(a) g is semi-simple. (b) The radical r of g is zero. (c) The Killing form $\beta$ of g is non-degenerate.

Moreover, a semi-simple Lie algebra is equal to its derived ideal.

(a) $\Rightarrow$ (b): for if $r \neq \{0\}$, the last non-zero derived algebra of r is a commutative ideal of g.

(b) $\Rightarrow$ (c): this follows from Proposition 5 (b) of § 5, no. 5 (which at the same time proves the last assertion of the theorem).

(c) $\Rightarrow$ (a): this follows from Proposition 6 (b) of § 4, no. 4.

The linear form $x \mapsto \operatorname{Tr} \rho(x)$ ($x \in g$) is zero when x is of the form $[y, z]$ ($y \in g, z \in g$) and hence on $\mathscr{D}g = g$.

The orthogonal of $g$ with respect to this form is a solvable ideal ($§ 5$, no. 4, Theorem 2) and hence is zero.

Let $\beta'$ be the restriction of $\beta$ to $a$: it is the bilinear form associated with the representation $x \mapsto \operatorname{ad}_g x$ of $a$ on the space $g$. This representation is faithful and hence $\beta'$ is non-degenerate (Proposition 1). Hence $h$ is supplementary to $a$ in $g$. On the other hand, if $x, y$ are in $a$ and $z \in h$, then

since $[x, y] \in a$, and hence $[y, z] \in h$, which proves that $[a, h] \subset h$. If $a$ is an ideal of $g$, we know that $h$ is an ideal of $g$ ($§ 3$, Proposition 7) and $g$ is identified with $a \times h$. As the centre of $a$ is zero, the centralizer of $a$ in $g$ is $h$.

This follows immediately from Corollary 1.

$\operatorname{ad} g$ is isomorphic to $g$ and hence semi-simple and is an ideal of the Lie algebra $\delta$ of derivations of $g$ ($§ 1$, Proposition 1). If $D \in \delta$ commutes with the elements of $\operatorname{ad} g$, then, for all $x \in g$, $\operatorname{ad} D(x) = [D, \operatorname{ad} x] = 0$, whence $D(x) = 0$; hence $D = 0$. Corollary 3 then follows from Corollary 1.

Let $a$ be an ideal of $g$. The orthogonal $b$ of $a$ in $g$ with respect to the Killing form is an ideal of $g$ and $a \cap b$ is a commutative ideal ($§ 3$, no. 6, Proposition 7) and hence zero. Hence $b$ is supplementary to $a$ in $g$. Moreover, as the Killing form of $g$ is non-degenerate, so are its restrictions to $a$ and $b$ (Algebra, Chapter IX, $§ 4$, no. 1, Corollary to Proposition 1) and hence $a$ and $b$ are semi-simple (no. 1, Theorem 1 and $§ 3$, no. 6, Proposition 9).

Clearly (a) implies (b). Suppose that (b) holds. Let $\sigma$ be a finite-dimensional representation of $g$ on a vector space M and N a vector subspace which is stable under $\sigma(g)$. Let $\mu$ be the representation of $g$ on $\mathscr{L}(M)$ canonically derived from $\sigma$ (§ 3, no. 3): recall that $\mu(x) = \mathrm{ad}_{\mathscr{L}(M)} \rho(x)$. Let V (resp. W) be the subspace of $\mathscr{L}(M)$ consisting of the linear mappings of M into N whose restriction to N is a homothety (resp. zero); then W is of codimension 1 in V and $\mu(x)(V) \subset W$ for all $x \in g$. By condition (b), there exists $u \in V$ which is annihilated by $\mu(x)$ for all $x \in g$ and whose restriction to N is a non-zero homothety. By multiplying $u$ by a suitable scalar, it can be assumed that $u$ is a projector of M onto N. To say that $\mu(x) \cdot u = 0$ means that $u$ is permutable with $\sigma(x)$. Hence the kernel of $u$ is a supplement of N in M which is stable under $\sigma(x)$ for all $x \in g$. Hence $\sigma$ is semi-simple.

