$\sigma^2=$ Id$_{\mathfrak{a}}, \sigma (\lambda x+\mu y) = \overline{\lambda \sigma}(x) + \overline{\mu\sigma}(y),[\sigma (x), \sigma (y)] =\sigma [x, y]$ (2)

for $\lambda , \mu\in \mathbf{C},x, y\in \mathfrak{a}$. An element $x$ of $\mathfrak{a}$ belongs to $\mathfrak{g}$ if and only if $\sigma (x) =x$.

b) Let $\sigma :\mathfrak{a}\rightarrow \mathfrak{a}$ be a map satisfying (2). Then the set $\mathfrak{g}$ of fixed points of $\sigma$ is a real form of $\mathfrak{a}$, and $\sigma$ is the conjugation of $\mathfrak{a}$ relative to $\mathfrak{g}$.

The proof is immediate.

Note that if B denotes the Killing form of $\mathfrak{a}$, and if $\mathfrak{g}$ is a real form of $\mathfrak{a}$, the restriction of B to $\mathfrak{g}$ is the Killing form of $\mathfrak{g}$; in particular, B is real-valued on $\mathfrak{g}\times \mathfrak{g}$. Assume that $\mathfrak{a}$ is reductive; then the real Lie algebra $\mathfrak{g}$ is compact if and only if the restriction of B to $\mathfrak{g}$ is negative (§1, no. 3). In that case we say that $\mathfrak{g}$ is a compact real form of $\mathfrak{a}$.

real form of $\mathfrak{a}$, of which $\mathfrak{h}_0$ is a Cartan subalgebra. The pair $(\mathfrak{a}_0,\mathfrak{h}_0)$ is a split semi-simple real Lie algebra, of which $(X_{\alpha})$ is a Chevalley system.

b) Let $\sigma$ be the conjugation of $\mathfrak{a}$ relative to $\mathfrak{a}_0$. Then $\sigma \circ \theta =\theta \circ \sigma$. The set of fixed points of $\sigma \circ \theta$ is a compact real form $\mathfrak{a}_u$ of $\mathfrak{a}$, of which $i\mathfrak{h}_0$ is a Cartan subalgebra.

Part a) follows immediately from the preceding. We prove b). Since $\sigma \circ \theta$ and $\theta \circ \sigma$ are two semi-linear maps from $\mathfrak{a}$ to $\mathfrak{a}$ that coincide on $\mathfrak{a}_0$, they coincide. Now $\sigma \circ \theta$ satisfies conditions (2) of no. 1, hence is the conjugation of $\mathfrak{a}$ relative to the real form $\mathfrak{a}_u$ consisting of the $x\in \mathfrak{a}$ such that $\sigma \circ \theta (x) =x$ (Prop. 1). For all $\alpha \in R$ put

Then the $\mathbf{R}$-vector space $\mathfrak{a}_u$ is generated by $i\mathfrak{h}_0$, the $u_{\alpha}$ and the $v_{\alpha}$. More precisely, if we choose a chamber C of R, then

It is clear that $i\mathfrak{h}_0$ is a Cartan subalgebra of $\mathfrak{a}_u$, and it remains to prove that the restriction of B to $\mathfrak{a}_u$ is negative. Now $i\mathfrak{h}_0$ and the different subspaces of the form $\mathbf{R}u_{\alpha}\oplus \mathbf{R}v_{\alpha}$ are orthogonal with respect to B, by (5); the restriction of B to $i\mathfrak{h}_0$ is negative and

hence the conclusion.

Let $\tau$ be the conjugation of $\mathfrak{a}$ relative to $\mathfrak{g}$. By hypothesis $\tau (x) =x$ for $x\in i\mathfrak{h}_0$, so $\tau (x) =-x$ for $x\in \mathfrak{h}_0$. Thus, for all $\alpha \in R$ and all $h\in \mathfrak{h}_0$,

it follows that $[h, \tau (X_{\alpha})] =-\alpha (h)\tau (X_{\alpha})$ for all $h\in \mathfrak{h}_0$, hence also for all $h\in \mathfrak{h}$. Hence there exists $c_{\alpha}\in \mathbf{C}^*$ such that $\tau (X_{\alpha}) =c_{\alpha}X_{-\alpha}$. Since $[X_{\alpha}, X_{-\alpha}]\in \mathfrak{h}_0$, we have $[\tau (X_{\alpha}), \tau (X_{-\alpha})] =-[X_{\alpha}, X_{-\alpha}]$, so $c_{\alpha}c_{-\alpha}$ = 1; similarly, formulas (3) and (4) give $c_{\alpha+\beta}=c_{\alpha}c_{\beta}$ if $\alpha , \beta , \alpha +\beta$ are roots. By Chap. VI, §1, no. 6, Cor. 2 of Prop. 19, there exists a homomorphism $\delta : Q(R)\rightarrow \mathbf{C}^*$ such that $\delta (\alpha ) =c_{\alpha}$ for all $\alpha \in R$.

