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This question should be extremely elementary but it stumped me somewhat. I know the definition of a (topological/smooth) manifold but I seem to have trouble when it comes to deciding whether a given subset of $\mathbb{R}^n$ is a manifold. In particular I was recently asked to say whether the following were manifolds:

  1. The set $M_1 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1, z \geq 0\}$
  2. The set $M_2 = \{(x, y) \in \mathbb{R}^2 : |x| + |y| = 1\}$

My intuition is that because the set $M_1$ has a "boundary" (the unit circle in the $xy$ plane) it isn't a manifold, but I'm not really sure of how to say taht precisely. In general I'm not exactly sure why having a boundary is a problem. On the other hand the set $M_2$ is just a square (I think), which is topologically the same as $S^1$, and so it should be a manifold since $S^1$ is. But I thought manifolds weren't supposed to have "sharp corners"? At least I remember hearing words to that effect. (To be honest I don't know what the precise definition of a "sharp corner" is).

Can anyone help me with this? Obviously I'm stilled a bit unclear about the concept of a manifold (even though I guess I understand the definition).

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2 Answers 2

up vote 3 down vote accepted

$M_1$ is what is known as a "manifold with boundary". If you haven't been introduced to those, yet, then the desired answer is probably that $M_1$ isn't a manifold at all.

You're right that $M_2$ is a square. This is then topologically equivalent to $S^1$, so is a topological manifold. However, it is not a smooth manifold, precisely because of those "sharp" corners.

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Ok, thanks. Can you explain why exactly the sharp corners prevent $M_2$ from being a smooth manifold? I know it has to do with a function not being continuous at those points, but I'm not sure of the details. –  rt93 Nov 5 '12 at 17:35
    
@rt93 It's not that a function isn't continuous there, but that it isn't smooth there. –  Matt Pressland Nov 5 '12 at 17:42
    
@rt93: Intuitively, it's because the tangent vector to $M_2$ does not vary continuously, so certainly doesn't vary smoothly. –  Cameron Buie Nov 5 '12 at 17:43
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@CameronBuie: I'm confused about something here. $M_2$ as I understand it, doesn't have a "natural" smooth structure. (If I understand it correctly, though, it can be equipped with a "non-natural" smooth structure, which does make it a smooth manifold and is unique up to diffeomorphism.) Here's what bothers me: don't you have to have a smooth structure to even speak of tangent vectors? Am I missing something? (Should I post this as a separate question?) –  Dejan Govc Nov 5 '12 at 18:12
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Right you are, Dejan. We can't really talk about the tangent vectors at the corners, so I suppose it's inaccurate to talk about the tangent vectors failing to vary continuously. More accurately, we can't even define tangent vectors at the corners in such a way that the tangent vectors vary continuously. –  Cameron Buie Nov 5 '12 at 18:24

The first example, $M_1$ is the Northern hemisphere. This is what we call a manifold with boundary.

You are right to say that $M_1$ is not a manifold. A necessary condition for $M_1$ to be a two dimensional manifold is that each open neighbourhood is diffeomorphic to an open region of the plane. However, consider the points on the equator. You can construct open neighbourhood of these equatorial points which are not diffeomorphic to an open region of the plane. In fact, you will find that they're diffeomorphic to the (non-open) half plane $\mathbb{H} = \{ (x,y) \in \mathbb{R}^2 : y \ge 0 \}.$

In this case, the boundary $\partial M_1$ is a circle which is a one dimensional manifold, while $M_1$ with the boundary deleted is a two dimensional manifold. This is one of the conditions of a manifold with boundary: both $\partial M$ and $M \backslash \partial M$ are to be manifolds, with some conditions on how you glue them.

Consider $M_2 = \{ (x,y) \in \mathbb{R}^2 : |x| + |y| = 1 \}.$ This is a square with vertices at $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$. This is clearly a (compact) topological manifold: it is homeomorphic to the circle. However, it fails to be a smooth manifold. If it were a smooth manifold then its tangent line would vary continuously. Consider the first quadrant where $x>0$ and $y > 0$. The tangent line to $M_2$ is the line $y=1-x$ for all points of $M_2$ in this quadrant. Next, consider the tangent line to $M_2$ in the second quadrant where $x<0$ and $y > 0$. The tangent line to $M_2$ is the line $y=x+1$ for all points of $M_2$ in this quadrant. Now, consider the point $(0,1)$ of $M_2$. Approaching this point along $M_2$ with $x < 0$, gives $y = x+ 1$ as the limiting tangent line. Approaching this point along $M_2$ with $x > 0$, gives $y = 1-x$ as the limiting tangent line. The two limits diagree, so the tangent line does not vary continuously in a neighbourhood of $(0,1)$, which means $M_2$ is not a smooth manifold.

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Thanks for this comprehensive answer. –  rt93 Nov 5 '12 at 18:02
    
You're welcome. I'm sorry it wasn't the best. –  Fly by Night Nov 5 '12 at 18:31
    
Well I suppose it is the better of the two answers by virtue of its level of detail, but I had already accepted the other answer before yours came on the scene, as it essentially resolved the question for me. I don't know whether it's kosher to change the accepted answer after the fact. –  rt93 Nov 5 '12 at 18:39
    
Patience is a virtue ;-) –  Fly by Night Nov 5 '12 at 18:58
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A word for the wise: a lot of people don't even read questions that have had an accepted answer. (The person who set the question is so happy s/he's effectively closed the discussion.) I was already writing my reply when you accepted your best answer. I usually try to wait 24 hours before accepting. It encourages replies. –  Fly by Night Nov 5 '12 at 19:01

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