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I've a doubt about compact manifolds and orientability.

I know that Compact Manifolds in $\mathbb{R^3}$ are orientable.

My questions is:

The statement above is valid only for compact manifolds without boundary (in this case, closed manifolds)?

I'm asking this because I'd read that the Möbius-Strip with its boundary is a compact manifold.

Can someone explain me this?

Thanks.

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2 Answers 2

up vote 3 down vote accepted

There are two notions here: manifold, and manifold-with-boundary. If you add the boundary circle to a Möbius strip, you get a manifold-with-boundary, which is not a manifold. The statement that all compact manifolds embedded in $\mathbb R^3$ is true. For $0$ and $1$-manifolds, it follows because all $0$ and $1$-manifolds are orientable. For surfaces it follows because any connected compact surface in $\mathbb R^3$ divides $\mathbb R^3$ into two pieces, which takes a little work to prove. (I like Alexander duality as a proof.) Then you can orient the surface by taking an outward-pointing normal vector at each point and using the right-hand rule to orient the tangent-plane at each point.

The statement is false for $\mathbb R^4$. You can embed the projective plane in $\mathbb R^4$.

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The (nice) proof you mention is perfectly fine for surfaces as 2-manifolds without boundary, but fails if you consider 2-manifolds with boundary. For example, both an "orientable strip" and a Möbius strip do not divide $\Bbb R^3$ in two distinct regions. –  A.P. Mar 28 '13 at 15:38
    
Ok, but is the mobius-strip a compact surface? And, the definition of compact surface is that which divides $\mathbb{R^3}$ in two pieces? –  DiegoMath Mar 28 '13 at 15:38
    
The user A.P. is right, the afirmation fail if we consider 2-manifolds WITH boundary, in this case, how is the assertion? –  DiegoMath Mar 28 '13 at 15:41
    
@A.P.:That is correct. A manifold by definition has no boundary. After manifolds-with-boundary are introduced, it is common practice to drop the "with boundary," but this is sloppy terminology. –  Grumpy Parsnip Mar 28 '13 at 15:54
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@user64437: You are answering your own question. The statement is false for manifolds-with-boundary because you can embed the Möbius strip in $\mathbb R^3$. –  Grumpy Parsnip Mar 28 '13 at 15:55
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It is pretty simple, actually. For some mathematicians the definition of manifold covers only those without boundary, while others distinguish between manifolds with or without boundary.

As you correctly observed, the Möbius strip is a non-orientable manifold with boundary which can be embedded in $\Bbb R^3$. Moreover it is a compact topological space, since for example it is a closed subspace of $\Bbb P^2_{\Bbb R}$.

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So, the statement about compact manifolds is valid only with compact manifolds without boundary, in this case, closed manifolds? –  DiegoMath Mar 28 '13 at 15:44
    
Yes, although I would prefer to state it for compact 2-manifolds without boundary. –  A.P. Mar 28 '13 at 15:48
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