Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Theorem 3.2.1. Let \( \mathfrak{g},\mathfrak{G} \), and \( \gamma \) be as defined above. Then \( \left( {\mathfrak{G},\gamma }\right) \) is a universal enveloping algebra of \( \mathfrak{g} \). If \( \left( {\mathfrak{G},{\gamma }^{\prime }}\right) \) is another universal enveloping algebra of \( \mathfrak{g} \), ther... | Proof. We have already observed that \( \gamma \left\lbrack \mathfrak{g}\right\rbrack \) generates \( \mathfrak{G} \). Since \( \gamma \left( {u}_{X, Y}\right) = 0 \) , we have\n\n\[ \gamma \left\lbrack {X, Y}\right\rbrack ) = \gamma \left( X\right) \gamma \left( Y\right) - \gamma \left( Y\right) \gamma \left( X\right)... | Yes |
Corollary 3.2.4. There is a unique antiautomorphism \( a \mapsto {a}^{t} \) of \( \mathfrak{G} \) such that\n\n\[ \n{X}^{t} = - X\;\left( {X \in \mathfrak{g}}\right) ,\n\]\nand this antiautomorphism is involutive. | Proof. Since \( X \mapsto - X \) is an antiautomorphism of \( \mathfrak{g} \), the first assertion follows from the theorem above. Since \( a \mapsto {\left( {a}^{t}\right) }^{t} \) is an automorphism of \( \mathfrak{G} \) which is the identity on \( \mathfrak{g},{\left( {a}^{t}\right) }^{t} = a \) for all \( a \in \ma... | Yes |
Let \( \mathfrak{g} \) be a Lie algebra over \( k,\mathfrak{G} \) its universal enveloping algebra. If \( \mathfrak{a} \) is a subalgebra of \( \mathfrak{g} \) and \( \mathfrak{A} \) the subalgebra of \( \mathfrak{G} \) generated by \( \mathfrak{a} \) , then \( \mathfrak{A} \), together with the identity map of \( \mat... | Let \( {\mathfrak{A}}^{\prime } \) be the universal enveloping algebra of the subalgebra \( \mathfrak{a} \) of \( \mathfrak{g} \) . We denote the product operation in \( {\mathfrak{A}}^{\prime } \) by \( \cdot \) . Then there is a homomorphism \( \xi \) of \( {\mathfrak{A}}^{\prime } \) into \( \mathfrak{G} \) such tha... | Yes |
Corollary 3.2.6. Let \( \mathfrak{g} \) be a Lie algebra over \( k,\mathfrak{G} \) its universal enveloping algebra. Then \( \mathfrak{G}\mathfrak{g}\mathfrak{G} = \mathfrak{G}\mathfrak{g} = \mathfrak{g}\mathfrak{G}\left( { = {\mathfrak{G}}^{ + }\text{, say }}\right) \) ; and \( \mathfrak{G} \) is the direct sum of \( ... | Proof. Let \( {\mathfrak{G}}^{ + } = \mathfrak{G}\mathfrak{g}\mathfrak{G} \) . Then \( {\mathfrak{G}}^{ + } \) is a proper two-sided ideal in \( \mathfrak{G} \) and \( \mathcal{G}/{\mathcal{B}}^{ + } \) is the universal enveloping algebra of \( \mathfrak{g}/\mathfrak{g} = 0 \) . So \( \dim \left( {\mathcal{G}/{\mathcal... | Yes |
Corollary 3.2.7. Let \( \mathfrak{a} \) and \( \mathfrak{b} \) be subalgebras of \( \mathfrak{g} \) such that \( \mathfrak{g} \) is their vectorial direct sum. Let \( \mathfrak{A} \) (resp. \( \mathfrak{B} \) ) be the subalgebra of \( \mathfrak{G} \) generated by \( \mathfrak{a} \) (resp. b). Then the left (resp. right... | Proof. Considering an ordered basis for \( \mathrm{g} \) which consists of an ordered basis for \( \mathfrak{b} \) followed by an ordered basis for \( \mathfrak{a} \), and using the Poincaré- Birkhoff-Witt theorem, we conclude that the map \( \left( {b, a}\right) \mapsto {ba} \) of \( \mathfrak{B} \times \mathfrak{A} \... | Yes |
Theorem 3.2.8. \( \mathfrak{g} \) is a free Lie algebra generated by \( {X}_{i}\left( {i \in I}\right) \) . Moreover \( \mathfrak{G} \), together with the identity map of \( \mathfrak{g} \) into it, is the universal enveloping algebra of \( \mathfrak{g} \) . | Proof. Let \( \mathfrak{h} \) be a Lie algebra over \( k,\mathfrak{H} \supseteq \mathfrak{h} \) the universal enveloping algebra of \( \mathfrak{h} \) . Suppose \( {X}_{i}^{\prime }\left( {i \in I}\right) \) are elements of \( \mathfrak{h} \) . We denote by \( \xi \) the homomorphism of \( \mathfrak{G} \) into \( \math... | Yes |
Theorem 3.2.9. (i) For any integer \( n \geq 0 \) let \( {\mathfrak{G}}_{n} \) be the subspace of homogeneous elements \( \dagger \) of \( \mathfrak{G} \) of degree \( n \), and let \( {\mathfrak{g}}_{n} = {\mathfrak{G}}_{n} \cap \mathfrak{g} \) . Then \( {\mathfrak{g}}_{n} \) is the linear span of \( \psi \left( {{X}_... | Proof. \( {\mathfrak{G}}_{n} \) is the linear span of the \( {X}_{{i}_{1}}\cdots {X}_{{i}_{n}}\left( {{i}_{1},\ldots ,{i}_{n} \in I}\right) \), and we have\n\n(3.2.16)\n\n\[ \psi \left( {{X}_{{i}_{1}}\cdots {X}_{{i}_{n}}}\right) = \left\lbrack {{X}_{{i}_{1}},\psi \left( {{X}_{{i}_{2}}\cdots {X}_{{i}_{n}}}\right) }\righ... | Yes |
Theorem 3.3.1. (i) \( {\mathfrak{G}}^{\left( 0\right) } = {k1};{\mathfrak{G}}^{\left( 1\right) } \) is the direct sum of \( {\mathfrak{G}}^{\left( 0\right) } \) and \( \mathfrak{g} \) ; and for any \( n \geq 0,{\mathcal{B}}^{\left( n\right) } \) is the linear span of 1 and all elements of the form \( {Z}_{1}\cdots {Z}_... | Proof. The relation \( {\mathcal{B}}^{\left( 0\right) } = k \cdot 1 \) is obvious. Let \( n \geq 1 \) be arbitrary. Then \( {\mathfrak{J}}^{\left( n\right) } \) is spanned by 1 and all tensors of the form \( {Z}_{1} \otimes \cdots \otimes {Z}_{s} \) with \( 1 \leq s \leq n \) and \( {Z}_{1},\ldots ,{Z}_{s} \in \mathfra... | Yes |
Lemma 3.3.3. Let notation be as above. Then 5 is the direct sum of \( \mathfrak{L} \) and \( \bar{3} \) . Moreover, if \( \gamma \) is the natural map of \( \mathfrak{I} \) onto \( \mathfrak{G},\gamma \) is a linear isomorphism of \( \mathop{\sum }\limits_{{0 \leq q \leq p}}{\overline{\mathfrak{J}}}_{q} \) onto \( {\ma... | Proof. \( \gamma \) is an isomorphism of \( {\overline{\mathfrak{I}}}_{0} = {\mathfrak{I}}_{0} \) onto \( {\mathfrak{G}}^{\left( 0\right) } \) . Let \( p \geq 1 \) be arbitrary. If \( {X}_{1},\ldots ,{X}_{p} \in \mathfrak{g},1 \leq r \leq p - 1 \), and \( s \) is the permutation of \( \{ 1,\ldots, p\} \) that interchan... | Yes |
Theorem 3.3.4 (i) \( \\lambda \) is a linear isomorphism of \( \\mathcal{S} \) onto \( \\mathfrak{G} \) and of \( {\\mathcal{S}}^{\\left( p\\right) } \) onto \( {\\mathcal{B}}^{\\left( p\\right) } \) for \( p \\geq 0 \) . If \( {X}_{1},\\ldots ,{X}_{p} \\in \\mathfrak{g} \) , | Proof. Only (3.3.13) remains to be proved in (i). We may assume that \( p \\geq 1 \) . Let \( {X}_{1},\\ldots ,{X}_{p} \\in \\mathfrak{g} \) and let \( t = {X}_{1} \\otimes \\cdots \\otimes {X}_{p} \) Then from (3.3.6), \( \\gamma \\left( {{Q}_{p}\\left( t\\right) }\\right) = \\left( {1/p!}\\right) \\mathop{\\sum }\\li... | Yes |
Corollary 3.3.5. For \( X \in \mathfrak{g} \), let \( {D}_{X} \) be the derivation of \( \mathcal{S} \) that extends the endomorphism ad \( X \) of \( \mathfrak{g} \) ; then\n\n(3.3.16)\n\n\[ \lambda \left( {{D}_{X}\left( u\right) }\right) = {X\lambda }\left( u\right) - \lambda \left( u\right) X\;\left( {u \in \mathbb{... | Proof. Take \( D = \) ad \( X \) in (3.3.15). | No |
Corollary 3.3.6. Let \( \mathfrak{g} \) be the direct sum of the subspaces \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{r} \) . Let \( {\mathcal{S}}_{i} \) be the subalgebra of \( \mathcal{S} \) generated by 1 and \( {\mathfrak{a}}_{i},{\mathcal{S}}_{i, d} \) the homogeneous subspace of \( {\mathcal{S}}_{i} \) of degr... | Proof. We begin by proving (3.3.17) using induction on \( p \) . We may assume \( p \geq 1 \) . Now, for \( {u}_{i} \in {\mathcal{S}}_{i,{d}_{i}} \) with \( {d}_{1} + \cdots + {d}_{r} \leq p \), we have from (3.3.14) (3.3.18) \[ \lambda \left( {{u}_{1}\cdots {u}_{r}}\right) \equiv \lambda \left( {u}_{1}\right) \cdots \... | Yes |
