integral curve starting at a zero of a vector field
$begingroup$
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
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
$endgroup$
add a comment |
$begingroup$
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
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
$endgroup$
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
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
asked Dec 13 '18 at 15:40
RagingBullRagingBull
465214
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
$endgroup$
You are basically done, because $c_p(t) = p = mathrm{id}(p)$, so $c(t)$ is the identity in $textrm{Diff}(M)$.
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
answered Dec 13 '18 at 15:46
GibbsGibbs
5,4383927
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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