Integrable system is not a level set: an example












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For a manifold $M$, let $f in C^{k}(M), k>0$, be a level set (i.e. $rank(df)=1$). I understand that this trivially implies that ${{k.df: k in C^{k}(M)}}$ forms an integrable regular Pfaffian system, and that the level sets of $f$ form the integral manifolds of such a Pfaffian system.



The converse, I know, is not true. The best one has is the following: Given a Pfaffian system generated by a one-form $alpha$ that is integrable, there exists an open nbd. $U$ for every $m in M$ on which there exists $f, g in C^k(U)$ s.t. $g.alpha |_U=df$ where $g$ is nowhere zero, and $f^{-1}(t_0)$ corresponds to the integral manifolds restricted to $U$.



Yet I still see physics references talking about a global level set prescription for such a manifold $M$. Does someone know any simple example where the manifold has smooth foliations from integrability, but lacks a level set prescription?










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  • 1




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    You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
    $endgroup$
    – Ted Shifrin
    Jan 8 at 17:23










  • $begingroup$
    Ah, yes, of course!
    $endgroup$
    – Sandesh Jr
    Jan 9 at 18:07
















0












$begingroup$


For a manifold $M$, let $f in C^{k}(M), k>0$, be a level set (i.e. $rank(df)=1$). I understand that this trivially implies that ${{k.df: k in C^{k}(M)}}$ forms an integrable regular Pfaffian system, and that the level sets of $f$ form the integral manifolds of such a Pfaffian system.



The converse, I know, is not true. The best one has is the following: Given a Pfaffian system generated by a one-form $alpha$ that is integrable, there exists an open nbd. $U$ for every $m in M$ on which there exists $f, g in C^k(U)$ s.t. $g.alpha |_U=df$ where $g$ is nowhere zero, and $f^{-1}(t_0)$ corresponds to the integral manifolds restricted to $U$.



Yet I still see physics references talking about a global level set prescription for such a manifold $M$. Does someone know any simple example where the manifold has smooth foliations from integrability, but lacks a level set prescription?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
    $endgroup$
    – Ted Shifrin
    Jan 8 at 17:23










  • $begingroup$
    Ah, yes, of course!
    $endgroup$
    – Sandesh Jr
    Jan 9 at 18:07














0












0








0





$begingroup$


For a manifold $M$, let $f in C^{k}(M), k>0$, be a level set (i.e. $rank(df)=1$). I understand that this trivially implies that ${{k.df: k in C^{k}(M)}}$ forms an integrable regular Pfaffian system, and that the level sets of $f$ form the integral manifolds of such a Pfaffian system.



The converse, I know, is not true. The best one has is the following: Given a Pfaffian system generated by a one-form $alpha$ that is integrable, there exists an open nbd. $U$ for every $m in M$ on which there exists $f, g in C^k(U)$ s.t. $g.alpha |_U=df$ where $g$ is nowhere zero, and $f^{-1}(t_0)$ corresponds to the integral manifolds restricted to $U$.



Yet I still see physics references talking about a global level set prescription for such a manifold $M$. Does someone know any simple example where the manifold has smooth foliations from integrability, but lacks a level set prescription?










share|cite|improve this question









$endgroup$




For a manifold $M$, let $f in C^{k}(M), k>0$, be a level set (i.e. $rank(df)=1$). I understand that this trivially implies that ${{k.df: k in C^{k}(M)}}$ forms an integrable regular Pfaffian system, and that the level sets of $f$ form the integral manifolds of such a Pfaffian system.



The converse, I know, is not true. The best one has is the following: Given a Pfaffian system generated by a one-form $alpha$ that is integrable, there exists an open nbd. $U$ for every $m in M$ on which there exists $f, g in C^k(U)$ s.t. $g.alpha |_U=df$ where $g$ is nowhere zero, and $f^{-1}(t_0)$ corresponds to the integral manifolds restricted to $U$.



Yet I still see physics references talking about a global level set prescription for such a manifold $M$. Does someone know any simple example where the manifold has smooth foliations from integrability, but lacks a level set prescription?







differential-geometry pde differential-topology foliations






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asked Jan 8 at 14:23









Sandesh JrSandesh Jr

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  • 1




    $begingroup$
    You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
    $endgroup$
    – Ted Shifrin
    Jan 8 at 17:23










  • $begingroup$
    Ah, yes, of course!
    $endgroup$
    – Sandesh Jr
    Jan 9 at 18:07














  • 1




    $begingroup$
    You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
    $endgroup$
    – Ted Shifrin
    Jan 8 at 17:23










  • $begingroup$
    Ah, yes, of course!
    $endgroup$
    – Sandesh Jr
    Jan 9 at 18:07








1




1




$begingroup$
You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
$endgroup$
– Ted Shifrin
Jan 8 at 17:23




$begingroup$
You're requiring $fcolon MtoBbb R$. So take the foliation of $S^1times S^1$ by circles ${p}times S^1$. (You can do this with a $0$-dimensional foliation of $S^1$, of course, but I figured you'd prefer a higher-dimensional example.)
$endgroup$
– Ted Shifrin
Jan 8 at 17:23












$begingroup$
Ah, yes, of course!
$endgroup$
– Sandesh Jr
Jan 9 at 18:07




$begingroup$
Ah, yes, of course!
$endgroup$
– Sandesh Jr
Jan 9 at 18:07










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