# Higher dimensional Jordan-Brower separation theorem

Is there an established higher dimensional version of the Jordan-Brower theorem? I would like some statement like this:

Let $M$ be an $n$-dimensional closed, compact and connected variety and suppose that there exists an embedding $M\hookrightarrow{\Bbb R}^{n+1}$. Then ${\Bbb R}^{n+1}-M$ has exactly two connected components.

Mind that I do not want to assume $M$ orientable in the first place.

If such a statement is true, then one can show that $M$ is orientable as follows. If not, let $\gamma$ be a closed path in $M$ starting at $P$ which realizes the non-trivial monodromy of the normal line bundle. By compactness, there is a tubular neighborhood $T$ of $\gamma$ such that $T\cap M\simeq\gamma\times[-1,1]$. Starting from $P$ choose compatibly (i.e. continuously) a normal vector $\vec{n}_Q$ contained in $T$ at each point $Q$ of $\gamma$.

The endpoint of $\vec{n}_Q$ describes a curve in $T$ which never meets $M$. It can be closed to a loop $\gamma^\prime$ moving along the normal line at $P$. The closed curve $\gamma^\prime$ meets $M$ only at $P$ and they intersect transversally there.

This would contradict the higher dimensional Jordan-Brower theorem, since a closed curve intersecting $M$ transversally must intersect an even number of times.

Motivation : I'm looking for an "easy" (maybe to be read: "highly intuitive") proof of the fact that the projective plane cannot be embedded in ${\Bbb R}^3$ which can be appreciated by a person without a solid background in mathematics (but she has a fairly good understanding of what the projective plane is). In this case, of course, the curve $\gamma$ can be taken as the "mid section" of a Möbius band in the projective plane.

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In the neighbourhood of a point of $M$, $M$ becomes nothing but $\mathbb R^n$ separating $\mathbb R^{n+1}$ in two half-spaces. Take two points $x_-$ and $x_+$, one in each of those “local connected components” of $\mathbb R^{n+1} \setminus M$. It is relatively easy to see that $\mathbb R^{n+1} \setminus M$ is connected if you can join $x_-$ and $x_+$ through a path avoiding $M$, and that it has 2 pathwise connected component otherwise.
So now, assume $\mathbb R^{n+1} \setminus M$ is pathwise connected, so that there exists such a path $\gamma : I \to \mathbb R^{n+1} \setminus M$. You can assume that $\gamma$ is injective (either by “generic position” arguments or because you can easily prove that pathwise connected components and arcwise connected components are the same). I will also assume $\gamma$ to be smooth.
I can also take the segment that goes from $x_-$ to $x_+$ while remaining in the neighbourhood I started with (in which $M$ looks like $\mathbb R^n$): it crosses $M$ transversely in exactly one point. Now I can glue together $\gamma$ and this arc. By still more (easy) general position arguments, what I get can be assumed to be an embedded circle $\widehat \gamma$ in $\mathbb R^{n+1}$ intersecting $M$ transversely in exactly one point.
I've thus obtained homology classes $[\widehat \gamma] \in H_1(\mathbb R^{n+1}), [M] \in H_n(\mathbb R^{n+1})$ such that $[\widehat \gamma] \cdot [M] = \pm 1$. This is absurd, because $H_1(\mathbb R^{n+1}) = H_n(\mathbb R^{n+1}) = 0$.
Well, it seems that this question did not attract much attention. Anyway, thanks for your argument which proves that the complement is not connected. Still, would like to have, for didactical purposes, an elementary (non cohomological?) proof that $\Bbb P^2(\Bbb R)$ cannot be embedded in $\Bbb R^3$. – Andrea Mori Dec 12 '11 at 11:05
I don't understand the last step: what is $[\tilde{\gamma}]\cdot [M]$ and why is it $\pm 1$? What you use looks like some kind of Poincare duality, but I can't see the details and to be honest, am not fully persuaded.. – Peter Franek May 22 '14 at 19:08