§ 3. Representations

1. Representations

Definition 1

Let $g$ be a Lie algebra over $K$ and $M$ a $K$ -module. A homomorphism of $g$ into the Lie Algebra $gl(M)$ is called a representation of $g$ on the module $M$ . An injective representation is called faithful. If $K$ is a field, the dimension (finite or infinite) of $M$ over $K$ is called the dimension of the representation. The representation $x \mapsto \operatorname{ad} x$ of $g$ on the $K$ -module $g$ is called the adjoint representation of $g$ .

A representation of $g$ on $M$ is thus a $K$ -linear mapping $\rho$ of $g$ into the endomorphism module of $M$ such that

\rho([x, y]) \cdot m = \rho(x) \rho(y) \cdot m - \rho(y) \rho(x) \cdot m

for all $x \in g, y \in g, m \in M$ .

Example

Let G be a real Lie group, g its Lie algebra and θ an analytic representation of G on a finite-dimensional real vector space E. Then the corresponding homomorphism of g into gl(E) is a representation of g on E.

Let U be the enveloping algebra of g. Proposition 1 of § 2, no. 1 defines a one-to-one correspondence between the set of representations of g on M and the set of representations of U on M. On the other hand we know (Algebra, Chapter VIII, § 13, no. 1) that there is an equivalence between the notion of representation of the associative algebra U and that of left U-module.

Definition 2

Let g be a Lie algebra over K and U its enveloping algebra. A unitary left module over U is called a left g-module, or simply a g-module.

If M is a g-module and x ∈ U, x_M will denote the homothety of M defined by x (cf. Algebra, Chapter VIII, § 1, no. 2).

A unitary right module over U is called a right g-module. Such a module is identified with a left U^0-module, that is (§ 2, no. 4) with a left g^0-module.

Let φ be the principal antiautomorphism of U. If M is a right g-module, a left g-module structure is defined on M by writing a.m = m.φ(a) for m ∈ M and a ∈ U.

The notions and results of the theory of modules can be translated into the language of representations:

(1) Two representations ρ and ρ' of g on M and M' are called similar or isomorphic if the g-modules M and M' are isomorphic. For this it is necessary and sufficient that there exist an isomorphism u of the K-module M onto the K-module M' such that

\rho'(x) = u \circ \rho(x) \circ u^{-1}

for all x ∈ g.

(2) For all i ∈ I, let ρ_i be a representation of g on M_i. Let M be the g-module the direct sum of the g-modules M_i. There is a corresponding representation ρ of g on M, called the direct sum of the ρ_i and denoted by $\sum_{i \in I} \rho_i$ (or $\rho_1 + \cdots + \rho_n$ in the case of n representations $\rho_1, \ldots, \rho_n$ ). If $m = (m_i)_{i \in I}$ is an element of M and x ∈ g, then $\rho(x).m = (\rho_i(x).m_i)_{i \in I}$ .

(3) A representation ρ of g on M is called simple or irreducible if the associated g-module is simple. It amounts to the same to say that there exists no sub-K-module of M (other than {0} and M) stable under all the $\rho(x), x \in g$ . A class of simple g-modules (Algebra, Chapter VIII, § 3, no. 2) defines a class of simple representations of g.

(4) A representation ρ of g on M is called semi-simple or completely reducible if the associated g-module is semi-simple. It amounts to the same to say that ρ is similar to a direct sum of simple representations or that every sub-K-module of

M stable under the $\rho(x)$ ( $x \in g$ ) has a supplement stable under the $\rho(x)$ ( $x \in g$ ) (cf. Algebra, Chapter VIII, § 3, no. 3).

(5) Let $\delta$ be a class of simple representations of $g$ corresponding to a class C of simple $g$ -modules. On the other hand let $\rho$ be a representation of $g$ on M. The isotypical component $M_C$ of species C of the $g$ -module M (Algebra, Chapter VIII, § 3, no. 4) is also called the isotypical component of M of species $\delta$ . This component is the sum of the sub-K-modules of M stable under the $\rho(x)$ and on which the $\rho(x)$ induce a representation of class $\delta$ ; it is the direct sum of certain of these submodules; if $M_C$ is of length $n$ , $\rho$ is said to contain $\delta$ n times. The sum of the different $M_C$ is direct; it is equal to M if and only if $\rho$ is semi-simple.

(6) Let $\rho, \rho'$ be two representations of $g$ . $\rho'$ is called a subrepresentation (resp. quotient representation) of $\rho$ if the module of $\rho'$ is a submodule (resp. quotient module) of the module of $\rho$ .

Let M be a K-module. The zero representation of $g$ on M defines on M a $g$ -module structure. With this structure M is called a trivial $g$ -module.

Let M be a $g$ -module. The quotient $g$ -modules of the sub- $g$ -modules of M are also the sub- $g$ -modules of the quotient modules of M: they are obtained by considering two sub- $g$ -modules U, $U'$ of M such that $U \supset U'$ and forming the $g$ -module $U/U'$ . Then if all the simple modules of the above type are isomorphic to a given simple $g$ -module N, M is called a pure $g$ -module of species N. If $\rho$ and $\sigma$ are the representations of $g$ corresponding to M and N, we also say that $\rho$ is pure of species $\sigma$ .

