Exercise 16, § 13

Let $\mathfrak{g}$ be an absolutely simple Lie algebra of type $A_l(l\geq 2), \overline{k}$ an algebraic closure of $k$, and $\pi :$ Gal($\overline{k}/k$)$\rightarrow$ Aut( $\overline{R},\overline{B}$) the homomorphism defined in Exerc. 8 of §5.

a) Assume that $\pi$ is trivial. Show that there exist exactly two two-sided ideals $\mathfrak{m}$ and $\mathfrak{m}'$ of $U(\mathfrak{g})$ such that $D = U(\mathfrak{g})/\mathfrak{m}$ and $D'= U(\mathfrak{g})/\mathfrak{m}'$ are central simple algebras of dimension $(l+ 1)^2$ (use Exerc. 8 of §7). The principal anti-automorphism of $U(\mathfrak{g})$ interchanges $\mathfrak{m}$ and $\mathfrak{m}'$; in particular, $D'$ is isomorphic to the opposite of D. The composite $\mathfrak{g}\rightarrow U(\mathfrak{g})\rightarrow D$ identifies $\mathfrak{g}$ with the Lie subalgebra $\mathfrak{s}\mathfrak{l}_D$ of D consisting of the elements of trace zero. Moreover, $\mathfrak{g}$ is splittable (and hence isomorphic to $\mathfrak{s}\mathfrak{l}(l+ 1, k)$) if and only if D is isomorphic to $\mathbf{M}_{l+1}(k)$.

Conversely, if $\Delta$ is a central simple algebra of dimension $(l+ 1)^2$, the Lie algebra $\mathfrak{s}\mathfrak{l}_{\Delta}$ is absolutely simple of type $A_l$, and the corresponding homomor-

$^{21}$For more details on this exercise, see: V. W. GUILLEMIN, D. QUILLEN and S.

STERNBERG, The classification of the irreducible complex algebras of infinite

type, J. Analyse Math., Vol. XVII (1967), pp. 107-112. phism $\pi$ is trivial. Two such algebras $\mathfrak{s}\mathfrak{l}_{\Delta}$ and $\mathfrak{s}\mathfrak{l}_{\Delta'}$ are isomorphic if and only if $\Delta$ and $\Delta '$ are isomorphic or anti-isomorphic.

b) Assume that $\pi$ is non-trivial. Since Aut( $\overline{R},\overline{B}$) has two elements, the kernel of $\pi$ is an open subgroup of Gal($\overline{k}/k$) of index 2, which corresponds by Galois theory to a quadratic extension $k_1$ of $k$. Denote the non-trivial involution of $k_1$ by $x \rightarrow \overline{x}$.

Show that there exists a unique two-sided ideal $\mathfrak{m}$ of $U(\mathfrak{g})$ such that D = $U(\mathfrak{g})/\mathfrak{m}$ is a simple algebra of dimension $2(l+ 1)^2$ whose centre is a quadratic extension of $k$ (same method). The centre of D can be identified with $k_1$. The ideal $\mathfrak{m}$ is stable under the principal anti-automorphism of $U(\mathfrak{g})$; that defines by passage to the quotient an involutive anti-automorphism $\sigma$ of D such that $\sigma (x) = \overline{x}$ for all $x\in k_1$. The composite map $\mathfrak{g}\rightarrow U(\mathfrak{g})\rightarrow D$ identifies $\mathfrak{g}$ with the Lie subalgebra $\mathfrak{s}\mathfrak{u}_{D,\sigma}$ of D consisting of the elements $x$ such that $\sigma (x) =-x$ and Tr$_{D/k_1}(x) = 0$. Conversely, if $\Delta$ is a central simple $k_1$-algebra of dimension $(l+1)^2$, equipped with an involutive anti-automorphism $\sigma$ whose restriction to $k_1$ is $x \rightarrow \overline{x}$, the Lie algebra $\mathfrak{s}\mathfrak{u}_{\Delta ,\sigma}$ is absolutely simple of type $A_l$, and the corresponding homomorphism $\pi$ is that associated to $k_1$. Two such algebras $\mathfrak{s}\mathfrak{u}_{\Delta ,\sigma}$ and $\mathfrak{s}\mathfrak{u}_{\Delta',\sigma'}$ are isomorphic if and only if there exists a $k$-isomorphism $f:\Delta \rightarrow \Delta '$ such that $\sigma '\circ f=f\circ \sigma$.

Show that, when $D =\mathbf{M}_{l+1}(k_1)$, there exists an invertible hermitian matrix $H$ of degree $l+ 1$, unique up to multiplication by an element of $k^*$, such that

for all $x\in \mathbf{M}_{l+1}(k_1)$; the algebra $\mathfrak{g}$ can then be identified with the algebra $\mathfrak{s}\mathfrak{u}(l+ 1, H)$ consisting of the matrices $x$ such that $x.H+H.^t\overline{x}= 0$ and Tr($x$) $= 0$.