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Corollary 11.1.3. Let \( G \) be a linear algebraic group. The following are equivalent:\n\n1. \( G \) is a connected topological space in the Zariski topology.\n\n2. \( G \) is irreducible as an affine algebraic set.\n\n3. The ring \( \mathcal{O}\left\lbrack G\right\rbrack \) has no zero divisors. | Proof. Apply Theorem 11.1.2 and Lemma A.1.10. | No |
Lemma 11.1.4. Let \( K \) be a subgroup of \( G \) . Then the closure \( \bar{K} \) of \( K \) in the Zariski topology is a group, and hence \( \bar{K} \) is an algebraic subgroup of \( G \) . Furthermore, if \( K \) contains a nonempty open subset of \( \bar{K} \) then \( K \) is closed in the Zariski topology. | Proof. Let \( x \in K \) . Then \( K = {xK} \subset x\bar{K} \) . Since left multiplication by \( x \) is a homeomorphism in the Zariski topology, we know that \( x\bar{K} \) is closed. Hence \( \bar{K} \subset x\bar{K} \), giving \( {x}^{-1}\bar{K} \subset \bar{K} \) . Thus \( K \cdot \bar{K} \subset \bar{K} \) . Repe... | Yes |
Theorem 11.1.5. Let \( \varphi : G \rightarrow H \) be a regular homomorphism of linear algebraic groups. Then \( F = \operatorname{Ker}\left( \varphi \right) \) is a closed subgroup of \( G \) and \( \varphi \left( G\right) \) is a closed subgroup of \( H \) . Hence \( \varphi \left( G\right) \) is an algebraic group.... | Proof. Since \( \varphi \) is continuous in the Zariski topology, it is clear that \( \operatorname{Ker}\left( \varphi \right) \) is closed. Set \( K = \overline{\varphi \left( G\right) } \) . Then \( K \) is an algebraic subgroup of \( H \) . By Theorem A.2.8, \( \varphi \left( G\right) \) contains a nonempty open sub... | Yes |
Corollary 11.1.8. Let \( \varphi : G \rightarrow H \) be a regular homomorphism of linear algebraic groups. Assume that \( G \) and \( H \) are connected in the Zariski topology and that \( \mathrm{d}\varphi : \operatorname{Lie}\left( G\right) \rightarrow \operatorname{Lie}\left( H\right) \) is surjective. Then \( \var... | Proof. By Corollary 11.1.3, \( G \) and \( H \) are irreducible affine algebraic sets. Since the differential of \( \varphi \) maps \( T{\left( H\right) }_{I} \) onto \( T{\left( G\right) }_{ }_{I} \), Theorem A.3.4 implies that \( \varphi \left( H\right) \) is Zariski dense in \( G \) . But \( \varphi \left( G\right) ... | Yes |
Proposition 11.1.9. Let \( G \) and \( H \) be algebraic subgroups of \( \mathbf{GL}\left( {n,\mathbb{C}}\right) \). Then the algebraic group \( G \cap H \) has Lie algebra \( \operatorname{Lie}\left( G\right) \cap \operatorname{Lie}\left( H\right) \). | Proof. Write \( \mathfrak{g} = \operatorname{Lie}\left( G\right) \) and \( \mathfrak{h} = \operatorname{Lie}\left( H\right) \). By Corollary 1.5.5 (1), \( \operatorname{Lie}\left( {G \cap H}\right) \subset \) \( \mathfrak{g} \cap \mathfrak{h} \). Let \( X = G \times H \) (as an affine algebraic set) and define \( \varp... | Yes |
Theorem 11.1.10. Suppose \( G \) is a connected algebraic group with Lie algebra \( \mathfrak{g} \) . Let \( \left( {\pi, V}\right) \) be a regular representation of \( G \) . If \( W \subset V \) is a linear subspace such that \( \mathrm{d}\mathbf{\pi }\left( X\right) W \subset W \) for all \( X \in \mathfrak{g} \) th... | Proof. We proved this result in Chapter 2 using the exponential map (Theorem 2.2.7); now we give a purely algebraic proof. Replacing \( G \) by \( \pi \left( G\right) \) and using Theorem 11.1.5, we may take \( G \subset \mathbf{{GL}}\left( V\right) \) . Set \( P = \{ h \in \mathbf{{GL}}\left( V\right) : {hW} \subset W... | Yes |