Let $M'$ be a sub- $g$ -module of M. For M to be pure of species N, it is necessary and sufficient that $M'$ and $M/M'$ be pure of species N. For the condition is obviously necessary. Suppose that it holds and let U, $U'$ be sub- $g$ -modules of M such that $U' \subset U$ and $U/U'$ is simple; let $\phi$ be the canonical homomorphism of M onto $M/M'$ ; if $\phi(U) \neq \phi(U')$ , $U/U'$ is isomorphism to $\phi(U)/\phi(U')$ and hence isomorphic to N; if $\phi(U) = \phi(U')$ , then $U \subset U' + M'$ , hence $U/U'$ is isomorphic to a simple submodule of $(U' + M')/U'$ and the latter module is itself isomorphic to $M'/(U' \cap M')$ ; hence $U/U'$ is again isomorphic to N, so that M is pure of species N.

Henceforth let M be a $g$ -module and suppose that the set of sub- $g$ -modules of M which are pure of species N admits a maximal element $M'$ . Then every submodule $M''$ of M which is pure of species N is contained in $M'$ . For $M''/(M' \cap M'')$ and $M'$ are pure of species N, hence $M' + M''$ is pure of species N by the above and hence $M' + M'' \subset M'$ .

Suppose that the $g$ -module M admits a Jordan-Hölder series $(M_i)_{0 \leq i \leq n}$ . For M to be pure of species N, it is necessary and sufficient that $M_0/M_1, M_1/M_2, \ldots, M_{n-1}/M_n$ be isomorphic to N; for the condition is obviously necessary and its sufficiency follows immediately by induction on $n$ from what we have seen above.

Proposition 1

Let g be a Lie algebra over K and a an ideal of g. Let M be a g-module* and $N$ a simple $a$ -module. Consider $M$ as an $a$ -module and suppose that the set of sub- $a$ -modules of $M$ which are pure of species $N$ admits a maximal element $M'$ . Then $M'$ is a sub- $g$ -module of $M$ .

Let $y \in g$ . Let $\phi$ be the canonical mapping of $M$ onto $M/M'$ and $f$ the mapping $m \mapsto \phi(y_M \cdot m)$ of $M'$ into $M/M'$ . It suffices to show that $f(M') = \{0\}$ . Let $x \in a$ . Then, for $m \in M$ ,

x_{M/M'} \cdot f(m) = \phi(x_M y_M \cdot m) = \phi(y_M x_M \cdot m) + \phi([x, y]_M \cdot m).

Now $[x, y] \in a$ , whence $\phi([x, y]_M \cdot m) = 0$ ; on the other hand,

\phi(y_M x_M \cdot m) = f(x_M \cdot m).

Hence $x_{M/M'} \cdot f(m) = f(x_M \cdot m)$ . It follows that $f(M')$ is a sub- $a$ -module of $M/M'$ isomorphic to a quotient of $M'$ and hence pure of species $N$ ; hence $f(M') = \{0\}$ .

Corollary

Let $g$ be a Lie algebra over $K$ and $a$ an ideal of $g$ . Let $M$ be a simple $g$ -module, of finite length as a $K$ -module. There exists a simple $a$ -module $N$ such that $M$ is a pure $a$ -module of species $N$ .

Since the $a$ -module $M$ is of finite length, there exists a minimal element $N$ in the set of sub- $a$ -modules of $M$ : it is a simple sub- $a$ -module of $M$ . The largest sub- $a$ -module of $M$ which is pure of species $N$ is therefore $\neq \{0\}$ and is a sub- $g$ -module of $M$ (Proposition 1) and is therefore identical with $M$ .

2. Tensor product of representations

We have defined in no. 1 the direct sum of a family of representations of $g$ . We shall now define other operations on representations.

Let $g_1, g_2$ be two Lie algebras over $K$ and $M_i$ a $g_i$ -module ( $i = 1, 2$ ). Let $U_i$ be the enveloping algebra of $g_i$ and $\sigma_i$ the canonical mapping of $g_i$ into $U_i$ . Then $M_i$ is a left $U_i$ -module and hence $M_1 \otimes_K M_2$ has a canonical left $(U_1 \otimes_K U_2)$ -module structure. Now $U_1 \otimes_K U_2$ is the enveloping algebra of $g_1 \times g_2$ and the mapping $(x_1, x_2) \mapsto \sigma_1(x_1) \otimes 1 + 1 \otimes \sigma_2(x_2)$ is the canonical mapping of $g_1 \times g_2$ into this enveloping algebra (§ 2, no. 2). Hence there exists a $(g_1 \times g_2)$ -module structure on $M = M_1 \otimes_K M_2$ such that:

\begin{align*} (1) \quad (x_1, x_2)_M \cdot (m_1 \otimes m_2) &= (\sigma_1(x_1) \otimes 1 + 1 \otimes \sigma_2(x_2)) \cdot (m_1 \otimes m_2) \\ &= ((x_1)_{M_1} \cdot m_1) \otimes m_2 + m_1 \otimes ((x_2)_{M_2} \cdot m_2). \end{align*}

This structure defines a representation of $g_1 \times g_2$ on $M$ .

If now $g_1 = g_2 = g$ , the homomorphism $x \mapsto (x, x)$ of $g$ into $g \times g$ , composed with the above representation, defines a representation of $g$ on $M$ and hence a $g$ -module structure on $M$ such that:

(2) \quad x_M \cdot (m_1 \otimes m_2) = (x_{M_1} \cdot m_1) \otimes m_2 + m_1 \otimes (x_{M_2} \cdot m_2).

