Existence of a Tubular neighborhood of a hypersurface












0












$begingroup$


Suppose $H$ is a co-dimension 1 embedded submanifold of $M$. Let $X$ be a vector field on $M$ such that $forall x in H$, $T_xM = T_xH oplus X_x$. Now I want to show that there exixts an open set $U$ such that $N subset U$, $U$ diffeo to $H times (-epsilon, epsilon)$.



The proof is supposed to go as follows:



Let $phi(x,t): M times (-a,a) to M$ be an integral flow of $X$, such that $phi(x,0) = id_M$, $phi(cdot,t)$ is a diffeo for each $t$, and $frac{partial}{partial t} phi = X$. Then by the inverse function theorem, $phi$ is a diffeo on a neighborhood of $U times V$ of each $(x,0)$.



How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?






differential-geometry differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:09










  • $begingroup$
    Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:19










  • $begingroup$
    Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:20












  • $begingroup$
    I see, may I ask whether you have any idea why does a local diffeo solve the question?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:21










  • $begingroup$
    Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:22


















0












$begingroup$


Suppose $H$ is a co-dimension 1 embedded submanifold of $M$. Let $X$ be a vector field on $M$ such that $forall x in H$, $T_xM = T_xH oplus X_x$. Now I want to show that there exixts an open set $U$ such that $N subset U$, $U$ diffeo to $H times (-epsilon, epsilon)$.



The proof is supposed to go as follows:



Let $phi(x,t): M times (-a,a) to M$ be an integral flow of $X$, such that $phi(x,0) = id_M$, $phi(cdot,t)$ is a diffeo for each $t$, and $frac{partial}{partial t} phi = X$. Then by the inverse function theorem, $phi$ is a diffeo on a neighborhood of $U times V$ of each $(x,0)$.



How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?






differential-geometry differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:09










  • $begingroup$
    Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:19










  • $begingroup$
    Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:20












  • $begingroup$
    I see, may I ask whether you have any idea why does a local diffeo solve the question?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:21










  • $begingroup$
    Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:22
















0












0








0





$begingroup$


Suppose $H$ is a co-dimension 1 embedded submanifold of $M$. Let $X$ be a vector field on $M$ such that $forall x in H$, $T_xM = T_xH oplus X_x$. Now I want to show that there exixts an open set $U$ such that $N subset U$, $U$ diffeo to $H times (-epsilon, epsilon)$.



The proof is supposed to go as follows:



Let $phi(x,t): M times (-a,a) to M$ be an integral flow of $X$, such that $phi(x,0) = id_M$, $phi(cdot,t)$ is a diffeo for each $t$, and $frac{partial}{partial t} phi = X$. Then by the inverse function theorem, $phi$ is a diffeo on a neighborhood of $U times V$ of each $(x,0)$.



How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?






differential-geometry differential-topology







share|cite|improve this question









$endgroup$




Suppose $H$ is a co-dimension 1 embedded submanifold of $M$. Let $X$ be a vector field on $M$ such that $forall x in H$, $T_xM = T_xH oplus X_x$. Now I want to show that there exixts an open set $U$ such that $N subset U$, $U$ diffeo to $H times (-epsilon, epsilon)$.



The proof is supposed to go as follows:



Let $phi(x,t): M times (-a,a) to M$ be an integral flow of $X$, such that $phi(x,0) = id_M$, $phi(cdot,t)$ is a diffeo for each $t$, and $frac{partial}{partial t} phi = X$. Then by the inverse function theorem, $phi$ is a diffeo on a neighborhood of $U times V$ of each $(x,0)$.



How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?






differential-geometry differential-topology




differential-geometry differential-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 0:05









KeithKeith

415316




415316












  • $begingroup$
    What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:09










  • $begingroup$
    Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:19










  • $begingroup$
    Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:20












  • $begingroup$
    I see, may I ask whether you have any idea why does a local diffeo solve the question?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:21










  • $begingroup$
    Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:22




















  • $begingroup$
    What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:09










  • $begingroup$
    Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:19










  • $begingroup$
    Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:20












  • $begingroup$
    I see, may I ask whether you have any idea why does a local diffeo solve the question?
    $endgroup$
    – Keith
    Dec 9 '18 at 0:21










  • $begingroup$
    Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 0:22


















$begingroup$
What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:09




$begingroup$
What is the derivative of $phi$ at $(x,0)$ for any $xin H$?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:09












$begingroup$
Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
$endgroup$
– Keith
Dec 9 '18 at 0:19




$begingroup$
Must it be decomposed into a vector tangent to $H$ and a vector in $X$?
$endgroup$
– Keith
Dec 9 '18 at 0:19












$begingroup$
Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:20






$begingroup$
Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:20














$begingroup$
I see, may I ask whether you have any idea why does a local diffeo solve the question?
$endgroup$
– Keith
Dec 9 '18 at 0:21




$begingroup$
I see, may I ask whether you have any idea why does a local diffeo solve the question?
$endgroup$
– Keith
Dec 9 '18 at 0:21












