If a Lie group is abelian, its Lie algebra has Lie bracket = zero

I futzed around with this yesterday and, while proving this on the other definitions of bracket is easy, doing it in the direct way, straight from the manifold definitions - without any recourse to flows, one parameter subgroups, exp map, the Lie derivative, adjoint map, etc., seems rather cumbersome - and I'm still not done. That's because looking at the flows, Lie derivative, etc, naturally provide a place to exploit commutivity, whereas the direct manifold-vector type defintions don't reveal anything about the Lie multiplication until you've drilled in deep. I'll leave it to you to complete this, if you like. Basically, I drilled down far enough to where the abelian-ness is obviously the cause of the brackets being 0. It looks like a short argument from there, but I'm done thinking about it.

I suspect it might be easier to prove that your definition of Lie brackets is equivalent to one of the many other definitions, and then prove that abelian=>[,]=0 from that other definition.

Of course, it's certainly possible that there's an easier way, still keeping just to the definitions, that I've simply overlooked. And of course, I certainly could've made a mistake somewhere.

$\text{Fix a chart } (U, \phi = (x_1, x_2, ... x_n) \) \text{about } e \in G.$

$\text{Define left translation by }g \in G \text{ as } L_g : G \rightarrow G \text{ by } L_g(h) = gh.$

$\text{Define smooth functions } c_{i, j} \text{ on } U \text{ by } T_eL_g(\partial_i|_e) = \sum_j c_{i, j}(g) \partial_j|_g \in T_gG.$

$\text{For any vector } \tilde{y} = \sum_i \tilde{y_i} \partial_i|_e \in T_eG \text{, there's a unique left invariant vector field having that value in } T_eG \text{, given by }$

$Y(g) = T_eL_g(\tilde{y}) = \sum_i \tilde{y_i} T_eL_g(\partial_i|_e) \in T_gG. \text{ Since } L_e = id_G, \text{ get } Y(e) = \tilde{y}.$

$\text{On } U \text{, } Y|U = \sum_i y_i \partial_i \in \Gamma(TU) \text{, and the functions } y_i \text{ are entirely determined by } \tilde{y}\text{ via: }$

$Y(g) = T_eL_g(\tilde{y}) = \sum_i \tilde{y_i} T_eL_g(\partial_i|_e) = \sum_{i} \tilde{y_i} \left\{\sum_j c_{i, j}(g) \partial_j|_g \right\}$

$= \sum_{j} \left\{\sum_{i} \tilde{y_i} c_{i, j}(g) \right\} \partial_j|_g = \sum_{j} y_j(g) \partial_j|_g \text{ so that }$

$y_j(g) = \sum_{i} \tilde{y_i} c_{i, j}(g) \text{, or simply } y_j = \sum_{i} \tilde{y_i} c_{i, j}$ .

$\text{Now, compute the bracket of left invariant vector fields } Y, W \text{ at } e \text{ based on their values at } T_eG:$

$[Y, W]|_e = \sum_j \left\{ \sum_i ( \ y_i(e) \partial_i|_e (w_j) - w_i(e) \partial_i|_e (y_j) \ ) \right\} \partial_j_|e$

$= \sum_{i, j} \left\{ \ y_i(e) \partial_i|_e \left(\sum_{k} w_k(e) c_{k, j}\right) - w_i(e) \partial_i|_e \left(\sum_{k} y_k(e) c_{k, j}\right) \ \right\} \ \partial_j|_e$

$= \sum_{i, j, k} \left\{ \ y_i(e) w_k(e) \partial_i|_e ( c_{k, j}) - y_k(e) w_i(e) \partial_i|_e ( c_{k, j}) \right\} \ \partial_j|_e$

$= \sum_{i, j, k} \left\{ \ [ \ y_i(e) w_k(e) - y_k(e) w_i(e) \ ] \ \partial_i|_e ( c_{k, j}) \ \right\} \ \partial_j|_e$

$= \sum_{j} \left\{ \sum_{i, k} \ [ \ y_i(e) w_k(e) - y_k(e) w_i(e) \ ] \ \partial_i|_e ( c_{k, j}) \right\} \ \partial_j|_e.$

