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?

  • $\begingroup$ 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

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 $X=\frac{\partial}{\partial x}$.


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.


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.