$begingroup$
Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:22






$begingroup$
Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 0:22












1 Answer
1






active

oldest

votes


















1












$begingroup$

Here's a sketch of the argument that when $H$ is compact, we can find $epsilon>0$ so that $phi$ is injective on $Xtimes (-epsilon,epsilon)$. Suppose not. Then for each $ninBbb N$ we have $x_n,x'_nin H$ and $|t_n|,|t'_n|<1/n$ so that $phi(x_n,t_n)=phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}to x$ and $x'_{n_k}to x'$. Then from $phi(x_{n_k},t_{n_k})=phi(x'_{n_k},t'_{n_k})$ we infer that $phi(x,0) = phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
    $endgroup$
    – Keith
    Dec 9 '18 at 16:37












  • $begingroup$
    This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 16:54










  • $begingroup$
    When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
    $endgroup$
    – Matt
    Dec 11 '18 at 8:35











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Here's a sketch of the argument that when $H$ is compact, we can find $epsilon>0$ so that $phi$ is injective on $Xtimes (-epsilon,epsilon)$. Suppose not. Then for each $ninBbb N$ we have $x_n,x'_nin H$ and $|t_n|,|t'_n|<1/n$ so that $phi(x_n,t_n)=phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}to x$ and $x'_{n_k}to x'$. Then from $phi(x_{n_k},t_{n_k})=phi(x'_{n_k},t'_{n_k})$ we infer that $phi(x,0) = phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
    $endgroup$
    – Keith
    Dec 9 '18 at 16:37












  • $begingroup$
    This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 16:54










  • $begingroup$
    When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
    $endgroup$
    – Matt
    Dec 11 '18 at 8:35
















1












$begingroup$

Here's a sketch of the argument that when $H$ is compact, we can find $epsilon>0$ so that $phi$ is injective on $Xtimes (-epsilon,epsilon)$. Suppose not. Then for each $ninBbb N$ we have $x_n,x'_nin H$ and $|t_n|,|t'_n|<1/n$ so that $phi(x_n,t_n)=phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}to x$ and $x'_{n_k}to x'$. Then from $phi(x_{n_k},t_{n_k})=phi(x'_{n_k},t'_{n_k})$ we infer that $phi(x,0) = phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
    $endgroup$
    – Keith
    Dec 9 '18 at 16:37












  • $begingroup$
    This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 16:54










  • $begingroup$
    When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
    $endgroup$
    – Matt
    Dec 11 '18 at 8:35














1












1








1





$begingroup$

Here's a sketch of the argument that when $H$ is compact, we can find $epsilon>0$ so that $phi$ is injective on $Xtimes (-epsilon,epsilon)$. Suppose not. Then for each $ninBbb N$ we have $x_n,x'_nin H$ and $|t_n|,|t'_n|<1/n$ so that $phi(x_n,t_n)=phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}to x$ and $x'_{n_k}to x'$. Then from $phi(x_{n_k},t_{n_k})=phi(x'_{n_k},t'_{n_k})$ we infer that $phi(x,0) = phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.






share|cite|improve this answer









$endgroup$



Here's a sketch of the argument that when $H$ is compact, we can find $epsilon>0$ so that $phi$ is injective on $Xtimes (-epsilon,epsilon)$. Suppose not. Then for each $ninBbb N$ we have $x_n,x'_nin H$ and $|t_n|,|t'_n|<1/n$ so that $phi(x_n,t_n)=phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}to x$ and $x'_{n_k}to x'$. Then from $phi(x_{n_k},t_{n_k})=phi(x'_{n_k},t'_{n_k})$ we infer that $phi(x,0) = phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 16:21









Ted ShifrinTed Shifrin

63.2k44489




63.2k44489












  • $begingroup$
    What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
    $endgroup$
    – Keith
    Dec 9 '18 at 16:37












  • $begingroup$
    This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 16:54










  • $begingroup$
    When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
    $endgroup$
    – Matt
    Dec 11 '18 at 8:35


















  • $begingroup$
    What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
    $endgroup$
    – Keith
    Dec 9 '18 at 16:37












  • $begingroup$
    This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
    $endgroup$
    – Ted Shifrin
    Dec 9 '18 at 16:54










  • $begingroup$
    When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
    $endgroup$
    – Matt
    Dec 11 '18 at 8:35
















$begingroup$
What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
$endgroup$
– Keith
Dec 9 '18 at 16:37






$begingroup$
What an elegant proof! Thank you so much. Just one more philosophical question. How do you come up with? Is there any similar construction in point-set topology?
$endgroup$
– Keith
Dec 9 '18 at 16:37














$begingroup$
This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 16:54




$begingroup$
This is a standard sort of proof technique. For example, this approach is the usual approach to prove that a continuous (say, real-valued) function on a compact metric space is uniformly continuous.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 16:54












$begingroup$
When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
$endgroup$
– Matt
Dec 11 '18 at 8:35




$begingroup$
When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…
$endgroup$
– Matt
Dec 11 '18 at 8:35


















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