Lie algebra of $\Bbb{R}^n$ I don't really understand the Lie algebra of the Euclidean space (the bold part):
Left translation by an element $b\in\Bbb{R}^n$ is given by the affine map $L_b(x)=b+x$, whose pushforward $(L_b)_*$is represented by the identity matrix in standard coordinates. Thus a vector field $V^i \partial /\partial x^i$ is left invariant iff its coefficients $V^i$ are constants.
So $x\partial/\partial x+y\partial/\partial y$ would be a counterexample but I don't see why this is not left invariant. Clearly $(L_b)_*(x\partial/\partial x+y\partial /\partial y)(x_0)=(x\partial /\partial x+y\partial /\partial y)(b+x_0)$ for any $b$, right?
 A: A vector field $X$ is left-invariant in $\mathbb R^2$ if
$$(L_v)_*X_w=X_{w+v},$$
for all $v$ and $w\in\mathbb R^2$. I think you forgot $w$ in your consideration.
If
$$X=X^1\frac{\partial}{\partial x} + X^2\frac{\partial}{\partial y}$$
then, for $v$, $w\in\mathbb R^2$ and any smooth function $f\colon\mathbb R^2\to\mathbb R$,
$$\begin{array}{rcl}
(L_v)_*X_w(f) & = & X_w(f\circ L_v) \\
& = & X^1(w)\frac{\partial L_v\circ f}{\partial x}(w) + X^2(w)\frac{\partial L_v\circ f}{\partial y}(w) \\
& = & X^1(w)\frac{\partial f}{\partial x}(w+v) + X^2(w)\frac{\partial f}{\partial y}(w+v),
\end{array}$$
and
$$X_{w+v}(f)=X^1(w+v)\frac{\partial f}{\partial x}(w+v) + X^2(w+v)\frac{\partial f}{\partial y}(w+v).$$
So, taking
$$X=x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y},$$
we have, for $v=(1,0)$ and $w=(x,y)\in\mathbb R^2$, that
$$\begin{array}{rcl}
(L_v)_*X_w(x) & = & x(w)\frac{\partial x}{\partial x}(w+v) + y(w)\frac{\partial x}{\partial y}(w+v) \\
& = & x\cdot 1+y\cdot 0
\end{array}$$
and
$$\begin{array}{rcl}
X_{w+v}(x) & = & x(w+v)\frac{\partial x}{\partial x}(w+v) + y(w+v)\frac{\partial f}{\partial y}(w+v) \\
& = & (x+1)\cdot 1 + y\cdot 0.
\end{array}$$
Thus, $X$ is not left-invariant.
