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Let \( \mathrm{L} \) be an associative algebra over \( \mathrm{K} \). The bracket \( \left\lbrack {x, y}\right\rbrack = {xy} - {yx} \) is a bilinear function of \( x \) and \( y \). | It is easily verified that the law of composition \( \left( {x, y}\right) \mapsto \left\lbrack {x, y}\right\rbrack \) on the K-module L makes L into a Lie algebra over K. | Yes |
Example 2. In Example 1 choose L to be the associative algebra of endomorphisms of a K-module E. We obtain the Lie algebra of endomorphisms of E, denoted by \( \mathfrak{{gl}}\left( \mathrm{E}\right) \) . (If \( \mathrm{E} = {\mathrm{K}}^{n} \), the Lie algebra \( \mathfrak{{gl}}\left( \mathrm{E}\right) \) is denoted b... | Every Lie subalgebra of \( \mathfrak{{gl}}\left( \mathrm{E}\right) \) is a Lie algebra over \( \mathrm{K} \) . In particular:\n\n(1) If \( \mathrm{E} \) is given a (not necessarily associative) algebra structure, the derivations of \( \mathrm{E} \) form a Lie algebra over \( \mathrm{K} \) .\n\n(2) If \( \mathrm{E} \) a... | No |
Let \( \mathfrak{g} \) be a Lie algebra. For all \( x \in \mathfrak{g} \) , \( \mathrm{{ad}}x \) is a derivation. The mapping \( x \mapsto \) ad \( x \) is a homomorphism of the Lie algebra \( \mathfrak{g} \) into the Lie algebra \( \mathfrak{d} \) of derivations of g. If \( \mathrm{D} \in \mathfrak{d} \) and \( x \in ... | Identity (4) can be written:\n\n\[ \left( {\operatorname{ad}x}\right) .\left\lbrack {y, z}\right\rbrack = \left\lbrack {\left( {\operatorname{ad}x}\right) .y, z}\right\rbrack + \left\lbrack {y,\left( {\operatorname{ad}x}\right) .z}\right\rbrack \]\n\nor:\n\n\[ \left( {\mathrm{{ad}}\left\lbrack {x, y}\right\rbrack }\rig... | Yes |
Proposition 2. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{a} \) an ideal (resp. a characteristic ideal) of \( \mathfrak{g} \) and \( \mathfrak{b} \) a characteristic ideal of \( \mathfrak{a} \) . Then \( \mathfrak{b} \) is an ideal (resp. a characteristic ideal) of \( \mathfrak{g} \) . | Every inner derivation (resp. every derivation) of \( g \) leaves \( \mathfrak{a} \) stable and induces on \( \mathfrak{a} \) a derivation and hence leaves \( \mathfrak{b} \) stable. | Yes |
Proposition 3. If \( \mathfrak{a} \) and \( \mathfrak{b} \) are ideals (resp. characteristic ideals) of \( \mathfrak{g},\left\lbrack {\mathfrak{a},\mathfrak{b}}\right\rbrack \) is an ideal (resp. a characteristic ideal) of \( \mathfrak{g} \) . | Let \( \mathrm{D} \) be an inner derivation (resp. a derivation) of \( \mathfrak{g} \) . If \( x \in \mathfrak{a} \) and \( y \in \mathfrak{b} \), then\n\n\[ \mathrm{D}\left( \left\lbrack {x, y}\right\rbrack \right) = \left\lbrack {\mathrm{D}x, y}\right\rbrack + \left\lbrack {x,\mathrm{D}y}\right\rbrack \in \left\lbrac... | Yes |
Proposition 4. Let \( \mathfrak{g} \) and \( \mathfrak{h} \) be two Lie algebras over \( \mathrm{K} \) and \( f \) a homomorphism of \( \mathfrak{g} \) onto \( \mathfrak{h} \) . Then \( f\left( {{\mathcal{D}}^{p}\mathfrak{g}}\right) = {\mathcal{D}}^{p}\mathfrak{h}, f\left( {{\mathcal{C}}^{p}\mathfrak{g}}\right) = {\mat... | If \( \mathfrak{a} \) and \( \mathfrak{b} \) are submodules of \( \mathfrak{g} \), it follows immediately that\n\n\[ f\left( \left\lbrack {\mathfrak{a},\mathfrak{b}}\right\rbrack \right) = \left\lbrack {f\left( \mathfrak{a}\right), f\left( \mathfrak{b}\right) }\right\rbrack \]\n\nThe proposition is then immediate by in... | No |
Proposition 5. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathfrak{a} \) an ideal (resp. a characteristic ideal) of \( \mathfrak{g} \). The centralizer \( {\mathfrak{a}}^{\prime } \) of \( \mathfrak{a} \) in \( \mathfrak{g} \) is an ideal (resp. a characteristic ideal) of \( \mathfrak{g} \). | Let \( \mathrm{D} \) be an inner derivation (resp. a derivation) of \( \mathfrak{g} \). If \( x \in {\mathfrak{a}}^{\prime } \) and \( y \in \mathfrak{a} \), then\n\n\[ \left\lbrack {\mathrm{D}x, y}\right\rbrack = \mathrm{D}\left( \left\lbrack {x, y}\right\rbrack \right) - \left\lbrack {x,\mathrm{D}y}\right\rbrack = 0 ... | Yes |
Proposition 6. Let\n\n\\[ \n a\\overset{\\lambda }{ \\rightarrow }g\\overset{\\mu }{ \\rightarrow }b \n\\]\n\nbe an extension of \\( \\mathfrak{b} \\) by \\( \\mathfrak{a} \\) and \\( \\mathfrak{n} \\) its kernel.\n\n(a) If there exists a subalgebra \\( \\mathfrak{m} \\) of \\( \\mathfrak{g} \\) supplementary to \\( \\... | The assertions of (a) are immediate. | No |
Let \( t \) be a Lie algebra over \( K \) . A linear mapping \( \theta \) of \( t \) into \( {af}\left( M\right) \) can be written \( x \mapsto (\left( {\zeta \left( x\right) ,\eta \left( x\right) }\right) \), where \( \zeta \) is a linear mapping of \( t \) into \( \mathrm{M} \) and \( \eta \) a linear mapping of \( t... | \[ \theta \left( \left\lbrack {x, y}\right\rbrack \right) = \left\lbrack {\theta \left( x\right) ,\theta \left( y\right) }\right\rbrack \] that is \[ \left( {\zeta \left( \left\lbrack {x, y}\right\rbrack \right) ,\eta \left( \left\lbrack {x, y}\right\rbrack \right) }\right) = \left\lbrack {\left( {\zeta \left( x\right)... | Yes |
Proposition 1. Let \( \sigma \) be an α-mapping of \( \mathfrak{g} \) into the associative algebra \( \mathbf{L} \) with unit element. There exists one and only one homomorphism \( \tau \) of \( \mathbf{U} \) into \( \mathbf{L} \), mapping 1 to 1, such that \( \sigma = \tau \circ {\sigma }_{0} \), where \( {\sigma }_{0... | Let \( {\tau }^{\prime } \) be the unique homomorphism of \( \mathrm{T} \) into \( \mathrm{L} \) which extends \( \sigma \) and maps 1 to 1 . Then, for \( x, y \) in \( g \) ,\n\n\[ \n{\tau }^{\prime }\left( {x \otimes y - y \otimes x - \left\lbrack {x, y}\right\rbrack }\right) = \sigma \left( x\right) \sigma \left( y\... | Yes |
Proposition 2. The homomorphism \( \phi \) is an algebra isomorphism | The mapping \( {\sigma }^{\prime } : \left( {{x}_{1},{x}_{2}}\right) \mapsto {\sigma }_{1}\left( {x}_{1}\right) \otimes 1 + 1 \otimes {\sigma }_{2}\left( {x}_{2}\right) \left( {{x}_{1} \in {\mathfrak{g}}_{1},{x}_{2} \in {\mathfrak{g}}_{2}}\right) \) is an \( \alpha \) -mapping of \( \mathfrak{g} \) into \( {\mathrm{U}}... | Yes |
Proposition 3. Let \( \mathfrak{h} \) be an ideal of \( \mathfrak{g},\mathfrak{p} \) the canonical homomorphism of \( \mathfrak{g} \) onto \( \mathfrak{g}/\mathfrak{h} \) and W the enveloping algebra of \( \mathfrak{g}/\mathfrak{h} \) . The homomorphism:\n\n\[ \widetilde{p} : \mathrm{U} \rightarrow \mathrm{W} \]\n\ndef... | Let \( {\sigma }^{\prime \prime } \) be the canonical mapping of \( \mathfrak{g}/\mathfrak{h} \) into W. The commutative diagram:\n\n\n\nproves that \( \widetilde{p} \) is zero on \( \sigma \left( \mathfrak{h}\right) \... | Yes |
Proposition 4. The homomorphism \( \phi \) is an isomorphism of \( \mathbf{V} \) onto \( {\mathbf{U}}^{0} \) . | There exists a homomorphism \( {\phi }^{\prime } \) of \( U \) into \( {V}^{0} \) mapping 1 to 1 and such that \( {\sigma }_{0} = {\phi }^{\prime } \circ \sigma .{\phi }^{\prime } \) can be considered as a homomorphism of \( {\mathrm{U}}^{0} \) into \( \mathrm{V} \) . Then \( {\sigma }_{0} = {\phi }^{\prime } \circ \ph... | Yes |