We now show that each $c_{\alpha}$ is strictly positive. Indeed, $c_{\alpha}B(X_{\alpha}, X_{-\alpha}) =$ $B(X_{\alpha}, \tau (X_{\alpha}))$, and since $B(X_{\alpha}, X_{-\alpha})$ is negative, it suffices to show that $B(z, \tau (z))<0$ for every non-zero element $z$ of $\mathfrak{a}$; but every element of $\mathfrak{a}$ can be written as $x+iy$, with $x$ and $y$ in $\mathfrak{g}$, and

hence the stated assertion, the restriction of B to $\mathfrak{g}$ being negative and separating by hypothesis.

It follows that the homomorphism $\delta$ takes values in $\mathbf{R}^*_+$; hence there exists a homomorphism $\gamma : Q(R)\rightarrow \mathbf{R}^*_+$ such that $\delta =\gamma^{-2}$. Then $f(\gamma )^{-1}(\mathfrak{g})$ is a real form of $\mathfrak{a}$; the corresponding conjugation is $\tau '=f(\gamma )^{-1}\circ \tau \circ f(\gamma )$. For all $\alpha \in R$, we have

and $\tau '(h) =\tau (h) =h$ for $h\in i\mathfrak{h}_0$; it follows that $\tau '$ is the conjugation with respect to $\mathfrak{a}_u$, and hence that $f(\gamma )^{-1}(\mathfrak{g}) =\mathfrak{a}_u$.

a$)\mathfrak{a}$ has compact (resp. splittable) real forms.

b) The group Int($\mathfrak{a}$) operates transitively on the set of compact (resp. splittable) real forms of $\mathfrak{a}$.

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{a}$. Then $(\mathfrak{a},\mathfrak{h})$ is split (Chap. VIII, §2, no. 1, Remark 2), and has a Chevalley system $(X_{\alpha})$ (Chap. VIII, §4, no. 4, Cor. of Prop. 5). Part a) now follows from Prop. 2. Let $\mathfrak{g}$ be a compact real form of $\mathfrak{a}$; we show that there exists $v\in$ Int($\mathfrak{a}$) such that $v(\mathfrak{a}_u) =\mathfrak{g}$. Let $\mathfrak{t}$ be a Cartan subalgebra of $\mathfrak{g}$; then $\mathfrak{t}_{(\mathbf{C})}$ is a Cartan subalgebra of $\mathfrak{a}$; since Int($\mathfrak{a}$) operates transitively on the set of Cartan subalgebras of $\mathfrak{a}$ (Chap. VII, §3, no. 2, Th. 1), we are reduced to the case in which $\mathfrak{t}_{(\mathbf{C})}=\mathfrak{h}$. Since $\mathfrak{g}$ is a compact form, the eigenvalues of the endomorphisms ad $h$, for $h\in \mathfrak{t}$, are purely imaginary (§1, no. 3, Prop. 1), so the roots $\alpha \in R$ map $\mathfrak{t}$ to $i\mathbf{R}$; this implies that $\mathfrak{t}=i\mathfrak{h}_0$. Then, by Prop. 3 (no. 2), there exists $v\in$ Int($\mathfrak{a}$) such that $v(\mathfrak{a}_u) =\mathfrak{g}$, hence b) in the case of compact forms. Finally, let $\mathfrak{m}_1$ and $\mathfrak{m}_2$ be two splittable real forms of $\mathfrak{a}$. There exist framings $(\mathfrak{m}_1,\mathfrak{h}_1,B_1,(X_{\alpha}^1))$ and $(\mathfrak{m}_2,\mathfrak{h}_2,B_2,(X_{\alpha}^2))$ (Chap. VIII, §4, no. 1). These extend in an obvious way to bases $e_1$ and $e_2$ of $\mathfrak{a}$. An automorphism of $\mathfrak{a}$ that maps $e_1$ to $e_2$ maps $\mathfrak{m}_1$ to $\mathfrak{m}_2$; thus, it suffices to apply Prop. 5 of Chap. VIII, §5, no. 3, to obtain the existence of an element $u$ of Aut$_0(\mathfrak{a}) =$ Int($\mathfrak{a}$) such that $u(\mathfrak{m}_1) =\mathfrak{m}_2$.

The condition is clearly necessary. Conversely, assume that $\mathfrak{g}_{(\mathbf{C})}$ and $\mathfrak{g}'_{(\mathbf{C})}$ are isomorphic. Let $\mathfrak{c}$ (resp. $\mathfrak{c}'$) be the centre of $\mathfrak{g}$ (resp. $\mathfrak{g}'$) and $\mathfrak{s}$ (resp. $\mathfrak{s}'$) the derived algebra of $\mathfrak{g}$ (resp. $\mathfrak{g}'$). Then $\mathfrak{c}_{(\mathbf{C})}$ and $\mathfrak{c}'_{(\mathbf{C})}$ are the centres of $\mathfrak{g}_{(\mathbf{C})}$ and $\mathfrak{g}'_{(\mathbf{C})}$, respectively, and hence are isomorphic; it follows that the commutative algebras $\mathfrak{c}$ and $\mathfrak{c}'$ are isomorphic. Similarly, $\mathfrak{s}_{(\mathbf{C})}$ and $\mathfrak{s}'_{(\mathbf{C})}$ are isomorphic, hence $\mathfrak{s}$ and $\mathfrak{s}'$, which are compact real forms of two isomorphic complex semi-simple Lie algebras, are isomorphic by Th. $1b)$.

