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The question: Let $L$ be a three dimensional simple Lie algebra over $\mathbb{C}$. Say that $x \in L$, $x \neq 0$, and that $ad_x$ has an eigenvalue $\lambda \neq 0$. Show that $L \cong \mathfrak{sl}(2, \mathbb{C})$.

Things I've tried: We can say $[h,e] = 2e$, $[h,f] = -2f$, and $[e,f] = h$. If I say $ad_x(y) = [x,y] = \lambda y$ then maybe I can have a basis $\{x,y,z\}$. I tried to just create a map $\phi(\frac{2}{\lambda}x)=h$, $\phi(y)=e$, and $\phi(z)=f$. Then I think $x$ and $y$ work but I'm not sure how to ensure $z$ works the way I want.

eternalmothra
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  • Can you use the fact that any simple Lie algebra over $\mathbb{C}$ is semisimple? – Жека Feb 17 '19 at 14:57
  • I should have said, actually $L$ in this problem is simple. – eternalmothra Feb 17 '19 at 16:48
  • I think your idea of extending those $\lbrace x, y\rbrace$ to a basis and then manipulating those basis vectors is good. You might need more than just scaling (namely, linear combinations) along the road to bring them to $h,e,f$. And you will have to use somewhere that $L$ is simple, since the statement is false otherwise. – Torsten Schoeneberg Feb 17 '19 at 20:24

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