Proposition 5. The mapping \( \phi \) of \( \mathrm{T} \) onto \( \mathrm{G} \) is an algebra homomorphism mapping 1 to 1 and is zero on the two-sided ideal generated by the tensors \( x \otimes y - y \otimes x\left( {x \in \mathfrak{g}, y \in \mathfrak{g}}\right) \) . | If \( t \in {\mathrm{T}}^{n} \) and \( {t}^{\prime } \in {\mathrm{T}}^{p} \), then \( \phi \left( t\right) \phi \left( {t}^{\prime }\right) = \phi \left( {t{t}^{\prime }}\right) \) by definition of the multiplication on G. Hence \( \phi \) is an algebra homomorphism and clearly \( \phi \left( 1\right) = 1 \) . If \( x,... | Yes |
Proposition 6. If \( \mathrm{K} \) is Noetherian and \( \mathrm{g} \) is a finitely generated module, the ring \( \mathrm{U} \) is right and left Noetherian. | \( \mathrm{S} \) is a finitely generated algebra over \( \mathrm{K} \) and hence a Noetherian ring (Commutative Algebra, Chapter III, § 2, no. 10, Corollary 3 to Theorem 2). Hence \( \mathrm{G} \), which is isomorphic to a quotient ring of \( \mathrm{S} \), is Noetherian. Hence \( \mathrm{U} \) is right and left Noethe... | Yes |
For every integer \( p \geq 0 \), there exists a unique homomorphism \( {f}_{p} \) of the \( \mathrm{K} \) - module \( \mathrm{g}{ \otimes }_{\mathrm{K}}{\mathrm{P}}_{p} \) into the \( \mathrm{K} \) -module \( \mathrm{P} \) satisfying the following conditions:\n\n\( \left( {\mathrm{A}}_{p}\right) \;{f}_{p}\left( {{x}_{... | The last assertion follows from the others since the restriction of \( {f}_{p} \) to \( \mathfrak{g}{ \otimes }_{\mathrm{K}}{\mathrm{P}}_{p - 1} \) satisfies conditions \( \left( {\mathrm{A}}_{p - 1}\right) ,\left( {\mathrm{B}}_{p - 1}\right) \) and \( \left( {\mathrm{C}}_{p - 1}\right) \) . We shall prove the existenc... | Yes |
Lemma 2. There exists an \( \alpha \) -mapping \( \sigma \) of \( \mathfrak{g} \) into \( {\mathcal{L}}_{\mathrm{K}}\left( \mathrm{P}\right) \) such that:\n\n(1) \( \sigma \left( {x}_{\lambda }\right) {z}_{\mathrm{M}} = {z}_{\lambda }{z}_{\mathrm{M}} \) for \( \lambda \leq \mathrm{M} \) ;\n\n(2) \( \sigma \left( {x}_{\... | By Lemma 1 there exists a homomorphism \( f \) of the K-module \( \mathfrak{g}{ \otimes }_{\mathbf{K}}{\mathrm{P}}_{p} \) into \( \mathrm{P} \) satisfying, for all \( p \), conditions \( \left( {\mathrm{A}}_{p}\right) ,\left( {\mathrm{B}}_{p}\right) ,\left( {\mathrm{C}}_{p}\right) \) (where \( {f}_{p} \) is replaced by... | Yes |
Lemma 3. Let \( t \) be a tensor in \( {\mathrm{T}}_{n} \cap \mathrm{J} \) . The homogeneous component \( {t}_{n} \) of \( t \) of order \( n \) is in the kernel I of the canonical homomorphism \( \mathrm{T} \rightarrow \mathrm{S} \) . | We write \( {t}_{n} \) in the form \( \mathop{\sum }\limits_{{i = 1}}^{r}{x}_{{\mathrm{M}}_{i}} \), where the \( {\mathrm{M}}_{i} \) are sequences of \( n \) elements of \( \Lambda \) . The mapping \( \sigma \) extends to a homomorphism of the algebra \( \mathrm{T} \) into the algebra \( {\mathcal{L}}_{\mathrm{K}}\left... | Yes |
Corollary 1. Suppose that \( \mathfrak{g} \) is a free \( \mathrm{K} \) -module. Let \( \mathrm{W} \) be a sub- \( \mathrm{K} \) -module of \( {\mathrm{T}}^{n} \) . If, in the notation of diagram (3), the restriction of \( {\tau }_{n} \) to \( \mathrm{W} \) is an isomorphism of \( \mathrm{W} \) onto \( {\mathrm{S}}^{n}... | The restriction to \( \mathrm{W} \) of \( {\omega }_{n} \circ {\tau }_{n} \) is a bijection of \( \mathrm{W} \) onto \( {\mathrm{G}}^{n} \) ; so is the restriction \( {\theta }_{n} \circ {\psi }_{n} \) to W. Hence the corollary. | Yes |
Corollary 2. If \( \mathfrak{g} \) is a free \( \mathrm{K} \) -module, the canonical mapping of \( \mathfrak{g} \) into its enveloping algebra is injective. | This follows from Corollary 1 taking \( \mathrm{W} = {\mathrm{T}}^{1} \). | No |
Corollary 3. If \( \mathfrak{g} \) admits a totally ordered basis \( {\left( {x}_{\lambda }\right) }_{\lambda \in \Lambda } \), the elements \( {x}_{{\lambda }_{1}}{x}_{{\lambda }_{2}}\ldots {x}_{{\lambda }_{n}} \) of the enveloping algebra \( \mathrm{U} \), where \( \left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right... | Let \( {\Lambda }_{n} \) be the set of increasing sequences of \( n \) elements of \( \Lambda \) . For \( \mathrm{M} = \left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \in {\Lambda }_{n} \), let \( {y}_{\mathrm{M}} = {x}_{{\lambda }_{1}} \otimes {x}_{{\lambda }_{2}} \otimes \cdots \otimes {x}_{{\lambda }_{n}} \) .... | Yes |
Corollary 4. Let \( {\mathrm{S}}^{\prime n} \subset {\mathrm{T}}^{n} \) be the set of homogeneous symmetric tensors of order \( n \) . Suppose that \( \mathrm{K} \) is a field of characteristic 0 . Then the composite mapping of the canonical mappings\n\n\[ \n{\mathrm{S}}^{n} \rightarrow {\mathrm{S}}^{\prime n} \rightar... | This follows from Corollary 1 taking \( \mathrm{W} = {\mathrm{S}}^{\prime n} \) . | No |
Corollary 5. Let \( \mathfrak{h} \) be a subalgebra of the Lie algebra \( \mathfrak{g} \) and \( {\mathbf{U}}^{\prime } \) its enveloping algebra. Suppose that the \( \mathrm{K} \) -modules \( \mathfrak{h} \) and \( \mathfrak{g}/\mathfrak{h} \) are free (for example if \( \mathrm{K} \) is a field). Let \( {\left( {x}_{... | We give \( \mathrm{L} \cup \mathrm{M} \) a total ordering such that every element of \( \mathrm{L} \) is less than every element of \( \mathrm{M} \) . The elements \( {x}_{{\alpha }_{1}}{x}_{{\alpha }_{2}}\ldots {x}_{{\alpha }_{p}} \) calculated in \( {\mathrm{U}}^{\prime } \) (where \( \left. {{\alpha }_{1} \leq \cdot... | Yes |
Corollary 6. Suppose that the K-module \( \mathfrak{g} \) is the direct sum of subalgebras \( {\mathfrak{g}}_{1},{\mathfrak{g}}_{2},\ldots \) , \( {\mathfrak{g}}_{n} \) and that each \( {\mathfrak{g}}_{i} \) is a free \( \mathrm{K} \) -module. Let \( {\mathrm{U}}_{i} \) be the enveloping algebra of \( {\mathfrak{g}}_{i... | Let \( {\left( {x}_{\lambda }^{i}\right) }_{\lambda \in {\mathrm{L}}_{1}} \) be a basis of \( {\mathfrak{g}}_{i} \) . We totally order \( {\mathrm{L}}_{1} \cup \cdots \cup {\mathrm{L}}_{n} \) so that every element of \( {\mathrm{L}}_{i} \) exceeds every element of \( {\mathrm{L}}_{j} \) for \( i \geq j \) . Then the el... | Yes |
Lemma 4. Let \( \mathrm{V} \) be a \( \mathrm{K} \) -module and \( \mathrm{T} \) the tensor algebra of \( \mathrm{V} \) . Let \( u \) be an endomorphism of \( \mathrm{V} \) . There exists one and only one derivation of \( \mathrm{T} \) which extends \( u \) . This derivation commutes with the symmetry operators on \( \... | Let \( \mathrm{F} = \mathrm{V} \times \mathrm{V} \times \cdots \times \mathrm{V} \) ( \( n \) factors). The mapping\n\n\[ \left( {{x}_{1},\ldots ,{x}_{n}}\right) \mapsto u{x}_{1} \otimes {x}_{2} \otimes \cdots \otimes {x}_{n} \]\n\n\[ + {x}_{1} \otimes u{x}_{2} \otimes \cdots \otimes {x}_{n} + \cdots + {x}_{1} \otimes ... | Yes |
