Infinitesimal Generator of Local Group of Transformation

Question

I am stuck on concept of infinitesimal generator. I am reading Olver and i quote definitions from there

Given a local group of transformation G acting on Manifold M via $g.x= \Psi(g,x)$ for $(g,x)\in \mathbb{U} \subset G\times M$. Then for every $v\in \mathbb{g}$ (the lie algebra of G) we define a vector field on M $$\begin{equation} \psi(v)|_{x} = \frac{d}{d\epsilon}\Bigg{|}_{\epsilon=0} \Psi(exp(\epsilon v),x) \tag{1} \end{equation}$$ Then it goes on to say that we identify $\mathbb{g}$ with $\psi{(\mathbb{g})}$ and this forms a Lie algebra of vector fields on M. In this language, "we recover $\mathbb{g}$ from group transformations by the basic formula $$\begin{equation} v|_{x} = \frac{d}{d\epsilon}\Bigg{|}_{\epsilon=0}exp(\epsilon v)x \tag{2} \end{equation}$$ A vector field $v$ in $\mathbb{g}$ is called an infinitesimal generator of the group action G.

From what I understand, when it says that we recover $\mathbb{g}$ from group transformation is that we obtain that subset of vectorfields on M which is an image of lie algebra of G (determined by number of parameters specified in group transformation) under $\psi$. Is that correct ?

For example if the given transformation is $(x,y) \mapsto (x+c\epsilon, y + \epsilon) $ then the underlying group is $R$. Eq (2) will give me $c\partial_x + \partial_y$ if i consider $exp(\epsilon v)(x,y)= (x+c \epsilon,y+\epsilon)$ with $v=\textbf{1}.\frac{d}{d\epsilon}$

i. This is the image of vector field "$\textbf{1}.\frac{d}{d\epsilon}$" on $R$ under $\psi$ by our first equation. So for a 1-d Group action can one say that image of the vector field "$\textbf{1}.\frac{d}{d\epsilon}$" is called infinitesimal generator.

ii. What is the infinitesimal generator of $(x,y)\mapsto(\epsilon x, \epsilon^3 y)$? As per equation (2) i get $\partial_x$ but i think that is not the case.

iii. how do we solve for infinitesimal generator if tranformation group has two parameters? say $(x,y)\mapsto(x+\mu + \epsilon, y + \epsilon)$. Here should i look for the image of the vector fields "$\textbf{1}.\frac{d}{d\epsilon}$" and "$\textbf{1}.\frac{d}{d\mu}$" ?

Answer 1

Ok, first a little bit of geometry.

Let $M,N$ be smooth manifolds, and let $\Phi: M\times N\to N$ be a smooth mapping. Suppose for some $p\in M$, we have that $\Phi(p,x) = x$ for every $x\in N$.

Since $\Phi$ is smooth, $\mathrm{d}\Phi$, evaluated at the point $(p,x)$, pushes forward the tangent space $T_{(p,x)} (M\times N)$ to $T_x N$. So for any vector $v\in T_p M$, and for any $x\in N$, this gives a vector in $T_x N$. Since $x$ is arbitrary, you've obtained a vector field.

In the case $M$ is a Lie group, and $\Phi$ is a group action, then the identity element $e\in M$ by definition has the property that $\Phi(e,x) = x$. So $\mathrm{d}\Phi$ induces a mapping from the tangent space $T_e M$ to sections of $TN$ (ie. vector fields on $N$).

So far nothing about groups is used except that it has an identity and that group actions must require $\Phi(e,x) = x$.

Lastly observe that $T_e M$ for $M$ being a Lie group is precisely its Lie algebra.

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Next, there's something strange about what you quoted. Does Olver assume that group actions are faithful? Otherwise, the trivial action $\Phi(p,x) = x$ for all $p$ will not allow you to identify the lie algebra $\mathbb{g}$ with the algebra of the induced vector fields (which in this case is just the $0$ vector field). But this is a minor aside.

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In terms of recovery: what Olver is saying is simply that "Lie algebras and Lie groups come in pairs" (up to some technical constraints. This is known modernly as Lie's third theorem.) So: if you know the (faithful) group action, you can compute the Lie algebra. If you know the Lie algebra, you can compute the group action.

The latter is given by the $\exp$ map you obtain by integrating along the vector field. The former you get by differentiating the group action along the group parameter. (This sounds very much like the fundamental theorem of calculus, and for good reasons.)

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All these prelude is to say that: yes, you are right in the first question.

For the third question the generators will be the constant vector fields $(1,1)$ and $(1,0)$, which are the images as you specified.

Answer 2

There may be no substitute for direct computation. Just do it.

ii. Confirm that, for $a=\ln \epsilon$, so the identity is at a=0, $$ e^{a(x\partial_x +3y \partial_y) } ~ (x,y)= (\epsilon x, \epsilon^3 y)~, $$ the standard double dilation of numerous physical systems. (You appear to have mis-identified the expansion parameter.)

iii. $$ e^{(\mu+\epsilon)\partial_x + \epsilon \partial_y} ~(x,y)= (x+\mu +\epsilon , y+\epsilon). $$ In this case, $\nu\equiv \mu+\epsilon$ may be a less confusing parameter.