By an analogous argument we see that:

Proposition 2

Let $g$ be a Lie algebra over $K$ and $M_i$ a $g$ -module ( $1 \leq i \leq n$ ). On the tensor product $M_1 \otimes_K M_2 \otimes \cdots \otimes M_n$ , there exists one and only one $g$ -module structure such that

x_M \cdot (m_1 \otimes \cdots \otimes m_m) = \sum_{t=1}^n m_1 \otimes \cdots \otimes (x_{M_t} \cdot m_i) \otimes \cdots \otimes m_n

for all $x \in g, m_1 \in M_1, \ldots, m_n \in M_n$ .

The corresponding representation is called the tensor product of the given representations of $g$ on the $M_i$ .

In particular, if $M$ is a $g$ -module, Proposition 2 defines a $g$ -module structure on each $M_p = \bigotimes^p M$ and hence on the tensor algebra $T$ of $M$ .

Formula (3) shows that, for all $x \in g, x_T$ is the unique derivation of the algebra $T$ which extends $x_M$ . We know (§ 2, no. 8) that $x_T$ defines on passing to the quotient a derivation of the symmetric algebra $S$ of $M$ . Hence $S$ can be considered as a quotient $g$ -module of $T$ and the $x_S$ are derivations of $S$ .

Still more particularly, consider $g$ as a $g$ -module by means of the adjoint representation of $g$ . Let $U$ be the enveloping algebra of $g$ . By Proposition 7 of § 2, $x_M$ defines on passing to the quotients a derivation of $U$ which is just the inner derivation defined by $\sigma(x)$ ( $\sigma$ denoting the canonical mapping of $g$ into $U$ ). Then $U$ can be considered as a quotient $g$ -module of $T$ . If $K$ is a field of characteristic 0, the canonical isomorphism of $S$ onto $U$ is a $g$ -module isomorphism (§ 2, no. 8).

3. Representations on homomorphism modules

Again let $g_1$ and $g_2$ be two Lie algebras over $K$ and $M_i$ a $g_i$ -module ( $i = 1, 2$ ). Let $U_i$ be the enveloping algebra of $g_i$ and $\sigma_i$ the canonical mapping of $g_i$ into $U_i$ . Then $M_i$ is a left $U_i$ -module and hence $\mathscr{L}_K(M_1, M_2)$ has a canonical left $(U_1^0 \otimes U_2)$ -module structure. Now $U_1^0 \otimes_K U_2$ is the enveloping algebra of $g_1^0 \times g_2$ and the mapping

(x_1, x_2) \mapsto \sigma_1(x_1) \otimes 1 + 1 \otimes \sigma_2(x_2)

is the canonical mapping of $g_1^0 \times g_2$ into this enveloping algebra. Hence there exists a $(g_1^0 \times g_2)$ -module structure on $M = \mathscr{L}_K(M_1, M_2)$ such that

((x_1, x_2)_M \cdot u) \cdot m_1 = ((\sigma_1(x_1) \otimes 1 + 1 \otimes \sigma_2(x_2)) \cdot u) \cdot m_1 \\ = u((x_1)_{M_1} \cdot m_1) + (x_2)_{M_2} \cdot u(m_1)

for all $u \in \mathscr{L}_K(M_1, M_2), m_1 \in M_1$ . This structure defines a representation of $g_1^0 \times g_2$ on $M$ .

If now $g_1 = g_2 = g$ , the homomorphism $x \mapsto (-x, x)$ of $g$ into $g^0 \times g$ , composed with the above representation, defines a representation of $g$ on $M$ and hence a $g$ -module structure on $M$ such that

(x_M \cdot u) \cdot m_1 = x_{M_2} \cdot u(m_1) - u(x_{M_1} \cdot m_1)

x_M \cdot u = x_{M_2} u - u x_{M_1}.

Combining these results with Proposition 2, we see that:

Proposition 3

Let $g$ be a Lie algebra over $K$ and $M_i$ a $g$ -module ( $1 \leq i \leq n + 1$ ). Let $N$ be the $K$ -module $\mathscr{L}_K(M_1, \ldots, M_n; M_{n+1})$ of multilinear mappings of $\prod_{i=1}^n M_i$ into $M_{n+1}$ . There exists one and only one $g$ -module structure on $N$ such that

(x_N \cdot u)(m_1, \ldots, m_n) = - \sum_{i=1}^n u(m_1, \ldots, x_{M_i} \cdot m_i, \ldots, m_n)

+ x_{M_{n+1}} \cdot u(m_1, \ldots, m_n)

for all $x \in g, u \in N$ and $m_i \in M_i$ ( $1 \leq i \leq n$ ).

In particular, let $g$ be a Lie algebra over $K$ and $M$ a $g$ -module and consider $K$ as a trivial $g$ -module. Proposition 3 defines a $g$ -module structure on $\mathscr{L}_K(M, K) = M^*$ . The corresponding representation is called the dual representation of the representation $x \mapsto x_M$ . We have:

(x_{M^*} \cdot f)(m) = -f(x_M \cdot m)

for all $x \in g, f \in M^*, m \in M$ . In other words:

x_{M^*} = -{}^t x_M.

When $K$ is a field and $M$ is finite-dimensional, the $g$ -module $M$ is simple (resp. semi-simple) if and only if the $g$ -module $M^*$ is simple (resp. semi-simple).

Proposition 4

Let $M_1, M_2$ be two $g$ -modules. The canonical $K$ -linear mappings* (Algebra, Chapter II, § 4, no. 2, Proposition 2 and no. 1, Proposition 1):

M_1^* \otimes_K M_2 \xrightarrow{\phi} \mathscr{L}_K(M_1, M_2), \quad \mathscr{L}_K(M_1, M_2^*) \xrightarrow{\psi} (M_1 \otimes_K M_2)^*

(where the second is bijective) are $g$ -module homomorphisms.

