Thanks for your hint, but I have already remarked this, even though I don't see where it is useful in the above approach. Or is the approach wrong?

@user123234 I don't see where in your post you say this. In any case, I'll elaborate further.

No not in the post, I remarked this on my notes.

@user123234 See my edit.

but couldn't I use that $H$ and $\overline{H}$ are commutative because my assistant told me that I could deduce this from the fact that $H$ and $\overline{H}$ are commutative

@user123234 I don't see any obvious way forward on that basis... do you have any results in your notes or textbook that apply to commutative groups/algebras?

my first problem in your approach is that you have never used that $\mathfrak{h}$ is maximal

I know that if $H$ is a lie subgroup then $H$ is a normal subgroup of $\overline{H}$

I also know $H$ is generated by $\exp(X)$ for $X\in \mathfrak{h}$

That's a fair point; I'm not sure if h being maximal is an important detail here.

me neither but that's why I'm a bit worried

Right and that last fact is what we show

What I show in the beginning I mean

Well not just generated by, actually equal to exp(X)

(the set of all exp(X) for X in h)

But I couldn't use this fact somehow to work out something with my $\overline{\mathfrak{h}}$ which I wanted to define in my solution?

I think there's an issue with my inverse map by the way

OH actually that's definitely the way to go

hmm okey could you maybe specify this a bit more?

will you rewrite the answer?

I'm getting stuck where you got stuck though... how to show that h-bar is commutative

But you think this is the right approach?

Maybe I need that since $\mathfrak{h}$ is commutative also $\overline{H}$ is?

Well we already know that H-bar is commutative

Oh that's what you mean

Yes I think that's the key. Show that non-commutativity of h-bar would induce non-abelian-ness of H.

*of H-bar

maybe but I don't see how

Just put my answer up

I saw it thanks. But you don't assume by contradiction that $H\neq \overline{H}$ right?

Right, I don't make that assumption

ah and what do you then assume to do your contradiction?

You could work by contradiction if you prefer

But I was thinking you know that H is a subset (not necessarily proper) of H-bar. You use this to show that H-bar is commutative. You use this to show that the algebra of H-bar is equal to h. Now conclude that H-bar is the Lie group associated with h, i.e. H-bar = H.

I mean I don't see what your attempt is, since you constructed $\overline{\mathfrak{h}}$ which is abelian and contain $\mathfrak{h}$ so $\mathfrak{h}$ is not maximal but is this thenn a contradiction to which assumption?

Again, there's no need to work by contradiction

but I somehow don't get your direct way.

What step are you having trouble with?

The main idea what we want to show.

We want to show $\overline{H}=H$ right?

Yes. We show this by showing that the Lie algebra of \overline H is equal to h, the Lie algebra of H.

and why is this enough?

Because we've just shown that both H and \overline H are the Lie group generated by the same Lie algebra

So they're the same group. They're equal.

ah okey so if h-bar=h then H-bar=H

got it.

Exactly

but I mean that h is contained in h-bar is clear

but how do you use commutativity to show the other direction?

That's what it means for h to be a "maximal" commutative subalgebra: it contains any commutative subalgebra, including h-bar

ah if we can show h-bar is abelian then we know by maximality that h-bar is contained in h.

so we are done. I see

Does it all make sense then?

but now one last question to the argument that h-bar is abelian. You have shown that $e^{sX}$ commutes with $Y$ for all $s$ but then you use that $Y$ is in H-bar?

No, I've only assumed that Y is in h-bar (the algebra not the group).

Hmm so I don't get the argument why $e^{sX}$ commutes with $Y$ and why $Y$ commutes with $X$

Use the derivative, as I suggest. Use the definition of a derivative.

But I mean you know $\exp(sX)\exp(tY)=\exp(tY)\exp(sX)$ and then if you derive you need to use the product rule on both sides

or do you first derive wrt $t$ and then $s$

because on my first attempt I used $t=s$ then I needed to apply the product rule

but here I see you used $t$ and $s$ different so you would not need it

As I wrote in the answer, the first derivative is with respect to t, holding s constant

but why doesn't it work if I take $t=s$

Why should that work? If that worked, I wouldn't have bothered with separate s and t variables

ah okey I think I got it

thanks a lot

It seems that you really have a knowledge about this topic. If I have further questions about analysis on matrix groups, is there a way to contact you so that you see that I asked a question or maybe ask you directly

You can email me at benwgrossmann (at) gmail (dot) com

okey perfect thanks a lot!