For all $x \in g$ let $\sigma(x)$ be the restriction of $\rho(x)$ to W. Suppose first that $\sigma$ is simple. If $\sigma = 0$, then $\rho(x)\rho(y) = 0$ for all $x, y$ in $g$, hence $\rho(g) = \rho(\mathscr{D}g) = \{0\}$ and our assertion is obvious. If $\sigma \neq 0$, let n be the kernel of $\sigma$ and let m be a supplementary ideal of n in $g$ (Lemma 1); then $m \neq \{0\}$ and the restriction of $\sigma$ to m is faithful; the restriction to m of the bilinear form associated with $\sigma$ is non-degenerate (Proposition 1) and hence the Casimir element c associated with m and $\sigma$ can be formed. By Proposition 12 of § 3, no. 7, $\sigma(c)$ is an automorphism of W. On the other hand, $\rho(c)(V) \subset W$. Hence the kernel Z of $\rho(c)$ is a supplementary line of W; since c belongs to the centre of the enveloping algebra of $g$, $\rho(c)$ is permutable with $\rho(x)$ for all $x \in g$ and hence Z is stable under $\rho(g)$.

In the general case we argue by induction on the dimension of V. Let T be a minimal non-zero stable subspace of W. Let $\rho'$ be the quotient representation on $V' = V/T$. Then, for all $x \in g$, $\rho'(x)(V') \subset W'$, where $W' = W/T$ is of codimension 1 in $V'$. By the induction hypothesis there exists a line $Z'$ which is supplementary to $W'$ and stable under $\rho'(g)$. Its inverse image Z in V is stable under $\rho(g)$, contains T as subspace of codimension 1, $Z \cap W = T$, and hence $\rho(x)(Z) \subset T$ for all $x \in g$. By what was proved above, there exists a supplementary line of T in Z which is stable under $\rho(g)$; this line is supplementary to W in V, which completes the proof.

This follows from Lemmas 2 and 3.

The condition is sufficient (no. 1, Remark 3). Conversely, suppose that g is semi-simple. Since the adjoint representation of g is semi-simple, g is the direct sum of minimal non-zero ideals $a_1, \ldots, a_m$. Then g is identified with the product algebra of the $a_i$ (§ 1, no. 1). Every ideal of $a_i$ is then an ideal of g and hence zero or equal to $a_i$. On the other hand $a_i$ is non-commutative. Hence the $a_i$ are simple Lie algebras.

$g = a_1 \times \cdots \times a_m$, where the $a_i$ are simple. As the centre of $a_i$ is zero, the centralizer of $a_i$ in g is the product of the $a_j$ for $j \neq i$. Then let a be an ideal of g. If it does not contain $a_i$, then $a \cap a_i = \{0\}$, hence $[a, a_i] = \{0\}$ and a is contained in the product of the $a_j$ for $j \neq i$. It follows that a is the product of certain of the $a_i$. Hence the simple ideals of g are precisely the $a_i$.

The simple ideals of a semi-simple Lie algebra are called the simple components of g.

As $f(r)$ is solvable, $f(r) \subset r'$. On the other hand, $g/r$ is semi-simple (§ 5, no. 2, Proposition 3), hence $g'/f(r)$, which is isomorphic to a quotient of $g/r$, is semi-simple (Lemma 1) and hence $f(r) \supset r'$ (§ 5, no. 2, Proposition 3).

If $K_1$ is an extension of K, the Killing form of $g_{(K_1)}$ is the extension to $g_{(K_1)}$ of that of $g$ (§ 3, no. 8) and hence is non-degenerate; therefore $g_{(K_1)}$ is semi-simple. It therefore suffices to prove Proposition 3 when the base field is algebraically closed, which we shall henceforth assume to be the case.

For every subspace N of M, let $g_N$ be the subalgebra of $\mathfrak{gl}(M)$ consisting of the elements which leave N stable and whose restriction to N has trace zero. As $g = \mathscr{D}g$, $g \subset g_N$ if N is stable under $g$. Then let $g^*$ be the intersection of the normalizer of $g$ in $\mathfrak{gl}(M)$ and the algebras $g_N$ where N runs through the set of subspaces of M which are stable under g. As the semi-simple (resp. nilpotent) component s (resp. n) of $x \in gl(M)$ is a polynomial in x with no constant term and ad s (resp. ad n) is the semi-simple (resp. nilpotent) part of ad x ($§ 5$, no. 4, Lemma 2), clearly $x \in g^*$ implies $s \in g^*$ and $n \in g^*$; it therefore suffices to show that $g^* = g$. Since g is a semi-simple ideal of $g^*$, $g^* = a \times g$ (no. 1, Corollary 1 to Proposition 1). Let $a \in a$ and let N be a subspace which is minimal among the non-zero subspaces of M which are stable under g. The restriction of a to N is a scalar multiple of the identity by Burnside’s Theorem, has trace zero by construction and hence is zero since K is of characteristic 0. As M is the direct sum of subspaces such as N, it follows that $a = 0$ and hence $g^* = g$.