We write

N = M_1^* \otimes M_2, \quad P = \mathscr{L}(M_1, M_2), \quad Q = \mathscr{L}(M_1, M_2^*), \quad R = (M_1 \otimes M_2)^*.

Then, for $x \in g, f \in M_1^*, m_1 \in M_1, m_2 \in M_2$ ,

((\phi x_N)(f \otimes m_2)) \cdot m_1 = (\phi(x_{M_1^*} f \otimes m_2 + f \otimes x_{M_2} m_2)) \cdot m_1 \\ = \langle x_{M_1^*} f, m_1 \rangle m_2 + \langle f, m_1 \rangle x_{M_2} m_2

((x_P \phi)(f \otimes m_2)) \cdot m_1 = x_{M_2}(\phi(f \otimes m_2) \cdot m_1) - \phi(f \otimes m_2)(x_{M_1} m_1) \\ = \langle f, m_1 \rangle x_{M_2} m_2 - \langle f, x_{M_1} m_1 \rangle m_2

and hence $\phi x_N = x_P \phi$ . On the other hand, for $x \in g, u \in \mathscr{L}(M_1, M_2^*), m_1 \in M_1, m_2 \in M_2$ :

(\psi x_Q u)(m_1 \otimes m_2) = \langle (x_Q u) \cdot m_1, m_2 \rangle = \langle x_{M_2^*} u m_1 - u x_{M_1} m_1, m_2 \rangle \\ (x_R \psi u)(m_1 \otimes m_2) = -\langle \psi u, x_{M_1} m_1 \otimes m_2 + m_1 \otimes x_{M_2} m_2 \rangle \\ = -\langle u x_{M_1} m_1, m_2 \rangle - \langle u m_1, x_{M_2} m_2 \rangle

and hence $\psi x_Q = x_R \psi$ , which completes the proof.

The $g$ -modules $\mathscr{L}(M_1, M_2^*)$ and $(M_1 \otimes M_2)^*$ are identified under the isomorphism $\psi$ . If $M_1$ and $M_2$ have finite bases, $\phi$ is an isomorphism (Algebra, Chapter II, § 4, no. 2, Proposition 2), which allows us to identify the $g$ -modules $M_1^* \otimes M_2$ and $\mathscr{L}(M_1, M_2)$ ; in that case, we can therefore identify the $g$ -modules $M_1^* \otimes M_2^*, \mathscr{L}(M_1, M_2^*)$ and $(M_1 \otimes M_2)^*$ .

4. Examples

Example 1

Let $g$ be a Lie algebra over $K$ and $M$ a $g$ -module. The $g$ -module structure on $M$ and the trivial $g$ -module structure on $K$ define a $g$ -module structure on the $K$ -module $N = \mathscr{L}(M, M; K)$ of bilinear forms on $M$ . Then

(x_N \cdot \beta)(m, m') = -\beta(x_M \cdot m, m') + \beta(m, x_M \cdot m')

for all $x \in g, m, m'$ in $M, \beta \in N$ . If $\beta$ is a given element of $N$ , the set of $x \in g$ such that $x_N \cdot \beta = 0$ is a subalgebra of $g$ .

Let $M$ be a $K$ -module and $\beta$ a bilinear form on $M$ . By the above, the set of $x \in gl(M)$ such that

\beta(x \cdot m, m') + \beta(m, x \cdot m') = 0

for all $m \in M$ and $m' \in M$ is a Lie subalgebra of $gl(M)$ . Suppose that $K$ is a field, $M$ is finite-dimensional and $\beta$ is non-degenerate. Then every $x \in gl(M)$ admits a left adjoint $x^*$ (relative to $\beta$ ) which is everywhere defined and the subalgebra in question is the set of $x \in gl(M)$ such that $x^* = -x$ . By this process we can construct two important examples of Lie algebras:

(a) Take $M = K^n$ and

\beta((\xi_1, \ldots, \xi_n), (\eta_1, \ldots, \eta_n)) = \xi_1 \eta_1 + \cdots + \xi_n \eta_n.

We canonically identify $gl(K^n)$ with $M_n(K)$ . Then the Lie algebra obtained is the Lie algebra of skew-symmetric matrices. *(When $K = \mathbf{R}$ , this algebra is the Lie algebra of the orthogonal group $O(n, \mathbf{R})$ ).

(b) Take $M = K^{2m}$ and

\beta((\xi_1, \ldots, \xi_{2m}), (\eta_1, \ldots, \eta_{2m})) = \xi_1 \eta_{m+1} - \eta_1 \xi_{m+1} + \cdots + \xi_m \eta_{2m} - \eta_m \xi_{2m}.

The matrix of

\beta

with respect to the canonical basis of

K^{2m}

is the matrix

\begin{pmatrix} 0 & I_m \\ -I_m & 0 \end{pmatrix}.

Let

U = \begin{pmatrix} A & B \\ C & D \end{pmatrix}

be the matrix with respect to the canonical basis of

K^{2m}

of an element

u

\mathrm{gl}(M)

(

A, B, C, D

lying in

\mathbf{M}_m(K)

). By formula (50) of Algebra, Chapter IX, § 1, no. 10,

u^*

has with respect to the same basis the matrix

\begin{pmatrix} 0 & -I_m \\ I_m & 0 \end{pmatrix} \begin{pmatrix} tA & tC \\ tB & tD \end{pmatrix} \begin{pmatrix} 0 & I_m \\ -I_m & 0 \end{pmatrix} = \begin{pmatrix} tD & -tB \\ -tC & tA \end{pmatrix}.

