Proposition 12

Let $\mathfrak{g}$ be a Lie algebra, $a$ an automorphism of $\mathfrak{g}$.

(i) For $\lambda , \mu\in k,[\mathfrak{g}^{\lambda}(a),\mathfrak{g}^\mu(a)]\subset \mathfrak{g}^{\lambda \mu}(a)$; in particular, $\mathfrak{g}^1(a)$ is a subalgebra of $\mathfrak{g}$.

(ii) If B is a symmetric bilinear form on $\mathfrak{g}$ invariant under $a,\mathfrak{g}^{\lambda}(a)$ and $\mathfrak{g}^\mu(a)$ are orthogonal relative to B for $\lambda \mu\not= 1$. Assume that B is non-degenerate. Then, if $\lambda \not= 0$, the restriction of B to $\mathfrak{g}^{\lambda}(a)\times \mathfrak{g}^{1/\lambda}(a)$ is non-degenerate.

Assertion (i) and the first half of (ii) follow from Prop. 2 (iii) applied to the composition law $\mathfrak{g}\times \mathfrak{g}\rightarrow \mathfrak{g}$ and the bilinear form B. To prove the second half of (ii), we can assume that $k$ is algebraically closed. Then $\mathfrak{g}=\bigoplus_{\nu\in k}\mathfrak{g}^{\nu}(a)$. In view of the above, $\mathfrak{g}^{\lambda}(a)$ is orthogonal to $\mathfrak{g}^{\nu}(a)$ if $\lambda \nu \not= 1$; since B is non-degenerate, it follows that its restriction to $\mathfrak{g}^{\lambda}(a)\times \mathfrak{g}^{1/\lambda}(a)$ is also.