Let s (resp. n) be the semi-simple (resp. nilpotent) component of $x \in g$. Then $s \in g$ and $n \in g$ (Proposition 3). Then $ad_g s$ (resp. $ad_g n$) is the semi-simple (resp. nilpotent) component of $ad_g x$, by Lemma 2 of $§ 5$, no. 4. If x is semi-simple (resp. nilpotent) so then is $ad_g x$. If now $ad_g x$ is semi-simple (resp. nilpotent), it is equal to $ad_g s$ (resp. $ad_g n$) and hence $x = s$ (resp. $x = n$) since the adjoint representation of g is faithful.

If $\rho$ is a representation of g', $\rho \circ f$ is a representation of g, whence the first assertion. If f is surjective, there exists a homomorphism g of g' into g such that $f \circ g$ is the identity homomorphism of g' (no. 1, Corollary 2 to Proposition 1) and the second assertion then follows from the first.

(a) Let $x \in g$. If there exists a faithful representation $\rho$ of g such that $\rho(x)$ is a semi-simple (resp. nilpotent) endomorphism, then x is semi-simple (resp. nilpotent).

(b) Every element of g can be written uniquely as the sum of a semi-simple element and a nilpotent element which commute with one another.

Suppose that the hypothesis of (a) holds. Let $\sigma$ be a representation of g, b the supplementary ideal of the kernel of $\sigma$ and $\alpha$ the projection of g onto b. Then $ad_g x$ is semi-simple (resp. nilpotent) by the Corollary to Proposition 3 and hence $ad_b \alpha(x)$ is semi-simple (resp. nilpotent). As $\sigma(x) = \sigma(\alpha(x))$, the first assertion follows from the Corollary to Proposition 3. The second result then follows from Proposition 3 applied to a faithful representation.

(a) $\Rightarrow$ (b): if the adjoint representation of $g$ is semi-simple, $g$ is a direct sum of minimal non-zero ideals $a_i$ and hence $g$ is isomorphic to the product of the $a_i$; and $a_i$ has no ideals other than $\{0\}$ and $a_i$ and hence is simple or commutative of dimension 1. Therefore $\mathscr{D}g$ is equal to the product of those $a_i$ which are simple and hence is semi-simple.

(b) $\Rightarrow$ (c): if $\mathscr{D}g$ is semi-simple, $g$ is isomorphic to the product of $\mathscr{D}g$ by a Lie algebra $h$ (no. 1, Corollary 1 to Proposition 1); $h$ is isomorphic to $g/\mathscr{D}g$ and hence is commutative.

(c) $\Rightarrow$ (d): let $g_1$ and $g_2$ be two Lie algebras, $\rho_i$ a finite-dimensional representation of $g_i$ and $\beta_i$ the bilinear form on $g_i$ associated with $\rho_i$ ($i = 1, 2$); $\rho_1$ and $\rho_2$ can be considered as representations of $g = g_1 \times g_2$; let $\rho$ be their direct sum. Clearly the bilinear form on $g$ associated with $\rho$ is the direct sum of $\beta_1$ and $\beta_2$ and hence is non-degenerate if $\beta_1$ and $\beta_2$ are non-degenerate. Then to prove the implication (c) $\Rightarrow$ (d) it suffices to consider the 2 following cases: (1) $g$ is semi-simple; then the adjoint representation admits as associated form the Killing form, which is non-degenerate; (2) $g = K$; then the identity representation of $g$ on $K$ has an associated bilinear form which is non-degenerate.

(d) $\Rightarrow$ (e): let $\rho$ be a finite-dimensional representation of $g$ and $\beta$ the associated bilinear form; by Proposition 4 of § 4, no. 3, there exists a finite-dimensional semi-simple representation $\sigma$ of $g$ such that the kernel $n$ of $\sigma$ is orthogonal to $g$ with respect to $\beta$. If $\beta$ is non-degenerate, then $n = \{0\}$ and hence $\sigma$ is faithful.

