How to compute one-parameter group and corresponding vector fields I have two related questions to ask - 
$1)$ Let $\rho : \mathbb{R} \rightarrow G$ be a one-parameter group. ($\mathbb{R}$ and $G$ are Lie groups). If we take $G = S^1$ then the left invariant vector fields form a $1-$D vector space generated by $X = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$  . 
Now the image of $\frac{d}{dt}$ under the map $t \mapsto (\cos at , \sin at)$ is compued to be $aX$. Hence it is the one-parameter group of the vector field $aX$. Can someone please help me in understanding how did we landed on $aX$ here? In general, how do we compute one-parameter groups?
I have taken the fist example from the book - Global Calculus by S.Ramanan. It's given in Remark $3.16$. And second example from same book  - Example $3.1$ 
$2)$ Given a one-parameter group of diffeomorphims, how do we compute the vector fields associated with it? Say for example, $\phi_t(x) = x+t , (t \in \mathbb{R})$ be a one-parameter group of diffeomorphims. What is the vector field associated with it?
 A: The two question are very much related; I hope this goes some way to help you. I have been suitably vague at points to hopefully get the ideas across but happy to try and answer any further concerns (to the best of my ability!).
1) Given a vector field $X$, the question arises: does this give rise to a family of curves? That is, can we find a family of curves whose tangent vectors are precisely $X$.
Let $X=\xi^{1}(x,y)\frac{\partial}{\partial x} + \xi^{2}(x,y)\frac{\partial}{\partial y}$ for some smooth functions $\xi^{i}(x,y)$ and define a curve on $\mathbb{R}^{2}$ by:
$$\begin{align}
      \gamma:[a,b] &\longrightarrow \mathbb{R}^{2} \\
               t   &\longmapsto (x,y) = (\gamma^{1}(t), \gamma^{2}(t)).
\end{align}$$
The vector field along $\gamma$ is then
$$V^{\gamma} = \gamma_{*}\frac{d}{dt} = \frac{d\gamma^{1}}{dt}\frac{\partial}{\partial x}+ \frac{d\gamma^{2}}{dt}\frac{\partial}{\partial y} $$
where $\gamma_{*}$ denotes the push-forward or tangent map. Thus $X$ is a tangent vector to the curve if the components of $X$ and $V^{\gamma}$ coincide on $\gamma$: 
$$\frac{d\gamma^{1}}{dt} = \xi^{1}(\gamma(t)), \quad \frac{d\gamma^{2}}{dt} = \xi^{2}(\gamma(t)).$$
From the theory of ordinary differential equations, solutions to this system always exist and are uniquely determined by initial conditions: $\gamma^{1}(0)=x_{0}$ and $\gamma^{2}(0)=y_{0}$. The curve is then said to start at the point $p=(x_{0},y_{0})$. With the map $\gamma$ starting at $p=\gamma(0)$ defined (the integral curves of the vector field $X$), we may then define
$$\begin{align}
      \varphi_{t}:\mathbb{R}^{2} &\longrightarrow \mathbb{R}^{2} \\
                     p &\longmapsto \varphi_{t}(p) = \gamma(t)
  \end{align}$$
which can be shown to be a local one-parameter family of diffeomorphisms with the group structure you are asking for. 
If you try this with your vector field $aX=ax\frac{\partial}{\partial y} - ay\frac{\partial}{\partial x}$, you get precisely the map $\rho:t\mapsto (\cos(at),\sin(at))$ provided the initial conditions $\rho^{1}(0)=1, \rho^{2}(0)=0$ are used. This then leads to a one-parameter family of diffeomorphisms $\varphi_{t}$ (a one-parameter group) as outlined above.
2) The second question is now the converse of the first: given a one-parameter family of diffeomorphisms, how do you find its associated vector field? 
In your example, you are working purely on $\mathbb{R}$, so let us stay there (hopefully extensions to higher dimensions should be obvious). 
We have a one-parameter family of diffeomorphisms:
$$\begin{align}
      \varphi_{t}: \mathbb{R} &\longrightarrow \mathbb{R} \\
                  x  &\longmapsto y = \varphi_{t}(x) 
  \end{align}$$
This induces a curve $\varphi_{x}$ on $\mathbb{R}$:
$$\begin{align}
      \varphi_{x}: [0,a] &\longrightarrow \mathbb{R} \\
                  t  &\longmapsto y = \varphi_{x}(t) = \varphi_{t}(x) 
  \end{align}$$
which starts at $x$ (i.e. $\varphi_{x}(0)=\varphi_{0}(x)=x$). Thus, we may define a curve starting at every $p\in\mathbb{R}$. Given the set of all such curves we may then define a tangent vector at each point in $\mathbb{R}$.  The tangent vector to $\varphi_{x}$ is simply
$$X = \varphi_{x*}\frac{d}{dt} = \left.\frac{d\varphi_{x}}{dt}\frac{\partial}{\partial x}\right|_{t=0}$$
Notice that the result is restricted to $t=0$ as this is where the curve is defined to start. 
Using this method with $\varphi_{x}(t)=x+t$ yields the vector $X=\frac{\partial}{\partial x}$. 
A: Let me respond to your second question first (since it is easier).
Let $\xi_a$ be a one-parameter family of diffeomorphisms (so, for each
value of the real parameter $a$, $\xi_a$ is a diffeomorphism), say, on
some manifold $M$.  Let $f:M \to \mathbb{R}$  be any smooth function on $M$.
Then consider $f$ (composed with) $\xi_a$.  For each $a$, this is also
a smooth function on $M$.  So (since $a$ can be varied) it is a
one-paramter family of smooth functions.  Call it $f_a$.  Now
consider $d f_a/da|$ (at $a = 0$).  This, again, is a smooth function on
$M$.  We thus acquire a mapping from smooth functions to smooth
functions.  One checks that it:


*

*is additive;

*satisfies the
Leibnitz rule; and

*annihilates the constant functions. 


Hence, there is some tangent vector field, $\psi$, on $M$ such that that
$\psi(f)$ is this function.  This is the vector field associated with
the family $\xi_a$ of diffeomorphisms.
As for the first question, I'm not quite sure what you are asking.
We have the Lie groups $\mathbb{R}$ (additive reals) and $G$ (circle group); and
we have this mapping $\rho$ from $\mathbb{R}$ to $G$ ("wrap the line around the
circle"). Further, the $X$ that you give is indeed a left-invariant
vector field on $G$; and every left-invariant vector field is a
constant multiple of this $X$.  We also have the vector field $d/dt$
on $\mathbb{R}$; and its image, under $\rho$, is indeed $X$.  Now (I believe$\ldots$) you want to consider a new map $\mathbb{R} \to G$.  Fix a number $a$, and let the map send $t$ in $\mathbb{R}$ to $(\cos at, \sin at)$ in $G$.  Now, you want to take the image of this same old vector field, $d/dt$, under this
new map.  This image is indeed the vector field $aX$ on $G$.  There
are a number of ways to see this.  One (which you may not find
very satisfying) is to invoke the following fact.  For $\gamma(t)$
a curve on a manifold $M$ (so $t$ in reals; $\gamma(t)$ in $M$), then
the tangent vector to the curve $\gamma(at)$ (where $a$ is a number)
is precisely $a$ times the tangent vector to the original curve.
Probably, this is not what you are asking for$\ldots$
Let me know if you have any further questions.
