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If a hypersurface in a manifold separates the ambient space into two disconnected pieces, is the surface necessarily orientable? This seems to be true when one considers the Jordan Brouwer theorem which implies the sphere $S^n$ embedded in $\mathbb{R}^n$ separates space into two disconnected components. But does the requirement of orientability extend to hypersurfaces in any manifold? A counter-example would show a non-orientable hypersurface separating the ambient space into two disconnected regions.

(edit: statement on Jordan Brouwer theorem refined per George Lowther's comment)

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A manifold cannot be separated by a connected non-orientable hypersurface. [Edit: A connected non-orientable hypersurface can't separate an orientable manifold] –  George Lowther Sep 29 '11 at 0:57
    
Can you please elaborate on a construction? Or is the proof standard and is there a reference I can look up? Thank you. –  snel Sep 29 '11 at 1:02
    
Actually, a closed nonorientable $n$-manifold can't even be embedded in an orientable $n+1$-manifold. –  George Lowther Sep 29 '11 at 1:06
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The line "...which implies that the hypersurface must be homeomorphic to $S^n$ embedded in $\mathbb{R}^n$ to guarantee separation" does not seem correct. A torus separates $\mathbb{R}^3$ for example. –  George Lowther Sep 29 '11 at 1:08

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A nonorientable surface can certainly separate a nonorientable manifold. Here is an example. Note that the Klein bottle $K$ is the boundary of a $3$-manifold $M$, which is gotten by identifying the two ends of a solid cylinder in an orientation reversing way. Now consider the $3$-manifold $(K\times [0,1])\cup M_0\cup M_1$ where $M_i$ is homeomorphic to $M$, and $\partial M_i$ is identified with $K\times\{i\}$. This is a nonorientable $3$-manifold. The middle slice $K\times\{1/2\}\subset K\times [0,1]$ separates the manifold into two pieces.

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Nice example! I guess I was thinking about non-compact manifolds that aren't $\mathbb{R}^n$, say for example $\mathbb{R}^3 \times M$. Are there non-orientable hypersurfaces that cause separation in such a case? Thanks. –  snel Sep 29 '11 at 0:42
    
@measure_noob: you can make the example noncompact in a stupid way by removing a point. –  Grumpy Parsnip Sep 29 '11 at 0:48
    
I'd have to think more about $\mathbb R^3\times M$. –  Grumpy Parsnip Sep 29 '11 at 0:50
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@measure_noob: If your ambient manifold is orientable, then no non-orientable surface can separate it. That's because the separating surface would be the boundary of one half of the manifold, and the boundary of an orientable manifold must always be orientable. –  Grumpy Parsnip Sep 29 '11 at 0:52
    
Well, if we consider $\mathbb{R}^3 \times RP^3$ and embed the non orientable product space $RP^4 \times S^1$ as a hypersurface, I do not understand the statement 'the separating surface would be the boundary of one half of the manifold'. –  snel Sep 29 '11 at 1:10

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