(e) $\Rightarrow$ (f): this is obvious.

(f) $\Rightarrow$ (g): if the nilpotent radical of $g$ is zero, $\mathscr{D}g \cap r$ is zero ($§ 5$, no. 3, Theorem 1); as $[g, r] \subset \mathscr{D}g \cap r$, $r$ is the centre of $g$.

(g) ⇒ (a): if r is the centre of g, the adjoint representation of g is identified with a representation of g/r, which is a semi-simple Lie algebra (§ 5, no. 2, Proposition 3); this representation is therefore semi-simple (Theorem 2).

Assertion (a) follows for example from condition (c) of Proposition 5.

Suppose that g is reductive. Let a be an ideal of g. Since the adjoint representation of g is semi-simple, a has a supplementary ideal b and g is identified with a × b. For all $x \in g$, let $\rho(x)$ be the restriction of $\mathrm{ad}_g x$ to a. Then $\rho$ is a semi-simple representation of g which is zero on b and defines on passing to the quotient the adjoint representation on a. Hence a is reductive. Similarly, g/a and b, which are isomorphic, are reductive. Finally, let d, d' be the centres of a and b; then $a = d \times \mathscr{D}a, b = d' \times \mathscr{D}b, d \times d' = c, \mathscr{D}a \times \mathscr{D}b = \mathscr{D}g$; hence $a = (a \cap c) + (a \cap \mathscr{D}g)$.

Clearly $[g, r] \subset \mathscr{D}g \cap r$. Now $\mathscr{D}g \cap r = s$ by Theorem 1 of § 5, no. 3. Let $g' = g/[g, r]$ and f be the canonical homomorphism of g onto $g'$; then $f(r)$ is the radical $r'$ of $g'$ (Corollary 3 to Proposition 2, no. 2), hence $[g', r'] = \{0\}$ and $r'$ is the centre of $g'$; therefore (Proposition 5) $g'$ has a finite-dimensional faithful semi-simple representation, whence $s \subset [g, r]$. This proves (a).

Let t be the intersection of the orthogonals of g with respect to the bilinear forms associated with the finite-dimensional representations of g. Then $s \subset t$ (§ 4, no. 3, Proposition 4 (d)). On the other hand, $g/s$ has a finite-dimensional faithful semi-simple representation and hence (Proposition 5) a finite-dimensional representation $\rho$ such that the associated bilinear form is non-degenerate; considered as a representation of g, $\rho$ has an associated bilinear form $\beta$ on g and the orthogonal of g with respect to $\beta$ is s, whence $t \subset s$. Hence $t = s$.

Even if $s \neq \{0\}$ there may exist non-degenerate symmetric bilinear forms on g (Exercise 18 (c)). Such forms, of course, are not associated with any representation of g.

If $r, r'$ are the radicals of $g, g'$, then $s' = [g', r'] = [f(g), f(r)] = f([g, r]) = f(s)$. Assertion (b) is an immediate consequence of (a).

(a) $\Rightarrow$ (b): if $\rho$ is semi-simple, $g'$ is reductive (Proposition 5); the associative algebra generated by 1 and $g'$ is semi-simple (\emph{Algebra}, Chapter VIII, § 5, no. 1, Proposition 3), hence its centre is semi-simple (\emph{loc. cit.}, § 5, no. 4, Proposition 12) and hence the elements of this centre are semi-simple (\emph{loc. cit.}, § 9, no. 1, Proposition 2).

(b) $\Rightarrow$ (c): if $g'$ is reductive, its centre is equal to its radical, that is $r'$, whence the implication (b) $\Rightarrow$ (c).

(c) $\Rightarrow$ (d): suppose that $r'$ consists of semi-simple endomorphisms. As $[g', r']$ consists of nilpotent endomorphisms (no. 4, Proposition 6), $[g', r'] = \{0\}$. Then the implication (c) $\Rightarrow$ (d) follows from \emph{Algebra}, Chapter VIII, § 9, no. 2, Theorem 1.

