Lie Algebras: [H1, H2] = 0 => h1h2 = h2h1

[Sajet]

Hi! I have the following problem:

Let G be a connected Lie Group, $\mathfrak g$ its Lie algebra. Let $\mathfrak h_1, \mathfrak h_2 \subset \mathfrak g$ be subalgebras of $\mathfrak g$ with $[\mathfrak h_1, \mathfrak h_2] = 0$, $H_1, H_2$ the subgroups belonging to them. Show: $h_1h_2 = h_2h_1$ for all $h_1 \in H_1, h_2 \in H_2$.

I tried to solve this by using the adjoint represantation and although it looks very promising, I can't really get it to work.

If I define $C_x(y) := xyx^{-1}$ as the conjugation, then what needs to be shown is: $C_{h_1}h_2 = h_2$.

I'm using these symbols: $Ad: G \rightarrow GL(\mathfrak g), Ad(g) := C_{g*e}$ and $ad_X Y = [X, Y]$.

My idea is as follows: $[\mathfrak h_1, \mathfrak h_2] = 0$ implies $ad_{h_1}ad_{h_2}=ad_{h_2}ad_{h_1} \Rightarrow Ad_{*e}(h_1)Ad_{*e}(h_2) = Ad_{*e}(h_2)Ad_{*e}(h_1) \Rightarrow (Ad(h_1)Ad(h_2))_{*e} = (Ad(h_2)Ad(h_1))_{*e}$

$\Rightarrow (Ad(h_1h_2))_{*e} = (Ad(h_2h_1))_{*e}$. Since $G$ is connected this implies $Ad(h_1h_2) = Ad(h_2h_1) \Rightarrow C_{h_1h_2*e} = C_{h_2h_1*e} \Rightarrow C_{h_1h_2}= C_{h_2h_1} \Leftrightarrow C_{h_1} = C_{h_2h_1h_2^{-1}}$ but I'm not getting any further from here.

Maybe this is the wrong path altogether but I don't see how else I could use the connectedness of G as well as the information on the Lie bracket to make a statement on the group elements themselves.

 

[Dick]

How do you define the Lie subgroup associated the Lie subalgebra h1? I'm going to guess that you want to use the exponential map here.

 

[Sajet]

"If $\mathfrak h \subset \mathfrak g$ is a Lie subalgebra, there exists a unique connected Lie subgroup $H \subset G$ with Lie algebra $\mathfrak h$."

I'm sure this unique connected subgroup is meant in the assignment.

 

[Dick]

"If $\mathfrak h \subset \mathfrak g$ is a Lie subalgebra, there exists a unique connected Lie subgroup $H \subset G$ with Lie algebra $\mathfrak h$."

I'm sure this unique connected subgroup is meant in the assignment.

I'm sure it is too. Look at how they proved that statement. I'm guessing you can use exp(A)exp(B)=exp(B)exp(A) if [A,B]=0.

 

[Sajet]

Hi! Thank you for your response.

The proof for this theorem uses distributions and leafs. But it is also shown that exp(A)exp(B) = exp(B)exp(A) if [A, B] = 0 at a different point in our script. But are you sure this fact can be used here? Is the exponential map necessarily surjective on a connected Lie group?

I'm a little confused. I tried to look this up and while I found two sources stating

"If G is a Lie group and G0 is the connected component of 1, then G0 is generated by exp(g)."

which would apply to a connected Lie group (right?), I found other sources stating:

"Not even on connected Lie groups is the exponential map necessarily surjective. Example: SL(2,IR)."

And if it's not surjective, how I could I use your idea if $h_1$ or $h_2$ are not included in the image of exp?

edit: Ok, I think I get it now. The first statement is just about a generator of the Lie group. I'll take a closer look at it now.

edit 2: Thank you, I think this works!