The condition

u^* = -u

is therefore equivalent to the conditions

D = -tA \quad B = tB \quad C = tC.

*When

K = \mathbf{R}

, the Lie algebra obtained is the Lie algebra of the symplectic group

\mathbf{Sp}(2m, \mathbf{R})

Example 2

We preserve the notation of Example 1.

The $g$ -module structure on $M$ defines on the $K$ -module $P = \mathscr{L}_K(M, M)$ of endomorphisms of $M$ a $g$ -module structure. By (6), for all $x \in g$ and $u \in P$ :

(11)

x_P \cdot u = [x_M, u] = (\mathrm{ad}\, x_M) \cdot u

where

\mathrm{ad}\, x_M

denotes the image of

x_M

under the adjoint representation of

\mathrm{gl}(M)

. In other words:

(12)

x_P = \mathrm{ad}\, x_M

\mathscr{L}(\mathscr{L}(M, M)) = \mathscr{L}(\mathrm{gl}(M))

5. Invariant elements

Definition 3

Let $g$ be a Lie algebra and $M$ a $g$ -module. An element $m \in M$ is called invariant (with respect to the $g$ -module structure on $M$ or with respect to the corresponding representation of $g$ ) if $x_M \cdot m = 0$ for all $x \in g$ .

Let $G$ be a connected real Lie group, $g$ its Lie algebra, $\theta$ an analytic representation of $G$ on a finite-dimensional real vector space $E$ and $\rho$ the corresponding representation of $g$ on $E$ . Let $m \in E$ . The element $m$ is invariant with respect to $\rho$ if and only if $\theta(g) \cdot m = m$ for all $g \in G$ . This justifies the use of the word "invariant".

Example 1

Let $M, N$ be two $g$ -modules and $P = \mathscr{L}_K(M, N)$ . For an element $f$ of $P$ to be invariant, it is necessary and sufficient, by (6), that $f$ be a homomorphism of the $g$ -module $M$ into the $g$ -module $N$ . In particular, if $M = N$ and $x_M = x_N$ for all $x \in g$ , $f$ is invariant if and only if $f$ is permutable with the $x_M$ .

Example 2

Let M be a K-module with a finite basis. If M has a g-module structure, $\mathscr{L}(M, M)$ and $M^* \otimes M$ have g-module structures and the canonical mapping of $M^* \otimes M$ into $\mathscr{L}(M, M)$ is a g-module isomorphism (Proposition 4). As $1 \in \mathscr{L}(M, M)$ is obviously an invariant (cf. Example 1), the corresponding element $u$ of $M^* \otimes M$ is an invariant. If $(e_i)_{1 \leq i \leq n}$ is a basis of M and $(e_i^*)_{1 \leq i \leq n}$ is the dual basis, we have $u = \sum_{i=1}^n e_i^* \otimes e_i$ .

Example 3

Let M be a g-module. Let $\beta$ be a bilinear form on M and f the corresponding element of $\mathscr{L}(M, M^*)$ . For $\beta$ to be invariant, it is necessary and sufficient that f be a g-module homomorphism (Proposition 4 and Example 1). Suppose that K is a field and that $\dim_K M < +\infty$ . A non-degenerate invariant bilinear form $\beta$ on M defines an isomorphism of the g-module M onto the g-module $M^*$ and hence an isomorphism of the g-module $M \otimes M$ onto the g-module $M^* \otimes M$ . Thus, by Example 2, giving $\beta$ defines canonically an invariant element c in the g-module $M \otimes M$ , which can be constructed as follows: let $(e_i)_{1 \leq i \leq n}$ be a basis of M and $(e'_i)_{1 \leq i \leq n}$ the basis of M such that $\beta(e_i, e'_j) = \delta_{ij}$ ; then $c = \sum_{i=1}^n e_i \otimes e'_i$ .

Proposition 5

Let g be a Lie K-algebra, h an ideal of g, $\rho$ a representation of g on M and $\rho'$ the restriction of $\rho$ to h. Then the set N of elements of M invariant with respect to $\rho'$ is stable under $\rho(g)$ .

Let $n \in N$ and $y \in g$ ; for all $x \in h, [x, y] \in h$ and hence

\rho(x)\rho(y)n = \rho([x, y])n + \rho(y)\rho(x)n = 0;

hence

\rho(y)n \in N

Proposition 6

Let M be a semi-simple g-module. Then the submodule $M_0$ of invariant elements of M admits one and only one supplement stable under the $x_M$ , namely the submodule $M_1$ generated by the $x_M.m$ ( $x \in g, m \in M$ ).

Let $M'$ be a submodule of M which is stable under the $x_M$ and a supplement of $M_0$ in M. For all $m \in M, m = m_0 + m'$ with $m_0 \in M_0, m' \in M'$ , and hence $x_M m = x_M m' \in M'$ . Hence $M_1 \subset M'$ . Let $M_2$ be a submodule of $M'$ stable under the $x_M$ and supplementary to $M_1$ in $M'$ . For all $m \in M_2$ ,

x_M m \in M_2 \cap M_1 = \{0\}

for all

x \in g

, hence

m \in M_0

and hence

m = 0

. Hence

M_2 = \{0\}

, which proves that

M_1 = M'

6. Invariant bilinear forms

Let g be a Lie algebra over K. The adjoint representation of g on g and the zero representation of g on K define a g-module structure on the K-module

$N = \mathscr{L}(g, g; K)$ of bilinear forms on $g$ . Briefly we say that a bilinear form $\beta$ on $g$ is invariant if it is invariant under the representation $x \mapsto x_N$ . By formula (10) the necessary and sufficient condition for this to be so is that:

\beta([x, y], z) = \beta(x, [y, z])

for all $x, y, z$ in $g$ .

