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Is the unit square $\partial I^2$ (i.e. the square with vertices $(0,0), (0,1), (1,0), (1,1) \in \mathbb R^2$) a smooth manifold?

I guess it shouldn't be smooth because it has "corners", but i have trouble actually finding an explicit atlas which "makes sense" and which contains two coordinate charts which are not compatible.

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  • $\begingroup$ A related question. $\endgroup$ Feb 7, 2012 at 0:49
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    $\begingroup$ Strictly speaking, it does not make sense to ask whether a topological space "is" a smooth manifold, in the same way that it does not make sense to ask whether a set "is" a group. Being a smooth manifold is a structure that one puts on a topological space (which needs to have the property of being a topological manifold first) in the same way that being a group is a structure that one puts on a set (that is, specifying it requires extra data). $\endgroup$ Feb 7, 2012 at 0:52
  • $\begingroup$ @QiaochuYuan Aren't both of those question valid if there is a parent object that already has (standard) structure? $\endgroup$
    – 2'5 9'2
    Feb 7, 2012 at 1:02
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    $\begingroup$ @alex: this is subtle. Certainly it makes sense to ask whether a subset of a group is a subgroup. But the naive way to endow a subspace of a smooth manifold with a smooth structure requires that the subspace be open. $\endgroup$ Feb 7, 2012 at 1:05
  • $\begingroup$ @Qiaochu I see. Using the specifics of this example, I thought maybe the local homeomorphism to $\mathbb{R}$ could be taken to be a restriction of a linearization of the local homeomorphism to $\mathbb{R}^2$. I see now that how to "linearize" is problematic. $\endgroup$
    – 2'5 9'2
    Feb 7, 2012 at 1:23

1 Answer 1

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I initially thought this question was about $I^2$, but I can give a definite answer for $\partial I^2$, which is that this question doesn't make sense. Note that as a topological space, $\partial I^2$ is homeomorphic to the unit circle $S^1$ (in particular, it is a topological manifold!), which can be equipped with a smooth structure in a fairly straightforward way (e.g. using the exponential map $e^{ix} : \mathbb{R} \to S^1$). So it's not clear what we would mean by the statement that $\partial I^2$ isn't smooth.

One way to make this intuition precise is to think of $\partial I^2$ as the image of $S^1$ under a continuous map $S^1 \to \mathbb{R}^2$. Then the statement you want is this: no such map can be an injective immersion. (Edit, 12/10/15: An earlier version of this answer claimed that no such map can be smooth. In fact this is false; a counterexample can be constructed by slowing down as you hit each corner using a bump function.)

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  • $\begingroup$ Thanks! Out of curiosity, would $I^2 \setminus \partial I^2 $ be a smooth submanifold of $\mathbb R^2$? $\endgroup$
    – Benno
    Feb 7, 2012 at 9:02
  • $\begingroup$ @Benno: sure. As I said above, any open subspace of a smooth manifold naturally inherits a smooth structure. $\endgroup$ Feb 7, 2012 at 9:05
  • $\begingroup$ Hi, I'm slightly confused as to whether one needs $\delta I^2$ to be diffeomorphic to $S^1$ (not just homeomorphic) for the first paragraph to work? Then again, your second paragraph shows this is not even possible... $\endgroup$
    – user71815
    Feb 6, 2014 at 21:38
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    $\begingroup$ @user: the point of the first paragraph is that the statement "$\partial I^2$ is diffeomorphic to $S^1$" doesn't even make sense a priori because $\partial I^2$ doesn't come with a distinguished smooth structure, in the same way that the statement "brown is bigger than yellow" doesn't even make sense a priori because colors don't come with a distinguished ordering. $\endgroup$ Feb 6, 2014 at 21:43
  • $\begingroup$ @user: in other words, the first paragraph claims that the OP is committing a type error (qchu.wordpress.com/2013/05/28/the-type-system-of-mathematics). $\endgroup$ Feb 6, 2014 at 21:44

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