(d) $\Rightarrow$ (a): let $s$ be the nilpotent radical of $g$ and $\rho'$ the restriction of $\rho$ to $r$. The elements of $\rho(s)$ are nilpotent and hence $s$ is contained in the largest nilpotency ideal of $\rho'$. As $\rho'$ is semi-simple, $\rho'(s) = \{0\}$ and $g'$ is reductive (Corollary to Proposition 6), so that $g' = a' \times r'$ with $a'$ semi-simple (Proposition 5). Let $A'$ (resp. $R'$) be the associative algebra generated by 1 and $a'$ (resp. $r'$). It is semi-simple (\emph{Algebra}, Chapter VIII, § 5, no. 1, Proposition 3), hence $A' \otimes_K R'$ is semi-simple (\emph{loc. cit.}, § 7, no. 6, Corollary 4 to Theorem 3) and hence the associative algebra generated by 1 and $g'$, which is a quotient of $A' \otimes_K R'$, is semi-simple, which proves that $\rho$ is semi-simple.

Let $r$ be the radical of $g$. For $x \in r$, $\rho(x)$ and $\rho'(x)$ are semi-simple (Theorem 4), hence $\rho(x) \otimes 1 + 1 \otimes \rho'(x)$ is semi-simple (\emph{Algebra}, Chapter VIII, § 9, Corollary to Theorem 1) and hence the tensor product of $\rho$ and $\rho'$ is semi-simple (Theorem 4).

Let $T^n$ be the subspace of $T$ consisting of the homogeneous tensors of order $n$. This subspace is stable under $\sigma_T$ and the representation defined by $\sigma_T$ on $T^n$ is semi-simple (Corollary 1). Hence the corollary for $\sigma_T$ and therefore for $\sigma_S$, which is a quotient representation of $\sigma_T$.

The $g$-module $\mathscr{L}_K(M, M')$ is canonically identified with the $g$-module $M^* \otimes_K M'$ (§ 3, no. 3, Proposition 4), so that Corollary 3 follows from Corollary 1.

Passing to the quotient by the kernel of $\rho$, $\rho$ can be assumed to be faithful. Then $g$ is reductive. Let $g = g_1 \times g_2$, where $g_1$ is the centre of $g$ and $g_2$ is semi-simple. Then $a = a_1 \times a_2$, where $a_1 \subset g_1$, $a_2 \subset g_2$ and $a_1$ is the centre of $a$. The elements of $\rho(g_1)$, and in particular those of $\rho(a_1)$, are semi-simple (Theorem 4) and hence $\rho'$ is semi-simple (Theorem 4). Hence (a). Assertion (b) follows from (a), using § 3, no. 1, Corollary to Proposition 1.

Let $W_0$ be a finite-dimensional simple sub-$h$-module of $V$. We need to prove that $\rho(x)(W_0) \subset W$ for all $x \in g$. Let $M$ denote the vector space $g$ considered as an $h$-module by means of the representation $x \mapsto \operatorname{ad}_g x$ of $h$ on $g$. Then $M \otimes_K W_0$ is a semi-simple $h$-module (Corollary 1 to Theorem 4). Let $\theta$ be the K-linear mapping of $M \otimes_K W_0$ into V defined by $\theta(x \otimes w) = \rho(x)w$. This is an $\mathfrak{h}$-module homomorphism, for if $y \in \mathfrak{h}$ then:

Hence $\theta(M \otimes_K W_0)$ is a finite-dimensional semi-simple $\mathfrak{h}$-module. Hence $\theta(M \otimes_K W_0) \subset W$, that is $\rho(x)(W_0) \subset W$ for all $x \in g$.

It suffices to consider the case where $\rho$ is simple. We adopt the notation V, W of Proposition 4. Let $W_1$ be a subspace of V minimal among the non-zero subspaces stable under $\rho(\mathfrak{h})$. Then $W_1 \subset W$, hence $W \neq \{0\}$ and hence $W = V$.

The representation $x \mapsto \mathrm{ad}_g x$ of $\mathfrak{h}$ on $g$ is semi-simple and hence its restriction to $\mathfrak{k}$ is semi-simple (Corollary 1).

