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I'm reading a paper which includes the following line, and I can't find a reference anywhere to the result the authors mention: "Let M be a compact orientable embedded minimal hypersurface of a compact orientable Riemannian manifold N. Suppose we know that the first Betti number is zero. Then using that M,N are both orientable and chasing through exact sequences of homology groups, it is easy to see that M divides N into two components $\Omega_1$ and $\Omega_2$ such that $\partial \Omega_1=M=\partial \Omega_2$."

It would be great if someone could help me with this - I can't imagine the argument is too complicated but I can't see where to go. Thanks.

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The argument the author has in mind is probably via Poincare-Lefschetz duality. Here is a slightly different argument, using the intersection numbers. Assume that the complement $N - M$ is connected. Then you can find an embedded $S^1 \cong S \subset N$ which meets $M$ transversally at precisely one point (if $dim (N) \geq 3$). So the intersection number $S \cdot M=\pm$, whence $S$ has to represent a nonzero element in $H_1 (N;\mathbb{Q})$, contradicting the assumption on the Betti number. So $N-M$ is disconnected and has precisely two components. The statement on the boundaries is clear if everything is smooth.

In dimension $2$, you still get a smooth map $\bS^1 \to M$ and the same argument applies.

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  • $\begingroup$ Naively, I would like to say that because $M, N$ are orientable, $M$ has dimension $n$, $N$ represents an element in $H_{n-1}(M)$. $H^1(M)$ is zero by the universal coefficient theorem; then by Poincare duality, so is $H_{n-1}(M)$, thus $N$ must be a boundary. $\Omega_1$ and $\Omega_2$ are both submanifolds it bounds. But that last part is not rigorous; this is mostly an intuitive analogue of why separating curves are homologically trivial. $\endgroup$ – Elizabeth S. Q. Goodman Dec 3 '10 at 2:50
  • $\begingroup$ I should also note that the universal coefficient theorem, $H^1(M)$ is $H_1(M)$ mod torsion, so it avoids the trap where $H_1(M)$ could be nontrivial. $\endgroup$ – Elizabeth S. Q. Goodman Dec 3 '10 at 2:55
  • $\begingroup$ Thanks for these suggestions. Would you know of any reference to the proof of the statement (or similar) using Poincaré-Lefschetz duality? $\endgroup$ – Michael Coffey Dec 3 '10 at 18:36
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    $\begingroup$ Poincare-Lefschetz duality is Theorem 8.3 in Bredon, ''Topology and Geometry'', page 350. From that $H_0(N-M)=H^n(N,M)$, which fits into the exact sequence $0=H_1 (N) = H^{n-1}(N) \to Z=H^{n-1} (M) \to H^{n}(N,M) \to H^n (N) =Z\to H^n (M)=0$. Thus $H^n (N,M)=Z^2$. B.t.w.: there is another essential assumption to be made: $M$ has to be connected. $\endgroup$ – Johannes Ebert Dec 3 '10 at 18:55

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