Proposition 7. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathrm{U} \) its enveloping algebra, \( \sigma \) the canonical mapping of \( \mathfrak{g} \) into \( \mathrm{U} \) and \( \mathrm{D} \) a derivation of \( \mathfrak{g} \) . (a) There exists one and only one derivation \( {\mathrm{D}}_{\mathrm{U}} \) of \( \ma... | Let \( {\mathrm{D}}_{\mathrm{T}} \) be the derivation of the tensor algebra \( \mathrm{T} \) of \( \mathrm{g} \) which extends \( \mathrm{D} \) (Lemma 4). The two-sided ideal \( \mathrm{J} \) of \( \mathrm{T} \) generated by the \[ x \otimes y - y \otimes x - \left\lbrack {x, y}\right\rbrack \] \( \left( {x, y\text{in}... | Yes |
Proposition 1. Let \( \mathfrak{g} \) be a Lie algebra over \( \mathrm{K} \) and \( \mathfrak{a} \) an ideal of \( \mathfrak{g} \) . Let \( \mathrm{M} \) be a \( \mathfrak{g} \) -module and \( \mathrm{N} \) a simple \( \mathfrak{a} \) -module. Consider \( \mathrm{M} \) as an \( \mathfrak{a} \) -module and suppose that ... | Let \( y \in \mathfrak{g} \) . Let \( \phi \) be the canonical mapping of \( \mathrm{M} \) onto \( \mathrm{M}/{\mathrm{M}}^{\prime } \) and \( f \) the mapping \( m \mapsto \phi \left( {{y}_{\mathrm{M}} \cdot m}\right) \) of \( {\mathrm{M}}^{\prime } \) into \( \mathrm{M}/{\mathrm{M}}^{\prime } \) . It suffices to show... | Yes |
Proposition 2. Let \( \mathfrak{g} \) be a Lie algebra over \( \mathrm{K} \) and \( {\mathrm{M}}_{i} \) a \( \mathfrak{g} \) -module \( \left( {1 \leq i \leq n}\right) \) . On the tensor product \( {\mathrm{M}}_{1}{ \otimes }_{\mathrm{K}}{\mathrm{M}}_{2} \otimes \cdots \otimes {\mathrm{M}}_{n} \), there exists one and ... | The corresponding representation is called the tensor product of the given representations of \( \mathfrak{g} \) on the \( {\mathrm{M}}_{i} \) . | No |
Proposition 4. Let \( {\mathrm{M}}_{1},{\mathrm{M}}_{2} \) be two \( \mathfrak{g} \) -modules. The canonical \( \mathrm{K} \) -linear mappings (Algebra, Chapter II, § 4, no. 2, Proposition 2 and no. 1, Proposition 1): \[ {\mathrm{M}}_{1}^{ * }{ \otimes }_{\mathrm{K}}{\mathrm{M}}_{2}\overset{\Phi }{ \rightarrow }{\mathc... | We write \[ \mathrm{N} = {\mathrm{M}}_{1}^{ * } \otimes {\mathrm{M}}_{2},\;\mathrm{P} = \mathcal{L}\left( {{\mathrm{M}}_{1},{\mathrm{M}}_{2}}\right) ,\;\mathrm{Q} = \mathcal{L}\left( {{\mathrm{M}}_{1},{\mathrm{M}}_{2}^{ * }}\right) ,\;\mathrm{R} = {\left( {\mathrm{M}}_{1} \otimes {\mathrm{M}}_{2}\right) }^{ * }. \] The... | Yes |
The \( \mathfrak{g} \) -module structure on \( \mathrm{M} \) defines on the \( \mathrm{K} \) -module \( \mathrm{P} = {\mathcal{L}}_{\mathrm{K}}\left( {\mathrm{M},\mathrm{M}}\right) \) of endomorphisms of \( \mathrm{M} \) a \( \mathfrak{g} \) -module structure. By (6), for all \( x \in \mathfrak{g} \) and \( u \in \math... | (11)\n\n\[ \n{x}_{\mathrm{P}}.u = \left\lbrack {{x}_{\mathrm{M}}, u}\right\rbrack = \left( {\operatorname{ad}{x}_{\mathrm{M}}}\right) .u \n\]\n\nwhere ad \( {x}_{\mathrm{M}} \) denotes the image of \( {x}_{\mathrm{M}} \) under the adjoint representation of \( \mathfrak{{gl}}\left( \mathrm{M}\right) \) . In other words:... | Yes |
Proposition 5. Let \( \mathfrak{g} \) be a Lie K-algebra, \( \mathfrak{h} \) an ideal of \( \mathfrak{g} \) , \( \mathfrak{g} \) a representation of \( \mathfrak{g} \) on \( \mathfrak{M} \) and \( {\mathfrak{p}}^{\prime } \) the restriction of \( \mathfrak{p} \) to \( \mathfrak{h} \) . Then the set \( \mathrm{N} \) of ... | Let \( n \in \mathrm{N} \) and \( y \in \mathfrak{g} \) ; for all \( x \in \mathfrak{h},\left\lbrack {x, y}\right\rbrack \in \mathfrak{h} \) and hence\n\n\[ \rho \left( x\right) \rho \left( y\right) n = \rho \left( \left\lbrack {x, y}\right\rbrack \right) n + \rho \left( y\right) \rho \left( x\right) n = 0; \]\n\nhence... | Yes |
Proposition 6. Let \( \mathrm{M} \) be a semi-simple g-module. Then the submodule \( {\mathrm{M}}_{0} \) of invariant elements of \( \mathrm{M} \) admits one and only one supplement stable under the \( {x}_{\mathrm{M}} \), namely the submodule \( {\mathbf{M}}_{1} \) generated by the \( {x}_{\mathbf{M}} \cdot m\left( {x... | Let \( {\mathrm{M}}^{\prime } \) be a submodule of \( \mathrm{M} \) which is stable under the \( {x}_{\mathrm{M}} \) and a supplement of \( {\mathrm{M}}_{0} \) in \( \mathrm{M} \) . For all \( m \in \mathrm{M}, m = {m}_{0} + {m}^{\prime } \) with \( {m}_{0} \in {\mathrm{M}}_{0},{m}^{\prime } \in {\mathrm{M}}^{\prime } ... | Yes |
Proposition 7. Let \( \mathfrak{g} \) be a Lie algebra, \( \beta \) an invariant symmetric bilinear form on \( \mathfrak{g} \) and \( \mathfrak{a} \) an ideal of \( \mathfrak{g} \) . (a) The orthogonal \( {\mathfrak{a}}^{\prime } \) of \( \mathfrak{a} \) with respect to \( \beta \) is an ideal of \( \mathfrak{g} \) . | Let \( \mathrm{D} \) be a derivation of \( \mathfrak{g} \) . Suppose that \( \mathfrak{a} \) is stable under \( \mathrm{D} \) and that \( \beta \left( {\mathrm{D}x, y}\right) + \beta \left( {x,\mathrm{D}y}\right) = 0 \) for \( x, y \) in \( \widetilde{\mathfrak{g}} \) . Then \( z \in {\mathfrak{a}}^{\prime } \) implies... | No |
Proposition 8. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathrm{M} \) a \( \mathfrak{g} \) -module. Supose that \( \mathrm{M} \), considered as a K-module, admits a finite basis. The bilinear form associated with \( \mathrm{M} \) is invariant. For \( x, y, z \) in \( \mathfrak{g} \), we have: | \[ \operatorname{Tr}({\left\lbrack x, y\right\rbrack }_{\mathrm{M}}{z}_{\mathrm{M}} = \operatorname{Tr}\left( {{x}_{\mathrm{M}}{y}_{\mathrm{M}}{z}_{\mathrm{M}}}\right) - \operatorname{Tr}\left( {{y}_{\mathrm{M}}{x}_{\mathrm{M}}{z}_{\mathrm{M}}}\right) = \operatorname{Tr}\left( {{x}_{\mathrm{M}}{y}_{\mathrm{M}}{z}_{\mat... | Yes |
Proposition 10. Suppose that \( \mathrm{K} \) is a field and that the Lie algebra \( \mathrm{g} \) is finite-dimensional over \( \mathrm{K} \) . The Killing form \( \beta \) of \( \mathrm{g} \) is completely invariant. | Let \( \mathrm{D} \) be a derivation of \( \mathfrak{g} \) . There exists a Lie algebra \( {\mathfrak{g}}^{\prime } \) containing \( \mathfrak{g} \) as an ideal of codimension 1 and an element \( {x}_{0} \) of \( {g}^{\prime } \) such that \( \mathrm{D}x = \left\lbrack {{x}_{0}, x}\right\rbrack \) for all \( x \in \mat... | Yes |
Proposition 11. Let \( \mathfrak{g} \) be a Lie algebra over a field \( \mathrm{K} \) , \( \mathrm{U} \) its enveloping algebra, \( \mathfrak{h} \) a finite-dimensional ideal of \( \mathfrak{g} \) and \( \beta \) an invariant bilinear form on \( \mathfrak{g} \), whose restriction to \( \mathfrak{h} \) is non-degenerate... | For \( x \in \mathfrak{g} \) let \( {x}_{\mathfrak{h}} \) be the restriction to \( \mathfrak{h} \) of \( {\operatorname{ad}}_{\mathfrak{g}}x \) . Then \( x \mapsto {x}_{\mathfrak{h}} \) is a representation of \( \mathfrak{g} \) on the vector space \( \mathfrak{h} \) and the restriction \( {\beta }^{\prime } \) of \( \b... | Yes |
