Proving that $I \times I$ is not a manifold

Question

I want to prove that $I \times I$ is not a manifold where $I=[0,1]$. But I believe this is false. Here we will denote closed unit disc as $\overline{D}^2$.Since we can take a homeomorphism $\phi : I \times I \to \overline{D}^2$ and then if $\{\psi _{\alpha},U_{\alpha}\}$ are the charts for $\overline{D}^2$ then $\{\psi_{\alpha} \circ \phi,\phi^{-1} U_{\alpha}\}$ are the required charts. Also the transition maps are $$\psi_{\alpha} \circ \phi \circ \phi^{-1} \circ \psi_{\beta}^{-1}=\psi_{\alpha} \circ \psi_{\beta}^{-1}$$ which is a diffeomorphism. Can somebody please tell me where I am going wrong? Thanks.

Answer 1

First things first, $I^2$ is not even a topological manifold (without boundary), so it has little chance of being a differentiable manifold. To see this, do as Berni Waterman suggests: a point of $\partial I$ (say, $(0,0)$ has no neighborhood homeomorphic to an open subset of $\mathbb{R}^2$, much less diffeomorphic.

However $I^2$ is a topological manifold with boundary, and it is indeed homeomorphic to the closed unit disk $\bar{D}^2$. So it is possible to do as you suggest and transport the differentiable structure of $\bar{D}^2$ to $I^2$ and obtain a differentiable structure on $I^2$. However, if you do this, the inclusion $I^2 \hookrightarrow \mathbb{R}^2$ will not be smooth: the corners at $(0,0)$, $(1,0)$... are troublesome. So this is not very satisfactory.

So what is $I^2$? It's a manifold with corners. There are several possible definitions of manifolds with corners, and they are not all equivalent; but one thing is sure, is that $I^2$ is one. The basic idea is that you have charts in $I^2$ that are homeomorphisms $\mathbb{R}^2 \to U \subset I^2$ (manifold), $\mathbb{R} \times [0,\infty) \to U \subset I^2$ (manifold with boundary), and $[0,\infty)^2 \to I^2$ (manifold with corners), and everything plays nice.

Remark. Manifolds with corners is a purely differentiable notion. Indeed, $[0,\infty)^2$ is homeomorphic to $\mathbb{R} \times [0,\infty)$, so from a topological point of view you can always "straighten up" corners. In the realm of topology one instead considers stratified spaces.

Answer 2

You dont even have to look at transition maps to show that $I \times I$ is not a manifold. I give you an idea: Let $p \in \partial (I \times I)$ be any point of the boundary (for example, one could simply take $p = (0,0)$ ) and let $U \ni p$ be a nice neighborhood of your choice in $I \times I$. Show that $U$ cannot be homeomorphic to any open set in $\mathbb R^2$.