Corollary 3.3.7. Let \( \mathfrak{a} \) be a subalgebra of \( \mathfrak{g} \) and \( \mathfrak{b} \) a subspace such that \( \mathfrak{g} \) is the direct sum of \( \mathfrak{a} \) and \( \mathfrak{b} \) . Let \( S\left( \mathfrak{b}\right) \) be the subalgebra of \( \mathcal{S} \) generated by 1 and b. Then\n\n(3.3.20... | Proof. Let \( S\left( \mathfrak{a}\right) \) and \( \mathfrak{A} \) be the respective subalgebras of \( \mathcal{S} \) and \( \mathfrak{G} \) generated by 1 and \( \mathfrak{a} \) . Since we can canonically identify \( S\left( \mathfrak{a}\right) \) with the symmetric algebra of \( \mathfrak{a} \), and \( \mathfrak{A} ... | Yes |
For any \( X \in {\mathfrak{g}}_{c} \), let \( \partial \left( X\right) \) be the differential operator \( f \mapsto {Xf}\left( {f \in {C}^{\infty }\left( G\right) }\right) \) . Then the map \( X \mapsto \partial \left( X\right) \) extends uniquely to an isomorphism \( \partial \left( {a \mapsto \partial \left( a\right... | We have, for all \( X, Y \in {\mathfrak{G}}_{c} \), \[ \partial \left( \left\lbrack {X, Y}\right\rbrack \right) = \partial \left( X\right) \partial \left( Y\right) - \partial \left( Y\right) \partial \left( X\right) \] Let \( \mathfrak{D} \) be the algebra (over \( \mathbf{C} \) ) of all left-invariant analytic differe... | Yes |
Theorem 3.4.2. Fix \( y \in G \) . For each \( a \in {\mathfrak{G}}_{c} \), let \( {\tau }_{a} \) be the element of \( {T}_{yc}^{\left( \infty \right) }\left( G\right) \) induced by the linear function \( f \mapsto f\left( {y;a}\right) \) on \( {C}^{\infty }\left( G\right) \) . Then \( \tau \left( {a \mapsto {\tau }_{a... | Proof. Clearly, \( {\tau }_{1} = {1}_{y} \) . If \( a = {Y}_{1}\cdots {Y}_{s}\left( {1 \leq s \leq r,{Y}_{1},\ldots ,{Y}_{s} \in \mathcal{G}}\right) \) , and \( f = {f}_{1}\cdots {f}_{r + 1} \), where \( {f}_{1},\ldots ,{f}_{r + 1} \) are \( {C}^{\infty } \) functions on \( G \) that vanish at \( y \), then \( f\left( ... | Yes |
Theorem 3.5.1. Let \( \mathfrak{g} \) be a Lie algebra over \( k,{k}^{\prime } \) an extension field of \( k \) , and \( {\mathfrak{g}}^{{k}^{\prime }} \) the extension of \( \mathfrak{g} \) to \( {k}^{\prime } \) . Then \( \mathfrak{g} \) is nilpotent if and only if \( {\mathfrak{g}}^{{k}^{\prime }} \) is. If \( \math... | Proof. Since \( \mathfrak{g} \subseteq {\mathfrak{g}}^{{k}^{\prime }},{\mathfrak{g}}^{{k}^{\prime }} \) nilpotent \( \Rightarrow \mathfrak{g} \) nilpotent. Suppose that \( \mathfrak{g} \) is nilpotent, and for any integer \( r \geq 1 \), let \( {\mathfrak{g}}_{r}\left( X\right) = \operatorname{tr}{\left( \operatorname{... | Yes |
Theorem 3.5.3. Let \( \mathfrak{g} \) be a nilpotent Lie algebra over \( k,\rho \) a nil representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( V \) over \( k \) . Define \( {V}_{0} = 0 \) and for \( i \geq 1 \), let \( {V}_{i} = \left\{ {v : v \in V,\rho \left( X\right) v \in {V}_{i - 1}\text{ fo... | Proof. It is clear by induction on \( i \) that the \( {V}_{i} \) are well-defined subspaces of \( V \) invariant for the representation \( \rho \), and \( {V}_{0} \subseteq {V}_{1} \subseteq \cdots \) . Suppose \( i \geq 0 \) and \( {V}_{i} \neq V \) . Then the quotient representation in \( V/{V}_{i} \) is also a nil ... | Yes |
Theorem 3.5.4. Let \( \mathfrak{g} \) be a nilpotent Lie algebra over \( k \) ; then:\n\n(i) Let \( {\mathfrak{g}}_{i}\left( {i \geq 0}\right) \) be defined inductively as follows. \( {\mathfrak{g}}_{0} = 0 \), and for \( i \geq 1 \), \[ {\mathfrak{g}}_{i} = \left\{ {X : X \in \mathfrak{g},\left\lbrack {X,\mathfrak{g}}... | Proof. (i) and (ii) are immediate consequence of Theorem 3.5.3 applied to the adjoint representation of \( \mathfrak{g} \). Note that (3.5.3) displays the fact that the \( {\mathfrak{g}}_{i} \) are ideals. To prove (iii), let \( {m}_{i} = \dim \left( {\mathfrak{g}}_{i}\right) \left( {0 \leq i \leq s,{m}_{s} = m}\right)... | Yes |
Let \( \mathfrak{g} \) be a Lie algebra over \( k \), and let \( {\mathfrak{C}}^{0}\mathfrak{g},{\mathfrak{C}}^{1}\mathfrak{g},\ldots \) be defined inductively as follows: \( {\mathcal{C}}^{0}\mathfrak{g} = \mathfrak{g} \), and for \( q \geq 1,{\mathcal{C}}^{q}\mathfrak{g} = \left\lbrack {\mathfrak{g},{\mathcal{C}}^{q ... | Write \( {\mathfrak{L}}_{q} = {\mathfrak{C}}^{q}\mathfrak{g}, q \geq 0 \) . If \( q \geq 0 \) and \( {\mathfrak{L}}_{q} \) is an ideal, then \( {\mathfrak{L}}_{q + 1} = \) \( \left\lbrack {\mathfrak{g},{\mathfrak{L}}_{q}}\right\rbrack \subseteq {\mathfrak{L}}_{q} \) and \( \left\lbrack {\mathfrak{g},{\mathfrak{L}}_{q +... | Yes |
Corollary 3.5.6. Let \( \mathfrak{g} \) be nilpotent, \( m = \dim \mathfrak{g} \) . Then there are ideals \( {\mathfrak{h}}_{i} \) of \( \mathfrak{g} \) such that (i) \( \dim {\mathfrak{h}}_{i} = m - i \) for \( 0 \leq i \leq m \) ,(ii) \( {\mathfrak{h}}_{0} = \mathfrak{g} \supseteq {\mathfrak{h}}_{1} \supseteq \cdots ... | Proof. Let \( {\mathfrak{g}}_{0} = 0,{\mathfrak{g}}_{1},\ldots ,{\mathfrak{g}}_{s} = \mathfrak{g} \) be the ideals defined by the equation (3.5.3). If \( \mathfrak{a},\mathfrak{b} \) are any two linear subspaces of \( \mathfrak{g} \) with \( {\mathfrak{g}}_{i} \subseteq \mathfrak{b} \subseteq \mathfrak{a} \subseteq {\m... | Yes |
Lemma 3.5.7. Let \( \mathfrak{g} \) be a nilpotent Lie algebra over \( k \) and let \( {\rho }_{i} \) be a \( {\lambda }_{i} \) - representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( {V}_{i}\left( {i = 1,2,{\lambda }_{i} \in {\mathfrak{g}}^{ * }}\right) \) . Then \( {\rho }_{1} \otimes {\rho }_{... | Proof. Let \( {1}_{j} \) be the identity endomorphism of \( {V}_{j}\left( {j = 1,2}\right) \) . Write \( V = {V}_{1} \otimes {V}_{2},\rho = {\rho }_{1} \otimes {\rho }_{2},\lambda = {\lambda }_{1} + {\lambda }_{2} \) . Then \( \rho \left( X\right) = {\rho }_{1}\left( X\right) \otimes {1}_{2} + {1}_{1} \) \( \otimes {\r... | Yes |
Theorem 3.5.8. Let \( \mathfrak{g} \) be a nilpotent Lie algebra, \( V \) a finite-dimensional vector space, both over \( k \) . Let \( \rho \) be a representation of \( \mathfrak{g} \) in \( V \) . Then the weight subspaces of \( \rho \) corresponding to distinct weights are linearly independent. If \( k \) is algebra... | Proof. We have already proved the first assertion. Let \( k \) be algebraically closed. We prove the second assertion by induction on dim \( V \) . Suppose that for each \( X \in \mathfrak{g},\rho \left( X\right) \) has exactly one eigenvalue, say \( \lambda \left( X\right) \) . Then, as we saw above, \( \lambda \) is ... | Yes |
Theorem 3.6.1. (i) There exists a polynomial mapping \( {}^{4}P \) of \( \mathfrak{g} \times \mathfrak{g} \) into 9 such that\n\n(3.6.1)\n\n\[ \exp X\exp Y = \exp P\left( {X : Y}\right) \;\left( {X, Y \in \mathfrak{g}}\right) . \] | Proof. We prove (i) using the results of \( §{2.15} \) . Let \( {c}_{n}\left( {n \geq 1}\right) \) be the maps of \( \mathfrak{g} \times \mathfrak{g} \) into \( \mathfrak{g} \) such that \( {c}_{1}\left( {X : Y}\right) = X + Y\left( {X, Y \in \mathfrak{g}}\right) \) and the recursion formulae (2.15.15) are satisfied. E... | Yes |
