In DoCarmo’s Riemannian Geometry book, a smooth vector field $X$ on a Lie group $G$ is called left invariant if $d(L_x)_yX=X$ for all $x,y\in G$.
I want to show that such a vector field $X$ is completely determined by its value at a single point, say $m$. I go as follows:
For any element $g\in G$, we have $X(g)=d(L_{gm^{-1}})_mX(m)$. Is it exactly what the author means?
Best Answer
Yes this is indeed exactly what is meant!
Perhaps unnecessary, but let me say what happens here in a more word-y, less notation-y way. The 'left multiplication'-action of $G$ on itself maps points of $G$ to points of $G$. 'Hence' it also maps tangent vectors to tangent vectors. As my physicist friends like to say: 'just like how a potato field is at every point a potato a vector field is at every point [of $G$] a vector'. So from the above it is obvious that the left action also maps vector fields to vector fields: viewing $X$ as a collection of vectors, the image of $X$ under a particular left-multiplication map $L_x$ consists just of all the images of all these vectors.
Being left-invariant means that as a collection of vectors, $X$ is mapped to itself under each $L_x$. Sure, individual vectors are mapped all over the place, but the total collection stays the same. In particular, if some $L_x$ maps $m$ to $g$ then the value of $X$ at $g$ is determined by the value of $X$ at $m$ since the vector in the tangent space at $g$ that $X(m)$ is mapped to, must belong to $X$ and $X$ only has one vector in that space. Very nice. Turning our heads 180 degrees we conclude that the last sentence means that $X(m)$ determines the value $X(g)$ for all $g$ for which there is some $x$ such that $L_xm = g$.
What then remains to be shown is that for each $g$ there exists an $x$ such that $L_xm = g$ and you achieved that marvelously by setting $x = gm^{-1}$.
EDIT: Now we also see what the more general case looks like. Instead of the smooth manifold being the group itself we might take it to be any manifold $M$ on which $G$ acts smoothly by a left-action $L$. In this case one would speak of $L$-invariant rather than left-invariant (except in sentences where left is used as a verb, like '$X$ is left invariant by $L$').
Anyway, the point is that if and only if the action of $G$ on $M$ is transitive, the same property holds, with the same proof. When the action is not necessarily transitive then the value of $X$ on each $G$-orbit is determined by the value at a single point (in that orbit).