# Infinitesimal Generator of Local Group of Transformation

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 $$$$\psi(v)|_{x} = \frac{d}{d\epsilon}\Bigg{|}_{\epsilon=0} \Psi(exp(\epsilon v),x) \tag{1}$$$$ Then it goes on to say that we identify $\mathbb{g}$ with $\psi{(\mathbb{g})}$ and this forms a Lie algebra of vector fields on M. In this language, "we recover $\mathbb{g}$ from group transformations by the basic formula $$$$v|_{x} = \frac{d}{d\epsilon}\Bigg{|}_{\epsilon=0}exp(\epsilon v)x \tag{2}$$$$ 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 "$\textbf{1}.\frac{d}{d\mu}$" ?

• For question 2: look at the definition of a group "acting" on a manifold. In particular the identity element of $R$ should act trivially on your set. The identity element is $0$. But your specified a "action" which is emphatically not the identity morphism (it sends all of ${R}^2$ to $(0,0)$). So you don't have a group action to start with. So you can't speak of the infinitesimal generators. – Willie Wong Nov 12 '15 at 15:14
• Incidentally, how much differential geometry do you know? If I use the word "pushforward" in my answer, would you be able to understand what I am talking about? – Willie Wong Nov 12 '15 at 15:16
• yes, i think i should be able to understand that. – Himanshu Nov 12 '15 at 15:21
• For the second question, the book mentions that for group action $(x,y)\mapsto (\lambda x,\lambda^{\alpha}y)$ where $\lambda > 0$ and $\alpha$ constant the infinitesimal generator is $x\partial_{x} + \alpha y \partial_{y}$. And i realize that the group transformation is for $(R,*)$ so identity element is 1. this works. thanks. – Himanshu Nov 12 '15 at 15:28
• need ur comment on first question. i am making some conceptual error there. – Himanshu Nov 12 '15 at 15:32

## 2 Answers

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.

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.

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.)

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.

• if by faithful you mean $\Phi(e,x)=x$ then yes, that is assumed as a part of definition of Local group of transformations. The third lie's theorem :Any finite-dimensional Lie algebra over $R$ is the Lie algebra of some analytic Lie group. Are we here talking of more general Lie algebra over $R$ than left invariant vector fields which are just 1-d? Welll olver has a theorem without proof saying " if we know the infinitesimal generators, which form a basis for a Lie Algebra, then we can find a local group of transformation whose Lie algebra coincides." Is this a corollary to Lie's third theorem? – Himanshu Nov 12 '15 at 16:52
• Faithful: $\Phi(a,x) = \Phi(b,x)$ for all $x$ if and only if $a = b$. Not $\Phi(e,x) =x$. // For your second follow-up: more or less. I don't have Olver's book in front of me so all I wrote are based on your quotes of him. – Willie Wong Nov 12 '15 at 17:28
• Another thing I should add is that the general nonsense above assumes that you have a group action, not a local group action. Most of the discussion survives in the local case, since we are only really interested in what happens in a neighbourhood of $e$. – Willie Wong Nov 12 '15 at 17:31
• i checked and i didn't find the assumption of faithfulness. in the example you gave $\Phi(p,x)=x$ for all $p$. The image of $\mathbb{g}$ as you said is 0 vector field. I understand that this is not an identification as a injective homomorphism but nevertheless 0 vector field is a sub-lie algebra of vector fields over M. so i think faithful constraint isn't that crucial ? or there's more that can go wrong ? – Himanshu Nov 13 '15 at 8:22
• Given two groups $A,B$. Generally we don't try to identify them as the same when there is only a homomorphism $A\to B$; generally we try to only identify them as the same if there is an isomorphism. – Willie Wong Nov 13 '15 at 16:05

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.