Theorem 3.6.2. Let \( G \) be simply connected. Then exp is an analytic diffeomorphism of \( \mathfrak{g} \) onto \( G \) . If \( H \) is an analytic subgroup of \( G \) and \( \mathfrak{h} \) is the corresponding subalgebra of \( \mathfrak{g} \), then \( H \) is closed in \( G \), is simply connected, and is equal to ... | Proof. Since \( \mathfrak{g} \) is a covering manifold of \( G \) with exp as the covering map, \( D \) must be \( \{ 0\} \) when \( G \) is simply connected. Hence exp is an analytic diffeomorphism of \( \mathfrak{g} \) onto \( G \) . Let \( H \) be an analytic subgroup of \( G,\mathfrak{h} \) the corresponding subalg... | No |
Corollary 3.6.4. The center of any nilpotent analytic group is connected. | Proof. Let \( H \) be a nilpotent analytic group with Lie algebra \( \mathfrak{h}, Z \) the center of \( H \) . Applying Theorem 3.6.3 to the case when \( V = \mathfrak{h} \) and \( \overline{\mathfrak{g}} = \operatorname{ad}\left\lbrack \mathfrak{h}\right\rbrack \) , we find that \( H/Z \cong \operatorname{Ad}\left( H... | Yes |
Lemma 3.6.5. Let \( A \) be an analytic manifold, \( H \) a Lie group acting analytically on \( A \) via the action \( \left( {h, a}\right) \mapsto h \cdot a\left( {h \in H, a \in A}\right) \). For any analytic function \( \varphi \) on \( A \) and \( h \in H \), let \( {\varphi }^{h}\left( a\right) = \varphi \left( {{... | Proof. Since \( {\varphi }_{i}^{h{h}^{\prime }} = {\left( {\varphi }_{i}^{{h}^{\prime }}\right) }^{h}\left( {h,{h}^{\prime } \in H}\right) \), it is clear that \( V \) is invariant under all the \( \pi \left( h\right) \). Let \( \left\{ {{\psi }_{1},\ldots ,{\psi }_{r}}\right\} \) be a basis for \( V \). Then there are... | Yes |
Theorem 3.7.1 (i) If \( \mathfrak{h} \) is an ideal in \( \mathfrak{g} \), the \( {\mathfrak{D}}^{p}\mathfrak{h} \) are ideals of \( \mathfrak{g} \) for all \( p \geq 0 \) . If \( D \) is a derivation of \( \mathfrak{g} \) that leaves \( \mathfrak{h} \) invariant, then \( D \) leaves each \( {\mathfrak{D}}^{p}\mathfrak... | Proof. (i) If \( X,{X}^{\prime } \in \mathfrak{l},\;Y \in \mathfrak{g} \), then \( \left\lbrack {Y,\left\lbrack {X,{X}^{\prime }}\right\rbrack }\right\rbrack = - \left\lbrack {X,\left\lbrack {{X}^{\prime }, Y}\right\rbrack }\right\rbrack - \left\lbrack {{X}^{\prime },\left\lbrack {Y, X}\right\rbrack }\right\rbrack \) .... | Yes |
Theorem 3.7.2. (i) \( \mathfrak{g} \) is solvable if and only if we can find ideals \( {\mathfrak{g}}_{0} = \mathfrak{g} \) , \( {\mathfrak{g}}_{1},\ldots ,{\mathfrak{g}}_{s + 1} = 0 \) such that \( {\mathfrak{g}}_{i} \supseteq {\mathfrak{g}}_{i + 1} \) and \( {\mathfrak{g}}_{i}/{\mathfrak{g}}_{i + 1} \) is abelian, fo... | Proof. (i) If \( \mathfrak{g} \) is solvable, \( {\mathfrak{g}}_{p} = {\mathfrak{D}}^{p}\mathfrak{g}\left( {p \geq 0}\right) \) have all the required properties. Conversely, let \( {\mathfrak{g}}_{0},{\mathfrak{g}}_{1},\ldots \) be as in (i). Since \( {\mathfrak{g}}_{i}/{\mathfrak{g}}_{i + 1} \) is abelian, \( \mathfra... | Yes |
Corollary 3.7.4. Let assumptions be as in the above theorem. If \( \rho \) is irreducible, then \( \dim V = 1 \) . | Proof. Obvious. | No |
Corollary 3.7.5. Let \( \mathfrak{g} \) be a solvable Lie algebra over \( k \) . Then we can find subalgebras \( {\mathfrak{g}}_{1} = {\mathfrak{g}}_{2},{\mathfrak{g}}_{2},\ldots ,{\mathfrak{g}}_{m + 1} = 0 \) such that (i) \( {\mathfrak{g}}_{i + 1} \subseteq {\mathfrak{g}}_{i} \) and \( {\mathfrak{g}}_{i + 1} \) is an... | Proof. We have seen that we can select an ideal \( {\mathfrak{g}}_{2} \) of \( {\mathfrak{g}}_{1} = \mathfrak{g} \) such that \( \dim \left( {{g}_{1}/{g}_{2}}\right) = 1 \) . The first assertion is now immediate by induction in \( \dim \left( g\right) \) . Suppose now that \( k \) is algebraically closed. Applying the ... | Yes |
Theorem 3.7.6. Let \( \mathfrak{g} \) be a solvable Lie algebra over \( k \) . Let \( \rho \) be a representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( V \) . Then the set of all \( X \in \mathfrak{g} \) with \( \rho \left( X\right) \) nilpotent is an ideal in \( \mathfrak{g} \) that contains \(... | Proof. First, assume that \( k \) is algebraically closed. Let \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) be a basis for \( V \) and \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) elements of \( {\mathfrak{g}}^{ * } \) such that\n\n(3.7.10)\n\n\[ \rho \left( X\right) {v}_{j} \equiv {\lambda }_{j}\left( X\right) {v}_{... | Yes |
Theorem 3.8.1. (i) If \( {k}^{\prime } \) is an extension field of \( k \), then \( {\left( \operatorname{rad}\mathfrak{g}\right) }^{{k}^{\prime }} = \operatorname{rad}{\mathfrak{g}}^{{k}^{\prime }} \) . | Proof. To prove (i), let \( \bar{k} \) be an algebraic closure of \( {k}^{\prime } \) . Let \( \mathfrak{q},{\mathfrak{q}}^{\prime },\overline{\mathfrak{q}} \) be the respective radicals of \( \mathfrak{g},{\mathfrak{g}}^{{k}^{\prime }},{\mathfrak{g}}^{k} \) . Since \( {\mathfrak{q}}^{k} \) is a solvable ideal of \( {\... | Yes |
Lemma 3.8.2. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) and \( \rho \) a representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( V \) over \( k \) . Let \( \mathfrak{G} \) be the universal enveloping algebra of \( \mathfrak{g} \) and let \( \sigma \) be the extension of \( \rho \) to a re... | Proof. We begin with the following simple result. Let \( \tau \) be an irreducible representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( W,\mathfrak{m} \) an ideal of \( \mathfrak{g} \) such that \( \tau \left( X\right) \) is nilpotent for all \( X \in \mathfrak{m} \) ; then \( \tau \left\lbrack ... | Yes |
Theorem 3.8.3. (i) If \( {k}^{\prime } \) is an extension field of \( k \), then (nil rad \( \mathfrak{g}{)}^{{k}^{\prime }} = \) nil rad \( {\mathfrak{g}}^{{k}^{\prime }} \) . | Proof. (i) is proved exactly as the assertion (i) of Theorem 3.8.1. Moreover, as in that theorem, we can prove that \( \mathfrak{n} \) is invariant under all derivations of \( g \) . | No |
Corollary 3.8.4. If \( \mathfrak{g} \) is solvable, nil rad \( \mathfrak{g} \) is the set of all \( X \in \mathfrak{g} \) for which ad \( X \) is nilpotent, and \( \mathfrak{{Dg}} \subseteq \) nil rad \( \mathfrak{g} \) . | Proof. Follows from (iii) above. | No |
Theorem 3.9.1. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) . Then \( \mathfrak{g} \) is solvable if and only if\n\n(3.9.15)\n\n\[ \langle X,\left\lbrack {Y, Z}\right\rbrack \rangle = 0\;\left( {X, Y, Z \in \mathfrak{g}}\right) . \]\n\nIn particular, if the Cartan-Killing form of \( \mathfrak{g} \) is identical... | Proof. Let \( \mathfrak{g} \) be solvable, \( X \in \mathfrak{g},{X}^{\prime } \in \mathfrak{{Dg}} \) . By Theorem 3.7.6, \( \mathfrak{{Dg}} \) is a nilpotent ideal of \( \mathfrak{g} \), and hence \( \mathfrak{D}\mathfrak{g} \subseteq \) nil rad \( \mathfrak{g} \) . Consequently, by Lemma 3.8.2, ad \( X \) ad \( {X}^{... | Yes |
