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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 unity –  user8268 Apr 21 '11 at 17:16
    
@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. –  Eric O. Korman Apr 21 '11 at 17:17
    
@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 $M$) –  user8268 Apr 21 '11 at 17:38
    
@user8268: It looks like in your construction $S = H^{-1}(0)$? That can't work in general. –  Eric O. Korman Apr 21 '11 at 17:49
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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 $M$. –  user8268 Apr 21 '11 at 18:22
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Perhaps my idea is silly, but could not you take $H=f\circ pr_2$? where $pr_2$ is your projection on the second factor and $f$ is a real-valued smooth function on $]-\epsilon,+\epsilon[$ such that: its support is compact and $f(0)\neq 0$ is one of its regular values. For example you could take f(x)=exp(1/(x−2ϵ/3))exp(−1/(x+ϵ/3)) for x∈]−ϵ/3,2ϵ/3[ and f(x)=0 otherwise.

The proposed $H$, defined on $N$, has compact support $S\times\mathrm{supp}(f)$, so it prolongs to the smooth function on $M$ vanishing outside $N$. In such a way, you get a smooth function $H$ on $M$ with a constant value $f(0)$ on $S$ which is not one of its singular values.

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Do you really mean composition? $pr_2$ is not defined on all of $M$, only on an open set. If I multiply $pr_2$ by a bump function to extend it to $M$ then 0 will not be a regular value (since the function will be identically zero off of $N$). –  Eric O. Korman Apr 21 '11 at 17:12
    
I mean the composition by $\circ$. Yes, $f\circ pr_2$ is defined only on the tubular neighborhood $N$, but its support is the compact $S\times\mathrm{supp}(f)$, so it extends to the smooth function on $M$ which vanishes outside $N$. By construction $f\circ pr_2$ is constant on $S$ with a value $f(0)\neq 0$ which is not singular. –  Giuseppe Tortorella Apr 21 '11 at 17:22
    
take for example $f(x)=\exp(1/(x-2\epsilon/3))\exp(-1/(x+\epsilon/3))$ for $x\in]-\epsilon/3,2\epsilon/3[$ and $f(x)=0$ otherwise. –  Giuseppe Tortorella Apr 21 '11 at 17:31
    
Say $H(S) = c$. Then in general $H^{-1}(c)$ will have to contain more than $S$. How can I guarantee other points in $H^{-1}(c)$ are not critical? –  Eric O. Korman Apr 21 '11 at 17:52
    
Sure $H^{-1}(c)$ could contain more than $S$. It is properly $pr_2^{-1}(f^{-1}(f(0)))$ by constuction, and does not contain singular points because $pr_2$ is non singular and because I requested $f(0)$ is not a singular value for $f$( so in particular $c=f(0)\neq 0$. –  Giuseppe Tortorella Apr 21 '11 at 18:53
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