Generalized Jordan-Brower separation theorem

Question

Suppose $M^{n+1}$ is a closed connected smooth manifold and $N^n$ is a closed connected smooth embedded submanifold of $M^{n+1}$.

What's the weakest condition under which the Jordan-Brower separation theorem hold?

Namely, $M^{n+1}\backslash N^n$ has exactly two connected components, that share $N^n$ as their common boundary. I think a simple connectedness condition suffices. But does a weaker set of conditions work? How about $H_1(M^{n+1},Z)=0$? And could you give me some reference? Many thanks!

Answer 1

For any closed connected submanifold $N \subset M$ of codimension 1 to separate $M$, it suffices that $H^1(M;\Bbb Z/2) = 0$. To avoid annoying difficulties I will assume $\text{dim } M \geq 2$.

For pick a closed submanifold $N$. Then there is a homomorphism $\varphi_N: \pi_1(M) \to \Bbb Z/2$ given by taking a loop and sending it to its mod-2 intersection number with $N$. That $N$ separates $M$ is equivalent to this homomorphism being zero.

For in one direction, if $N$ does not separate $M$, pick a point in $N$ and two points $x$ and $y$ in $M \setminus N$ that locally look to be on 'opposite sides' of $N$ (so that there is a path from one to the other that passes through $N$ exactly once). Then pick a path in the complement of $N$ from $x$ to $y$. Joining this with the path in the above parenethetical we get a loop that's tranvserse to $N$ and passes through it exactly once; so it has nonzero mod-2 intersection number with $N$. So $\varphi_N$ is nonzero.

Conversely suppose $\varphi_N$ is nonzero. Suppose $M$ was disconnected by $N$. Pick points $x, y$ in the two different components. Now pick an immersed loop with nonzero intersection number with $N$. Modify it so that it passes once each through $x$ and $y$, and so that it's immersed and, of course, transverse to $N$. Then as we follow the path from $x$ to $y$ the intersection number increases precisely when we pass between components. So going from $x$ to $y$ and then back to $x$ we must cross $N$ an even number of times! This is a contradiction to our choice of path. So $M \setminus N$ is connected.

So $H^1(M;\Bbb Z/2) = 0$ (or equivalently by Poincare duality $H_{n-1}(M;\Bbb Z/2) = 0$) suffices to know that every closed connected submanifold disconnects $M$. I hope you agree that this is a very weak condition, much weaker than $\pi_1(M) = 0$. It's true, for instance, of every lens space $L(2p+1,q)$ (or as you say if $H_1(M;\Bbb Z) = 0$). But even if you don't have this, it still suffices to check whether or not $\varphi_N = 0$.

I suspect there is a manifold $M$ with $H^1(M;\Bbb Z/2)$ nonzero but such that any closed connected submanifold still separates $M$. I don't have an example, though.

Edit: as pointed out by Daniel Valenzuela in the comments, if one has a nontrivial cohomology class $\xi \in H^1(M;\Bbb Z/2)$, consider it as a real line bundle over $M$, and take a generic section of $\xi$ (in the sense that it intersects the zero section transversely). Let $N$ be the zero set of this generic section, and $\iota$ its inclusion map; then one can check that the normal bundle of $N$ is $\iota^* \xi$. This immediately implies that the complement of $N$ is connected, and thus we have constructed a non-separating manifold. (Alternatively one can show that $\varphi_N$ is the same homomorphism as $\xi$; same trick.) Note that $N$ may not be connected, but just restrict to one of its connected components and one has a connected separating submanifold.

So $H^1(M;\Bbb Z/2) = 0$ is equivalent to the separation theorem: that any closed submanifold of $M$ of codimension 1 separates $M$ into two components.

(As far as I can tell everything should be perfectly valid for $M$ noncompact and $N$ closed (in the point-set sense, not in the compact without boundary sense) submanifolds.)

Answer 2

A properly embedded connected compact submanifold $N \hookrightarrow M$ gives you a map $H_1(M;Z) \to Z/2$ by counting mod 2 intersections. By counting those you see (as explained by Mike) that we have equivalent conditions: $N$ seperates $M$ into exactly 2 components (as we have a trivial normal bundle) $\Leftrightarrow$ the induced homomorphism $H_1(M;Z) \to Z/2$ is trivial.

Now this characterizes those submanifolds pretty well. In particular we see that if $H_1(M;Z)=0$ every such homomorphism induced by any submanifold is necessarily trivial. Or if $H_1(N;Z) \to H_1(M;Z)$ is an isomorphism as we see that every homology class can be represented as a curve on $N$ which can be pushed off $N$, which makes the intersection trivial.