# Orientability implies separation of space?

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
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

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.
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
I'd have to think more about $\mathbb R^3\times M$. –  Grumpy Parsnip Sep 29 '11 at 0:50
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