Smooth frame for the tangent bundle $TM$

Question

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 ?

Answer 1

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 $$ C^k_{ij}=\sum_{\mu\nu}\theta^k_\nu(e^\mu_i\partial_\mu e^\nu_j-e^\mu_j\partial_\mu e^\nu_i). $$

Answer 2

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.