Proposition 12. Let \( \mathfrak{g} \) be a Lie algebra over a field \( \mathrm{K},\mathfrak{h} \) an ideal of \( \mathfrak{g} \) of finite dimension \( n \) and \( \mathrm{M} \) a \( \mathrm{g} \) -module of finite dimension over \( \mathrm{K} \) . Let \( c \) be the Casimir element (assumed to exist) associated with ... | In the notation of Proposition 11,\n\n\[ \operatorname{Tr}\left( {c}_{\mathrm{M}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\operatorname{Tr}\left( {{\left( {e}_{i}\right) }_{\mathrm{M}}{\left( {e}_{i}^{\prime }\right) }_{\mathrm{M}}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\beta \left( {{e}_{i},{e}_{i}^{\prime }... | Yes |
Proposition 1. Let \( \mathfrak{g} \) be a Lie algebra. The following conditions are equivalent:\n\n(a) \( \mathfrak{g} \) is nilpotent;\n\n(b) \( {\mathcal{C}}^{k}\mathfrak{g} = \{ 0\} \) for sufficiently large \( k \) ;\n\n(c) \( {\mathcal{C}}_{k}\mathfrak{g} = \mathfrak{g} \) for sufficiently large \( k \) ;\n\n(d) ... | If \( {\mathcal{C}}^{k}\mathfrak{g} = \{ 0\} \) (resp. \( {\mathcal{C}}_{k}\mathfrak{g} = \mathfrak{g} \) ), clearly the sequence \( {\mathcal{C}}^{1}\mathfrak{g},\ldots ,{\mathcal{C}}^{k}\mathfrak{g} \) (resp. \( \left. {{\mathcal{C}}_{k}\mathfrak{g},{\mathcal{C}}_{k - 1}\mathfrak{g},\ldots ,{\mathcal{C}}_{0}\mathfrak... | Yes |
Corollary 2. The Killing form of a nilpotent Lie algebra is zero. | For all \( x \) and \( y \) in a nilpotent Lie algebra ad \( x \circ \) ad \( y \) is nilpotent and hence of zero trace. | Yes |
Proposition 2. Subalgebras, quotient algebras and central extensions of a nilpotent Lie algebra are nilpotent. A finite product of nilpotent Lie algebras is a nilpotent Lie algebra. | Let \( \mathfrak{g} \) be a Lie algebra, \( {\mathfrak{g}}^{\prime } \) a subalgebra of \( \mathfrak{g},\mathfrak{h} \) an ideal of \( \mathfrak{g},\mathfrak{k} = \mathfrak{g}/\mathfrak{h} \) and \( \phi \) the canonical mapping of \( \mathfrak{g} \) onto \( \mathfrak{k} \) . If \( \mathfrak{g} \) is nilpotent, then \(... | Yes |
Proposition 3. Let \( \mathfrak{g} \) be a nilpotent Lie algebra and \( \mathfrak{h} \) a subalgebra of \( \mathfrak{g} \) distinct from \( \mathfrak{g} \) . The normalizer of \( \mathfrak{h} \) in \( \mathfrak{g} \) is distinct from \( \mathfrak{h} \) . | Let \( k \) be the greatest integer such that \( {\mathcal{C}}^{k}\mathfrak{g} + \mathfrak{h} \neq \mathfrak{h} \) . Then\n\n\[ \left\lbrack {{\mathcal{C}}^{k}g + \mathfrak{h},\mathfrak{h}}\right\rbrack \subset {\mathcal{C}}^{k + 1}g + \mathfrak{h} \subset \mathfrak{h} \]\n\nand hence the normalizer of \( \mathfrak{h} ... | Yes |
Lemma 1. Let \( \mathrm{V} \) be a vector space over \( \mathrm{K} \) . If \( x \) is a nilpotent endomorphism of \( \mathrm{V} \), the mapping \( y \mapsto \left\lbrack {x, y}\right\rbrack \) of \( \mathcal{L}\left( \mathrm{V}\right) \) into \( \mathcal{L}\left( \mathrm{V}\right) \) is nilpotent. | If \( f \) denotes this mapping, \( {f}^{m}\left( y\right) \) is a sum of terms of the form \( \pm {x}^{l}y{x}^{j} \) with \( i + j = m \) . If \( {x}^{k} = 0 \), then \( {f}^{{2k} - 1}\left( y\right) = 0 \) for all \( y \) . | Yes |
For a Lie algebra \( \mathfrak{g} \) to be nilpotent, it is necessary and sufficient that, for all \( x \in \mathfrak{g} \), ad \( x \) be nilpotent. | The condition is necessary (Proposition 1). Suppose that its sufficiency has been proved for Lie algebras of dimension \( < n\left( {n \neq 0}\right) \) . Let \( \mathfrak{g} \) be an \( n \) -dimensional Lie algebra such that, for all \( x \in \mathfrak{g} \), ad \( x \) is nilpotent. Theorem 1, applied to the set of ... | No |
Corollary 2. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathfrak{h} \) an ideal of \( \mathfrak{g} \) . Suppose that \( \mathfrak{g}/\mathfrak{h} \) is nilpotent and that, for all \( x \in \mathfrak{g} \), the restriction of \( \operatorname{ad}x \) to \( \mathfrak{h} \) is nilpotent. Then \( \mathfrak{g} \) is ni... | Let \( x \in \mathfrak{g} \) . As \( \mathfrak{g}/\mathfrak{h} \) is nilpotent, there exists an integer \( k \) such that \( {\left( \operatorname{ad}x\right) }^{k}\left( \mathfrak{g}\right) \subset \mathfrak{h} \) . By hypothesis there exists an integer \( {k}^{\prime } \) such that \( {\left( \operatorname{ad}x\right... | Yes |
Corollary 3. Let \( \mathrm{V} \) be a vector space and \( \mathrm{g} \) a finite-dimensional subalgebra of \( \mathfrak{{gl}}\left( \mathrm{V}\right) \) whose elements are nilpotent endomorphisms of \( \mathrm{V} \) . Then \( \mathrm{g} \) is a nilpotent Lie algebra. | This follows immediately from Lemma 1 and Corollary 1. | Yes |
Lemma 2. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{a} \) an ideal of \( \mathfrak{g} \) and \( \mathrm{M} \) a simple \( \mathfrak{g} \) -module. If, for all \( x \in \mathfrak{a},{x}_{\mathrm{M}} \) is nilpotent, then \( {x}_{\mathrm{M}} = 0 \) for all \( x \in \mathfrak{a} \) . | Let \( \mathrm{N} \) be the subspace of \( \mathrm{M} \) consisting of the \( m \in \mathrm{M} \) such that \( {x}_{\mathrm{M}} \cdot m = 0 \) for all \( x \in \mathfrak{a} \) . By Theorem \( 1,\mathrm{N} \neq \{ 0\} \) . On the other hand, for all \( y \in \mathfrak{g},\mathrm{N} \) is stable under \( {y}_{\mathrm{M}}... | Yes |
Lemma 3. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{a} \) an ideal of \( \mathfrak{g},\mathrm{M} \) a \( \mathfrak{g} \) -module of finite dimension over \( \mathrm{K} \) and \( {\left( {\mathrm{M}}_{i}\right) }_{0 \leq i \leq n} \) a Jordan-Hölder series of the \( \mathrm{g} \) -module \( \mathrm{M} \) . Th... | (b) \( \Rightarrow \) (a): as A is finite-dimensional over K, the Jacobson radical of A is a nilpotent ideal (Algebra, Chapter VIII,§ 6, no. 4, Theorem 3) and hence every element of this radical is nilpotent.\n\n\( \left( a\right) \Rightarrow \left( c\right) : \mathrm{{each}}\;{Q}_{i} = {M}_{i}/{\overset{˜}{M}}_{i + 1}... | Yes |
Proposition 4. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathrm{M} \) a \( \mathfrak{g} \) -module of finite dimension over \( \mathrm{K} \) and \( \mathrm{A} \) the associative algebra generated by 1 and the set of \( {x}_{\mathrm{M}}\left( {x \in \mathfrak{g}}\right) \) .\n\n(a) The ideals \( \mathfrak{a} \) of \(... | The set of \( x \in \mathfrak{g} \) such that \( {x}_{\mathrm{M}} \) belongs to the Jacobson radical of \( \mathrm{A} \) is obviously an ideal of \( \mathfrak{g} \) . The proposition then follows immediately from Lemma 3. | No |
Proposition 1. Subalgebras and quotient algebras of a solvable Lie algebra are solvable. Every extension of a solvable algebra by a solvable algebra is solvable. Every finite product of solvable algebras is solvable. | Let \( \mathfrak{g} \) be a Lie algebra, \( {\mathfrak{g}}^{\prime } \) a subalgebra, \( \mathfrak{h} \) an ideal of \( \mathfrak{g},\mathfrak{k} = \mathfrak{g}/\mathfrak{h} \) and \( \phi \) the canonical mapping of \( \mathfrak{g} \) onto \( \mathfrak{k} \) . If \( \mathfrak{g} \) is solvable then \( {\mathcal{D}}^{k... | Yes |