$\text{Now, since } (y_i(e) w_k(e) - y_k(e) w_i(e)) \text{ is skew-symmetric in } i \text{ and } k, \text{if } \partial_i|_e ( c_{k, j})$

$\text{were symmetric in } i \text{ and } k \text{, then the product would be skew symmetric, and so the double sum would be } 0.$

$\text{Thus if } \partial_i|_e ( c_{k, j}) = \partial_k|_e ( c_{i, j}) \ \forall i, j, k \in \{1, 2, ... n\}, \text{ then }$

$(y_i(e) w_k(e) - y_k(e) w_i(e)) \partial_i|_e ( c_{k, j}) = -(y_k(e) w_i(e) - y_i(e) w_k(e)) \partial_k|_e ( c_{i, j}) \ \forall i, j, k,$

$\text{and so } \sum_{i, k} \ (y_i(e) w_k(e) - y_k(e) w_i(e)) \partial_i|_e ( c_{k, j}) \} = 0 \ \forall j \in \{1, 2, ... n\},$

$\text{and so } [Y, W]|_e = 0.$

$\text{So if can show that } G \text{ abelian implies } \partial_i|_e ( c_{k, j}) = \partial_k|_e ( c_{i, j}) \ \forall i, j, k \in \{1, 2, ... n\},$

$\text{then will have proven that } G \text{ abelian implies that the brackets at } T_eG \text{ of any left invariant vector fields are } 0.$

$\text{Now } c_{i,s}(g) = \sum_j c_{i, j}(g) \delta_j^s = \sum_j c_{i, j}(g) \partial_j|_g [x_s] = ( \ T_eL_g(\partial_i|_e) \ )[x_s]\text{, so }$

$c_{i,s}(g) = \partial_i|_e[x_s \circ L_g]. \text{ Thus } \ \forall g \in U, \ \forall i, j, \ c_{i,j}(g) = \partial_i|_e[x_j \circ L_g].$

$\text{Thus want to show that } G \text{ abelian implies }$

$\partial_k|_e \{ g \mapsto \partial_i|_e[x_j \circ L_g] \} = \partial_i|_e \{ g \mapsto \partial_k|_e[x_j \circ L_g] \} \ \forall i, j, k \in \{1, 2, ... n\}.$

$\text{(I've written it as } \partial_k|_e \{ g \mapsto \partial_i|_e[x_j \circ L_g] \} \text{ to emphasize which function } \partial_k|_e \text { acts on.)}$

$\text{Thus want to show that } G \text{ abelian implies }$

$\partial_k|_e \partial_i|_e[x_j \circ L_g] = \partial_i|_e \partial_k|_e[x_j \circ L_g] \ \forall i, j, k \in \{1, 2, ... n\}.$

$\text{Define } \phi : G \times G \rightarrow G \text{ by } \phi(u_1, u_2) = u_2u_1 = L_{u_2}u_1.$

$\text{Then } (x_j \circ L_g)(u_1) = x_j(L_g(u_1)) = (x_j \circ \phi) (u_1, g),$

$\text{so } \partial_i|_e [x_j \circ L_g] = \partial_i|_{u_1=e} [x_j \circ L_g(u_1)] = \partial_i|_{u_1=e}[(x_j \circ \phi) (u_1, g)],$

$\text{so } \partial_k|_e \{ g \mapsto \partial_i|_e [x_j \circ L_g] \} = \partial_k|_{u_2 = e} \partial_i|_{u_1=e}[(x_j \circ \phi) (u_1, u_2)],$

$\text{So, } \ \forall i, j, k \in \{1, 2, ... n\}, \text{ define }b_{i, j, k} =\partial_k|_{u_2 = e} \partial_i|_{u_1=e}[(x_j \circ \phi) (u_1, u_2)].$

$\text{The problem thus reduces to showing that: }$

$G \text{ abelian implies } b_{i, j, k} = b_{k, j, i} \forall i, j, k \in \{1, 2, ... n\}.$

If I've made no mistake, this does "drive" the problem down to where commutivity makes the brackets 0. I'll let you figure that out. I'm sick of this. It looks like you just use that partials commute, then use that phi is symmetric, and you're done. But I'm done - I'll leave that for you to think about.