Let $G$ be a Lie group, $g\in G$ and $L_g$ be left translation by $g$. I want to compute the differential $dL_g|_0$ of $L_g$ at $0$.
Attempt:
Let $v\in T_0G$ be a tangent vector at $0$. Let $x:\mathbb R^n\to G$ be chart around $0$, such that $x(0)=0$. Let $a:\mathbb R\to \mathbb R^n:t\to vt$ be a straight line with gradient $v$.
Then $dL_g(v)=v(L_g)=\frac{d}{dt}|_0[L_g(x(a(t)))]=\frac{d}{dt}|_0[gx(vt)]=gx'(0)v$
As $g$ is a group element, and $x'(0)v$ is a tangent vector, we cannot multiply these together. Therefore, the last term above is meaningless. How should we remedy this problem? How can we compute the matrix of $dL_g|_0$?
EDIT: Thank you for your comment, Jorkug. I am worried because by following the definitions I arrive at the meaningless symbol $gx'(0)v$.
$\newcommand{\Reals}{\mathbf{R}}$To be invariant about it, if $v$ is a tangent vector at $e$, then the one-parameter subgroup $x(t) = \exp(tv)$ is a curve through $e$ with $x'(0) = v$, and $$ dL_{g}(v) = \frac{d}{dt}\bigg|_{t=0} L_{g}\exp(tv), $$ just as you say.
To get more detailed and/or explicit information, you need to invoke specifics about the group structure. If $G$ is a Lie subgroup of $GL(n, \Reals)$, the group operation is the restriction of matrix multiplication. For each $g$ in $G$, left multiplication $L_{g}(x) = gx$ is the restriction of a linear transformation $\Reals^{n \times n} \to \Reals^{n \times n}$, and may be identified with its differential: $$ dL_{g}(v) = gv. $$
For example, if $G = SO(n)$, then $g$ is a rotation matrix, a tangent vector $v$ at the identity is a skew-symmetric matrix, and $$ dL_{g}(v) = gv $$ is an ordinary matrix product, viewed as an element of $T_{g} SO(n)$.