Theorem 3.9.2. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) . Then \( \mathfrak{g} \) is semisimple if and only if the Cartan-Killing form of \( \mathfrak{g} \) is nonsingular. | Proof. Suppose \( \mathfrak{g} = \operatorname{rad}\mathfrak{g} \neq 0 \) . Let \( p \geq 0 \) be such that \( \mathfrak{a} = {\mathfrak{D}}^{p}\mathfrak{g} \neq 0 \) but \( \mathfrak{{Da}} = 0 \) . Then \( \mathfrak{a} \) is abelian and is an ideal of \( \mathfrak{g} \) by (i) of Theorem 3.7.1. Suppose \( X \in \mathf... | Yes |
Corollary 3.9.3. Let \( \mathfrak{g} \) admit no ideals other than 0 and \( \mathfrak{g} \) . Then \( \mathfrak{g} \) is either of dimension 1 or semisimple. | Proof. Let \( \mathfrak{m} \) be as in (3.9.16). Then \( \mathfrak{m} \) is an ideal. If \( \mathfrak{m} = 0,\langle \cdot , \cdot \rangle \) is nonsingular, so \( \mathfrak{g} \) is semisimple. If \( \mathfrak{m} = \mathfrak{g},\mathfrak{g} \) is solvable. In this case, \( \mathfrak{D}\mathfrak{g} \neq \mathfrak{g} \)... | Yes |
Corollary 3.9.4. Let \( \mathfrak{g} \) be a Lie algebra over \( k,\left\{ {{X}_{1},\ldots ,{X}_{m}}\right\} \) a basis for \( \mathfrak{g} \) . Then \( \mathfrak{g} \) is semisimple if and only if\n\n(3.9.17)\n\n\[ \det {\left( \left\langle {X}_{i},{X}_{j}\right\rangle \right) }_{1 \leq i, j \leq m} \neq 0. \] | Proof. The relation (3.9.17) is the criterion for \( \langle \cdot , \cdot \rangle \) to be nonsingular. The second statement follows trivially from the first. | No |
Lemma 3.9.5. Let \( \mathfrak{g} \) be a Lie algebra over \( k \), and \( \rho \) a representation of \( \mathrm{g} \) in a finite dimensional vector space \( V \) over \( k \) . Let \( {B}^{\rho } \) be as in (3.9.11) and let \( \mathfrak{p} \) be defined by\n\n(3.9.19)\n\n\[ \mathfrak{p} = \left\{ {X : X \in \mathfra... | Proof. The relation (3.9.14) implies at once that \( \mathfrak{p} \) is an ideal of \( \mathfrak{g} \) . Let \( X \in \left\lbrack {\mathfrak{p},\mathfrak{g}}\right\rbrack \) and write\n\n\[ X = \mathop{\sum }\limits_{{1 \leq i \leq r}}\left\lbrack {{Y}_{i},{Z}_{i}}\right\rbrack \;\left( {{Y}_{i} \in \mathfrak{p},{Z}_{... | Yes |
Theorem 3.10.1 Let \( \mathfrak{g} \) be semisimple. If \( \mathfrak{h} \) is an ideal of \( \mathfrak{g} \), then \( {\mathfrak{h}}^{ \bot } \) is also an ideal, \( \left\lbrack {\mathfrak{h},{\mathfrak{h}}^{ \bot }}\right\rbrack = 0 \), and \( \mathfrak{g} \) is the direct sum of \( \mathfrak{h} \) and \( {\mathfrak{... | Proof. That \( {\mathfrak{h}}^{ \bot } \) is an ideal follows from (3.9.9). Suppose \( \mathfrak{h} \cap {\mathfrak{h}}^{ \bot } \neq 0 \) . Then \( \left\langle {X,{X}^{\prime }}\right\rangle = 0 \) for \( X,{X}^{\prime } \in \mathfrak{h} \cap {\mathfrak{h}}^{ \bot } \) . So \( \langle \cdot , \cdot {\rangle }_{\mathf... | Yes |
Corollary 3.10.2. If \( \mathfrak{g} \) is scmisimple, then\n\n\[ \mathfrak{g} = \mathfrak{D}\mathfrak{g} \] | Proof. If \( \mathfrak{D}\mathfrak{g} \neq \mathfrak{g},\mathfrak{g}/\mathfrak{D}\mathfrak{g} \) will be nonzero, abelian, and semisimple all at once, which is impossible. | Yes |
Corollary 3.10.3. Let \( \mathfrak{g} \) be semisimple, \( \mathfrak{l} \) an ideal of \( \mathfrak{g} \), and \( \mathfrak{a} \) an ideal of \( \mathfrak{h} \). Then \( \mathfrak{a} \) is an ideal of \( \mathfrak{g} \). In particular, if \( \mathfrak{h} \) is a minimal element of the set of all ideals of \( \mathfrak{... | Proof. Since \( \mathfrak{g} = \mathfrak{h} + {\mathfrak{h}}^{ \bot } \) and \( \left\lbrack {\mathfrak{h},{\mathfrak{h}}^{ \bot }}\right\rbrack = 0 \), we have \( \left\lbrack {\mathfrak{a},\mathfrak{g}}\right\rbrack = \left\lbrack {\mathfrak{a},\mathfrak{h}}\right\rbrack \subseteq \mathfrak{a} \). | No |
Theorem 3.10.5. Let \( \mathfrak{h} \) be a Lie algebra over \( k,\mathfrak{q} \) the radical of \( \mathfrak{h} \) . If \( \mathfrak{a} \) is an ideal such that \( \mathfrak{h}/\mathfrak{a} \) is semisimple, then \( \mathfrak{q} \subseteq \mathfrak{a} \) . If \( \pi \) is a homomorphism of \( \mathfrak{h} \) onto a Li... | Proof. Let \( \tau \) be the natural map of \( \mathfrak{h} \) onto \( \mathfrak{h}/\mathfrak{a} \) . If \( \mathfrak{q} \nsubseteq \mathfrak{a},\tau \left\lbrack \mathfrak{q}\right\rbrack \) will be a nonzero solvable ideal of \( \mathfrak{h}/\mathfrak{a} \) . So we must have \( \mathfrak{q} \subseteq \mathfrak{a} \) ... | Yes |
Theorem 3.10.6. Let \( \mathfrak{g} \) be semisimple. If \( D \) is a derivation of \( \mathfrak{g} \), there is a unique \( X \in \mathfrak{g} \) such that \( D = {adX} \). Any \( X \in \mathfrak{g} \) can be written as \( S + N \), where \( \left\lbrack {S, N}\right\rbrack = 0 \), ad \( S \) is semisimple, and ad \( ... | Proof. Since \( \langle \cdot , \cdot \rangle \) is nonsingular, we can find \( X \in \mathfrak{g} \) such that \( \langle X, Y\rangle = \) \( \operatorname{tr}\left( {D\text{ad}Y}\right) \) for all \( Y \in \mathfrak{g} \). Let \( {D}^{\prime } = D - \) ad \( X \). Then \( {D}^{\prime } \) is a derivation of \( \mathf... | Yes |
Lemma 3.10.7. Let \( \mathfrak{g} \) be semisimple, \( \rho \) a finite-dimensional representation of \( \mathfrak{g} \) in a vector space \( V \), and \( {B}^{\rho } \) the bilinear form on \( \mathfrak{g} \times \mathfrak{g} \) defined by \( \rho \) . If \( \mathfrak{h} \) is the kernel of \( \rho \) and \( \mathfrak... | Proof. I) is an ideal and \( \mathfrak{g} \) is the direct sum of \( \mathfrak{h} \) and \( \mathfrak{p} \) by Theorem 3.10.1. Let\n\n\[ \mathfrak{m} = \left\{ {X : X \in \mathfrak{p},{B}^{\rho }\left( {X, Y}\right) = 0\text{ for all }Y \in \mathfrak{p}}\right\} .\n\]\n\nThen \( \mathfrak{m} \) is an ideal in \( \mathf... | Yes |
Theorem 3.10.8. Let \( G \) be a semisimple analytic group over \( k \) with Lie algebra \( \mathfrak{g} \) . Then \( \operatorname{Ad}\left\lbrack G\right\rbrack = \operatorname{Aut}{\left( \mathfrak{g}\right) }^{0} \), the component of 1 in \( \operatorname{Aut}\left( \mathfrak{g}\right) \), and \( \operatorname{Aut}... | Proof. It is obvious that \( \operatorname{Ad}\left\lbrack G\right\rbrack \subseteq \operatorname{Aut}{\left( \mathfrak{g}\right) }^{0} \) . To prove the first assertion it is therefore enough to prove that the Lie algebra of \( \operatorname{Aut}{\left( \mathfrak{g}\right) }^{0} \) is contained in ad[g]. But this is i... | Yes |
Theorem 3.11.1. The Casimir element \( \omega \) belongs to the center of \( \mathfrak{G} \) . Let \( \left\{ {{X}_{1},\ldots ,{X}_{m}}\right\} \) be a basis for \( \mathrm{g} \) and let \( \left\{ {{X}^{1},\ldots ,{X}^{m}}\right\} \) be the dual basis defined by\n\n(3.11.5)\n\n\[ \left\langle {{X}_{i},{X}^{j}}\right\r... | Proof. Since \( {D}_{x}\widetilde{\xi } = 0 \) for all \( X \in \mathfrak{g},\omega \) lies in the center of \( \mathcal{G} \) by Theorem 3.3.8. We now prove (3.11.6). Fix the basis \( \left\{ {{X}_{1},\ldots ,{X}_{m}}\right\} \) of \( \mathfrak{g} \) . The existence of the dual basis \( \left\{ {{X}^{1},\ldots ,{X}^{m... | Yes |
Theorem 3.11.2. Let \( \rho \) be a finite-dimensional representation of \( \mathfrak{g} \) and \( \mathfrak{p} = {\left( \operatorname{kernel}\rho \right) }^{ \bot } \) . Let \( {\omega }^{\rho } \) be defined as above. Then \( {\omega }^{\rho } \) lies in the center of \( \mathfrak{G} \) . Let \( \left\{ {{X}_{1},\ld... | Proof. Proceed as in the preceding theorem. Since \( \operatorname{tr}\rho \left( {{X}_{i}{X}^{i}}\right) = {B}^{\rho }\left( {{X}_{i},{X}^{i}}\right) \) \( = 1 \), we have (3.11.10). | Yes |
