topsquark said:
Okay, I did finally find this and the d is a differential operator.
Here's my problem in detail. Let G be a Lie group. Define [math]L_a(g): G \times G: g \mapsto a \cdot g,~\forall g \in G, \text{ for fixed } a \in G[/math].
Let G be embedded in a smooth manifold M. Then G inherits the differential structure of M. So we have a map [math]dL_a: G \to TG[/math] using this differential structure. In general [math]g \in G[/math] does not lie in TG (Unless perhaps we are taking the tangent space over the identity in G. Not sure on this point.)
Now, a vector field X over a Lie group G is left invariant if
[math](dL_g)X(x) = X(L_g(x)) = X(g \cdot x)[/math]
I would expect a relationship like [math](dL_g)X(x) = X(g \cdot dx)[/math]. Why is there no differential operation on the RHS of the left invariant definition?
-Dan
Hi Dan,
I'm having some issues with the contents of your question. Why do you embed $G$ in a smooth manifold $M$? The group $G$ is itself a smooth manifold, so it also has a differentiable structure. Also, you're not viewing $d$ correctly. This $d$ is the exterior derivative. Let $f : M \to N$ be a differentiable map between smooth manifolds $M$ and $N$. Then for each $x\in M$, the differential of $f$ at $x$ is a map $df_x : T_x M \to T_{f(x)} N$ such that
$$df_x(\xi):= \frac{d}{dt}|_{t = 0} f(c(t))$$
whenever $\xi \in T_xM$ and $c : [0,1] \to M$ is a curve with $c(0) = x$ and $\dot{c}(0) = \xi$. The total differential of $f$ is a map $df : TM \to TN$ such that $df(x,\xi) := (f(x), df_x(\xi))$ for all $x\in M$ and for all $\xi\in T_xM$.
Since $G$ is a Lie group, the left translation maps $L_g : G \to G$ are diffeomorphisms. So given a vector field $X$ on $G$, the expressions $dL_g X$ makes sense as a vector fields on $G$. The field $X$ is left-invariant if $dL_g X = X$ for all $g\in G$. Let's make sense of this. Let $h\in G$. Then $X_h$ is a vector tangent to $h$ in $G$. Thus for fixed $g$, $dL_g(X_h)$ is a vector tangent to $L_g(h) = gh$ in $G$. So for the equation $dL_g X = X$ to make sense, we must have $dL_g(X_h) = X_{gh}$ for all $h\in G$. Another way of stating this is that $X$ is $L_g$-related to itself.
It's important to note that if $X$ is left-invariant on $G$, then $X$ is completely determined by $X_e$, where $e$ denotes the identity of $G$. For given $g\in G$,
$$ X_g = X_{ge} = dL_g(X_e).$$
Conversely, given a vector $Y_e\in T_eG$, we can form a left-invariant vector field $X$ on $G$ by setting $X_g := dL_g(Y_e)$ for all $g\in G$. By the chain rule, we have
$$ dL_g(X_h) = dL_g(dL_h(Y_e)) = d(L_g \circ L_h)(Y_e) = dL_{gh}(Y_e) = X_{gh}$$
for all $g, h\in G$. So indeed, $X$ is left-invariant. This gives a correspondence between the tangent space to the identity at $G$, $T_eG$, and the space of left-invariant vector fields on $G$, $L(G)$. This correspondence is actually an isomorphism of vector spaces.
Try as an exercise to find all left-invariant vector fields on $\Bbb R$.