Proposition 11.1.11. Let \( G \) be a connected linear algebraic group with Lie algebra \( \mathfrak{g} \). Suppose \( \sigma : G \rightarrow \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) is a regular representation and \( H \subset \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) is a linear algebraic subgroup with Lie alg... | Proof. By the Hilbert basis theorem (Theorem A.1.2) there is a finite set \( {f}_{1},\ldots ,{f}_{r} \) of regular functions on \( \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) that generate the ideal \( {\mathcal{J}}_{H} \). Let \( V \subset {\mathcal{J}}_{H} \) be the subspace spanned by the right translates \( R\left... | Yes |
1. There exist a regular representation \( \left( {\pi, V}\right) \) of \( G \) and a one-dimensional subspace \( {V}_{0} \subset V \) such that \( H = \left\{ {g \in G : \pi \left( g\right) {V}_{0} = {V}_{0}}\right\} \) and \( \mathfrak{h} = \left\{ {X \in \mathfrak{g} : \mathrm{d}\pi \left( X\right) {V}_{0} \subset {... | (1): The defining ideal \( {\mathcal{J}}_{H} \subset \mathcal{O}\left\lbrack G\right\rbrack \) for \( H \) is finitely generated, so there is a finite-dimensional right \( G \) -invariant subspace \( L \subset \mathcal{O}\left\lbrack G\right\rbrack \) that contains a set of generators for \( {\mathcal{J}}_{H} \) . Let ... | Yes |
Lemma 11.1.14. Let \( M \) be a d-dimensional subspace of \( {\mathbb{C}}^{n} \). Let \( \pi \) be the representation of \( \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) on \( \mathop{\bigwedge }\limits^{d}{\mathbb{C}}^{n} \) and let \( N = \mathop{\bigwedge }\limits^{d}M \). 1. Suppose \( g \in \mathbf{{GL}}\left( {n,\... | Proof. (1): We may assume that \( M = \operatorname{Span}\left\{ {{e}_{1},\ldots ,{e}_{d}}\right\} \), where \( \left\{ {e}_{j}\right\} \) is the standard basis for \( {\mathbb{C}}^{n} \). Assume for the sake of contradiction that \( g \cdot M ⊄ M \). After performing row and column reductions of \( g \) by multiplying... | Yes |
Theorem 11.1.15. Suppose that \( G, K \), and \( M \) are algebraic groups and \( G \) is connected. Let \( \psi : G \rightarrow K \) and \( \varphi : G \rightarrow M \) be regular homomorphisms. Assume that \( \psi \) is surjective and \( \operatorname{Ker}\left( \psi \right) \subset \operatorname{Ker}\left( \varphi \... | Proof. Because \( \operatorname{Ker}\left( \psi \right) \subset \operatorname{Ker}\left( \varphi \right) \), we can define a homomorphism \( \mu \) of abstract groups satisfying the commutative diagram\n\n\n\nSince \... | Yes |
Corollary 11.1.16. Assume that \( G \) and \( K \) are connected algebraic groups and that \( \psi : G \rightarrow K \) is a bijective regular homomorphism. Then \( {\psi }^{-1} \) is regular, and hence \( \psi \) is an isomorphism of algebraic groups. | Proof. Take \( M = G \) and \( \varphi \) as the identity map in Theorem 11.1.15. | No |
