How to prove these Lie algebra relations

This is a bit of a basic computational question concerning Lie algebras, but I'm getting kind of bamboozled so I thought I'd post it.

I'm confused about how to perform some computations in Serre's Complex Semisimple Lie algebras. The first case is the following: Let $X,Y,H$ be a basis for $\mathfrak{sl}_2(\mathbb{C})$ satisfying the usual commutation relations: $[H,X] = 2X$, $[H,Y] = -2Y]$, and $[X,Y] = H$. Then the claim is that for any representation of $\mathfrak{sl}_2(\mathbb{C}))$, if

$$\Theta = e^Xe^{-Y}e^X$$

then

$$\Theta H = -H \Theta, \Theta X = - Y \Theta, \Theta Y = - X \Theta.$$

I started trying to force this out with the commutation relations, but I got $[e^X, H] = H - 2Xe^X$ so that

$$\Theta H = e^X e^{-Y} (He^X + -2Xe^X).$$

At this point, I became overwhelmed with the prospect of trying to commute $X$ with $e^Y$.

How do you do these computations???

Secondly, Serre also states the following. Let $\mathfrak{g}$ be a Lie algebra which decomposes under the action of a Cartan subalgebra $\mathfrak{h}$ as $$\mathfrak{g} = \mathfrak{h} \bigoplus_{\alpha} \mathfrak{g}^{\alpha}$$ and let $X_{\alpha} \in \mathfrak{g}^{\alpha} , Y_{\alpha} \in \mathfrak{g}^{-\alpha}, H_{\alpha} \in \mathfrak{h}$ be elements satisfying the commutation relations as above. Then $$e^{X_{\alpha}}e^{-Y_{\alpha}}e^{X_{\alpha}}$$ restricts on $\mathfrak{h}$ to be the usual reflection associated to $\alpha$, i.e. negating the root $\alpha$ and fixing the hyperplane determined by $\alpha$.

It's an easy computation to see that the above element does indeed fix the orthogonal hyperplane, but I don't know how to show the rest.

How can one see this?

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Since $X$, $Y$, $H$ are fixed elements of $sl_2(\mathbb{C})$, I don't understand the part where you say "in any representation of $sl_2(\mathbb{C})$". – Ted May 3 '12 at 7:10
For the second, shouldn't be $e^{\operatorname{ad}X_\alpha}e^{-\operatorname{ad}Y_\alpha}e^{\operatorname{ad}‌​X_\alpha}$? – Yuki Jul 20 '12 at 15:00

First Part

(I assume that really $X\equiv d(X)$ for some representation $d$; this makes no odds.)

You're apparently familiar with the fact that $[[H,Y],Y]=0$ gives (inductively) $$[H,Y^n]=nY^{n-1}[H,Y]$$

Another inductive argument (which can in fact be generalized) would show $$[X,Y^n]=n(H+(n-1))Y^{n-1}$$

However there is a more satisfying approach involving showing

$$\boxed{e^{A}Be^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\frac{1}{3!}[A,[A,[A,B]]]+\cdots}$$

(Wikipedia points out that this is essentially equivalent to $e^{\textrm{ad}(A)} B$ in the case of the fundamental representation.)

Proof: Write $f_B(\lambda)=e^{\lambda A}Be^{-\lambda A}$. Then $$f_B'(\lambda)=e^{\lambda A}(AB-BA)e^{-\lambda A} = f_{[A,B]}(\lambda)$$ Inductively, we find $$f_B^{(n)}(\lambda) = f_{[A,\cdots [A,B]\cdots]}(\lambda)$$ Setting $\lambda=1$ and using the Taylor expansion of $f_B(1)$ gives the result. QED.

This generalizes the identity you already used. It is useful because, setting $A=Y,B=X$, we find $[A,B]=-H$ and so $[A,[A,B]]=[Y,-H]=-2Y$, and finally $$\boxed{[A,[A,[A,B]]]=[Y,-2Y]=0}$$ so the series terminates. You should be able to use this to compute the result you want.

Second Part

To see that the root is negated, I think you're simply supposed to use the definition of the action of group elements on the Lie algebra to observe that $$\Theta H_\alpha \Theta^{-1} = -H_\alpha \Theta \Theta^{-1} = -H_\alpha$$ You can use the $X,Y$ as well if you like.

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