How to find in practise a $\mathfrak{sl}_2$-triple in a simple lie algebra?

Question

I want to justify the existence of a $\mathfrak{sl}_2$-triple in a simple lie algebra. I know there exists the Jacobson-Morozov theorem. It states that given a nilpotent element of the lie algebra, one can construct a $\mathfrak{sl}_2$-triple. I think it's a quite big theorem, is there other arguments to find in practise a $\mathfrak{sl}_2$-triple in a (semi)-simple lie algebra ?

Answer 1

$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\fg{\mathfrak{g}}\def\fa{\mathfrak{a}}$I am not sure in what sense you are "given" your Lie algebra, and I'm also not sure whether you are working over $\RR$ or $\CC$. But I'll walk through the approach implicit in the standard description. Our Lie algebra is called $\fg$.

Step 1 Find an abelian subalgebra $\fa$ which you hope is a Cartan.

Step 2 Let $\fa$ act on $\fg$ by $a \ast g = [a,g]$. Decompose $\fg$ (or $\fg \otimes \CC$ if you started with a real Lie algebra) into eigenspaces for this action: $\fg = \bigoplus_{\mu} \fg_{\mu}$, for various linear functionals $\mu : \fa \to \CC$. The summands $\fg_{\mu}$ are called ``weight spaces".

Step 3 Check your initial guess: If the $\fa$ action isn't diagonalizable, then $\fa$ isn't a Cartan; start again. (If we are working over $\RR$, and if the Killing form is negative-definite, this can't happen, but in general it is an issue. Consider $\fg= \mathfrak{sl}_2$, with the guess $\fa = \left[ \begin{smallmatrix} 0&\ast \\ 0&0 \end{smallmatrix} \right]$.) Start over if this happens.

If the action is diagonalizable, but $\fg_0 \supsetneq \fa$, then $\fa$ also isn't a Cartan, but in this case you can still use your data to make progress. Look for a new abelian $\fa'$ with $\fa \subsetneq \fa' \subseteq \fg_0$, and try again.

Step 4: Choose a nonzero weight space $\fg_{\mu}$ with $\mu \neq 0$. Then $[\ , \ ]$ is a bilinear map $\fg_{\mu} \times \fg_{- \mu} \longrightarrow \fa$. If this map is the zero map (in particular, if $\fg_{-\mu} = 0$, then your Lie algebra wasn't semi-simple. File a complaint with the appropriate authority :).

As a boring example of this failure; consider the Lie algebra $\left[ \begin{smallmatrix} \ast & \ast \\ 0 & \ast \end{smallmatrix}\right]$ with $\fa = \left[ \begin{smallmatrix} \ast & 0 \\ 0 & \ast \end{smallmatrix}\right]$. The action diagonalizes, with a weight space $\left[ \begin{smallmatrix} 0 & \ast \\ 0 & 0 \end{smallmatrix}\right]$, but the weight space with negative weight is $0$.

As a more interesting example, consider the Lie algebra $\left[ \begin{smallmatrix} a&0&x \\ 0&-a&y \\ 0&0&0& \end{smallmatrix} \right]$. Take $\fa = \left[ \begin{smallmatrix} a&0&0 \\ 0&a&0 \\ 0&0&0& \end{smallmatrix} \right]$. Then the $\fa$ action diagonalizes, with the nonzero weight spaces $\left[ \begin{smallmatrix} 0&0&\ast \\ 0&0&0 \\ 0&0&0& \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 0&0&0 \\ 0&0&\ast \\ 0&0&0& \end{smallmatrix} \right]$. Here we have weight spaces whose weights are negative of each other, but the bracket between them is $0$.

Otherwise, choose $e$ and $f$ in $\fg_{\mu}$ and $\fg_{-\mu}$ with $[e, f] \neq 0$. Put $h = [e, f]$.

Step 5: Look at $\mu(h)$. (In other words, the scalar $c$ such that $[h,e] = c e$.) If it is zero than, again, your Lie algebra wasn't semisimple; complain. I think an example of this issue should be $\fg = \left[ \begin{smallmatrix} 0&\ast&\ast \\ 0&\ast&\ast \\ 0&0&0 \\ \end{smallmatrix} \right]$ with $\fa = \left[ \begin{smallmatrix} 0&0&\ast \\ 0&\ast&0 \\ 0&0&0 \\ \end{smallmatrix} \right]$. Then we get nonzero weight spaces whose pairing is nonzero, and we can take $e = \left[ \begin{smallmatrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \\ \end{smallmatrix} \right]$ and $f= \left[ \begin{smallmatrix} 0&0&0 \\ 0&0&1 \\ 0&0&0 \\ \end{smallmatrix} \right]$, so $h = [e,f] = \left[ \begin{smallmatrix} 0&0&1\\ 0&0&0 \\ 0&0&0 \\ \end{smallmatrix} \right]$, but $[h,e] = 0$.

If $\mu(h)$ is not zero and you are working over $\CC$, rescale $e$ (or $f$) such that $\mu(h) = 2$. Then $(e,f,h)$ is an $\mathfrak{sl}_2$ triple.

If you are working over $\RR$ and $\mu(h) \neq 0$, then either $\mu$ was a real weight, meaning a linear map $\fg \to \RR$, or an imaginary weight, meaning a linear map $\fg \to \RR \sqrt{-1}$. In the former case, rescale $e$ such that $\mu(h) = 2$ and you have an $\mathfrak{sl}_2$ triple. In the latter case, rescale $e$ such that $\mu(h) = 2i$, and you get an $\mathfrak{su}_2$ triple instead.