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

• Oct 13th 2012, 10:58 AM
ModusPonens
If a Lie group is abelian, its Lie algebra has Lie bracket = zero
Hello

I'm looking for tips (and just tips) on how to prove the proposition in the title: If a Lie group is abelian, its Lie algebra has Lie bracket = zero. I know I have to prove that the Lie bracket of the left invariant vector fields (to which the Lie algebra is isomorphic) is zero. But that is as far as I can get. I can't find a way to relate the commutativity of the Lie group with the Lie bracket

• Oct 13th 2012, 02:53 PM
johnsomeone
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
It depends on how the brackets of the Lie Algebra is being defined.
If it's defined via the adjoint map ad = derivative of Ad, then note that, for all a, Ad_a is the identity for a commutative group.
If it's defined via the exponential map, then use that exp(v1)exp(v2) = exp(v2)exp(v1) for a commutative group, to show that the second order term of f in exp(v1)exp(v2) = exp(f(v1, v2)) that defines the bracket is both symmetric and skew symmetric (and bilinear), hence is 0.
• Oct 13th 2012, 03:44 PM
ModusPonens
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
The Lie bracket on the Lie algebra is determined by the isomorphism between the left invariant vector fields on the Lie group G, with the usual Lie bracket [X,Y], and the Lie algebra it self, which associates each V in the Lie algebra with $\displaystyle X_g^V(L_g)_1V$ in the set of the vector fields. So none of those aproaches I understand. But I'm very thankful anyway. But, if you have any sugestion on how to aproach the problem from this angle, please share.
• Oct 13th 2012, 04:02 PM
johnsomeone
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
Quote:

Originally Posted by ModusPonens
The Lie bracket on the Lie algebra is determined by the isomorphism between the left invariant vector fields on the Lie group G, with the usual Lie bracket [X,Y], and the Lie algebra it self, which associates each V in the Lie algebra with $\displaystyle X_g^V(L_g)_1V$ in the set of the vector fields.

If it's known (defined) by such an isomorphism, then you need only show that the bracket is always 0 in the simpler case to show that it's always 0 in the other case. Presumably, the simpler case is "left invariant vector fields on the Lie group G, with the usual Lie bracket [X,Y]". Unfortunately, what exactly "the usual Lie bracket [X,Y]" means there isn't clear. How is your book defining that?
Is it the operation of vector fields on test functions in the manifold sense? (i.e. [X,Y](f) = X(Y(f)) - Y(X(f)) ?)
• Oct 13th 2012, 04:17 PM
ModusPonens
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
$\displaystyle [X,Y]_p=(X.Y_1-Y.X_1)(p)(\partial/\partial x_1)_p + ... + X.Y_n-Y.X_n)(p)(\partial/\partial x_n)_p$ , so we're probably speaking of the same thing.
• Oct 14th 2012, 03:50 AM
ModusPonens
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
The dots mean composition, that is, $\displaystyle X.Y_1=X(Y_1)$ .
• Oct 15th 2012, 02:11 PM
johnsomeone
Re: 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.
• Oct 15th 2012, 03:38 PM
ModusPonens
Re: If a Lie group is abelian, its Lie algebra has Lie bracket = zero
Wow! Thanks a lot for your work! I really apreciate it.

I never thought it would be that hard. I'll read it when I have the time, since I'm close to going to bed.