(i) $\mathfrak{a}$ is reductive.

(ii) There exists a compact real Lie algebra $\mathfrak{g}$ such that $\mathfrak{a}$ is isomorphic to $\mathfrak{g}_{(\mathbf{C})}$.

(iii) There exists a compact Lie group G such that $\mathfrak{a}$ is isomorphic to $L(G)_{(\mathbf{C})}$.

By Def. 1 of §1, no. 3, conditions (ii) and (iii) are equivalent and imply (i). If $\mathfrak{a}$ is reductive, it is the direct product of a commutative algebra, which clearly has a compact real form, and a semi-simple algebra which has one by Th. $1a)$, hence (i) implies (ii).

Indeed, there exists a compact real form $\mathfrak{g}_1$ (resp. $\mathfrak{g}_2$) of $\mathfrak{a}_1$ (resp. $\mathfrak{a}_2$); then $\mathfrak{g}_1\times \mathfrak{g}_2$ is a compact real form of $\mathfrak{a}_1\times \mathfrak{a}_2$. The corollary now follows from Th. $1b)$, applied to $\mathfrak{a}_1,\mathfrak{a}_2$ and $\mathfrak{a}_1\times \mathfrak{a}_2$.

Note that it follows from Cor. 3 above that a compact real Lie algebra $\mathfrak{g}$ is simple if and only if the complex Lie algebra $\mathfrak{g}_{(\mathbf{C})}$ is simple. We say that $\mathfrak{g}$ is of type $A_n$, or $B_n,. .$., if $\mathfrak{g}_{(\mathbf{C})}$ is of type $A_n$, or $B_n,. .$. (Chap. VIII, §2, no. 2). By Cor. 1 above, two compact simple real Lie algebras are isomorphic if and only if they are of the same type.

Let G be an almost simple connected compact Lie group (Chap. III, §9, no. 8, Def. 3). We say that G is of type $A_n$, or $B_n,. .$., if its Lie algebra is of type $A_n$, or $B_n,. .$.. Two simply-connected almost simple compact Lie groups are isomorphic if and only if they are of the same type.

b) The algebras $\mathfrak{u}(\Phi )$ are the compact real forms of $\mathfrak{g}\mathfrak{l}(V)$.

Let $\Phi$ be a separating positive hermitian form on V. For all $x\in \mathfrak{g}\mathfrak{l}$(V), put $\sigma (x) =-x^*$ (where $x^*$ is the adjoint of $x$ relative to $\Phi$ ). Then $\sigma$ satisfies conditions (2) of Prop. 1 of no. 1, so the set $\mathfrak{u}(\Phi )$ (resp. $\mathfrak{s}\mathfrak{u}(\Phi )$) of fixed points of $\sigma$ on $\mathfrak{g}\mathfrak{l}(V)$ (resp. $\mathfrak{s}\mathfrak{l}(V)$) is a compact real form of $\mathfrak{g}\mathfrak{l}(V)$ (resp. $\mathfrak{s}\mathfrak{l}(V)$). Since $\mathbf{G}\mathbf{L}(V)$ operates transitively on the set of separating positive hermitian forms on V (Algebra, Chap. IX) and on the set of compact real forms of $\mathfrak{s}\mathfrak{l}(V)$ (no. 3, Th. 1 and Chap. VIII, §13, no. 1 (VII)), Prop. 4 is proved.

Indeed, every complex Lie algebra of type $A_n$ is isomorphic to $\mathfrak{s}\mathfrak{l}(n+ 1,\mathbf{C})$ (Chap. VIII, §13, no. 1).

The first assertion follows from the Cor. of Prop. 4 and Prop. 5. Since $\mathbf{S}\mathbf{U}(2,\mathbf{C})$ is homeomorphic to $\mathbf{S}_3($General Topology, Chap. VIII, §1, no. 4, Remark 4), hence is simply-connected (General Topology, Chap. XI, in preparation), every simply-connected compact semi-simple Lie group of rank 1 is isomorphic to $\mathbf{S}\mathbf{U}(2,\mathbf{C})$; every connected compact semi-simple Lie group of rank 1 that is not simply-connected is isomorphic to a quotient of $\mathbf{S}\mathbf{U}(2,\mathbf{C})$ by a subgroup of Z that does not reduce to the identity element, hence to $\mathbf{S}\mathbf{O}(3,\mathbf{R})$.