Proposition 2. Let \( \mathfrak{g} \) be a Lie algebra. The following conditions are equivalent:\n\n(a) \( \mathrm{g} \) is solvable;\n\n(b) there exists a decreasing sequence \( \mathfrak{g} = {\mathfrak{g}}_{0} \supset {\mathfrak{g}}_{1} \supset \cdots \supset {\mathfrak{g}}_{n} = \{ 0\} \) of ideals of \( \mathfrak{... | (a) \( \Rightarrow \) (b): it suffices to consider the sequence of derived ideals of \( \mathfrak{g} \).\n\n(b) \( \Rightarrow \) (c): this is obvious.\n\n(c) \( \Rightarrow \) (d): suppose that condition (c) holds; every vector subspace of \( {\mathfrak{g}}_{i}^{\prime } \) containing \( {\mathfrak{g}}_{i + 1}^{\prime... | Yes |
Proposition 3. The radical \( \mathfrak{r} \) of a Lie algebra \( \mathfrak{g} \) is the smallest ideal of \( \mathfrak{g} \) such that \( \mathfrak{g}/\mathfrak{r} \) has radical \( \{ 0\} \) . | Let \( \mathfrak{a} \) be an ideal of \( \mathfrak{g} \) and \( \phi \) the canonical mapping of \( \mathfrak{g} \) onto \( \mathfrak{g}/\mathfrak{a} \) . If the radical of \( \mathfrak{g}/\mathfrak{a} \) is zero, then \( \phi \left( \mathfrak{r}\right) \), which is a solvable ideal of \( \mathfrak{g}/\mathfrak{a} \), ... | Yes |
Proposition 4. Let \( {\\mathfrak{g}}_{1},\\ldots ,{\\mathfrak{g}}_{n} \) be Lie algebras. The radical \( \\mathfrak{r} \) of the product of the \( {\\mathfrak{g}}_{i} \) is the product of the radicals \( {\\mathfrak{r}}_{i} \) of the \( {\\mathfrak{g}}_{i} \) . | The product \( {\\mathfrak{r}}^{\\prime } \) of the \( {\\mathfrak{r}}_{i} \) is a solvable ideal (Proposition 1) and hence \( {\\mathfrak{r}}^{\\prime } \\subset \\mathfrak{r} \) . The canonical image of \( \\mathfrak{r} \) in \( {\\mathfrak{g}}_{i} \) is a solvable ideal of \( {\\mathfrak{g}}_{i} \) and hence is cont... | Yes |
Lemma 1. Let \( \mathrm{V} \) be a finite-dimensional vector space over \( \mathrm{K},\mathrm{\;g} \) a subalgebra of \( \mathrm{{gl}}\left( \mathrm{V}\right) \) such that \( \mathrm{V} \) is a simple \( \mathfrak{g} \) -module and \( \mathfrak{a} \) a commutative ideal of \( \mathfrak{g} \) . Then \( \mathfrak{a} \cap... | Let \( \mathrm{S} \) be the subalgebra of \( \mathcal{L}\left( \mathrm{V}\right) \) generated by 1 and \( \mathfrak{a} \). If \( \mathfrak{b} \) is an ideal of \( \mathfrak{g} \) contained in \( \mathfrak{a} \) such that \( \operatorname{Tr}{bs} = 0 \) for all \( b \in \mathfrak{b} \) and all \( s \in \mathbb{S} \), th... | Yes |
Let \( \mathfrak{g} \) be a solvable Lie algebra. The nilpotent radical of \( \mathfrak{g} \) is \( \mathcal{D}\mathfrak{g} \). If \( \rho \) is a finite-dimensional simple representation of \( \mathfrak{g},\rho \left( \dot{\mathfrak{g}}\right) \) is commutative and the associative algebra \( \mathrm{L} \) generated by... | Here \( \mathfrak{r} = \mathfrak{g}, \) whence \( \mathfrak{s} = \mathcal{D}\mathfrak{g}. \) Hence \( \mathfrak{p}\left( {\mathcal{D}\mathfrak{g}}\right) = \{ 0\} , \) which shows that \( {\mathfrak{g}}^{\prime } = \mathfrak{p}\left( \mathfrak{g}\right) \) is commutative. Every element \( \neq 0 \) of \( \mathrm{L} \) ... | Yes |
Corollary 2 (Lie’s Theorem). Let \( \mathfrak{g} \) be a solvable Lie algebra; suppose that \( \mathrm{K} \) is algebraically closed. Let \( \mathrm{M} \) be a \( \mathrm{g} \) -module of finite dimension over \( \mathrm{K} \) and let \( {\left( {\mathrm{M}}_{i}\right) }_{0 \leq i \leq r} \) be a Jordan-Hölder series o... | Let \( {\rho }_{i} \) be the representation of \( g \) on \( {M}_{i - 1}/{M}_{i} \) . The associative algebra \( {\mathrm{L}}_{i} \) generated by 1 and \( {\rho }_{i}\left( g\right) \) is a field, a finite extension of \( \mathrm{K} \) and therefore equal to \( \mathrm{K} \) ; and \( {\mathrm{M}}_{i - 1}/{\mathrm{M}}_{... | "No" |
Corollary 3. Suppose that \( \mathrm{K} \) is algebraically closed. If \( \mathfrak{g} \) is an r-dimensional solvable Lie algebra, every ideal of \( \mathfrak{g} \) is a term of a decreasing sequence of ideals of dimensions \( r, r - 1,\ldots ,0 \) . | Every ideal is part of a Jordan-Hölder series of \( \mathfrak{g} \), considered as the space of the adjoint representation (Algebra, Chapter I,§ 6, no. 14, Corollary to Theorem 8); then it suffices to apply Corollary 2. | Yes |
Corollary 4. Suppose that \( \mathrm{K} = \mathbf{R} \) . Let \( \mathfrak{g} \) be a solvable Lie algebra. Every simple representation of \( \mathfrak{g} \) is of dimension \( \leq 2 \) . Every ideal of \( \mathfrak{g} \) is a term of a decreasing sequence \( {\left( {\mathfrak{g}}_{i}\right) }_{0 \leq i \leq m} \) of... | This is proved in a similar way to that for Corollaries 2 and 3, using the fact that every algebraic extension of \( \mathbf{R} \) is of degree \( \leq 2 \) . | No |
Corollary 5. For a Lie algebra \( \mathfrak{g} \) to be solvable, it is necessary and sufficient that Dg be nilpotent. | The condition is necessary by Corollary 1. It is sufficient since \( \mathfrak{g}/\mathcal{D}\mathfrak{g} \) is commutative. | No |
Corollary 6. Let \( \rho \) be a finite-dimensional representation of a Lie algebra \( g \) . Let \( \mathfrak{r} \) be the radical of \( \mathfrak{g} \) . Every element \( x \in \mathfrak{r} \) such that \( \rho \left( x\right) \) is nilpotent belongs to the largest nilpotency ideal \( n \) of \( \rho \) . | Let \( \mathrm{V} \) be the space of \( \rho \) ; let \( {\left( {\mathrm{V}}_{i}\right) }_{0 \leq i \leq r} \) be a Jordan-Hölder series for the \( \mathfrak{r} \) -module structure on \( \mathrm{V} \) and let \( {\rho }_{i} \) be the representation of \( \mathfrak{r} \) with space \( {\mathrm{V}}_{i}/{\mathrm{V}}_{i ... | Yes |
Lemma 2. Let \( x \) be an endomorphism of a finite-dimensional vector space \( \mathrm{V} \) and \( s \) (resp. n) its semi-simple (resp. nilpotent) component (cf. Algebra, Chapter VIII, § 9, no. 4, Definition 4). Let ad \( x \), ad \( s \), ad \( n \) be the respective images of \( x, s, n \) in the adjoint represent... | We know that \( \operatorname{ad}x = \operatorname{ad}s + \operatorname{ad}n,\left\lbrack {\operatorname{ad}s,\operatorname{ad}n}\right\rbrack = 0 \) and \( \operatorname{ad}n \) is nilpotent ( \( §4 \), Lemma 1). We show that ad \( s \) is semi-simple. It suffices to do this when K is algebraically closed (cf. Algebra... | Yes |
Lemma 3. Let \( \mathrm{M} \) be a finite-dimensional vector space, \( \mathrm{A} \) and \( \mathrm{B} \) two vector subspaces of \( \mathfrak{{gl}}\left( \mathrm{M}\right) \) such that \( \mathrm{B} \subset \mathrm{A} \) and \( \mathrm{T} \) the set of \( t \in \mathfrak{{gl}}\left( \mathrm{M}\right) \) such that \( \... | It suffices to prove this when \( \mathrm{K} \) is algebraically closed, which we shall assume henceforth. Let \( s \) and \( n \) be the semi-simple and nilpotent components of \( z \) and let \( \left( {e}_{i}\right) \) be a basis of \( \mathrm{M} \) such that \( s\left( {e}_{i}\right) = {\lambda }_{i}{e}_{i}\left( {... | Yes |