The identity representation of $\mathrm{gl}(V)$ is simple, hence $\mathrm{gl}(V)$ is reductive and therefore $\mathrm{gl}(V)$ is the direct sum of its centre $c$ and its derived algebra $\mathscr{D}(\mathrm{gl}(V))$. The centre $c$ is the set of homotheties (Algebra, Chapter II, § 2, no. 5, Corollary 1 to Proposition 5). Clearly $\mathscr{D}(\mathrm{gl}(V)) \subset \mathrm{sl}(V)$. As $\mathrm{sl}(V) \cap c = \{0\}$, $\mathscr{D}(\mathrm{gl}(V)) = \mathrm{sl}(V)$. Hence $\mathrm{sl}(V)$ is semi-simple.

For all u ∈ gl(V) let u* denote its adjoint relative to β; then Tr(u) = Tr(u*) by Proposition 7 of Algebra, Chapter IX, § 1, no. 8. The condition

for all m, m' in V means that $u + u^* = 0$. In particular, if $v \in \mathrm{gl}(V)$ then $(v - v^*)^* = v^* - v$ and hence $v - v^* \in g$. Then let u be an element of g orthogonal to g with respect to the bilinear form φ associated with the identity representation of g. For all $v \in \mathrm{gl}(V)$, $\mathrm{Tr}\,u(v - v^*) = 0$, hence

and hence $\mathrm{Tr}(uw) = 0$. It follows that $u = 0$, so that φ is non-degenerate. Hence g is reductive (Proposition 5). It remains to show that the centre of g is zero (except when β is symmetric and $n = 2$). By extending the base field, we can assume that K is algebraically closed.

(a) When β is symmetric, it can be identified with the bilinear form on $K^n$ with matrix $I_n$ with respect to the canonical basis (Algebra, Chapter IX, § 6, Corollary 1 to Theorem 1). Under these conditions g is identified with the Lie algebra of skew-symmetric matrices ($§ 3$, no. 4, Example 1). Let $U = (u_{ij}) \in g$; we use the fact that U commutes with the matrix $(v_{ij}) \in g$ all of whose elements are zero except $v_{i_0j_0}$ and $v_{j_0i_0}$ ($i_0 \neq j_0$) which are equal respectively to 1 and −1. We find that $u_{i_0j} = u_{j_0i} = u_{i_0} = u_{j_0} = 0$ for $i \neq i_0, j_0$ and $j \neq i_0, j_0$. If $n > 2$, there exist, for all distinct indices $i_0$ and $j$, distinct indices i and $j_0$ such that $i \neq i_0, j_0 \neq j, j_0 \neq i_0$; hence $u_{i_0j} = 0$. This proves that an element of the centre of g is zero.

(b) When β is alternating and $n = 2m$, β can be identified with the bilinear form on $K^{2m}$ with matrix $\begin{pmatrix} 0 & I_m \\ -I_m & 0 \end{pmatrix}$ with respect to the canonical basis (Algebra, Chapter IX, § 5, Corollary 1 to Theorem 1). Under these conditions g is identified with the Lie algebra of matrices of the form $U = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ where $D = -{}^tA$, B and C are symmetric ($A, B, C, D$ in $\mathbf{M}_m(K)$) ($§ 3$, no. 4, Example 1). We use first the fact that U commutes with the matrix $\begin{pmatrix} X & 0 \\ 0 & -{}^tX \end{pmatrix}$, where $X \in \mathbf{M}_m(K)$. Then $AX = XA, CX = -{}^tXC, XB = -B\cdot{}^tX$; as these equalities must hold for all X, it follows that A is a scalar matrix $\lambda I_m$. We now use the fact that U commutes with the matrix $\begin{pmatrix} 0 & Y \\ 0 & 0 \end{pmatrix}$, where

Y is a symmetric matrix of $\mathbf{M}_m(\mathbf{K})$. Then $\lambda Y = YC = CY = 0$. This proves first that $\lambda = 0$. Moreover, for all $X \in \mathbf{M}_m(\mathbf{K})$, $X + {}^tX$ is symmetric and hence $XC = -{}^tXC$. Using the equation $CX = -{}^tXC$ obtained above, we see that $C$ commutes with every element of $\mathbf{M}_m(\mathbf{K})$ and hence that $C$ is a scalar matrix, necessarily zero since $YC = 0$. It is similarly shown that $B = 0$.

For $\beta$ symmetric and $n = 2$, $g$ is of dimension 1 and hence commutative. For the other cases cf. Exercises 25 and 26.