Lie algebra action

Particular case of this (I think): Lie algebra representation.
Definition
The action of a Lie algebra $g$ on a manifold $M$ is a Lie algebra homomorphism

A : g \to X (M)

such that the map

g \times M \to T M

defined by

(v, p) \mapsto A (v)_{p} \in T_{p} M

is smooth.
$◼$

Example
Given the group action of a Lie group $G$ on $M$ it induces a Lie algebra action via fundamental vector fields. But I think that the converse is not true. I guess there must be problems of "completeness". Maybe this is related to the notion of pseudogroup or local group of transformations (are they the same?).

Infinitesimal action

A Lie algebra action is also called an infinitesimal action. Why?
Let's consider the Lie algebra action induced by a Lie group action of $G$ over a manifold $M$ . Given $V \in g$ we can identify it, loosely speaking, with

e + δ V \in G

if $δ$ is small. In the same way, if $X_{V}$ is the fundamental vector field on $M$ then for $p \in M$ we can imagine the point $q$ that results of adding to $p$ a bit of $X_{V} (p)$ , that is,

q = p + δ X_{V} (p) = p (e + δ V)

We can think of all of this as the action on $M$ of elements of $G$ that live very near to the identity element, that of course when applied to $p \in M$ yields elements of $M$ that live near $p$ .

Recovering of the group action

Surprisingly, the infinitesimal action let us recover the group action. Intuitively it goes in this way. Define

T_{δ} (p) := p (e + δ V)

We can apply the idea of "adding a bit of $X_{V}$ " successively, with smaller and smaller $δ$ s. This way

T_{\frac{1}{N}} (T_{\frac{1}{N}} (T_{\frac{1}{N}} (. . . T_{\frac{1}{N}} (p)))) = p \cdot (e + \frac{1}{N} V)^{N} \equiv p \cdot e x p (V)

This is strongly related to exponential map#Motivation.

I think that this can be seen in the particular case of a matrix Lie group/ matrix Lie algebra as relation between the determinant and the trace.

Example

Let's see an example

When you have a relatively complicated active coordinate transformation, for example

\begin{aligned} x & \mapsto c o s (θ) x + s i n (θ) y \\ y & \mapsto - s i n (θ) x + c o s (θ) y \end{aligned}

we can change $θ$ by a small $δ$ and take approximation to first order in $δ$

So you can replace $c o s (δ)$ with $1$ and $s i n (δ)$ with $δ$ , first order approximation. So we obtain

\begin{aligned} x & \mapsto x + δ y \\ y & \mapsto - δ x + y \end{aligned}

That can be rewritten as

(x, y) ⟼ (x, y) + δ \cdot (y, - x)

This is justified because

d τ_{e} (\partial_{δ}) = X_{\partial_{δ}} = (y, - x)

To understand this a bit more, let us restrict to $R^{2}$ but with a general transformation $F_{λ} : R^{2} \mapsto R^{2}$ , and being $λ \in R$ a parameter. It could be think as an action of the group $(R, +)$ over $R^{2}$ . So, for every $λ$ :

(x, y) ⟼ F_{λ} (x, y) = (F_{λ}^{1} (x, y), F_{λ}^{2} (x, y))

being $F_{0} = i d$ .
Since $R^{2}$ is an affine space we can take the vector $(F_{λ}^{1} (x, y), F_{λ}^{2} (x, y)) - (x, y)$ . As a first order approximation we would assume:

(\frac{F_{λ}^{1} (x, y) - x}{λ}, \frac{F_{λ}^{2} (x, y) - y}{λ}) ≅ (\frac{d F_{λ}^{1} (x, y)}{d λ} |_{λ = 0}, \frac{d F_{λ}^{2} (x, y)}{d λ} |_{λ = 0})

so the transformation, for a small $λ$ , could be considered as

(x, y) ⟼ (x, y) + λ \cdot (\frac{d F_{λ}^{1} (x, y)}{d λ} |_{λ = 0}, \frac{d F_{λ}^{2} (x, y)}{d λ} |_{λ = 0})

The approximation taken in \ref{aproximacion} could be interpreted like a Taylor expansion, since

\begin{aligned} x & \mapsto F_{λ}^{1} (x, y) = F_{0}^{1} + \deriv F_{λ}^{1} λ |_{λ = 0} \cdot λ + \dots ≅ x + \deriv F_{λ}^{1} λ |_{λ = 0} \cdot λ \\ y & \mapsto F_{λ}^{2} (x, y) = F_{0}^{2} + \deriv F_{λ}^{2} λ |_{λ = 0} \cdot λ + \dots ≅ y + \deriv F_{λ}^{2} λ |_{λ = 0} \cdot λ \end{aligned}

as most physics books do.

Several considerations:

If $F$ depends on more parameters, we can choose what direction are we moving in inside the "parameter space". It corresponds to being in an multidimensional Lie group $G$ and choosing an specific element of $g$ .
If we are not in an affine space we cannot define the vector at this way. But anyway, $λ \mapsto F_{λ}$ defines a curve, and we can take the tangent vector at $λ = 0$ as usual and proceed.
$X (M)$ is a Lie algebra (more or less). Its subalgebras determine the Lie groups that can act over $M$ .
The iterative process $(e + δ V)^{N} = e x p (t V)$ could be formal if $G$ is inside an $R^{M}$ , likewise $p + δ X_{V} (p)$ would have sense if $M$ was inmersed in $R^{K}$ . But the fact is that $G$ is always inside a matrix space, and I think that $(e + δ V)^{N} = e x p (t V)$ is formalized with the exponential of matrices.
If $M$ is a vector space ( $R^{n}$ ) then linear automorphisms of $M$ are included in $D i f f (M)$ . It turns out that infinitesimal generators of subgroups of $A u t (M)$ corresponds to elements of the Lie algebra $Γ (T M)$ (vector fields) with linear coefficients. Indeed, given $X$ a vector field in $M$ such that

X = \sum [\sum a_{i j}] x_{i} \partial x_{j}

we can identify $X$ with a matrix also named $X$ and

e x p (δ X) \cdot m = e^{δ X} \cdot m

(the LHS is the general exponential map and the RHS is the matrix exponential map).
For $d i m (M) = 1$ we recover the usual exponential map in $R$ .

Given a function $F : M \to R$ we have that it is invariant by a group $G \subseteq D i f f (M)$ (that is $F (g (x)) = F (x)$ ) if every of its infinitesimal generator seen as a vector field $X$ satisfies:

X (F) = 0