13
$\begingroup$

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?

  • 0
    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 $sl_2(\mathbb{C})$".2012-05-03
  • 1
    For the second, shouldn't be $e^{\operatorname{ad}X_\alpha}e^{-\operatorname{ad}Y_\alpha}e^{\operatorname{ad}X_\alpha}$?2012-07-20

1 Answers 1