhood diffeomorphic to an open of $[0,\infty[^k$, any of its neighbourhood contains an infinite number of corners.

Of course if you are talking about topological manifold or if you don't want your manifold structure to be compatible with the differentiable structure induced from $\mathbb{R}^2$ then this is no longer a counter example: the things I defined is homeomorphic to the closed disk so there is a structure of $C^{\infty}$ manifold with boundaries compatible to its topology, so that is why I said that the question need to be made more precise.

| cite | improve this answer | |
$\endgroup$
14
$\begingroup$

Let $C \subset S^1 \subset \mathbb{R}^2$ be a Cantor set. Let $H_C$ be its convex hull, the smallest closed subset of $\mathbb{R}^2$ containing $C$. Then $H_C$ is not a manifold with corners. Its boundary $\partial H_C$ is $C^1$ at every point except for the countably many points which are endpoints of components of $S^1-C$, these points forming a dense subset of $C$. Also, $\partial H_C$ fails to be $C^2$ at every point of $C$. But $H_C$ is homeomorphic to the closed 2-disc, indeed there is an ambient homeomorphism $\mathbb{R}^2 \mapsto \mathbb{R}^2$ taking $H_C$ to the unit closed disc.

| cite | improve this answer | |
$\endgroup$
14
$\begingroup$

See de Rham's "cutting corners" curve.

picture

Picture from an MO answer by Bill Thurston; also see a description there.

The limiting curve is $C^1$ but not $C^2$. It has a tangent everywhere, but curvature zero almost everywhere. When you say "manifold with boundary" what do you require?

Plug: English translation of de Rham's paper is in my book Classics on Fractals

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.