What is an example and a non-example of a left-invariant vector field? I am going through a few different books reviewing Lie Groups and Lie Algebras and came across the notion of a left-invariant vector field. Given a diffeomorphism $L_ag : G \to G, L_a(g) = ag.$ We then have a map from $T_gG$ to $T_{ag}G$ such that $L_{a*} X |_g = X|_{ag}.$ Additionally, we have in coordinates that $$X^u(g) \frac{\partial x^v(ag)}{\partial x^u(g)}\frac{\partial}{\partial x^v} |_{ag} = X^v(ag)\frac{\partial}{\partial x^v} |_{ag}.$$ However I have two issues. The fist is this definition seems so intuitively obvious I cannot think of a non-example (Note: I've checked previous Stackexchange questions on this matter and the solutions amounted to short comments that didn't fully answer my question) and the second is I am a bit shaky as to how use the coordinate definition to check if a vector field is or isn't left invariant.
Therefore, I am hoping someone could provide two examples. The first being an example of a left invariant vector field and the second a non-example. If you could show why these examples satisfy and fail to satisfy the requirements respectively I would be greatly apppreciative!
 A: Here are two fundamental examples.
Take the vector field $X=\frac{d}{d\theta}$ on $S^1$, which we think of as laying in the complex plane $\Bbb{C}\cong \Bbb{R}^2$. Viewed geometrically, this is the vector field which sends $(x,y)\in S^1$ to the tangent vector $(y,-x)\in T_{(x,y)}S^1.$ You should draw a picture of this and it should become clear why it is left invariant. It is because $e^{i\theta}\in S^1$ acts by rotation by $\theta$ radians counterclockwise for $\theta \in [0,2\pi)$ (say), and it is a good exercise to check that the rotation by $\psi$, $R_\psi$, acts on $X_\theta$ by $R_\psi(X_\theta)=X_{\theta+\psi}.$
It is easy to construct non-invariant vector fields on $\Bbb{R}^2$. Note that left-invariance is equivalent to translation invariance under all $v\in \Bbb{R}^2$. Pictorally, this is a vector field that does not "vary" from point to point, like $X:=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$ or really any vector field of the form $Y=a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}$. To construct non-invariant vector fields, just introduce non-constant $C^\infty$ coefficient functions like $Z=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$ for instance. At least for this type of example it is easy to see non-invariance. Left multiplication with $v\in \Bbb{R}^2$ is $\mu_v(x,y)=(x,y)+v$ and taking the differential of this affine transformation gives $(\mu_v)_*=\text{Id}.$ So, a general vector field
$$
X_{(x,y)}=f(x,y)\partial_x+g(x,y)\partial_y
$$  is invariant if and only if for all $v$, $(\mu_v)_*X_{(x,y)}=X_{(x,y)+v}$. Writing out the equations and comparing coefficients of $\partial_x$ and $\partial_y$ we see that this is in turn equivalent to $f(x,y)$ and $g(x,y)$ being translation invariant for all $v$. I.e. for all $v$ $f((x,y)+v)=f(x,y)$. This is equivalent to $f$ and $g$ being constant and hence equivalent to $X=a\partial_x+b\partial_y$.
This idea generalizes to $\Bbb{R}^n$ quite easily.
A: If you take the case that $G=(\mathbb R^n,+)$ then vector fields on $G$ are smooth functions $G\to\mathbb R^n$. Since the derivative of a translation is the identity, the conditions of left invariance reads as $f(x+y)=f(x)$ for any $x,y\in\mathbb R^n$. Thus left invariant vector fields correspond to constant functions.
