This question was asked in my quiz on smooth manifolds and I was unable to solve it. I tried it again at home and still not able to solve it. Question is:
Let $\sigma$ be an integral curve for a vector field X on a manifold M, with $\sigma(0)=p$. If $X_p\neq 0$, show that there exists $\epsilon >0$ such that $\sigma : (-\epsilon , \epsilon) \to M $ is an immersion.
What I could prove is that: If $p\in M$ then there exists $\epsilon >0$ and an integral curve of X such that $\sigma : (-\epsilon ,\epsilon)\to M$ with $\sigma(0)=p$.
But How can I prove that a $\sigma$ also exists which is an immersion.
Can you please tell?
First of all one has to be clear what does it means integral curve, so it would be useful to remember this concept:
An integral curve of starting point $p\in M$ of a vector field $X\colon M\to TM$ is a curve $ \sigma \colon (-\epsilon, \epsilon)\to M$ such that
$\sigma(0)=p$
$\sigma’(t)=X(\sigma(t))$
The first important theorem is the following one:
There exists always a unique integral curve of starting point $p$ related to a vector field $X$
The proof of the theorem is quite easy, because in a coordinate chart $(U,\phi)$, the condition became
$(\sigma\circ \phi)^{-1}(0)=0$
$(\sigma\circ\phi^{-1})’(t)= \sum_i d\phi_{i,\sigma(t)}(X(\sigma(t))\frac{\partial}{\partial \phi_{i,\sigma(t)}}$
In other words
$(\sigma\circ \phi)^{-1}(0)=0$
$\frac{d}{dt}(\sigma\circ\phi^{-1})_i= d\phi_{i,\sigma(t)}(X(\sigma(t))$ for each $i=1,\dots n$
This is a Cauchy problem in $t,y_1=\sigma\circ\phi^{-1}_1,\cdots ,y_n= \sigma\circ \phi^{-1}_n$ variables and so by locally existence and uniqueness Cauchy-Lipschitz theorem (or equivalently Picard-Lindelöf theorem) there exists a unique solution $\gamma(t)\colon (-\epsilon, \epsilon)\to \phi(U)$ for an $\epsilon$ enough small.
Then our integral curve will be $\sigma:=\phi^{-1} \circ \gamma $ and we have done.
But now you have an important condition to your problem:
$X_p\neq 0$
So by continuity (X is a smooth map) of $X$ there exists an appropriate neighbourhood of $p$ such that $X_q\neq 0$. If you intersect your integral curve $\sigma$ with that neighbourhood then you will get $d\sigma_t(d/dt)=\sigma’(t)=X(\sigma(t))\neq 0$
But $d/dt$ is a basis of $T_t(-\epsilon,\epsilon)$ and so the differential of $\sigma$ in $t$, $d\sigma_t$, has to be injective, that is the definition of a smooth map to be an immersion.
Thus we have done. If there is some mistake or error please tell me and we will discuss