Theorem 2 (Cartan’s criterion). Let \( \mathfrak{g} \) be a Lie algebra, \( \mathrm{M} \) a finite-dimensional vector space, \( \rho \) a representation of \( \mathfrak{g} \) on \( \mathbf{M} \) and \( \beta \) the bilinear form on \( \mathfrak{g} \) associated with p. Then \( \rho \left( g\right) \) is solvable if and... | It can obviously be reduced to the case where \( g \) is a Lie subalgebra of \( \mathfrak{{gl}}\left( \mathbf{M}\right) \) and \( \rho \) is the identity mapping. If \( \mathfrak{g} \) is solvable, \( \mathcal{D}\mathfrak{g} \) is contained in the largest nilpotency ideal of the identity representation of \( g \) (Theo... | Yes |
Proposition 5. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathfrak{r} \) its radical.\n\n(a) If \( \rho \) is a finite-dimensional representation of \( \mathfrak{g} \) and \( \beta \) is the associated bilinear form, \( \mathfrak{r} \) and \( \mathcal{D}\mathfrak{g} \) are orthogonal with respect to \( \beta \) . | Let \( x, y \) be in \( \mathfrak{g}, z \in \mathfrak{r} \) . Then \( \left\lbrack {y, z}\right\rbrack \in \mathcal{D}\mathfrak{g} \cap \mathfrak{r} \) and hence\n\n\[\n\beta \left( {\left\lbrack {x, y}\right\rbrack, z}\right) = \beta \left( {x,\left\lbrack {y, z}\right\rbrack }\right) = 0\n\]\n\n(Theorem 1). Hence (a)... | Yes |
Corollary 1. Let \( \mathfrak{g} \) be a Lie algebra. Then \( \mathfrak{g} \) is solvable if and only if \( \mathcal{D}\mathfrak{g} \) is orthogonal to \( \mathrm{g} \) with respect to the Killing form. | This is an immediate consequence of Proposition 5 (b). | No |
Corollary 3. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{r} \) its radical and \( \mathfrak{a} \) an ideal of \( \mathfrak{g} \) . Then the radical of \( \mathfrak{a} \) is equal to \( \mathfrak{r} \cap \mathfrak{a} \) . | \( \mathfrak{r} \cap \mathfrak{a} \) is a solvable ideal of \( \mathfrak{a} \) and hence is contained in the radical \( {\mathfrak{r}}^{\prime } \) of \( \mathfrak{a} \) . Conversely, \( {\mathfrak{r}}^{\prime } \) is an ideal of \( \mathfrak{g} \) (Corollary 2 and \( §1 \), no. 4, Proposition 2) and hence \( {\mathfra... | Yes |
Proposition 6. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{r} \) its radical and \( \mathfrak{n} \) its largest nilpotent ideal. Every derivation of \( \mathfrak{g} \) maps \( \mathfrak{r} \) into \( \mathfrak{n} \) . | Let \( \mathrm{D} \) be a derivation of \( \mathfrak{g} \) . Let \( {\mathfrak{g}}^{\prime } = \mathfrak{g} + \mathrm{K}{x}_{0} \) be a Lie algebra in which \( \mathfrak{g} \) is an ideal of codimension 1 such that \( \mathrm{D}x = \left\lbrack {{x}_{0}, x}\right\rbrack \) for all \( x \in \mathfrak{g}(§1 \ | No |
Proposition 1. Let \( \mathfrak{g} \) be a semi-simple Lie algebra and \( \mathfrak{g} \) a finite-dimensional faithful representation of \( \mathfrak{g} \) . Then the bilinear form on \( \mathfrak{g} \) associated with \( \rho \) is non-degenerate. | The orthogonal of \( g \) with respect to this form is a solvable ideal \( (§5 \), no. 4, Theorem 2) and hence is zero. | No |
Let \( \mathfrak{g} \) be a Lie algebra, \( \beta \) its Killing form and \( \mathfrak{a} \) a semi-simple subalgebra of \( \mathfrak{g} \). The orthogonal \( \mathfrak{h} \) of \( \mathfrak{a} \) with respect to \( \mathfrak{g} \) is a supplementary subspace of \( \mathfrak{a} \) in \( \mathfrak{g} \) and \( \left\lbr... | Let \( {\beta }^{\prime } \) be the restriction of \( \beta \) to \( \mathfrak{a} \): it is the bilinear form associated with the representation \( x \mapsto {\operatorname{ad}}_{\mathfrak{g}}x \) of \( \mathfrak{a} \) on the space \( \mathfrak{g} \). This representation is faithful and hence \( {\beta }^{\prime } \) i... | Yes |
Corollary 2. Every extension of a semi-simple Lie algebra by a semi-simple Lie algebra is semi-simple and trivial. | This follows immediately from Corollary 1. | No |
Corollary 3. If \( \mathfrak{g} \) is semi-simple, every derivation of \( \mathfrak{g} \) is inner. | ad \( g \) is isomorphic to \( g \) and hence semi-simple and is an ideal of the Lie algebra \( \mathfrak{d} \) of derivations of \( \mathfrak{g}\left( {§1\text{, Proposition 1}}\right) \) . If \( \mathrm{D} \in \mathfrak{d} \) commutes with the elements of ad \( \mathfrak{g} \), then, for all \( x \in \mathfrak{g} \),... | Yes |
Lemma 1. Let \( \mathfrak{g} \) be a semi-simple Lie algebra. The adjoint representation of \( \mathfrak{g} \) is semisimple. Every ideal and every quotient algebra of \( \mathfrak{g} \) is semi-simple. | Let \( \mathfrak{a} \) be an ideal of \( \mathfrak{g} \) . The orthogonal \( \mathfrak{b} \) of \( \mathfrak{a} \) in \( \mathfrak{g} \) with respect to the Killing form is an ideal of \( \mathfrak{g} \) and \( \mathfrak{a} \cap \mathfrak{b} \) is a commutative ideal ( \( §3 \), no. 6, Proposition 7) and hence zero. He... | No |
Lemma 2. Let \( \mathfrak{g} \) be a Lie algebra. Then the following two conditions are equivalent:\n\n(a) All finite-dimensional linear representations of \( \mathfrak{g} \) are semi-simple.\n\n(b) Given a linear representation \( \circ \) of \( \mathfrak{g} \) on a finite-dimensional vector space \( \mathrm{V} \) and... | 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 \( \sigma \left( \mathfrak{g}\right) \) . Let \( \mu \) be the representation of \( \mathfrak{g} \) on \( \mathcal{L}\left( \m... | Yes |
Lemma 3. Let \( \mathfrak{g} \) be a semi-simple Lie algebra, \( \mathfrak{p} \) a linear representation of \( \mathfrak{g} \) on a finite-dimensional vector space \( \mathrm{V} \) and \( \mathrm{W} \) a subspace of \( \mathrm{V} \) of codimension 1 such that \( \rho \left( x\right) \left( \mathrm{V}\right) \subset \ma... | For all \( x \in \mathfrak{g} \) let \( \sigma \left( x\right) \) be the restriction of \( \rho \left( x\right) \) to W. Suppose first that \( \sigma \) is simple. If \( \sigma = 0 \), then \( \;\rho \left( x\right) \;\rho \left( y\right) \; = \;0\; \) for all \( \;x, y\; \) in \( \;\mathfrak{g} \), hence \( \;\rho \le... | Yes |
Proposition 2. For a Lie algebra \( \mathfrak{g} \) to be semi-simple, it is necessary and sufficient that it be a product of simple algebras. | The condition is sufficient (no. 1, Remark 3). Conversely, suppose that \( \mathfrak{g} \) is semi-simple. Since the adjoint representation of \( \mathfrak{g} \) is semi-simple, \( \mathfrak{g} \) is the direct sum of minimal non-zero ideals \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{m} \) . Then \( \mathfrak{g} \) ... | Yes |
Corollary 1. A semi-simple Lie algebra is the product of its simple ideals \( {\mathfrak{g}}_{i} \) . Every ideal of \( \mathfrak{g} \) is the product of certain of the \( {\mathfrak{g}}_{i} \) . | \( \mathfrak{g} = {\mathfrak{a}}_{1} \times \cdots \times {\mathfrak{a}}_{m} \), where the \( {\mathfrak{a}}_{i} \) are simple. As the centre of \( {\mathfrak{a}}_{i} \) is zero, the centralizer of \( {\mathfrak{a}}_{i} \) in \( \mathfrak{g} \) is the product of the \( {\mathfrak{a}}_{j} \) for \( j \neq i \) . Then le... | Yes |
Proposition 3. Let \( \mathrm{M} \) be a finite-dimensional vector space over \( \mathrm{K} \) and \( \mathrm{g} \) a semisimple subalgebra of \( \mathfrak{{gl}}\left( \mathbf{M}\right) \) . Then \( \mathfrak{g} \) contains the semi-simple and nilpotent components of its elements. | If \( {\mathrm{K}}_{1} \) is an extension of \( \mathrm{K} \), the Killing form of \( {\mathrm{g}}_{\left( {\mathrm{K}}_{1}\right) } \) is the extension to \( {\mathfrak{g}}_{\left( {\mathrm{K}}_{1}\right) } \) of that of \( \mathfrak{g}\left( {§3\text{, no. 8}}\right) \) and hence is non-degenerate; therefore \( {\mat... | Yes |
