Small question regarding left invariant vectorfields Usually, one writes out the left invariance of vector fields $X$ on a Lie group $G$ as $$(L_x)_*X=X$$for every $x$. 
However, I have trouble understanding this equality, since both sides does not map to the same domain. Evaluating the right hand side in $y$ gives an element in $T_yG$. Evaluating the left hand side in $y$ yields an element in $T_{xy}G$. Is there some canonical identification between $T_{xy}G$ and $T_yG$ going on, or did I make a mistake somewhere? Or, are we actually evaluating the right hand side in $xy$ instead of $y$?
Thank you. My apologies if this is duplicate. 
I already found this one: Definition of a left-invariant vector field, 
but I do not fully understand the answer. 
 A: Your instinct to keep track of the base point of a vector is a good one. Doing so here sheds a little light on the situation:
Each map $L_x : G \to G$, $x \in G$ is a diffeomorphism, and for any vector field $X$, $(L_x)_* X$ conveniently denotes the vector field whose value at $y \in G$ is $$((L_x)_* X)_y = ((L_x)_*)_{x^{-1} y} X_{x^{-1} y} .$$ To say that $X$ is left-invariant is just to say that this coincides with $X_y$ for all $x$. (This notation is obviously a little unwieldy, and the lower-star notation for the pushforward discourages writing down the base point explicitly, so I would prefer to write this as $$T_{x^{-1} y} L_x \cdot X_{x^{-1} y} ,$$ where $T_{x^{-1} y} L_x : T_{x^{-1} y} G \to T_y G$ is the differential of $L_x$ at $x^{-1} y$.) In practice we often take $y = x$ and so write the left-invariance condition as $$L_x \cdot X_e = X_x$$ (for all $x \in G$). (I encourage you to convince yourself that the two given conditions really are equivalent.)
More generally, we can talk about the pushforward of a vector field $Y \in \Gamma(TM)$ along a diffeomorphism $\phi : M \to N$, namely, $T\phi \cdot Y$, where
$$(T \phi \cdot Y)_q = T_{\phi^{-1}(q)} \phi\cdot Y_{\phi^{-1}(q)} .$$
One cannot, however, push forward a vector field by an arbitrary smooth map.
A: The derivative of the left multiplication map, $L_g(x)=gx$ gives an isomorphism of the tangent spaces.  See this.
A: Let $X$ be a vector field on $G$. For each $g\in G,$ the map $g\mapsto xg$ is a diffeomorphism of $G$ so the pushforward $(L_x)_*:T_gG\to T_{xg}G$ is well-defined.  
Now, $(L_x)_*$ maps vectors in $T_gG$ to vectors in $T_{xg}G$. And so if $X$ is a vector field on $G$, then $(L_x)_*X_g$ is $some$ vector $v_{xg}\in T_{xg}G$. 
$\textit{If}\ $  it turns out that $v_{xg}=X_{xg}$ for every $g\in G,$  we say that $X$ is left-invariant. Of course, this does not say that there are any such fields, but this is another matter. 
