1
$\begingroup$

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.

$\endgroup$
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, 2018 at 15:45

1 Answer 1

Reset to default
1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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
  • 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, 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.