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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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A related question. –  J. M. Feb 7 '12 at 0:49
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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). –  Qiaochu Yuan Feb 7 '12 at 0:52
    
@QiaochuYuan Aren't both of those question valid if there is a parent object that already has (standard) structure? –  alex.jordan Feb 7 '12 at 1:02
    
@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. –  Qiaochu Yuan Feb 7 '12 at 1:05
    
@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. –  alex.jordan Feb 7 '12 at 1:23
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1 Answer

up vote 2 down vote accepted

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 smooth. (Note that the smoothness of a continuous map between smooth manifolds is a genuine property as opposed to a structure.) This can be verified by showing that tangent vectors to $S^1$ can't have a well-defined image at the corners.

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Thanks! Out of curiosity, would $I^2 \setminus \partial I^2 $ be a smooth submanifold of $\mathbb R^2$? –  Benno Feb 7 '12 at 9:02
    
@Benno: sure. As I said above, any open subspace of a smooth manifold naturally inherits a smooth structure. –  Qiaochu Yuan Feb 7 '12 at 9:05
    
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... –  user71815 Feb 6 at 21:38
    
@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. –  Qiaochu Yuan Feb 6 at 21:43
    
@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). –  Qiaochu Yuan Feb 6 at 21:44
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