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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?

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  • $\begingroup$ I really can't make much sense of your terminology. Which might be due to the fact that I'm a physicist, NO mathematician by education. I've always thought that $\mathbb{R}$ are the reals, while you say it's a Lie group. What does $S^1$ mean? Is $X = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$ not just an operator, instead of an "1-D vector space"? It might be that your terminology comes from the nowadays standard books on the subject. In that case I'm giving up. $\endgroup$ – Han de Bruijn Feb 10 '17 at 16:20
  • $\begingroup$ @HandeBruijn $S^1 = \{ (x,y) \in \mathbb{R^2} : x^2 + y^2 = 1 \}$. For more reference - en.wikipedia.org/wiki/N-sphere . Yes, $X$ is an operator which generates vector space. $\endgroup$ – Dark_Knight Feb 12 '17 at 6:03
  • $\begingroup$ Existing Mathematics Stack Exchange answers + chain references are in here, starting with : Advanced beginners textbook on Lie theory from a geometric viewpoint . $\endgroup$ – Han de Bruijn Feb 28 '17 at 12:54
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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 $X=\frac{\partial}{\partial x}$.

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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:

  1. is additive;
  2. satisfies the Leibnitz rule; and
  3. 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.

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