Exercise 16, § 1

Let $\mathfrak{g}$ be a simple Lie algebra of dimension 3, and let $\Phi$ be its Killing form. Denote by $\mathfrak{o}(\Phi )$ the orthogonal algebra of $\Phi$, i.e. the subalgebra of $\mathfrak{g}\mathfrak{l}(\mathfrak{g})$ consisting of the elements leaving $\Phi$ invariant.

a) Show that ad$:\mathfrak{g}\rightarrow \mathfrak{o}(\Phi )$ is an isomorphism.

b) Prove the equivalence of the following properties:

(i) $\mathfrak{g}$ contains a non-zero isotropic vector (for $\Phi$ ).

(ii) $\mathfrak{g}$ contains a non-zero nilpotent element.

(iii) $\mathfrak{g}$ is isomorphic to $\mathfrak{s}$.

c) Show that there exists an extension $k_1$ of $k$, of degree $\leq 2$, such that $\mathfrak{g}_{(k_1)}=k_1\otimes_k\mathfrak{g}$ is isomorphic to $\mathfrak{s}_{(k_1)}$.

$^8$ For more details on Exerc. 12, 13 and 14, see: D. ARNAL and G. PINCZON,

Sur les représentations algébriquement irréductibles de l’algèbre de Lie $\mathfrak{s}\mathfrak{l}$(2), J.

Math. Phys., Vol. 15 (1974), pp. 350-359. d) Give the vector space $A =k\oplus \mathfrak{g}$ the unique algebra structure admitting 1 as unit element, and such that the product $\mathfrak{g}\times \mathfrak{g}\rightarrow A$ is given by the formula

Show that A is a quaternion algebra over $k$, and that $\mathfrak{g}$ is the Lie subalgebra of A consisting of the elements of trace zero (reduce, by extension of the base field, to the case where $\mathfrak{g}=\mathfrak{s}$, and show that A can then be identified with the matrix algebra $\mathbf{M}_2(k)$).

Conversely, if D is a quaternion algebra over $k$, the elements of D of trace zero form a simple Lie algebra of dimension 3, and the corresponding algebra A can be identified with D.

e) Prove the formulas

$\Phi (x, x) =-$8Nrd$_A(x),\Phi (x, y) =$ 4Trd$_A(xy)$

f) Show that the discriminant of $\Phi$ (with respect to any basis of $\mathfrak{g}$) is of the form $-2\lambda^2$, with $\lambda \in k^*$.

g) Let $n$ be an integer $\geq 0$. Show that the $\mathfrak{g}$-module $\mathbf{S}^n(\mathfrak{g})$ has a unique simple submodule of dimension $2n+ 1$, and that this module is absolutely simple (reduce to the case where $\mathfrak{g}=\mathfrak{s}$, and use Exerc. 9).

Show that $\mathfrak{g}$ has an absolutely simple module of dimension $2n$ only if $\mathfrak{g}$ is isomorphic to $\mathfrak{s}$, i.e. only if A is isomorphic to $\mathbf{M}_2(k)$. (Let V be such a module. If $n\geq 2$, show by means of Exerc. $6c$) that the $\mathfrak{g}$-module $V\otimes \mathfrak{g}$ has a unique absolutely simple submodule of dimension $2n-2$. Then reduce to the case $n= 1$, which is trivial.)