1

I struggle with a question I thought was apparently very simple.

Suppose I have a triangle in the plane and a circle in the plane.

How can I prove that the triangle is not a smooth manifold, and how can I prove is not diffeomorphic to a circle?

I looked around and I couldn't find much, I've tried using the definition of smooth manifold but I can't really get anywhere.

The diffeomorphic part I think it will be straightforward once I figure the first part.

Can you help?

user8469759
  • 5,285

1 Answers1

0

If it was a smooth manifold, it would be a submanifold of the plane and hence any point of the triangle, and in particular any vertex, would have a neighbourhood $V$, that, in some basis of the plane could be written $V= \{ (x,f(x)) \ , x \in (-1,1) \}$ where $f:(-1,1) \to \Bbb R$ is a smooth map and $(0,f(0))$ is the coordinates of the vertex in the chosen basis of the plane.

But such an $f$ cannot be smooth because it would also have to be piecewise linear.

Edit: There are many equivalent definitions of a submanifold of $\Bbb R^n$, I used one of them, you can find it here at (d) in proposition 3.2.1. When working with subsets of $\Bbb R^n$, it is often better to use one of these caracterisations instead of just charts, because it simplifies thing so much.

So if your triangle was a smooth submanifold of $\Bbb R^2$ it should be locally parametrized by a smooth function, and this is not the case at a vertex, since the only parametrizations are piecewise linear, and henceforth not smooth.

But note that it depends essentially on the fact that your triangle is a subset of $\Bbb R^2$: it inherits the topology of the plane but cannot inherit the differentiable structure because of the "corner". You can find much more details in here (the accepted answer and the comments under it should really help you out).

Nal
  • 465
  • 3
  • 8
  • What do you mean with "in some basis of the plane..." – user8469759 Feb 03 '20 at 09:38
  • The plane is a vector space, any couple of non colinear vectors gives a vector space basis. Take any vertex of your triangle, there exists a neighbourhood of the point in the triangle that looks like the graph of a function, meaning that you can choose a basis of the plane where the neighbourhood is explicitely the graph of a function $f$. (For example take a colinear vector to the bisection of the vertex, and the other one perpendicular, that should give you such a basis.) – Nal Feb 03 '20 at 10:01
  • You meant "basis" as a vector space basis, not as "basis" for a topology, also why does the triangle locally look like the "graph of a function"? – user8469759 Feb 03 '20 at 11:04
  • Yes exactly, a vector space basis. As to why it looks like the graph of a function: if you take a small enough neighbourhood of the vertex, it looks like a "v", hence it can only be described as the graph of a piecewise linear function. (And so no smooth function can describe it.) – Nal Feb 03 '20 at 13:13
  • Are you using the implicit function theorem to assert the graph part? – user8469759 Feb 03 '20 at 13:18
  • Well it depends on the caracterisation of a submanifold of $\Bbb R^n$ that you use. Here if you use the "parametrisation" definition it is direct, if you use the "implicit" definition it is equivalent to the parametrisation one by implicit function theorem yes. – Nal Feb 03 '20 at 13:35
  • Can you please expound your answer with more details? I can see the the graph has a non differentiable point, but I'd like to put all these pieces together to show that the manifold is not differentiable (like with charts and everything), I came across many example to show that a manifold is smooth, some examples that show that certain sets aren't manifold, but not a single one showing that we have a non smooth manifold. I'd really appreciate if you could work out the details. – user8469759 Feb 03 '20 at 13:39
  • I've edited my answer, I hope it will help you out. – Nal Feb 03 '20 at 14:46