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.
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$I\times I$ is a manifold with boundary, but it isn't a manifold without boundary. Also, there is no homeomorphism from $I\times I$ to the open disc $\mathbb{D}^2$ (I don't think you want $\mathbb{D}^1$). There is a homeomorphism from $I\times I$ to the closed disc $\bar{\mathbb{D}}^2$, though. – Batominovski Nov 06 '15 at 04:46
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$I \times I$ is by itself a manifold diffeomorphic to $D^2$, but it is not a submanifold of $\Bbb R^2$, because the inclusion immersion is not smooth. – Paul Sinclair Nov 06 '15 at 04:46
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2@Batominovski, $I\times I$ is not a manifold with boundary, it is a manifold with corners. – R_D Nov 06 '15 at 04:52
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Are you talking about topological manifolds or differentiable manifolds? – Batominovski Nov 06 '15 at 04:56
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@Batominovski I am talking about differentiable manifolds – happymath Nov 06 '15 at 04:57
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Then, your problem is assuming that $\phi$ is a differentiable map. – Batominovski Nov 06 '15 at 04:58
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Comments are not for extended discussion; this conversation has been moved to chat. – user642796 Nov 06 '15 at 08:04
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@ArthurFischer, there was no need to move the conversation to chat. Answers and explanations were there, and now the math mode in the chat is not usable. – Batominovski Nov 06 '15 at 12:02
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1@Batominovski Answers should be conveniently posted as answers (that's why those posts are called answers). (Also, you may use robjohn's ChatJax bookmarklet to allow MathJax rendering in chat.) – user642796 Nov 06 '15 at 12:05
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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.
Najib Idrissi
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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 $\mathbb R^2$.
H1ghfiv3
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