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I am reading Analysis on Manifolds by Munkres, and I am trying to understand better what manifolds are. Here are the relevant definitions:
Defn: For $k \in \mathbb{Z}^+$, a k-manifold in $\mathbb{R}^n$ of class $C^r$ is $M \subseteq \mathbb{R}^n$ s.t. $\forall p \in M$, $\exists V$ open in $M$, $p \in V$, $U$ open in either $\mathbb{R}^k$ or $\mathbb{H}^k$ and a bijection $\alpha:U \to V$ s.t.

  • $\alpha$ is of class $C^r$
  • $\alpha^{-1}:V \to U$ is continuous
  • $D \alpha (x)$ has rank k $\forall x \in U$

Here $\mathbb{H}^k=\{x \in \mathbb{R}^k:x_k \geq 0 \}$

My intuitive understanding of manifolds, for instance, a surface (aka 2-dimensional manifold) in $\mathbb{R}^3$ lives in 3-dimensional space, but really only requires 2 parameters to describe. Now $[0,1]^2 \times \{ 0 \} = \{x \in \mathbb{R}^3: 0 \leq x_1,x_2 \leq 1, x_3=0 \}$ is not a 2-manifold in $\mathbb{R}^3$ b/c of its corners, for instance at the origin $\textbf{0}_3$. While it can be shown to violate the definition, why is this a bad thing, in other words, why should the definition exclude cases like this? I am new to this subject so any intuition that can be provided is greatly appreciated.

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    What do you mean by the $\times 0$ part? – 2017-01-25
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    $[0,1]^2 \times 0= \{x \in \mathbb{R}^3: 0 \leq x_1,x_2 \leq 1, x_3=0 \}$ – 2017-01-25
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    That would usually be written as $\times \{ 0\} $, fyi. – 2017-01-25
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    Manifolds are usually not supplied with a smooth structure unless otherwise. So a square (with corners) is a manifold, but not a *smooth* manifold. – 2017-01-25
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    There is something called a "manifold with corners". It takes a bit of extra work to define them. I think authors focus on smooth manifolds "without corners" mainly for simplicity. – 2017-01-25
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    It's not a "bad thing". It's a "different thing". The behavior of manifolds without corners is slightly different from the behavior of manifolds with corners, and so must be dealt with. Also, since manifolds without corners are simpler objects from a mathematical perspective, and since they are more common objects, the literature has a strong tendency to focus on them. But, when you need manifolds with corners, then there are places to learn about them and their slighly different properties. – 2017-01-25
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    @littleO,Lee Mosher, Tac-Tics, and others. I greatly appreciate the comments/answers. To see if I understand correctly, Munkres tried to strike a balance between a relatively straightforward definition for manifold (even though I already find it hard to grasp intuitively, it can be made even more intricate by allowing for "corners"...), and something that would lead to a nontrivial theorem (Stokes' Thm). – 2017-01-25

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There's nothing necessarily wrong with such a manifold. What you have described is usually called a manifold with corners. The kind of manifold you're reading about is often called a manifold with boundary.

Having a boundary is necessary if you want to talk about Stoke's theorem. Adding in corners would only serve to complicate the theorem as it's usually stated. I don't know much about manifolds with corners, but I imagine it isn't hard to adapt Stoke theorem to that case.

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    Some writers have manifolds with boundaries as manifolds, some don't. Whilst they share structure, you can see that there are points without charts, so its not a manifold according to the definition. – 2017-01-25
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Think of an $n$-manifold as a place where, local to every point, the space is like a copy of $\mathbb{R}^n$. So, if you zoom in enough on the surface of a 3d ball, its kind of flat like a plane, and so that surface is a manifold called the sphere.

But what if we cut a region out of the ball, and cut out the corresponding region of the sphere, and then we stood on the boundary of that region. Well, its like a half plane, but its not like a plane anymore. Its boundary makes it not a manifold. Manifolds can't have those "half" places. So your closed square has a boundary, and its not a manifold.

If it was an open square, it would be a manifold again

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The reason for defining manifolds this way is that they are intended to be locally similar to euclidean space. The term locally means that near each point of the manifold there is the corresponding bijection to an open set in $\mathbb{R}^k$. This implies that topologically the manifold has the same properties than the corresponding euclidean space, at least locally.

In your example, the corners nor the points in the edges can't be seen in the manifold as one can see a point in euclidean space (of dimension $2$, i.e. a plane)