Smooth frame for the tangent bundle $TM$ Suppose that $e_1, \ldots, e_n$ is a smooth frame for the tangent bundle $TM$ of an $n$-dimensional manifold $M$. Then, at each $p\in M$, $\lbrace e_i(p) \rbrace$ is a basis for the tangent space $T_pM$. Being a tangent vector at $p$, $[e_i,e_j]_p$ is uniquely a combination 
\begin{align*}
[e_i,e_j]_p  = \sum_{k=1}^n C^k_{ij}(p) e_k(p).
\end{align*}
How can I determine the coefficients $C^k_{ij}$?. Any help please ?
 A: You need to know the expression of your frame with respect to a coordinate frame. Coordinate frames are preferred, because their brackets are zero, so you can use them as reference points.
If $e_i=\sum_{\mu=1}^ne^\mu_i\partial_\mu$, then $$ [e_i,e_j]=\sum_{\mu,\nu}[e^\mu_i\partial_\mu,e^\nu_j\partial_\nu]=\sum_{\mu,\nu}(e^\mu_i\partial_\mu e^\nu_j-e^\mu_j\partial_\mu e^\nu_i)\partial_\nu, $$ so now we let the $k$-th element of the dual frame of $e$ (written as $\theta^k$) act on the expression. In components we have $\theta^i_\mu$ being the inverse matrix of $e^\mu_i$ as $\sum_\mu\theta^i_\mu e^\mu_j=\delta^i_j$, so we have $$ C^k_{ij}=\sum_{\mu\nu}\theta^k_\nu(e^\mu_i\partial_\mu e^\nu_j-e^\mu_j\partial_\mu e^\nu_i). $$
A: Sans other information, the only way I know to detrtmine the $C_{ij}^k$ is via direct computation in a coordinate basis.  Granted the the $e_i$ are known in terms of a such a basis
$\partial_i = \dfrac{\partial}{\partial x_i}, \tag{1}$
that is,
$e_i = \sum_l a_i^l \partial_l, \tag{2}$
where the $a_i^j$ are smooth functions in a neighborhood $U(p)$ of $p$, we may proceed to evalutate the bracket $[e_i, e_j]$ on a smooth function $f: U \to \Bbb R$ as follows:  we have
$e_i[f] = \sum_l a_i^l \partial_l[f]; \tag{3}$
applying $e_j$ to $e_i[f]$ yields
$e_j[e_i[f]] = \sum_m a_j^m \partial_m[\sum_l a_i^l \partial_l[f]]; \tag{4}$
we evaluate $\partial_m[\sum_l a_i^l \partial_l[f]]$ using the Leibniz rule:
$\partial_m[\sum_l a_i^l \partial_l[f]] = \sum_l (\partial_m a_i^l \partial_l[f] + a_i^l \partial_m [\partial_l[f]]), \tag{5}$
whence
$e_j[e_i[f]] = \sum_m a_j^m(\sum_l (\partial_m a_i^l \partial_l[f] + a_i^l \partial_m [\partial_l[f]]))$
$= \sum_m (\sum_l (a_j^m \partial_m a_i^l \partial_l[f] + a_j^ma_i^l\partial_m[\partial_l[f]])) = \sum_{l, m} (a_j^m \partial_m a_i^l \partial_l[f] + a_j^ma_i^l\partial_m[\partial_l[f]]); \tag{6}$
likewise, reversing the roles of $i$ and $j$ we find
$e_i[e_j[f]] = \sum_{l, m} (a_i^m \partial_m a_j^l \partial_l[f] + a_i^ma_j^l\partial_m[\partial_l[f]]), \tag{7}$
whence,
$e_i[e_j[f]] - e_j[e_i[f]]$
$= \sum_{l, m} (a_i^m \partial_m a_j^l \partial_l[f] + a_i^ma_j^l\partial_m[\partial_l[f]]) - \sum_{l, m} (a_j^m \partial_m a_i^l \partial_l[f] + a_j^ma_i^l\partial_m[\partial_l[f]])$
$= \sum_{l, m} (a_i^m \partial_m a_j^l \partial_l[f] - a_j^m \partial_m a_i^l \partial_l[f]) + \sum_{l, m}(a_i^ma_j^l\partial_m[\partial_l[f]] - a_j^ma_i^l\partial_m[\partial_l[f]]); \tag{8}$