Proposition 4. Let \( \mathfrak{g},{\mathfrak{g}}^{\prime } \) be semi-simple Lie algebras and \( f \) a homomorphism of \( \mathfrak{g} \) into \( {\mathfrak{g}}^{\prime } \) . If \( x \in \mathfrak{g} \) is semi-simple (resp. nilpotent), so is \( f\left( x\right) \) . If \( f \) is surjective, every semisimple (resp.... | If \( \rho \) is a representation of \( {\mathfrak{g}}^{\prime },\rho \circ f \) is a representation of \( \mathfrak{g} \), whence the first assertion. If \( f \) is surjective, there exists a homomorphism \( g \) of \( {\mathfrak{g}}^{\prime } \) into \( \mathfrak{g} \) such that \( f \circ g \) is the identity homomo... | Yes |
Proposition 6. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{r} \) its radical and \( \mathfrak{s} \) its nilpotent radical.\n\n(a) \( \mathfrak{s} = \left\lbrack {\mathfrak{g},\mathfrak{r}}\right\rbrack = \mathcal{D}\mathfrak{g} \cap \mathfrak{r} \) . | Clearly \( \left\lbrack {\mathfrak{g},\mathfrak{r}}\right\rbrack \subset {\mathcal{D}}_{\mathcal{G}} \cap \mathfrak{r} \) . Now \( \mathcal{D}\mathfrak{g} \cap \mathfrak{r} = \mathfrak{s} \) by Theorem 1 of \( §5 \), no. 3. Let \( {\mathfrak{g}}^{\prime } = \mathfrak{g}/\left\lbrack {\mathfrak{g},\mathfrak{r}}\right\rb... | Yes |
Corollary 2. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{p} \) a semi-simple representation of \( \mathfrak{g} \) on a finite-dimensional vector space \( \mathrm{V},\mathrm{T} \) and \( \mathrm{S} \) the tensor and symmetric algebras of \( \mathrm{V} \) and \( {\sigma }_{\mathrm{T}} \) , \( {\sigma }_{\mathrm... | Let \( {\mathrm{T}}^{n} \) be the subspace of \( \mathrm{T} \) consisting of the homogeneous tensors of order \( n \) . This subspace is stable under \( {\sigma }_{\mathrm{T}} \) and the representation defined by \( {\sigma }_{\mathrm{T}} \) on \( {\mathrm{T}}^{n} \) is semi-simple (Corollary 1). Hence the corollary fo... | Yes |
Corollary 3. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathfrak{p} \) and \( {\mathfrak{p}}^{\prime } \) two finite-dimensional semi-simple representations of \( \mathrm{g} \) on spaces \( \mathrm{M} \) and \( {\mathrm{M}}^{\prime } \) . Then the representation of \( \mathrm{g} \) on \( {\mathcal{L}}_{\mathrm{K}}... | The \( \mathfrak{g} \) -module \( {\mathcal{L}}_{\mathrm{K}}\left( {\mathrm{M},{\mathrm{M}}^{\prime }}\right) \) is canonically identified with the \( \mathfrak{g} \) -module \( {\mathrm{M}}^{ * }{ \otimes }_{\mathrm{K}}{\mathrm{M}}^{\prime } \) (§ 3, no. 3, Proposition 4), so that Corollary 3 follows from Corollary 1. | Yes |
Corollary 4. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{a} \) an ideal of \( \mathfrak{g} \) and \( \mathfrak{g} \) a semi-simple representation of \( \mathfrak{g} \). (a) The restriction \( {\rho }^{\prime } \) of \( \rho \) to \( \mathfrak{a} \) is semi-simple. | Passing to the quotient by the kernel of \( \rho ,\rho \) can be assumed to be faithful. Then \( \mathfrak{g} \) is reductive. Let \( \mathfrak{g} = {\mathfrak{g}}_{1} \times {\mathfrak{g}}_{2} \), where \( {\mathfrak{g}}_{1} \) is the centre of \( \mathfrak{g} \) and \( {\mathfrak{g}}_{2} \) is semi-simple. Then \( \m... | Yes |
Proposition 7. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{h} \) a subalgebra reductive in \( \mathfrak{g} \) , \( \mathfrak{p} \) a representation of \( \mathfrak{g} \) on a vector space \( \mathrm{V} \) and \( \mathrm{W} \) the sum of the finite-dimensional subspaces of \( \mathrm{V} \) which are simple \( ... | Let \( {\mathrm{W}}_{0} \) be a finite-dimensional simple sub- \( \mathfrak{h} \) -module of \( \mathrm{V} \) . We need to prove that \( \rho \left( x\right) \left( {\mathrm{W}}_{0}\right) \subset \mathrm{W} \) for all \( x \in \mathfrak{g} \) . Let \( \mathrm{M} \) denote the vector space \( \mathfrak{g} \) considered... | Yes |
Corollary 1. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{h} \) a subalgebra reductive in \( \mathfrak{g} \) and \( \mathfrak{p} \) a finite-dimensional semi-simple representation of \( \mathfrak{g} \) . Then the restriction of \( \rho \) to \( \mathfrak{h} \) is semisimple. | It suffices to consider the case where \( \rho \) is simple. We adopt the notation \( \mathrm{V},\mathrm{W} \) of Proposition 4. Let \( {\mathrm{W}}_{1} \) be a subspace of \( \mathrm{V} \) minimal among the nonzero subspaces stable under \( \rho \left( \mathfrak{h}\right) \) . Then \( {\dot{W}}_{1} \subset W \), hence... | No |
Corollary 2. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{h} \) a subalgebra reductive in \( \mathfrak{g} \) and \( \mathfrak{k} \) a subalgebra of \( \mathfrak{h} \) reductive in \( \mathfrak{h} \) . Then \( \mathfrak{k} \) is reductive in \( \mathfrak{g} \) . | The representation \( x \mapsto {\operatorname{ad}}_{\mathfrak{g}}x \) of \( \mathfrak{h} \) on \( \mathfrak{g} \) is semi-simple and hence its restriction to \( \mathfrak{k} \) is semi-simple (Corollary 1). | "Yes" |
Proposition 8. Let \( \mathrm{V} \) be a finite-dimensional vector space. Then \( \mathfrak{{gl}}\left( \mathrm{V}\right) \) is reductive. its centre is the set of homotheties of \( \mathrm{V} \), its derived algebra is \( \mathfrak{{sl}}\left( \mathrm{V}\right) \) and the latter is semi-simple. | The identity representation of \( \mathfrak{{gl}}\left( \mathrm{V}\right) \) is simple, hence \( \mathfrak{{gl}}\left( \mathrm{V}\right) \) is reductive and therefore \( \mathfrak{{gl}}\left( \mathrm{V}\right) \) is the direct sum of its centre \( \mathfrak{c} \) and its derived algebra \( \mathcal{D}\left( {\mathfrak{... | Yes |
Proposition 9. Let \( \mathrm{V} \) be a vector space of finite dimension \( n \) over \( \mathrm{K} \) and \( \beta \) a nondegenerate symmetric (resp. alternating) bilinear form on \( \mathrm{V} \) . Let \( \mathfrak{g} \) be the Lie algebra consisting of the \( x \in \mathfrak{{gl}}\left( \mathrm{V}\right) \) such t... | For all \( u \in \mathfrak{{gl}}\left( V\right) \) let \( {u}^{ * } \) denote its adjoint relative to \( \beta \) ; then \( \mathrm{{Tr}}\left( u\right) \; = \mathrm{{Tr}}\left( {u}^{ * }\right) \) by Proposition 7 of Algebra, Chapter IX, § 1, no. 8. The condition\n\n\[ \beta \left( {{um},{m}^{\prime }}\right) + \beta ... | Yes |
Corollary 1. Let \( \mathfrak{s} \) be a Levi subalgebra of \( \mathfrak{g} \) and \( \mathfrak{h} \) a semi-simple subalgebra of \( \mathfrak{g} \) .\n\n(a) There exists a special automorphism of \( \mathfrak{g} \) mapping \( \mathfrak{h} \) onto a subalgebra of \( \mathfrak{s} \) . | Let \( \mathfrak{r} \) be the radical of \( \mathfrak{g} \) and \( \mathfrak{a} = \mathfrak{h} + \mathfrak{r} \), which is a subalgebra of \( \mathfrak{g} \) . Then \( \mathfrak{a}/\mathfrak{r} \) is semi-simple and \( \mathfrak{r} \) is solvable, hence \( \mathfrak{r} \) is the radical of \( \mathfrak{a} \) and \( \ma... | Yes |