Now let $\mathfrak{d}$ be the Lie algebra of derivations of $g$ . The identity representation of $\mathfrak{d}$ and the zero representation of $\mathfrak{d}$ on $K$ define a representation $D \mapsto D_N$ of $\mathfrak{d}$ on $N$ . Briefly we say that a bilinear form on $g$ is completely invariant if it is invariant under the representation $D \mapsto D_N$ . A completely invariant bilinear form is invariant. For a bilinear form $\beta$ on $g$ to be completely invariant, it is necessary and sufficient that:

\beta(Dx, y) + \beta(x, Dy) = 0

for all $x, y$ in $g$ and $D \in \mathfrak{d}$ .

Proposition 7

Let $g$ be a Lie algebra, $\beta$ an invariant symmetric bilinear form on $g$ and $a$ an ideal of $g$ .

(a) The orthogonal $a'$ of $a$ with respect to $\beta$ is an ideal of $g$ . (b) If $a$ is characteristic and $\beta$ is completely invariant, $a'$ is characteristic. (c) If $\beta$ is non-degenerate, $a \cap a'$ is commutative.

Let $D$ be a derivation of $g$ . Suppose that $a$ is stable under $D$ and that $\beta(Dx, y) + \beta(x, Dy) = 0$ for $x, y$ in $g$ . Then $z \in a'$ implies $Dz \in a'$ , since, for all $t \in a$ , $Dt \in a$ and hence $\beta(Dz, t) = -\beta(z, Dt) = 0$ . Thus $a'$ is stable under $D$ . This establishes (a) and (b).

Now let $b$ be an ideal of $g$ and suppose that the restriction of $\beta$ to $b$ is zero. For $x, y$ in $b$ and $z \in g$ , $\beta([x, y], z) = \beta(x, [y, z]) = 0$ , for $[y, z] \in b$ . Thus $[b, b]$ is orthogonal to $g$ . If $\beta$ is non-degenerate, $b$ is therefore commutative. This result applied to $a \cap a'$ proves (c).

Definition 4

Let $g$ be a Lie $K$ -algebra and $M$ a $g$ -module. Suppose that $M$ , considered as a $K$ -module, admits a finite basis. The bilinear form associated with the $g$ -module $M$ (or with the corresponding representation) is the symmetric bilinear form $(x, y) \mapsto \operatorname{Tr}(x_M y_M)$ on $g$ . If the representation in question is the adjoint representation, the associated bilinear form is called the Killing form of $g$ .

Proposition 8

Let $g$ be a Lie algebra and $M$ a $g$ -module. Suppose that $M$ , considered as a $K$ -module, admits a finite basis. The bilinear form associated with $M$ is invariant.

For $x, y, z$ in $g$ , we have:

\begin{align*} \operatorname{Tr}([x, y]_M z_M) &= \operatorname{Tr}(x_M y_M z_M) - \operatorname{Tr}(y_M x_M z_M) = \operatorname{Tr}(x_M y_M z_M) - \operatorname{Tr}(x_M z_M y_M) \\ &= \operatorname{Tr}(x_M [y, z]_M). \end{align*}

Proposition 9

Suppose that $K$ is a field and that the Lie algebra $g$ is finite-dimensional* over $K$ . *Let $a$ be an ideal of $g$ , $\beta$ the Killing form of $g$ and $\beta'$ the Killing form of $a$ . Then $\beta'$ is the restriction of $\beta$ to $a$ .

Let $u$ be an endomorphism of the vector space $g$ which leaves $a$ stable. Let $v$ be the restriction of $u$ to $a$ and $w$ the endomorphism of the vector space $g/a$ derived from $u$ when passing to the quotient. Then $\operatorname{Tr} u = \operatorname{Tr} v + \operatorname{Tr} w$ as is seen by taking a basis $(x_1, \ldots, x_n)$ of $g$ of which the first $p$ elements form a basis of $a$ . Then let $x \in a, y \in a$ and apply the above formula to the case where $u = (\operatorname{ad}_g x)(\operatorname{ad}_g y)$ . Then $v = (\operatorname{ad}_a x)(\operatorname{ad}_a y)$ and $w = 0$ . Hence $\beta(x, y) = \beta'(x, y)$ .

Proposition 10

Suppose that $K$ is a field and that the Lie algebra $g$ is finite-dimensional over $K$ . The Killing form $\beta$ of $g$ is completely invariant.

Let $D$ be a derivation of $g$ . There exists a Lie algebra $g'$ containing $g$ as an ideal of codimension 1 and an element $x_0$ of $g'$ such that $Dx = [x_0, x]$ for all $x \in g$ (§ 1, no. 8, Example 1). Let $\beta'$ be the Killing form of $g'$ . For $x, y$ in $g$ , $\beta'([x, x_0], y) = \beta'(x, [x_0, y])$ , that is $\beta'(Dx, y) + \beta'(x, Dy) = 0$ . Now the restriction of $\beta'$ to $g$ is $\beta$ (Proposition 9). Hence the proposition.

