Structure constants of Lie algebra Let $(x^i)$ be a local coordinates system near identity of a Lie group $G$ such that $x(e)=0$. Suppose the multiplication has local form
$$m(x_1,x_2)^k=x_1^k+x_2^k+\frac{1}{2}b_{ij}^k x^i_1 x^j_2+O_3(x_1,x_2).$$
Let $e_i=\partial_{i}\big|_{x=0}$ be the basis of Lie algebra (as left invariant vector fields) $T_e G=\mathfrak{g}$ induced by $x$. How can we that $c_{ij}^k=\frac{1}{2}(b_{ij}^k-b_{ji}^k)$ gives us the structure constants of $\mathfrak{g}$? That is how do we know
$$[e_i,e_j]=c_{ij}^ke_k?$$
If $(x^i)$ is given by (inverse of) the exponential map, this follows easily from CBH formula. How to prove this for a general coordinates? After a while consideration, I figured out that if it holds for coordinates $x$ then it holds for coordinates $y$. However I still prefer a direct proof.
 A: Let $U$ be an open neighborhood of $e$ in $G$ such that the local coordinates system $x=(x^i)$ is well defined on $U\cdot U$. Given $g\in G$, denote $L_g$ the left translation by $g$, i.e.
$$L_g:G\to G,\quad h\mapsto gh.$$
When $g,h\in U$, following your notations,
$$x^k(L_g(h))=m(x(g),x(h))^k=x^k(g)+x^k(h)+\frac{1}{2}b_{i,j}^kx^i(g)x^j(h)+O_3\big(x(g),x(h)\big),$$
so for fixed $g$, the Jacobi matrix of the map $x(h)\mapsto x(L_g(h))$ at $h=e$ is given by
$$\frac{\partial x^k(L_g(h))}{\partial x^j(h)}\Big|_{h=e}=\delta_j^k+\frac{1}{2}b_{i,j}^kx^i(g)+O_2(x(g)). \tag{1}$$
Let $(L_g)_*:TG\to TG$ be the tangent map of $L_g$, which maps $T_eG\to T_gG$. Since $e_j$ is a left invariant vector field with $e_j(e)=\partial_j(e)$, then by $(1)$,
$$e_j(g)=(L_g)_*\partial_j(e)=\frac{\partial x^k(L_g(h))}{\partial x^j(h)}\Big|_{h=e}\cdot \partial_k(g)=\partial_j(g)+\frac{1}{2}b_{i,j}^kx^i(g)\cdot\partial_k(g)+O_2(x(g)).$$
It follows that at $g=e$,
$$
[e_i,e_j]=[\partial_i+\frac{1}{2}b_{k,i}^lx^k\partial_l\,,\partial_j+\frac{1}{2}b_{m,j}^nx^m\partial_n]=\frac{1}{2}\left(\partial_i(x^m)b_{m,j}^n\partial_n-\partial_j(x^k)b_{k,i}^l\partial_l\right)=c_{ij}^ke_k.
$$
