Exercise 23, § 6

Let $\mathfrak{g}$ be a 3-dimensional non-commutative Lie algebra over a field $K$ of arbitrary characteristic.

(a) If $\mathfrak{g}$ has centre $\mathfrak{z} \neq \{0\}$, then $\dim \mathfrak{z} = 1$ (§ 1, Exercise 2) and $\dim \mathscr{D}\mathfrak{g} = 1$. If $\mathfrak{z} \neq \mathscr{D}\mathfrak{g}$, $\mathfrak{g}$ is the product of $\mathfrak{z}$ and the 2-dimensional non-commutative algebra. If $\mathfrak{z} = \mathscr{D}\mathfrak{g}$, $\mathfrak{g}$ is the 3-dimensional non-commutative nilpotent algebra (§ 4, Exercise 9 (a)).

(b) If $\mathfrak{z} = \{0\}$ but there exists in $\mathfrak{g}$ a 2-dimensional commutative subalgebra, this subalgebra is unique and is equal to $\mathscr{D}\mathfrak{g}$; for all $x \notin \mathscr{D}\mathfrak{g}$, the restriction $u$ of $\mathrm{ad}\,x$ to $\mathscr{D}\mathfrak{g}$ is a bijection of this vector space, determined to within a multiplicative constant. Conversely, every automorphism $u$ of a 2-dimensional vector space $Ka + Kb$ determines a Lie algebra structure on $g = Ka + Kb + Kc$ by the conditions $[a, b] = 0,\ [c, a] = u(a),\ [c, b] = u(b)$. For two Lie algebras thus defined by automorphisms $u_1, u_2$ to be isomorphic, it is necessary and sufficient that the matrices of $u_1$ and $u_2$ be similar to within a scalar factor.

(c) Suppose that there exists in $\mathfrak{g}$ no commutative subalgebra of dimension $> 1$. Show that there then exists $a \in \mathfrak{g}$ such that $a \notin (\mathrm{ad}\,a)\mathfrak{g}$ (assuming that an element $x$ does not have this property, show, using the Jacobi identity, that another element has the desired property). There then exists a basis $(a, b, c)$ of $\mathfrak{g}$ such that $[a, b] = c,\ [a, c] = \beta b,\ [b, c] = \gamma a$, with $\beta \neq 0$ and $\gamma \neq 0$, and $\mathfrak{g}$ is simple. Let $\phi$ be a canonical isomorphism of $\mathfrak{g}$ onto the exterior power $\bigwedge^2 \mathfrak{g}^*$ of the dual space of $\mathfrak{g}$, determined to within an invertible constant factor (Algebra, Chapter III, § 8, no. 5); let $u$ be the linear mapping of $\mathfrak{g}$ into $\mathfrak{g}^*$ defined by the condition $\langle [x, y], u(z) \rangle = \langle x \wedge y, \phi(z) \rangle$; the bilinear form $\Phi(x, y) = \langle x, u(y) \rangle$ on $\mathfrak{g} \times \mathfrak{g}$ is symmetric and non-degenerate and $2\Phi$ is the Killing form of $\mathfrak{g}$ to within a factor $\neq 0$. For two 3-dimensional simple Lie algebras over $K$ to be isomorphic, it is necessary and sufficient that the corresponding bilinear forms be equivalent to within a constant factor $\neq 0$.

(d) If $K$ is not of characteristic 2, the simple algebra $\mathfrak{g}$ defined in (c) is isomorphic to the Lie algebra $\mathfrak{o}(\Phi)$ of the formal orthogonal group $\mathbf{G}(\Phi)$ (§ 1, Exercise 26 (a)). For the existence in $\mathfrak{g}$ of vectors $x$ such that $\mathrm{ad}\,x$ admits eigenvectors not proportional to $x$, it is necessary and sufficient that $\Phi$ be of index > 0; g then admits a basis (a, b, c) such that [a, b] = b, [a, c] = -c, [b, c] = a.

(e) If K is of characteristic 2, show that there is no 2-mapping on g and therefore that g is not the Lie algebra of a formal group. Show that g admits derivations which are not inner.