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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.

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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. $\endgroup$ Dec 13, 2018 at 15:45

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

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  • $\begingroup$ 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, 2018 at 15:52
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    $\begingroup$ 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, 2018 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, 2018 at 16:01
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Gibbs
    Dec 13, 2018 at 16:02
  • $\begingroup$ Thanks a lot!!! $\endgroup$
    – RagingBull
    Dec 13, 2018 at 16:03

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