we examine the second sum on the right of (8):
$\sum_{l, m}(a_i^ma_j^l\partial_m[\partial_l[f]] - a_j^ma_i^l\partial_m[\partial_l[f]]) = \sum_{l, m}(a_i^ma_j^l\partial_m[\partial_l[f]] - a_i^l a_j^m\partial_m[\partial_l[f]])$
$= \sum_{l, m}a_i^ma_j^l\partial_m[\partial_l[f]] - \sum_{l, m}a_i^l a_j^m\partial_m[\partial_l[f]]) =$
$\sum_{l, m}a_i^ma_j^l\partial_m[\partial_l[f]] - \sum_{l, m}a_i^m a_j^l\partial_m[\partial_l[f]]) = 0, \tag{9}$
where we are able to affirm that 
$\sum_{l, m}a_i^l a_j^m\partial_m[\partial_l[f]]) = \sum_{l, m}a_i^m a_j^l\partial_m[\partial_l[f]]) \tag{10}$
by virtue of the fact that $l$ and $m$ are dummy indices; since every conceivable $l, m$ pair occurs in each sum, the individual terms all cancel out and we are left with $0$.  Equation (8) thus becomes
$[e_i, e_j][f] = e_i[e_j[f]] - e_j[e_i[f]$
$= \sum_{l, m} (a_i^m \partial_m a_j^l \partial_l[f] - a_j^m \partial_m a_i^l \partial_l[f]) = \sum_{l, m} (a_i^m \partial_m a_j^l \partial_l - a_j^m \partial_m a_i^l \partial_l)[f], \tag{11}$
and since (11) holds for every smooth $f:U \to \Bbb R$, we may abstract it away and affirm that
$[e_i, e_j] = (\sum_{l, m} a_i^m \partial_m a_j^l \partial_l - a_j^m \partial_m a_i^l \partial_l), \tag{12}$
which is the formula for $[e_i, e_j]$ in terms of the coefficients $a_k^m$.
(12) may be further refined to yield
$[e_i, e_j] = \sum_l (\sum_m  a_i^m \partial_m a_j^l  - a_j^m \partial_m a_i^l)\partial_l, \tag{13}$
which expresses $[e_i, e_j]$ in terms of the coordinate basis $\partial_l = \partial / \partial x_l$.  The process of expressing the commutators $[e_i, e_j]$ in terms of the $e_k$ themselves may be completed with the aid of (2); since both $\partial_i$ and the $e_j$ form bases of $T_pM$ for any $p \in M$, the matrix of coefficients $[a_i^l]$ occurring in (2) must be invertible; thus there exists a matrix $[b_k^l]$ such that
$\partial_l = \sum_p b_l^k e_k, \tag{14}$
with 
$\sum_k b_i^k a_k^j = \delta_i^j; \tag{15}$
we may use he formula (14) to express $[e_i, e_j]$ in terms of the $e_p$ themselves:
$[e_i, e_j] = \sum_l (\sum_m  a_i^m \partial_m a_j^l  - a_j^m \partial_m a_i^l)\sum_k b_l^k e_k$
$= \sum_{l, k} (\sum_m  a_i^m \partial_m a_j^l  - a_j^m \partial_m a_i^l)b_l^k e_k, \tag{16}$
so if we set
$C_{ij}^k = \sum_l (\sum_m  a_i^m \partial_m a_j^l  - a_j^m \partial_m a_i^l) b_l^k, \tag{17}$
we see that
$[e_i, e_j] = \sum_k C_{ij}^k e_k, \tag{18}$
as desired.
Nota Bene:  The importance of a coordinate basis to this computation cannot be overemphasized.  Indeed, since a manifold such as $M$ is essentially defined in terms of local coordinate patches, ultimately every geometric quantity must be referred to them.  So the need for coordinate expressions for the $e_i$, such as (2), should come as no surprise.
