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I'm stuck on the following problem: let $S$ be a compact orientable hypersurface in the symplectic manifold $(M,\omega)$. Prove that there exists a smooth function $H: M \to \mathbb R$ such that $0$ is a regular value of $H$ and $S \subset H^{-1}(0)$.

Since $S$ and $M$ are orientable, I can find a tubular neighborhood $N\simeq S \times (-\epsilon, \epsilon)$, which is open in $M$ and with $S$ corresponding to $S \times \{0\}$. Then $S$ is the inverse image of the regular value 0 under the projection onto the second factor $N \to (-\epsilon,\epsilon)$. Is there a clean way to see that I can extend this map to all of $M$ such that 0 remains a regular value?

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    use a partition of unity2011-04-21
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    @user8268: Can you be more specific? If I simply extend the map by multiplying by a bump function then 0 will no longer be a regular value.2011-04-21
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    @Eric: sorry, it was too sloppy. Since $M$ and $S$ are orientable, $S$ cuts $M$ to two pieces. You can take $H$ which is $+1$ on one piece, $-1$ on the other piece, and grows from $-1$ to $1$ on a tubular neighbourhood of $S$. (first construct that function on your tubular neighbourhood so that it's constant $\pm1$ away from $S\times(-\epsilon/2,\epsilon/2)$ and then extend it to $\pm1$ on the rest of $M$)2011-04-21
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    @user8268: It looks like in your construction $S = H^{-1}(0)$? That can't work in general.2011-04-21
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    ah, I'm on leave of absence from my mind :) But one more attempt: take a smooth function on $\mathbb{R}$ which is $1$ away from $(-\epsilon,\epsilon)$ and has two regular zeros, one at $0$ and the other at (say) $\epsilon/2$. This gives you a function on $S\times(-\epsilon,\epsilon)$ where $0$ is a regular value and $H^{-1}(0)$ are two copies of $S$. Extend this function by $1$ to the resto of $M$.2011-04-21

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