7. Casimir element

Proposition 11

Let $g$ be a Lie algebra over a field $K$ , $U$ its enveloping algebra, $h$ a finite-dimensional ideal of $g$ and $\beta$ an invariant bilinear form on $g$ , whose restriction to $h$ is non-degenerate. Let $(e_i)_{1 \leq i \leq n}, (e'_j)_{1 \leq j \leq n}$ be two bases of $h$ such that $\beta(e_i, e'_j) = \delta_{ij}$ . Then the element $c = \sum_{i=1}^n e_i e'_i$ of $U$ belongs to the centre of $U$ and is independent of the choice of basis $(e_i)$ .

For $x \in g$ let $x_h$ be the restriction to $h$ of $\operatorname{ad}_g x$ . Then $x \mapsto x_h$ is a representation of $g$ on the vector space $h$ and the restriction $\beta'$ of $\beta$ to $h$ is invariant under this representation. By no. 5, Example 3, the tensor $\sum_{i=1}^n e_i \otimes e'_i$ is independent of the choice of basis $(e_i)$ and is an invariant element of the tensor algebra of $h$ . It is also an element of the tensor algebra $T$ of $g$ , which is invariant for the representation derived from the adjoint representation of $g$ . Its canonical image in $U$ , that is $c$ , is therefore independent of the choice of basis $(e_i)$ and is an invariant for the representation of $g$ on $U$ considered at the end of no. 2. This element is therefore permutable with every element of $g$ and therefore belongs to the centre of $U$ .

When $\beta$ is the bilinear form associated with a $g$ -module $M$ , the element $c$ of Proposition 11 is called the Casimir element associated with $M$ (or with the corresponding representation). This element exists if the restriction of $\beta$ to $h$ is non-degenerate.

Proposition 12

Let $g$ be a Lie algebra over a field $K$ , $h$ an ideal of $g$ of finite dimension n and M a g-module of finite dimension over K. Let c be the Casimir element (assumed to exist) associated with M and h.

(a) $\operatorname{Tr}(c_M) = n$ . (b) If M is simple and n is not divisible by the characteristic of K, $c_M$ is an automorphism of M.

In the notation of Proposition 11,

\operatorname{Tr}(c_M) = \sum_{i=1}^n \operatorname{Tr}((e_i)_M(e'_i)_M) = \sum_{i=1}^n \beta(e_i, e'_i) = n.

Hence, if n is not divisible by the characteristic of K, $c_M \neq 0$ . On the other hand, as c belongs to the centre of U, $c_M$ is permutable with all the $x_M, x \in g$ . If further M is simple, $c_M$ is therefore invertible in $\mathscr{L}(M)$ (Algebra, Chapter VIII, § 4, no. 3, Proposition 2).

8. Extension of the base ring

Let $K_1$ be a commutative ring with unit element and $\phi$ a homomorphism of K into $K_1$ mapping 1 to 1. Let g be a Lie K-algebra, U its enveloping algebra and M a left g-module, that is a left U-module. Then $M_{(K_1)}$ has a canonical left $U_{(K_1)}$ -module structure and hence a left $g_{(K_1)}$ -module structure. Let $\rho$ and $\rho_{(K_1)}$ be the representations of g and $g_{(K_1)}$ corresponding to M and $M_{(K_1)}$ : $\rho_{(K_1)}$ is said to be derived from $\rho$ by extending the base ring and the results of Algebra, Chapter VIII, § 13, no. 4 can be applied. If $x \in g$ , $\rho_{(K_1)}(x)$ is just the endomorphism $\rho(x) \otimes 1$ of $M_{(K_1)} = M \otimes_K K_1$ .

Suppose that K is a field, that $K_1$ is an extension of K and that $\phi$ is the canonical injection of K into $K_1$ . Let V and V' be vector subspaces of M. Let a be the vector subspace of g consisting of the $x \in g$ such that $\rho(x)(V) \subset V'$ . Let $a'$ be the vector subspace of $g_{(K_1)}$ consisting of the $x' \in g_{(K_1)}$ such that $\rho_{(K_1)}(x')(V_{(K_1)}) \subset V'_{(K_1)}$ . Then $a' = a_{(K_1)}$ . For clearly $a_{(K_1)} \subset a'$ . Now let $x' \in a'$ . We may write $x' = \sum_{i=1}^n \lambda_i x_i$ , where the $x_i$ are in g and the $\lambda_i$ are elements of $K_1$ linearly independent over K. For all $u \in V$ , $\rho(x') \cdot u \in V'_{(K_1)}$ , that is $\sum_{i=1}^n \lambda_i \rho(x_i) \cdot u \in V'_{(K_1)}$ , whence $\rho(x_i) \cdot u \in V'$ , hence $x_i \in a$ and $x' \in a_{(K_1)}$ . This shows that $a' = a_{(K_1)}$ . In particular, the centre of $g_{(K_1)}$ is derived from the centre of g by extending K to $K_1$ : it suffices to apply the above to the adjoint representation of g. It follows that $\mathscr{C}_p(g_{(K_1)}) = (\mathscr{C}_p g)_{(K_1)}$ for all $p$ . Similarly, let h be a subalgebra of g and n the normalizer of h in g. Then the normalizer of $h_{(K_1)}$ in $g_{(K_1)}$ is $n_{(K_1)}$ .