Corollary 2. For a subalgebra \( \mathfrak{h} \) of \( \mathfrak{g} \) to be a Levi subalgebra of \( \mathfrak{g} \), it is necessary and sufficient that \( \mathfrak{h} \) be a maximal semi-simple subalgebra of \( \mathfrak{g} \) . | This follows immediately from Corollary 1. | No |
Corollary 3. Let \( \mathfrak{g} \) be a Lie algebra and \( \mathfrak{m} \) an ideal of \( \mathfrak{g} \) such that \( \mathfrak{g}/\mathfrak{m} \) is semisimple. Then \( \mathfrak{g} \) contains a subalgebra supplementary to \( \mathfrak{m} \) in \( \mathfrak{g} \) . In other words, every extension of a semi-simple L... | Let \( \mathfrak{s} \) be a Levi subalgebra of \( \mathfrak{g} \) (Theorem 5). Its canonical image in \( \mathfrak{g}/\mathfrak{m} \) is a Levi subalgebra and therefore equal to \( \mathfrak{g}/\mathfrak{m} \), hence \( \mathfrak{g} = \mathfrak{s} + \mathfrak{m} \) . Then an ideal of \( \mathfrak{s} \) supplementary in... | Yes |
Corollary 4. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{r} \) its radical, \( \mathfrak{s} \) a Levi subalgebra of \( \mathfrak{g} \) and \( \mathfrak{m} \) an ideal of \( \mathfrak{g} \) . Then \( \mathfrak{m} \) is the direct sum of \( \mathfrak{m} \cap \mathfrak{r} \) which is its radical and \( \mathfrak... | We know that \( m \cap s \) is the radical of \( m(§5 \), no. 5, Corollary 3 to Proposition 5). Let \( \mathfrak{h} \) be a Levi subalgebra of \( \mathfrak{m} \) and \( {\mathfrak{s}}^{\prime } \) a Levi subalgebra of \( \mathfrak{g} \) containing \( \mathfrak{h} \) (Corollary 1). The algebra \( \mathfrak{m} \cap {\mat... | Yes |
Lemma 4. Let \( \mathfrak{p},\sigma ,\tau \) be representations of \( \mathfrak{g} \) on vector spaces \( \mathrm{M},\mathrm{N},\mathrm{P} \) . Suppose that we are given a K-bilinear mapping \( \left( {m, n}\right) \mapsto m.n \) of \( \mathrm{M} \times \mathrm{N} \) into \( \mathrm{P} \) such that\n\n\[ \left( {\rho \... | (a) If \( {m}_{0} \in {\mathrm{M}}_{0} \), the mapping \( n \mapsto {m}_{0}.n \) is a \( \mathfrak{g} \) -module homomorphism.\n\n(b) If \( n \in {\mathrm{N}}_{\delta } \), then \( {m}_{0} \cdot n \in {\mathrm{P}}_{\delta } \).\n\n(c) If \( \mathrm{M} \) is a (not necessarily associative) algebra and the \( \wp \left( ... | No |
Lemma 5. Suppose further that \( \sigma \) and \( \tau \) are semi-simple and hence \( \mathrm{N} \) (resp. P) is the direct sum of the \( {\mathrm{N}}_{\delta } \) (resp. \( {\mathrm{P}}_{\delta } \) ). For all \( n \in \mathrm{N} \) (resp. \( p \in \mathrm{P} \) ), let \( {n}^{\natural } \) (resp. \( {p}^{\natural } ... | By linearity it suffices to consider the case where \( n \in {\mathbf{N}}_{\delta } \) . If \( \delta \neq {\delta }_{0},{n}^{\natural } = 0 \) and \( {m}_{0}.n \in {\mathrm{P}}_{\delta } \) (Lemma 4), hence \( {\left( {m}_{0}.n\right) }^{\natural } = 0 = {m}_{0}.{n}^{\natural } \) . If \( \delta = {\delta }_{0},{n}^{\... | Yes |
Lemma 1. Let \( \mathfrak{g} = {\mathfrak{g}}^{\prime } + \mathfrak{h} \) be a Lie algebra which is the sum of an ideal \( {\mathfrak{g}}^{\prime } \) and a subalgebra \( \mathfrak{h} \) . Let \( \sigma \) be a finite-dimensional representation of \( \mathfrak{g} \) . Suppose that \( \sigma \left( x\right) \) is nilpot... | Passing to the quotient by the kernel of \( \sigma ,\sigma \) may be assumed to be faithful. Then \( {\mathfrak{g}}^{\prime } \) and \( \mathfrak{h} \) are nilpotent and hence \( \mathfrak{g} \), which is an extension of a quotient of \( \mathfrak{h} \) by \( {\mathfrak{g}}^{\prime } \), is solvable. Then \( \mathfrak{... | No |
Proposition 1. Let \( \mathfrak{g} \) be a Lie algebra, \( \mathfrak{n} \) its largest nilpotent ideal, \( \mathfrak{a} \) a nilpotent ideal of \( \mathfrak{g} \) and \( \rho \) a finite-dimensional representation of \( \mathfrak{a} \) such that every element of \( \rho \left( \mathfrak{a}\right) \) is nilpotent. Then ... | Let \( \mathfrak{a} = {\mathfrak{n}}_{0} \subset {n}_{1} \subset \cdots \subset {n}_{p} = \mathfrak{n} \) be a sequence of subalgebras of \( \mathfrak{n} \) such that \( {n}_{i - 1} \) is an ideal of \( {n}_{i} \) of codimension 1 for \( 1 \leq i \leq p \) (§ 4, no. 1, Proposition 1 (e)). The algebra \( {\mathfrak{n}}_... | Yes |
Proposition 1. Every u-primitive element of \( \mathrm{E} \) belongs to \( {\mathrm{E}}^{ + } \) . | (2) implies \( x = \varepsilon \left( x\right) \cdot u + \varepsilon \left( u\right) \cdot x = \varepsilon \left( x\right) \cdot u + x \), whence \( \varepsilon \left( x\right) = 0 \) . | Yes |
Proposition 2. We have\n\n\[ \left( {{\pi }_{u} \otimes {\pi }_{u}}\right) \circ c = {c}_{u}^{ + } \circ {\pi }_{u} \] | Let \( x \) be in \( \mathrm{E} \) ; then\n\n\[ \left( {{\pi }_{u} \circ {\pi }_{u}}\right) \left( {c\left( x\right) }\right) = \left( {\left( {1 - {\eta }_{u}}\right) \otimes \left( {1 - {\eta }_{u}}\right) }\right) \left( {c\left( x\right) }\right) \]\n\n\[ = c\left( x\right) - \left( {1 \otimes {\eta }_{u}}\right) \... | Yes |
Proposition 3. If the cogebra (E, c) is coassociative (resp. cocommutative) (Algebra, Chapter III, \( §1\overline{1} \), no. 2), so is the cogebra \( \left( {{\mathrm{E}}^{ + },{c}_{u}^{ + }}\right) \) . | This follows from the following lemma:\n\nLemma 1. Let \( \pi : \) | No |
Lemma 1. Let \( \pi : \mathrm{E} \rightarrow {\mathrm{E}}^{\prime } \) be a surjective cogebra morphism. If \( \mathrm{E} \) is coassociative (resp. cocommutative), so is \( {\mathbf{E}}^{\prime } \) . | Let \( \mathrm{B} \) be an associative K-algebra; the mapping \( f \mapsto f \circ \pi \) is an injective algebra homomorphism of \( {\operatorname{Hom}}_{\mathrm{K}}\left( {{\mathrm{E}}^{\prime },\mathrm{B}}\right) \) into \( {\operatorname{Hom}}_{\mathrm{K}}\left( {\mathrm{E},\mathrm{B}}\right) \) . It then suffices ... | No |
Proposition 4. The set \( \mathrm{P}\left( \mathrm{E}\right) \) of primitive elements of \( \mathrm{E} \) is a Lie subalgebra of E. | If \( x, y \) are in \( \mathrm{P}\left( \mathrm{E}\right) \), then\n\n\[ c\left( {xy}\right) = c\left( x\right) c\left( y\right) = \left( {x \otimes 1 + 1 \otimes x}\right) \left( {y \otimes 1 + 1 \otimes y}\right) \]\n\n\[ = {xy} \otimes 1 + 1 \otimes {xy} + x \otimes y + y \otimes x, \]\n\nwhence\n\n\[ c\left( \left... | Yes |
Proposition 5. Let \( f : \mathrm{E} \rightarrow {\mathrm{E}}^{\prime } \) be a bigebra morphism. If \( x \) is a primitive element of \( \mathrm{E} \) , then \( f\left( x\right) \) is a primitive element of \( {\mathbf{E}}^{\prime } \) and the restriction of \( f \) to \( \mathbf{P}\left( \mathbf{E}\right) \) is a Lie... | Let \( c \) (resp. \( {c}^{\prime } \) ) be the coproduct of \( \mathrm{E} \) (resp. \( {\mathrm{E}}^{\prime } \) ). Since \( f \) is a cogebra morphism \( {c}^{\prime } \circ f = \left( {f \otimes f}\right) \circ c \), whence\n\n\[ \n{c}^{\prime }\left( {f\left( x\right) }\right) = \left( {f \otimes f}\right) \left( {... | Yes |
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