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well In warner book, page 36. A curve $\gamma:(a,b)\rightarrow M$ is integral curve iff

$$d\gamma(\frac{d}{dr}|_t)=X(\gamma(t))$$ Could anyone explain me about the left side in a detail and breaking into the coordinates?well, $X$ is a vector field on $M$.

I curve $\sigma$ is an integral in $M$ if $\dot{\sigma(t)}=X(\sigma(t)) \forall t \in \text{domain of $\sigma$ }$

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If $\gamma: N\to M$ is a smooth map between manifolds, $d\gamma: TN\to TM$ is a map between tangent spaces defined by $$ d\gamma(X_x)h := X_x(h\circ \gamma) $$ where $X_x$ is an element of the tangent space of $N$ at $x$ and $h$ is a smooth real function on $M$. Let's call $d\gamma$ the (total) tangent map. In your case $N$ is $(a, b)$, and $\gamma$ is an integral curve iff the induced tangent map sends the element of the tangent space of $(a, b)$ at $t$, $\left. \frac d {dr} \right|_t$, into the element of tangent space of $M$ at $\gamma(t)$, $X(\gamma(t))$, or in others words iff $$ \left. \frac d {dr} \right|_t h\circ\gamma = X(\gamma(t)) h $$ for each function $h$ on $M$.

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thank you very much.... –  El Angel Exterminador Jun 27 '12 at 13:08
    
I'd add that $d\gamma(\frac{d}{dr}|_t) = \gamma'(t)$, the velocity vector of $\gamma$ at $t$. –  Neal Jun 27 '12 at 14:23

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