Theorem 3.12.1. Let \( \mathfrak{g} \) be a semisimple Lie algebra over \( k,\rho \) a representation of \( \mathfrak{g} \) in a vector space \( F \) of finite dimension over \( k \) . Then\n\n\[ \n{H}^{1}\left( {\mathfrak{g},\rho }\right) = 0,\;{H}^{2}\left( {\mathfrak{g},\rho }\right) = 0.\n\] | Proof. To start with, we take up the proof that \( {H}^{1}\left( {\mathfrak{g},\rho }\right) = 0 \) . This is equivalent to proving that\n\n\[ \n{C}^{1}\left( {\mathfrak{g},\rho }\right) = {B}^{1}\left( {\mathfrak{g},\rho }\right)\n\]\n\nIn what follows, we write \( {C}^{i} \) and \( {B}^{i} \) for \( {C}^{i}\left( {\m... | Yes |
Corollary 3.14.3. Let \( \mathfrak{m} \) be a Levi subalgebra of \( \mathfrak{g} \) , \( \mathfrak{a} \) a semisimple subalgebra. Then there is \( y \in {G}_{\mathfrak{p}} \) such that \( {\mathfrak{a}}^{y} \subseteq \mathfrak{m} \) . In particular, a maximal semisimple subalgebra is a Levi subalgebra. | Proof. \( {g}_{0} = q + a \) is a subalgebra, and \( a \) is a Levi subalgebra of \( {g}_{0} \) , while \( \mathfrak{q} = \operatorname{rad}\left( {\mathfrak{g}}_{0}\right) \) . Let \( {\mathfrak{a}}^{\prime } = \mathfrak{m} \cap {\mathfrak{g}}_{0} \) . Then \( {\mathfrak{g}}_{0} = \mathfrak{q} + {\mathfrak{a}}^{\prime... | Yes |
Lemma 3.16.1. Let \( A \) be an associative algebra with unit 1 over \( k, S \) a subset of \( A \) such that \( S \) and 1 generate \( A \) . Let \( D \) be a derivation of \( A \) with the property that for each \( u \in S \), there is an integer \( n\left( u\right) > 0 \) such that \( {D}^{n\left( u\right) }u = 0 \)... | Proof. For \( u, v \in A \) and any integer \( m > 0 \), we have the Leibniz formula\n\n\[ \n{D}^{m}\left( {uv}\right) = \mathop{\sum }\limits_{{0 \leq r \leq m}}\left( \begin{matrix} m \\ r \end{matrix}\right) {D}^{r}u{D}^{m - r}v.\n\]\n\nLet \( {A}^{\prime } \) be the set of all \( u \in A \) with the property that f... | Yes |
Theorem 3.16.3. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) . Then the following statements are equivalent.\n\n(i) \( \mathrm{g} \) is reductive\n\n(ii) \( \mathfrak{g} \) has a faithful semisimple representation\n\n(iii) The adjoint representation of \( \mathfrak{g} \) is semisimple.\n\n(iv) \( \mathfrak{D}\m... | Proof. (i) \( \Leftrightarrow \) (ii) and (i) \( \Rightarrow \) (iv) by the previous theorem. If \( \mathfrak{{Dg}} \) is semisimple, it must be a Levi subalgebra of \( \mathfrak{g} \) by Theorem 3.14.1. So \( \mathfrak{g} = \mathfrak{q} + \) \( \mathfrak{D}\mathfrak{g} \) is a direct sum. But then \( \left\lbrack {\ma... | Yes |
Corollary 3.16.4. Let \( \mathfrak{n} \) be the nil radical of \( \mathfrak{g},\mathfrak{h} = \mathfrak{g}/\mathfrak{n} \), and let \( \gamma \) be the natural map of \( \mathfrak{g} \) onto \( \mathfrak{h} \) . Then \( \mathfrak{h} \) is reductive, and \( \gamma \left\lbrack \mathfrak{q}\right\rbrack \) is the center ... | Proof. By Theorem 3.10.5, \( \gamma \left\lbrack \mathfrak{q}\right\rbrack = \operatorname{rad}\mathfrak{h} \) . Since \( \left\lbrack {\mathfrak{q},\mathfrak{g}}\right\rbrack \subseteq \mathfrak{u} \), we have \( \left\lbrack {\gamma \left\lbrack q\right\rbrack ,\mathfrak{h}}\right\rbrack = 0 \) . This implies that \(... | "No" |
Theorem 3.16.6. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) and let \( \rho ,{\rho }_{1},\ldots ,{\rho }_{s} \) be finite-dimensional semisimple representations of \( \mathfrak{g} \) . Then the representations \( {\rho }^{ * } \) and \( {\rho }_{1} \otimes \cdots \otimes {\rho }_{s} \) are semisimple. | Proof. Let \( \mathfrak{q} = \operatorname{radg} \) . For \( X \in \mathfrak{q},\rho \left( X\right) \) is semisimple. Hence \( {\rho }^{ * }\left( X\right) \) is semisimple by Lemma 3.1.11. This proves that \( {\rho }^{ * } \) is semisimple. For the next assertion it is enough to consider the case \( s = 2 \) ; the ge... | Yes |
Lemma 3.17.1. Let \( \\mathfrak{a} \) be a Lie algebra over \( k \), and let \( \\mathfrak{A} \) be its universal enveloping algebra.\n\n(i) Suppose \( \\mathfrak{M} \) is a proper two-sided ideal of \( \\mathfrak{A} \) . Then in order that dim \( \\left( {\\mathfrak{A}/\\mathfrak{M}}\\right) < \\infty \), it is necess... | Proof. (i) Suppose \( \\dim \\left( {\\mathfrak{A}/\\mathfrak{M}}\\right) < \\infty \) and \( a \\in \\mathfrak{A} \) . Then for some integer \( N > 0,1, a,{a}^{2},\\ldots ,{a}^{N} \) are linearly dependent modulo \( \\mathfrak{M} \) . So \( p\\left( a\\right) \\in \\mathfrak{M} \) for some \( p \\in k\\left\\lbrack T\... | Yes |
Lemma 3.17.2. Let \( \mathfrak{a},\mathfrak{A} \) be as above. Suppose that \( \mathfrak{n} \) is an ideal of \( \mathfrak{a} \) and \( \rho \) is a finite-dimensional representation of \( \mathfrak{a} \) such that \( \rho \left( X\right) \) is nilpotent for all \( X \in \mathfrak{n} \). Let \( \sigma \) be the represe... | Proof. Let \( V \) denote the vector space on which \( \sigma \) acts. Let \( {V}_{i}\left( {0 \leq i \leq r}\right) \) be invariant subspaces for \( \sigma \) such that \( {V}_{0} = V \supseteq {V}_{1} \supseteq \cdots \supseteq {V}_{r} = 0 \) and such that the representation \( {\sigma }_{i} \) of \( \mathfrak{N} \) ... | Yes |
Lemma 3.17.3. Let notation be as above and let \( r \geq 1 \) be such that \( {\mathfrak{M}}_{r} \subseteq \) \( \mathfrak{K} \) . Let \( U = \mathfrak{A}/{\mathfrak{M}}_{r} \), let \( \gamma \) be the natural map of \( \mathfrak{A} \) onto \( U \), and let \( \xi \) be the natural representation of \( \mathfrak{A} \) ... | Proof. Since \( D \mapsto \widetilde{D} \) is a representation of \( \mathfrak{d} \) in \( \mathfrak{A} \) ,(i) is immediate. Suppose that \( a, b \in \mathfrak{A} \) . Then\n\n\[ \xi \left( {\widetilde{D}a}\right) \left( {\gamma \left( b\right) }\right) = \gamma \left( {\left( {\widetilde{D}a}\right) b}\right) \]\n\n\... | Yes |
Lemma 3.17.4. Let \( \mathfrak{g} \) be a Lie algebra over \( k \), and let \( \mathfrak{g} = \mathfrak{a} + \mathfrak{b} \), where \( \mathfrak{a} \) is an ideal and \( \mathfrak{b} \) is a subalgebra of \( \mathfrak{g} \), the sum being direct. Suppose \( \mathfrak{n} \) is an ideal of \( \mathfrak{a} \) such that \(... | Proof. We use the notation of the previous lemma. Let \( \widehat{x} = \operatorname{kernel}\left( \sigma \right) \) . For \( Y \in \mathfrak{b} \), let \( {D}_{Y} \) be the derivation \( X \mapsto \left\lbrack {Y, X}\right\rbrack \) of \( \mathfrak{a} \) . Then \( {D}_{Y} \in \mathfrak{d} \) for all \( Y \in \mathfrak... | Yes |
Corollary 3.17.5. Suppose \( \sigma \) is faithful on \( \mathfrak{a} \) . Then \( {\sigma }^{\prime } \) is faithful on \( \mathfrak{a} \) . | Proof. This follows at once from (i). | No |
Corollary 3.17.6. Suppose \( \mathfrak{g} \) is a nilpotent Lie algebra over \( k \) . Then \( \mathfrak{g} \) has a faithful finite-dimensional nil representation. | Proof. We use induction on \( \dim \left( \mathfrak{g}\right) \) . If \( \dim \left( \mathfrak{g}\right) = 1 \), this is obvious. For \( \dim \left( \mathfrak{g}\right) > 1 \), we select an ideal \( \mathfrak{a} \) of \( \mathfrak{g} \) such that \( \dim \left( \mathfrak{a}\right) = \dim \left( \mathfrak{g}\right) - 1 ... | Yes |
