0

I have read some articles about this and I think I have a good intuitive understanding now. It simply is "some space that locally is equivalent to an open set of $\mathbb{R}^n$ via a homeomorphism".

But in a lecture, we had this definition for a submanifold of $\mathbb{R}^n$:

Let $M \subset \mathbb{R}^n$. It's a differentiable submanifold of $\mathbb{R}^n$ of dimension $k$ if

$\forall a\in M: \exists \,\, \mathbb{R}^n = E^k \oplus E^{n-k}$ and surroundings $U' \subset E^k, \,U'' \subset E^{n-k}, \,\, \varphi \in C^\alpha(U', U''):$

$a=(a',a'')\in U' \times U''\,\, $ and $ \,\, M \,\cap\, (U' \times U'') = \{(x',x'') \in U' \times U'' \,\,|\,\, x''=\varphi(x')\}$

I understand this definition as: "M is locally Graph of a C-alpha function"

But I just don't see where we get that M "locally is equivalent to an open set of $\mathbb{R}^n$ via a homeomorphism".

Paul Frost
  • 87,968
mathematics-and-caffeine
  • 860
  • "locally is equivalent to an open set of ℝ via a homeomorphism" : that obviously won't be the case. Any set of "dimension $n-1$" in $\mathbb R^n$ won't be open in $\mathbb R^n$(for example, how ${(x,x)\in\mathbb R}\subset \mathbb R^2$ can be open in $\mathbb R^2$ ? – Surb Jan 21 '22 at 08:26
  • 2
    Your understanding ($M$ is locally a graph) is correct, you only have to get the dimensions right. In the example you cited you are talking about a $k$- dimensional manifold. The local homeomorphism is just the orthogonal projection onto the first $k$ coordinates. – Thomas Jan 21 '22 at 09:36
  • The equivalences of various definitions of "$k$-dimensional submanifold of $\Bbb{R}^n$" can be proved using the inverse/implicit function theorem; see this answer of mine for most of the details. – peek-a-boo Jan 25 '22 at 00:09

0 Answers0