integral curve starting at a zero of a vector field












0












$begingroup$


This is a question from Loring Tu's book "Introduction to manifolds" (Page-161 14.6(b))



Show that if X is the zero vector field on a manifold M, and ct(p) is the maximal integral curve of X starting at p, then the one-parameter group of diffeomorphisms c:R->Diff(M) is the constant map c(t)=1M.



From the previous part of this question I know that if X is a smooth vector field on a manifold M that vanishes at a point p in M then the integral curve of X with initial point p is the constant curve c(t)=p.



I am stuck and really don't know how to proceed.Thanks.






differential-geometry manifolds smooth-manifolds diffeomorphism







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
    $endgroup$
    – Hans Lundmark
    Dec 13 '18 at 15:45
















0












$begingroup$


This is a question from Loring Tu's book "Introduction to manifolds" (Page-161 14.6(b))



Show that if X is the zero vector field on a manifold M, and ct(p) is the maximal integral curve of X starting at p, then the one-parameter group of diffeomorphisms c:R->Diff(M) is the constant map c(t)=1M.



From the previous part of this question I know that if X is a smooth vector field on a manifold M that vanishes at a point p in M then the integral curve of X with initial point p is the constant curve c(t)=p.



I am stuck and really don't know how to proceed.Thanks.






differential-geometry manifolds smooth-manifolds diffeomorphism







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
    $endgroup$
    – Hans Lundmark
    Dec 13 '18 at 15:45














0












0








0





$begingroup$


This is a question from Loring Tu's book "Introduction to manifolds" (Page-161 14.6(b))



Show that if X is the zero vector field on a manifold M, and ct(p) is the maximal integral curve of X starting at p, then the one-parameter group of diffeomorphisms c:R->Diff(M) is the constant map c(t)=1M.



From the previous part of this question I know that if X is a smooth vector field on a manifold M that vanishes at a point p in M then the integral curve of X with initial point p is the constant curve c(t)=p.



I am stuck and really don't know how to proceed.Thanks.






differential-geometry manifolds smooth-manifolds diffeomorphism







share|cite|improve this question









$endgroup$




This is a question from Loring Tu's book "Introduction to manifolds" (Page-161 14.6(b))



Show that if X is the zero vector field on a manifold M, and ct(p) is the maximal integral curve of X starting at p, then the one-parameter group of diffeomorphisms c:R->Diff(M) is the constant map c(t)=1M.



From the previous part of this question I know that if X is a smooth vector field on a manifold M that vanishes at a point p in M then the integral curve of X with initial point p is the constant curve c(t)=p.



I am stuck and really don't know how to proceed.Thanks.






differential-geometry manifolds smooth-manifolds diffeomorphism




differential-geometry manifolds smooth-manifolds diffeomorphism






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 15:40









RagingBullRagingBull

465214




465214








  • 1




    $begingroup$
    Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
    $endgroup$
    – Hans Lundmark
    Dec 13 '18 at 15:45














  • 1




    $begingroup$
    Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
    $endgroup$
    – Hans Lundmark
    Dec 13 '18 at 15:45








1




1




$begingroup$
Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
$endgroup$
– Hans Lundmark
Dec 13 '18 at 15:45




$begingroup$
Aren't you almost done already? Since $X$ vanishes at every point, every integral curve is just a constant curve. So no point moves, which is just what the identity map does.
$endgroup$
– Hans Lundmark
Dec 13 '18 at 15:45










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 15:52






  • 1




    $begingroup$
    You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
    $endgroup$
    – Gibbs
    Dec 13 '18 at 15:58












  • $begingroup$
    i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:01










  • $begingroup$
    Yes, precisely.
    $endgroup$
    – Gibbs
    Dec 13 '18 at 16:02










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:03












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 15:52






  • 1




    $begingroup$
    You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
    $endgroup$
    – Gibbs
    Dec 13 '18 at 15:58












  • $begingroup$
    i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:01










  • $begingroup$
    Yes, precisely.
    $endgroup$
    – Gibbs
    Dec 13 '18 at 16:02










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:03
















1












$begingroup$

You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 15:52






  • 1




    $begingroup$
    You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
    $endgroup$
    – Gibbs
    Dec 13 '18 at 15:58












  • $begingroup$
    i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:01










  • $begingroup$
    Yes, precisely.
    $endgroup$
    – Gibbs
    Dec 13 '18 at 16:02










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:03














1












1








1





$begingroup$

You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.






share|cite|improve this answer









$endgroup$



You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 15:46









GibbsGibbs

5,4383927




5,4383927












  • $begingroup$
    Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 15:52






  • 1




    $begingroup$
    You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
    $endgroup$
    – Gibbs
    Dec 13 '18 at 15:58












  • $begingroup$
    i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:01










  • $begingroup$
    Yes, precisely.
    $endgroup$
    – Gibbs
    Dec 13 '18 at 16:02










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:03


















  • $begingroup$
    Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 15:52






  • 1




    $begingroup$
    You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
    $endgroup$
    – Gibbs
    Dec 13 '18 at 15:58












  • $begingroup$
    i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:01










  • $begingroup$
    Yes, precisely.
    $endgroup$
    – Gibbs
    Dec 13 '18 at 16:02










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – RagingBull
    Dec 13 '18 at 16:03
















$begingroup$
Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
$endgroup$
– RagingBull
Dec 13 '18 at 15:52




$begingroup$
Actually all i know is that one-parameter group of diffeomorphisms is a homomorphism c:R->Diff(M). This is all that this book says about it. But i think you have related it to the integral cures. How to do that?
$endgroup$
– RagingBull
Dec 13 '18 at 15:52




1




1




$begingroup$
You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
$endgroup$
– Gibbs
Dec 13 '18 at 15:58






$begingroup$
You said that from the previous part of the question you know that if $X$ is a smooth vector field on $M$ that vanishes at $p$, then the integral curve of $X$ with initial point $p$ is the constant curve $c_p(t)=p$. Now your $X$ vanishes at each point of the manifold, so $c_p(t)=p$ for any point on $M$ and any $t$ (which gives you the maximal integral curve).
$endgroup$
– Gibbs
Dec 13 '18 at 15:58














$begingroup$
i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
$endgroup$
– RagingBull
Dec 13 '18 at 16:01




$begingroup$
i think i got it. c is actually a homomorphism which is induced by the integral curves, right?
$endgroup$
– RagingBull
Dec 13 '18 at 16:01












$begingroup$
Yes, precisely.
$endgroup$
– Gibbs
Dec 13 '18 at 16:02




$begingroup$
Yes, precisely.
$endgroup$
– Gibbs
Dec 13 '18 at 16:02












$begingroup$
Thanks a lot!!!
$endgroup$
– RagingBull
Dec 13 '18 at 16:03




$begingroup$
Thanks a lot!!!
$endgroup$
– RagingBull
Dec 13 '18 at 16:03


















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