Theorem 3.17.7. Let \( \mathfrak{g} \) be a Lie algebra over \( k \) and \( \mathfrak{u} \) its nil radical. Then there exists a faithful finite-dimensional representation \( \rho \) of \( \mathfrak{g} \) such that \( \rho \left( X\right) \) is nilpotent for all \( X \in \mathfrak{n} \) . | Proof. We use induction on \( \dim \left( \mathfrak{g}\right) \) . Suppose first that \( \mathfrak{g} \) is solvable. In view of the preceding corollary, we may assume that \( \mathfrak{n} \neq \mathfrak{g} \) . Now \( \mathfrak{D}\mathfrak{g} \subseteq \mathfrak{n} \) ; hence if \( \mathfrak{a} \) is any linear subspa... | Yes |
Theorem 3.18.1. Let \( G \) be a simply connected analytic group with Lie algebra \( \mathfrak{g} \) . Let \( \mathfrak{a} \) be an ideal in \( \mathfrak{g}, A \) the analytic subgroup of \( G \) defined by \( \mathfrak{a} \) . Then \( A \) is a closed normal subgroup of \( G \) . | Proof. We need prove only that \( A \) is closed. By the global form of the third fundamental theorem of Lie, there exists an analytic group \( H \) whose Lie algebra \( \mathfrak{h} \) is isomorphic to \( \mathfrak{g}/\mathfrak{a} \) . Then there exists a homomorphism \( \lambda \) of \( \mathfrak{g} \) onto \( \mathf... | Yes |
Theorem 3.18.2. Let \( G \) be a simply connected analytic group and \( A \) a normal analytic subgroup. Then \( A \) is closed, \( A \) and \( G/A \) are both simply connected, and the coset space \( G/A \) admits a global analytic section. | Essentially, we follow Hochschild's method of proof [1]. We need some lemmas. Note that, by the preceding theorem, \( A \) is closed. | No |
Lemma 3.18.3. Let \( \mathfrak{h} \) be a Lie algebra over a field \( {k}^{\prime } \) of characteristic zero and \( \mathfrak{a} \subseteq \mathfrak{h} \) an ideal which is maximal among the ideals of \( \mathfrak{h} \) that are properly contained in \( \mathfrak{h} \) . Then there exists a subalgebra \( \mathfrak{b} ... | Proof. As \( \mathfrak{h}/\mathfrak{a} \) has no ideals other than 0 and \( \mathfrak{h}/\mathfrak{a} \), either \( \dim \left( {\mathfrak{h}/\mathfrak{a}}\right) = 1 \) or \( \mathfrak{h}/\mathfrak{a} \) is semisimple. If \( \dim \left( {\mathfrak{h}/\mathfrak{a}}\right) = 1 \), we can take \( \mathfrak{h} = {k}^{\pri... | Yes |
Lemma 3.18.4. Let \( H \) be a simply connected analytic group with Lie algebra \( \mathfrak{h} \) . Suppose that \( \mathfrak{a} \) is an ideal of \( \mathfrak{h} \) and that \( \mathfrak{b} \) is a subalgebra of \( \mathfrak{h} \) such that (3.18.1) is satisfied. Let \( A \) and \( B \) be the respective analytic sub... | Proof. For \( Y \in \mathfrak{b} \) let \( \sigma \left( Y\right) = \left( {\operatorname{ad}Y}\right) \mid \mathfrak{a} \) . Then the map \( \xi (\left( {X, Y}\right) \mapsto X + \) \( Y \) ) is an isomorphism of \( \mathfrak{a} \times {}_{\sigma }\mathfrak{b} \) onto \( \mathfrak{h} \) . Let \( {H}^{\prime } = A \tim... | Yes |
Lemma 3.18.5. Let \( G \) be a simply connected analytic group with Lie algebra \( \mathfrak{g} \) . Suppose \( \mathfrak{a},{\mathfrak{b}}_{1},\ldots ,{\mathfrak{b}}_{r} \) are subalgebras of \( \mathfrak{g} \) such that (i) \( \mathfrak{g} = \mathfrak{a} + {\mathfrak{b}}_{1} + \) \( \cdots + {\mathfrak{b}}_{r} \) is ... | Proof. For \( r = 1 \), this follows from the preceding lemma. We use induction on \( r \) . Assume that \( r \geq 2 \) . Let \( {H}_{r - 1} \) be the analytic subgroup of \( G \) defined by \( {\mathfrak{h}}_{r - 1} \) . By the previous lemma, \( {H}_{r - 1} \) and \( {B}_{r} \) are closed and simply connected in \( G... | Yes |
Theorem 3.18.8. For any \( r \geq 1,{D}^{r}G \) (resp. \( {C}^{r}G \) ) is the analytic subgroup of \( G \) defined by \( {\mathfrak{D}}^{r}\mathfrak{g} \) (resp. \( {\mathfrak{C}}^{r}\mathfrak{g} \) ). These are all normal, and if \( G \) is simply connected, they are all closed and simply connected. | Proof. This follows at once from Theorems 3.18.7 and 3.18.2. | No |
Theorem 3.18.11. Let \( G \) be a simply connected solvable analytic group with Lie algebra \( \mathfrak{g} \) . Suppose \( {}^{6}\left\{ {{X}_{1},\ldots ,{X}_{m}}\right\} \) is a basis of \( \mathfrak{g} \) with the following property: \( {\mathfrak{h}}_{i} = \mathop{\sum }\limits_{{1 \leq j \leq i}}k \cdot {X}_{j} \)... | Proof. Write \( {\mathfrak{b}}_{i} = k \cdot {X}_{i} \) . Let \( {B}_{i} \) be the analytic subgroup of \( G \) defined by \( {\mathfrak{v}}_{i} \) . Then Lemma 3.18.5 is applicable (with \( \mathfrak{a} = 0, r = m \) ), and we conclude that (i) the \( {B}_{i} \) are all closed and simply connected in \( G \), and (ii)... | Yes |
Theorem 3.18.13. Let \( G \) be an analytic group with Lie algebra \( \mathfrak{g} \), and \( Q \) (resp. \( N \) ) the radical (resp. nil radical) of \( G \) . Then \( Q \) and \( N \) are closed. Suppose that \( \mathfrak{g} = \mathfrak{q} + \mathfrak{m} \) is a Levi decomposition of \( \mathfrak{g} \) and that \( M ... | \[ {QM} = G,\;Q \cap M = \{ 1\} . \] | No |
Lemma 3.18.14. Let \( H \) be an analytic group and \( A \subseteq H \) an analytic subgroup. Suppose \( A \) is solvable (resp. nilpotent). Then \( {Cl}\left( A\right) \) is a solvable (resp. nilpotent) analytic group. | Proof. The arguments for the two cases are quite similar, so we treat only the solvable case. Let \( {A}_{0} = A \) and \( {A}_{p} = {D}^{p}A\left( {p \geq 1}\right) \) . Since \( A \) is solvable, we can find \( a \geq 0 \) such that \( {A}_{s + 1} = \{ 1\} \) . Let \( {B}_{p} = {Cl}\left( {A}_{p}\right) \) . Then \( ... | Yes |
Lemma 3.18.15. Let \( G \) be a nilpotent analytic group with Lie algebra \( \mathfrak{g} \) . Suppose \( \pi \) is a unipotent representation of \( G \) . Then\n\n(3.18.6)\n\n\[ \operatorname{kernel}\left( \pi \right) = \exp \left\lbrack {\operatorname{kernel}\left( {d\pi }\right) }\right\rbrack . \]\n\nIn particular,... | Proof. Let \( Z = \operatorname{kernel}\left( \pi \right) ,\mathfrak{z} = \operatorname{kernel}\left( {d\pi }\right) \) . Write \( N = \pi \left\lbrack G\right\rbrack ,\mathfrak{n} = \left( {d\pi }\right) \left\lbrack \mathfrak{g}\right\rbrack \) . Let \( V \) be the vector space on which \( \pi \) acts. Then \( N \) i... | Yes |
Lemma 4.1.1. Let \( \mathfrak{g} \) be a Lie algebra over \( k,\mathfrak{h} \) a CSA. Then: (i) \( \mathfrak{h} \) is maximal nilpotent. (ii) If \( \mathfrak{z} \) is any subalgebra of \( \mathfrak{g} \) such that \( \mathfrak{h} \subseteq \mathfrak{z} \subseteq \mathfrak{g} \), then \( \mathfrak{h} \) is a CSA of \( 3... | Proof. (i) Let \( \mathfrak{h} \) be a CSA of \( \mathfrak{g},\mathfrak{n} \neq \mathfrak{h} \) a nilpotent subalgebra containing \( \mathfrak{h} \) . By Engel’s theorem (cf. §3.5) applied to the nil representation \( H \mapsto {\left( \operatorname{ad}H\right) }_{\mathfrak{n}\left( \mathfrak{h}\right) } \) \( \left( {... | Yes |
