About separation property of hypersurface

Question

Let N be a complete Riemannian manifold and M be a complete hypersurface in N. M is said to have separation property if N\M is disjoint union of 2 connected open sets in N. Under what reasonable assumptions on N (like simply connected or vanishing homology or cohomology)can ensure that M must have separation property? Thank you.

Answer 1

Assuming $M$ and $N$ are orientable, it suffices for the first Betti number of $N$ to be zero.

To see this, let $U$ be a tubular neighborhood of $M$, let $V = N-M$, and consider the Mayer-Vietoris sequence in (say) reduced de Rham cohomology $$ \cdots \;\leftarrow\; H^1_{\scriptscriptstyle\mathrm{dR}}(N) \;\leftarrow\; \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(U\cap V) \;\leftarrow\; \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(U) \oplus \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(V) \;\leftarrow\; \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(N) $$ Since $M$ and $N$ are orientable, the intersection $U\cap V$ has two components, so $\tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(U\cap V)\cong\mathbb{R}$. Since $M$ and $N$ are connected, we also know that $\tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(U) = \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(N) = 0$. Thus, the exact sequence above simplifies to $$ \cdots \;\leftarrow\; H^1_{\scriptscriptstyle\mathrm{dR}}(N) \;\leftarrow\; \mathbb{R} \;\leftarrow\; \tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(V) \;\leftarrow\; 0 $$ But $V=N-M$ has two components if and only if $\tilde{H}\,\!^0_{\scriptscriptstyle\mathrm{dR}}(V) \cong \mathbb{R}$, and this is guaranteed as long as $H^1_{\scriptscriptstyle\mathrm{dR}}(N)=0$.