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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.

Benno
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  • A related question. – J. M. ain't a mathematician Feb 07 '12 at 00:49
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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 07 '12 at 00:52
  • @QiaochuYuan Aren't both of those question valid if there is a parent object that already has (standard) structure? – 2'5 9'2 Feb 07 '12 at 01:02
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    @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 07 '12 at 01:05
  • @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. – 2'5 9'2 Feb 07 '12 at 01:23
  • It's actually a tricky business to find manifolds which do not have any smooth structure. You have to go to dimension 4 to find the first example. – Cheerful Parsnip Feb 07 '12 at 03:13

1 Answers1

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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.)

Qiaochu Yuan
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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 07 '12 at 09:02
  • @Benno: sure. As I said above, any open subspace of a smooth manifold naturally inherits a smooth structure. – Qiaochu Yuan Feb 07 '12 at 09:05
  • 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... –  Feb 06 '14 at 21:38
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    @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 06 '14 at 21:43
  • @user: in other words, the first paragraph claims that the OP is committing a type error (http://qchu.wordpress.com/2013/05/28/the-type-system-of-mathematics/). – Qiaochu Yuan Feb 06 '14 at 21:44
  • Okay... but your points about $\delta I^2$ being homeomorphic to $S^1$ and $S^1$ having a smooth structure don't seem to relate, nor do they seem to illustrate your point about the question 'not making sense'. E.g. "'Is ${a,b}$ a group?' does not make sense. There is a bijection onto $\mathbf{Z}/2\mathbf{Z}$ and $\mathbf{Z}/2\mathbf{Z}$ is a group. So it's not clear." –  Feb 06 '14 at 21:54
  • @user: it's true that "is ${ a, b }$ a group?" does not make sense, but if ${ a, b }$ is naturally a subset of another group $G$, then "is ${ a, b }$ a subgroup of $G$" does make sense. Similarly, the unit square is naturally a subspace of $\mathbb{R}^2$, which does have a smooth structure, so it makes sense to ask whether it appears as the image of a smooth map $S^1 \to \mathbb{R}^2$. The answer is no, so in particular the unit square cannot even be an immersed submanifold of $\mathbb{R}^2$. – Qiaochu Yuan Feb 06 '14 at 21:56
  • @user: it's also still interesting to ask whether you can give a set which does not have some structure the structure you want, perhaps in a way which is compatible with other structure you already have. For example, it's easy to give any finite set a group structure, but giving all infinite sets a group structure turns out to be equivalent to the axiom of choice. In this case any subspace of a topological space is naturally a topological space (not that the analogous statement for smooth manifolds is false in a big way), so it was natural to keep the topological structure. Not all... – Qiaochu Yuan Feb 06 '14 at 21:58
  • ...topological spaces can be the underlying topological spaces of smooth manifolds, so it's a non-vacuous statement to observe that the unit square is such a topological space. – Qiaochu Yuan Feb 06 '14 at 22:00
  • In this case, though, the unit square is explicitly embedded in $\mathbf{R}^2$. There is certainly a standard way to define what it means for a map from an arbitrary subset of $\mathbf{R}^n$ into $\mathbf{R}^m$ to be smooth. Isn't this enough in this example? –  Feb 06 '14 at 22:08
  • @user: what is this standard definition in the case that the subset isn't open? If the definition is that it's the restriction to the subset of a smooth map on $\mathbb{R}^n$, then every subset of $\mathbb{R}^m$ is the image of a smooth map in this sense. Most of these are not in any reasonable sense manifolds. – Qiaochu Yuan Feb 06 '14 at 22:20
  • The definition is that this map is smooth if it is locally a restriction of a smooth map defined on an open set. Surely not many subsets of $\mathbf{R}^m$ are the diffeomorphic image of an open subset of $\mathbf{R}^n$? (I am reading the Milnor book and I still don't see a problem with subsets explicitly embedded in $\mathbf{R}^m$...) A smooth manifold has to be locally diffeomorphic to open subsets of $\mathbf{R}^n$...? –  Feb 06 '14 at 22:30
  • @user: I don't understand what point you're trying to make with any sentence in that comment except the first one. – Qiaochu Yuan Feb 06 '14 at 22:48
  • @user: no, we can't. Arbitrary subsets of a smooth manifold inherit at best a topology; they don't inherit a smooth structure in any reasonable sense (indeed they need not even be topological manifolds) in general. We can ask, for example, whether they're immersed submanifolds, but this is extra structure (namely the data of the immersion). – Qiaochu Yuan Feb 06 '14 at 23:09