Theorem 4.1.2. Let \( \mathfrak{g} \) be a Lie algebra over \( k \). For \( X \in \mathfrak{g} \), let\n\n\[ \mathfrak{h}_{X} = \left\{ Y : Y \in \mathfrak{g}, (\operatorname{ad}X)^s(Y) = 0 \text{ for some integer } s \geq 1 \right\} \].\n\nThen for any regular \( X, \mathfrak{h}_{x} \) is a CSA, and \( \dim \mathfrak{... | Proof. Let \( X \in \mathfrak{g} \) be regular and let us write \( \mathfrak{h} \) for \( \mathfrak{h}_{X} \). If \( Y \in \mathfrak{h} \), \( (\operatorname{ad}X)^s(Y) = \left\lbrack \underbrace{\operatorname{ad}X, \lbrack \operatorname{ad}X, \lbrack \cdots \lbrack \operatorname{ad}X, \operatorname{ad}Y \rbrack \cdots... | Yes |
Corollary 4.1.4. The dimension of any CSA is the rank of \( \mathfrak{g} \) . If \( \mathfrak{h} \) is any CSA | \[ \eta \left( Y\right) = \det {\left( \operatorname{ad}Y\right) }_{\mathfrak{g}/\mathfrak{h}}\;\left( {Y \in \mathfrak{h}}\right) . \] | No |
Lemma 4.2.1. Let \( p \) be an integer \( \geq 0 \) . Then we have the following identities in \( \& \) :\n\n(4.2.4)\n\n\[ X{Y}^{p + 1} = {Y}^{p + 1}X + \left( {p + 1}\right) {Y}^{p}\left( {H - p}\right) \]\n\n\[ Y{X}^{p + 1} = {X}^{p + 1}Y - \left( {p + 1}\right) {X}^{p}\left( {H + p}\right) . \] | Proof. We use induction on \( p \) . For \( p = 0 \), this is clear, since \( {XY} - {YX} \) \( = H \) . Assume (4.2.4) for some \( p \geq 0 \) . Since \( {HY} = Y\left( {H - 2}\right) \), \n\n\[ X{Y}^{p + 2} = \left( {{Y}^{p + 1}X + \left( {p + 1}\right) {Y}^{p}\left( {H - p}\right) }\right) Y \]\n\n\[ = {Y}^{p + 2}X ... | Yes |
Theorem 4.2.2. Let \( \rho \) be a finite-dimensional irreducible representation of \( \mathfrak{g} \) in a complex vector space \( V \) . Then \( \rho \left( H\right) \) is semisimple, its eigenvalues being all simple and integral. Moreover, there is an integer \( j \geq 0 \) and a basis \( \left\{ {{v}_{0},{v}_{1}}\r... | Proof. Let \( \rho \) be an irreducible representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( V \) . For any \( \lambda \in \mathbf{C} \), write \( {V}_{\lambda } \) for the eigenspace of \( \rho \left( H\right) \) corresponding to the eigenvalue \( \lambda \), if \( \lambda \) is an eigenvalue o... | Yes |
Corollary 4.2.3. Let \( \rho \) be a representation of \( \mathfrak{g} \). Then \( \rho \left( H\right) \) is semisimple, while \( \rho \left( X\right) \) and \( \rho \left( Y\right) \) are nilpotent. The eigenvalues of \( \rho \left( H\right) \) are all integers; and the set of eigenvalues of \( \rho \left( H\right) \... | It is enough to prove these assertions for irreducible \( \rho \); for such \( \rho \), they are immediate from (4.2.5). | No |
Corollary 4.2.4. Let \( \pi \) be a representation of \( \mathfrak{g} \) in a finite-dimensional vector space \( V \) . Assume that (i) all eigenvalues of \( \pi \left( H\right) \) are of multiplicity 1, and (ii) the difference between any two eigenvalues of \( \pi \left( H\right) \) is even. Then \( \pi \) is irreduci... | Proof. \( \pi \) is seen to be either irreducible or equivalent to the direct sum of \( {\pi }_{j} \) and \( {\pi }_{{j}^{\prime }} \) where \( \left| {j - {j}^{\prime }}\right| \) is odd, on noting that in the irreducible representation \( \rho \) corresponding to \( j \geq 0 \) the eigenvalues of \( \rho \left( H\rig... | Yes |
Corollary 4.2.7. Let \( \omega = {H}^{2} + {2H} + {4YX} + 1 \) . Then \( \omega \) lies in the center of \( \mathfrak{G} \), and\n\n\[{\pi }_{j}\left( \omega \right) = {\left( j + 1\right) }^{2} \cdot 1.\] | Proof. It is an easy verification that \( \omega \) lies in the center of \( \mathfrak{G} \) . So, by Schur’s lemma, \( {\pi }_{j}\left( \omega \right) = {c}_{j} \cdot 1 \) for some constant \( {c}_{j} \) . In particular, \( {\pi }_{j}\left( \omega \right) \cdot {v}_{0} = \) \( {c}_{j}{v}_{0} \) . But\n\n\[{\pi }_{j}\l... | Yes |
Lemma 4.3.1. Let \( \lambda ,\mu \in {\mathfrak{h}}^{ * } \) . Then \( \left\lbrack {{\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\mu }}\right\rbrack \subseteq {\mathfrak{g}}_{\lambda + \mu } \) ; if \( \lambda + \mu \neq 0,{\mathfrak{g}}_{\lambda } \bot {\mathfrak{g}}_{\mu } \) | Proof. Since ad \( H \) is a derivation of \( \mathfrak{g} \), we have, for \( X \in {\mathfrak{g}}_{\lambda } \) and \( Y \in \) \( {\mathcal{g}}_{\mu },\left\lbrack {H,\left\lbrack {X, Y}\right\rbrack }\right\rbrack = \left\lbrack {\left\lbrack {H, X}\right\rbrack, Y}\right\rbrack + \left\lbrack {X,\left\lbrack {H, Y... | Yes |
Corollary 4.3.2. \( {\mathfrak{g}}_{\lambda } \bot \mathfrak{h}\left( {\lambda \in \Delta }\right) \) . If \( \alpha \in \Delta \), then \( - \alpha \in \Delta \) . Moreover, \( \langle \cdot , \cdot \rangle \) is non-singular on \( {\mathfrak{g}}_{\alpha } \times {\mathfrak{g}}_{-\alpha } \) for any \( \alpha \in \Del... | Proof. The first relation is clear, since \( \lambda \neq 0 \) . If \( \alpha \in \Delta \) but \( - \alpha \notin \Delta \) , then \( {\mathfrak{g}}_{\alpha } \bot {\mathfrak{g}}_{\beta } \) for all \( \beta \in \Delta \) . So, since \( {\mathfrak{g}}_{\alpha } \bot \mathfrak{h} \) too, \( {\mathfrak{g}}_{\alpha } \bo... | Yes |
Lemma 4.3.3. \( \mathrm{C} \cdot \Delta = {\mathfrak{h}}^{ * } \) | Proof. For \( H,{H}^{\prime } \in \mathfrak{h} \) we have\n\n(4.3.4)\n\n\[ \left\langle {H,{H}^{\prime }}\right\rangle = \mathop{\sum }\limits_{{\alpha \in \Delta }}\dim \left( {\mathfrak{g}}_{\alpha }\right) \alpha \left( H\right) \alpha \left( {H}^{\prime }\right) .\n\]\n\nIf \( \mathbf{C} \cdot \Delta \neq {\mathfra... | Yes |
Lemma 4.3.4. Let \( \alpha \in \Delta, X \in {\mathfrak{g}}_{\alpha }, Y \in {\mathfrak{g}}_{-\alpha } \). Then\n\n(4.3.7)\n\n\[ \left\lbrack {X, Y}\right\rbrack = \langle X, Y\rangle {H}_{\alpha } \]\n\n\nIn particular,\n\n(4.3.8)\n\n\[ \left\lbrack {{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }}\right\rbrack = \... | Proof. By Lemma 4.3.1, \( \left\lbrack {{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }}\right\rbrack \subseteq \mathfrak{h} \). If \( X, Y \) are as above, then \( \langle H,\left\lbrack {X, Y}\right\rbrack \rangle = \alpha \left( H\right) \langle X, Y\rangle = \left\langle {H,\langle X, Y\rangle {H}_{\alpha }}\rig... | Yes |
Lemma 4.3.5. Let \( \alpha ,\beta \in \Delta \) . Then there is a rational number \( {q}_{\beta \alpha } \) such that\n\n\[ \langle \beta ,\alpha \rangle = {q}_{\beta \alpha }\langle \alpha ,\alpha \rangle \]\n\nMoreover, \( \langle \alpha ,\alpha \rangle \) is a rational number \( > 0 \) . | Proof. Let \( V = \mathop{\sum }\limits_{{k \in \mathbf{Z}}}{\mathfrak{g}}_{\beta + {k\alpha }} \) (the sum is finite). By Lemma 4.3.1, \( \left\lbrack {{\mathfrak{g}}_{\alpha }, V}\right\rbrack \) \( \subseteq V,\left\lbrack {{\mathfrak{g}}_{-\alpha }, V}\right\rbrack \subseteq V \) . Let \( {d}_{k} = \dim \left( {\ma... | Yes |
Corollary 4.3.6. Let\n\n(4.3.10)\n\n\[ \n{\left\lbrack \right\rbrack }_{\mathbf{R}} = \mathop{\sum }\limits_{{\alpha \in \Delta }}\mathbf{R} \cdot {H}_{\alpha } \]\n\nThen \( {\dim }_{\mathbf{R}}{\mathfrak{h}}_{\mathbf{R}} = l \) . Moreover, \( \langle \cdot , \cdot \rangle \) is a positive definite scalar product on \... | Proof. By (4.3.9) it is clear that each root is real-valued on \( {\mathfrak{h}}_{\mathbf{R}} \) . By (4.3.4),\n\n\[ \n\langle H, H\rangle = \mathop{\sum }\limits_{{\beta \in \Delta }}\dim \left( {\mathfrak{g}}_{\beta }\right) \beta {\left( H\right) }^{2}\;\left( {H \in {\mathfrak{h}}_{\mathbf{R}}}\right) .\n\]\n\nSinc... | Yes |