Let K, $K_1$ , g, $\rho$ , M be as in the last paragraph. Let b be a vector subspace of g and W a vector subspace of M. Let V be the vector subspace of M consisting of the $m \in M$ such that $\rho(b) \cdot m \subset W$ . Let V' be the vector subspace of $M_{(K_1)}$ consisting of the $m' \in M_{(K_1)}$ such that $\rho_{(K_1)}(b_{(K_1)}) \cdot m' \subset W_{(K_1)}$ . As above it is seen that $V' = V_{(K_1)}$ . In particular, the vector subspace of invariants of $M_{(K_1)}$ is derived from the vector subspace of invariants of $M$ by extending the base field from $K$ to $K_1$ .

Let $K, K_1$ and $\phi$ be as at the beginning of this no. Let $g$ be a Lie $K$ -algebra and $M$ and $N$ $g$ -modules. If $M$ and $N$ are isomorphic $g$ -modules, $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic $g_{(K_1)}$ -modules. Conversely:

Proposition 13

Let $K$ be a field, $K_1$ an extension of $K$ , $g$ a Lie $K$ -algebra and $M, N$ two $g$ -modules of finite dimension over $K$ . If $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic $g_{(K_1)}$ -modules, $M$ and $N$ are isomorphic $g$ -modules.

The proof is in two steps.

(1) Suppose first that $K_1$ is an extension of $K$ of finite degree $n$ . Let $U$ be the enveloping algebra of $g$ , so that the enveloping algebra of $g_{(K_1)}$ is $U_{(K_1)} = U \otimes_K K_1$ (§ 2, no. 9). As $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic as $U_{(K_1)}$ -modules they are a fortiori isomorphic as $U$ -modules; but as $U$ -modules they are respectively isomorphic to $M^n$ and $N^n$ . Now $M$ and $N$ are $U$ -modules of finite length; $M$ (resp. $N$ ) is therefore the direct sum of a family $(P_i^{r_i})_{1 \leq i \leq p}$ (resp. $(Q_j^{s_j})_{1 \leq j \leq q}$ ) of submodules such that the $P_i$ (resp. $Q_j$ ) are indecomposable and two $P_i$ (resp. $Q_j$ ) of different indices are not isomorphic (Algebra, Chapter VIII, § 2, no. 2, Theorem 1). Then $M^n$ (resp. $N^n$ ) is isomorphic to the direct sum of the $P_i^{n r_i}$ (resp. $Q_j^{n s_j}$ ); it follows (loc. cit.) that $p = q$ and that after permuting the $Q_j$ if necessary $n r_i = n s_i$ and $P_i$ is isomorphic to $Q_i$ for $1 \leq i \leq p$ , hence $M$ is isomorphic to $N$ .

(2) General case. Let $P$ be the $g$ -module $\mathscr{L}_K(M, N)$ and $Q$ the subspace of invariants of $P$ , that is the set of homomorphisms of the $g$ -module $M$ into the $g$ -module $N$ . In the $g_{(K_1)}$ -module $\mathscr{L}_{K_1}(M_{(K_1)}, N_{(K_1)}) = (\mathscr{L}_K(M, N))_{(K_1)}$ , the subspace of invariants is $Q_{(K_1)}$ . The hypothesis that $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic implies that $M$ and $N$ have the same dimension over $K$ and that there exists in $Q_{(K_1)}$ an element $g$ which is an isomorphism of $M_{(K_1)}$ onto $N_{(K_1)}$ . Let $(f_1, \ldots, f_d)$ be a basis of $Q$ over $K$ and choose bases of $M$ and $N$ over $K$ . If $\lambda_k \in K_1$ for $1 \leq k \leq d$ , the matrix of $f = \sum_{k=1}^d \lambda_k f_k$ with respect to these bases has determinant which is a polynomial $D(\lambda_1, \ldots, \lambda_d)$ with coefficients in $K$ . When $f = g$ , this determinant is non-zero and hence the coefficients of $D$ are not all non-zero. Therefore, if $\Omega$ is the algebraic closure of $K$ , there exists (since $\Omega$ is infinite) elements $\mu_k \in \Omega$ $(1 \leq k \leq d)$ such that $D(\mu_1, \ldots, \mu_d) \neq 0$ (Algebra, Chapter IV, § 2, no. 5, Proposition 8). If $K_2$ is the algebraic extension of $K$ generated by the $\mu_k$ $(1 \leq k \leq d)$ , it follows that $\sum_{k=1}^d \mu_k f_k$ is an isomorphism of $M_{(K_2)}$ onto $N_{(K_2)}$ ; but $K_2$ is of finite degree over $K$ (Algebra, Chapter V, § 3, no. 2, Proposition 5) and hence $M$ and $N$ are isomorphic by the first part of the argument.

Again let $K, K_1$ and $\phi$ be as the beginning of this no. Let $\rho$ be a representation of $g$ on a $K$ -module $M$ with a finite basis $(x_1, \ldots, x_n)$ . Then the bilinear form on $g_{(K_1)}$ associated with $\rho_{(K_1)}$ is derived from the bilinear form associated with $\rho$ by extending the base ring to $K_1$ (for, if $u \in \mathscr{L}_K(M)$ , $u$ has the same matrix with respect to $(x_1, \ldots, x_n)$ as $u \otimes 1$ with respect to $(x_1 \otimes 1, \ldots, x_n \otimes 1)$ and hence $u$ and $u \otimes 1$ have the same trace). In particular, if the $K$ -module $g$ has a finite basis, the Killing form of $g_{(K_1)}$ is derived from that of $g$ by extending the base ring to $K_1$ .

Exercises

See the exercises for § 3.