Lie algebra of $\mathbb{R}^{n}$ Until now the only example of lie groups I have seen are subgroups of $GL_n$. Today I had the idea, that also $G=(\mathbb R^n,+)$ must be a lie group ($(\mathbb R^n,+)$ is a group with the differentiable group operation $+$). Is it right that the lie algebra of this lie group is $\mathfrak g = \mathbb R^n$ with the exponential map $\exp : \mathfrak g \rightarrow G: x \mapsto x$? What is the lie bracket of $\mathfrak g$? Is $[x,y]$ always zero, because $+$ is commutative?
 A: Solution 1
The left translation maps have the form $L_a:\mathbb R^n \rightarrow \mathbb R^n: x\mapsto a+x$. So $D L_a(x)=\operatorname{id}$ for all $a,x\in\mathbb R^n$ ($D L_a$ is the total derivative of $L_a$). So the set of all left-invariant vector fields is the set of all constant vector fields $\mathfrak g=\{f:\mathbb R^n\rightarrow T\mathbb R^n: f = \operatorname{const}\}$ (which is isomorph to $\mathbb R^n$ because $\mathfrak g$ is $n$-dimensional).
Because $X\in \mathfrak g$ is constant, one has $[X,Y]=\mathcal L_X(Y)=0$ for any given vector field $Y$. 
Solution 2
$(\mathbb R^n, +)$ can be embedded into $GL_n$ via $$f:\mathbb R^n \rightarrow GL_n: (x_1,x_2,\ldots,x_n) \mapsto \left(\begin{matrix} e^{x_1} & & & & \\ & e^{x_2} & & \\ & & \ddots & \\ & & & e^{x_n}\end{matrix}\right)$$ $f$ is well defined because the determinant $e^{x_1 + x_2 + \ldots + x_n}$ is always positive and because of $e^{a+b}=e^a\cdot e^b$ the function $f$ is a group homomorphism.
Via $f$ one can show that $(\mathbb R^n,+)$ is isomorph to the set $D^+$ of diagonal matrices with positive entries on the diagonal. The lie algebra $\mathfrak g$ of $D^+$ is given by $$\mathfrak g = \{\dot\gamma(0): \gamma:(-\epsilon,\epsilon)\rightarrow D^{+}\}$$ which is also $D^{+}$ (as one can easily show). Because $D^{+}$ is in the center of $GL_n$ the lie bracket is always zero.
A: Yes and yes. (Exercise: exhibit $\mathbb{R}^n$ as a Lie subgroup of $\text{GL}_{n+1}$. If you get stuck, try the case $n = 1$ first.)  