Lemma 4.3.7. Let \( \alpha \in \Delta \) . Then \( \dim \left( {\mathrm{g}}_{\alpha }\right) = 1 \), and none of \( \pm {2\alpha }, \pm {3\alpha },\ldots \) , are roots. | Proof. Select \( X \in {\mathfrak{g}}_{\alpha }, Y \in {\mathfrak{g}}_{-\alpha } \) such that \( {H}_{\alpha } = \left\lbrack {X, Y}\right\rbrack \), and let \( V = \) C. \( Y + \mathbf{C} \cdot {H}_{\alpha } + \mathop{\sum }\limits_{{k \geq 1}}{g}_{k\alpha } \) . It is clear that \( V \) is invariant under both ad \( ... | Yes |
Lemma 4.3.8. Let \( \alpha ,\beta \in \Delta \) with \( \beta \neq \pm \alpha \) . Then there exist two integers \( p = p\left( {\alpha ,\beta }\right) \) and \( q = q\left( {\alpha ,\beta }\right) \), both \( \geq 0 \), such that for any integer \( k,\beta + {k\alpha } \in \) \( \Delta \) if and only if \( - q \leq k ... | Proof. \( \beta + {k\alpha } \neq 0, k = 0, \pm 1, \pm 2,\ldots \), so \( \dim \left( {\mathfrak{g}}_{\beta + {k\alpha }}\right) \leq 1 \) for all \( k \) . Let\n\n(4.3.15)\n\n\[ \n{\mathfrak{g}}_{\beta ,\alpha } = \mathop{\sum }\limits_{{k \in \mathbf{Z}}}{\mathfrak{g}}_{\beta + {k\alpha }}\n\]\n\n\( {\rho }_{\alpha }... | Yes |
Corollary 4.3.9. Let \( \alpha \in \Delta \) . If \( c \in \mathbf{C} \), then \( {c\alpha } \) is a root if and only if \( c = \pm 1 \) . | Proof. We may assume that \( c \neq 0 \) . Let \( \beta = {c\alpha } \) . Then \( {2c} = \beta \left( {\bar{H}}_{\alpha }\right) \) is an integer. Similarly, since \( \alpha = {c}^{-1}\beta ,2{c}^{-1} \) is an integer too. So \( c = \pm \frac{1}{2}, \pm 1 \) , or \( \pm 2 \) . If \( c \neq \pm 1 \), then either \( \alp... | Yes |
Corollary 4.3.10. Suppose \( \alpha ,\beta \in \Delta \) and \( \beta \neq \alpha \) . If \( \beta - \alpha \) is not a root, \( \langle \alpha ,\beta \rangle \leq 0 \) ; moreover, in this case, \( q = 0 \) and \( p = - \beta \left( {\bar{H}}_{\alpha }\right) \) . | Proof. If \( \beta - \alpha \) is not a root, \( {\mathfrak{g}}_{\beta ,\alpha } = {\mathfrak{g}}_{\beta } + {\mathfrak{g}}_{\beta + \alpha } + \cdots + {\mathfrak{g}}_{\beta + {p\alpha }} \) . So \( \beta \left( {\bar{H}}_{\alpha }\right) \) is the lowest eigenvalue of \( {\rho }_{\alpha }\left( {\bar{H}}_{\alpha }\ri... | No |
Corollary 4.3.11. Suppose that \( \alpha ,\beta \in \Delta \) and \( \alpha + \beta \in \Delta \) . Then \( \left\lbrack {{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }}\right\rbrack \) \( = {g}_{\alpha + \beta } \) . | Proof. Otherwise \( \left\lbrack {{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }}\right\rbrack = 0 \), so \( \mathop{\sum }\limits_{{-q \leq k \leq 0}}{\mathfrak{g}}_{\beta + {k\alpha }} \) is invariant under \( {\rho }_{\alpha } \) . This implies that \( p = 0 \) or \( \beta + \alpha \notin \Delta \), a contradictio... | No |
Corollary 4.3.12. Let \( \alpha ,\beta \in \Delta \), with \( \beta \neq \pm \alpha \), and let \( p, q \) be as in Lemma 4.2.8. Then \( p + q \leq 3 \) . Furthermore, the possible values of \( 2\langle \alpha ,\beta \rangle /\langle \alpha ,\alpha \rangle \) are \( 0, \pm 1, \pm 2, \pm 3 \) . | Proof. Let \( m = 2\langle \alpha ,\beta \rangle /\langle \alpha ,\alpha \rangle n = 2\langle \beta ,\alpha \rangle /\langle \beta ,\beta \rangle \) . Then both \( m \) and \( n \) are integers. Suppose \( m \neq 0 \) . Then \( n \neq 0 \) also. On the other hand, by Corollary 4.3.6, \( \langle \cdot , \cdot \rangle \)... | Yes |
Corollary 4.3.13. Let \( \\alpha ,\\beta \\in \\Delta \), with \( \\beta \\neq \\pm \\alpha \), and let \( p, q \) be as in Lemma 4.3.8. Suppose that \( {Z}_{\\alpha } \\in {\\mathfrak{g}}_{\\alpha } \) and \( {Z}_{-\\alpha } \\in {\\mathfrak{g}}_{-\\alpha } \), both \( \\neq 0 \). Then for any \( X \\in {\\mathfrak{g}... | Proof. Let \( {X}_{\\alpha } \\in {\\mathfrak{g}}_{\\alpha },{X}_{-\\alpha } \\in {\\mathfrak{g}}_{-\\alpha } \) be such that (4.3.12) is satisfied. Since \( {\\bar{H}}_{\\alpha } = \\left( {2/\\langle \\alpha ,\\alpha \\rangle }\\right) {H}_{\\alpha } \) and \( \\left\\lbrack {{X}_{\\alpha },{X}_{-\\alpha }}\\right\\r... | Yes |
Lemma 4.3.14. \( \mathfrak{w} \) is a finite subgroup of the orthogonal group of \( \mathfrak{h} \) with respect to \( \langle \cdot , \cdot \rangle \) . Each element of \( \mathfrak{w} \) induces a permutation of \( \mathbf{\Delta } \) . | Proof. If \( \alpha \in \Delta \), then for any \( \beta \in \Delta ,\beta - \beta \left( {\bar{H}}_{\alpha }\right) \alpha = \gamma \) is also in \( \Delta \) . But \( {H}_{\gamma } = {H}_{\beta } - \beta \left( {\bar{H}}_{\alpha }\right) {H}_{\alpha } = {H}_{\beta } - \alpha \left( {H}_{\beta }\right) {\bar{H}}_{\alp... | Yes |
Theorem 4.3.16. Let \( S = \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{l}}\right\} \) be a simple system of roots, and let \( P \) be the corresponding positive system. Then:\n\n(i) If \( 1 \leq i, j \leq l \) and \( i \neq j \), then \( {\alpha }_{i} - {\alpha }_{j} \) is not a root, and\n\n\[ \left\langle {{\alpha }_{i... | Proof. If \( {m}_{1}{\alpha }_{1} + \cdots + {m}_{l}{\alpha }_{l} \) is a root, then we must have either all \( {m}_{i} \geq 0 \) or all \( {m}_{i} \leq 0 \) . So \( {\alpha }_{i} - {\alpha }_{j} \) cannot be a root if \( i \neq j \) ; by Corollary 4.3.10, \( \left\langle {{\alpha }_{i},{\alpha }_{j}}\right\rangle \leq... | Yes |
Lemma 4.3.17. Let \( V \) be a real vector space with a positive definite scalar product \( \left( {\cdot , \cdot }\right) \) . Let \( {v}_{1},\ldots ,{v}_{s} \) be elements of \( V \) with the following two properties: (i) \( \left( {{v}_{i},{v}_{j}}\right) \leq 0 \) for \( i \neq j,1 \leq i, j \leq s \), and (ii) if ... | Proof. Suppose to the contrary. Let \( {c}_{1}{v}_{1} + \cdots + {c}_{s}{v}_{s} = 0 \) be a nontrivial relation between the \( {v}_{i}^{\prime }s \) . By throwing away the \( {v}_{i} \) for which \( {c}_{i} = 0 \) , we may assume that \( {c}_{i} \neq 0, i = 1,\ldots, s \) . By (ii) the \( {c}_{i} \) cannot all be of th... | Yes |
Corollary 4.3.19. Let \( P \) be a positive system, \( C \), the corresponding chamber. Then \( - P \) is a positive system with corresponding chamber \( - C \), and there is a unique element \( {s}_{0} \in \mathfrak{w} \) such that \( {s}_{0}P = - P \) . Moreover, \( {s}_{0}^{2} = 1 \) . | Proof. \( - P \) is obviously the positive system corresponding to \( - C \) . Existence and uniqueness of \( {s}_{0} \) are clear. Since \( {s}_{0}^{2}P = P,{s}_{0}^{2} = 1 \) . | No |
Corollary 4.3.20. Let \( P \) be any positive system, and let \( S \) be the set of roots in \( P \) which cannot be expressed as a sum of two elements of \( P \) . Then \( S \) is a simple system, and \( P \) is the positive system corresponding to \( S \) . | Proof. Obvious, since the result is valid when \( P = P\left( C\right) \) . | No |
Corollary 4.3.21. Let \( w \) be the order of \( \mathfrak{w} \) . Then there are exactly \( w \) chambers (resp. simple systems, positive systems). | Proof. Obvious. | No |
Lemma 4.3.22. Let \( {N}_{\alpha ,\beta } \) be defined as above. Then:\n\n(i) \( {N}_{\alpha ,\beta } = - {N}_{\beta ,\alpha } \) | Proof. (i) is trivial. | No |
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