Theorem 11.1.17. Let \( G \) be a connected linear algebraic group and \( H \) a normal algebraic subgroup. Choose a rational representation \( \varphi \) of \( G \) with \( \operatorname{Ker}\left( \varphi \right) = H \), and make \( G/H \) into a linear algebraic group by identifying it with \( \varphi \left( G\right... | Proof. Assertion (1) is immediate from Theorem 11.1.15 and Corollary 11.1.16. To prove (2), we see from the definition given above of \( \mathcal{O}\left\lbrack {G/H}\right\rbrack \) as \( {\mu }^{ * }\mathcal{O}\left\lbrack K\right\rbrack \) that \( {\pi }^{ * }\mathcal{O}\left\lbrack {G/H}\right\rbrack \subset \mathc... | Yes |
Corollary 11.1.18. Let \( G, H \), and \( K \) be linear algebraic groups with Lie algebras \( \mathfrak{g} \) , \( \mathfrak{h} \), and \( \mathfrak{k} \), respectively. Suppose that \( G \) is connected and \( H\overset{\varphi }{ \rightarrow }G\overset{\psi }{ \rightarrow }K \) is an exact sequence of regular homomo... | Proof. We know that \( \varphi \left( H\right) \) is closed in \( G \) by Theorem 11.1.5; since \( \varphi \) is injective, we also have \( H \cong \varphi \left( H\right) \) by Corollary 11.1.16, and \( \operatorname{Ker}\left( {\mathrm{d}\varphi }\right) = 0 \) . Identifying \( H \) with \( \varphi \left( H\right) \)... | Yes |
Theorem 11.2.1. Suppose \( G \subset \mathbf{GL}\left( V\right) \) is a commutative algebraic group.\n\n1. The set \( {G}_{s} \) of semisimple elements and the set \( {G}_{u} \) of unipotent elements are subgroups of \( G \) .\n\n2. There exists a basis for \( V \) such that the matrix \( \left\lbrack {g}_{ij}\right\rb... | Proof. To obtain (1), take \( x, y \in G \) . Since \( x \) and \( y \) commute, so do \( {x}_{s} \) and \( {y}_{s} \), by Theorem B.1.4. Hence \( {x}_{s}{y}_{s} \) is semisimple and is the semisimple factor of \( {xy} \) . This implies that \( {G}_{s} \) is a group. The same argument applies to \( {G}_{u} .\n\nFor the... | Yes |
Theorem 11.2.3 (Engel). Let \( G \) be an algebraic group and \( U \subset G \) a normal subgroup consisting of unipotent elements. Suppose \( \left( {\rho, V}\right) \) is a regular representation of \( G \) . Then there is a \( G \) -invariant flag of subspaces \[ V = {V}_{1} \supset {V}_{2} \supset \cdots \supset {V... | Proof. We may assume that \( U \) is Zariski closed. Indeed, if \( G \subset \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) as an algebraic subgroup, then every unipotent element \( u \) in \( G \) satisfies \( {\left( u - I\right) }^{n} = 0 \) . Hence the closure of \( U \) is a normal subgroup whose elements are all un... | Yes |
Lemma 11.2.4. \( {\operatorname{Rad}}_{u}\left( G\right) \) is a closed normal unipotent subgroup. | Proof. Let \( {U}_{1},{U}_{2} \subset G \) be normal unipotent subgroups of \( G \) . Set\n\n\[ W = {U}_{1} \cdot {U}_{2} = \left\{ {{u}_{1}{u}_{2} : {u}_{i} \in {U}_{i}}\right\} \]\n\nThen \( W \) is a normal subgroup of \( G \) . We will show that the elements of \( W \) are unipotent. We may assume \( G \subset \mat... | Yes |
Theorem 11.2.5. Let \( G \) be an (abstract) group and \( \left( {{\rho }_{i},{V}_{i}}\right) \) a finite-dimensional completely reducible representation of \( G \), for \( i = 1,2 \) . Then \( \left( {{\rho }_{1} \otimes {\rho }_{2},{V}_{1} \otimes {V}_{2}}\right) \) is a completely reducible representation of \( G \)... | Proof. It suffices to consider the case \( {\rho }_{1} = {\rho }_{2} \), since \( {\rho }_{1} \oplus {\rho }_{2} \) is completely reducible and \( {\rho }_{1} \otimes {\rho }_{2} \) is a subrepresentation of \( \left( {{\rho }_{1} \oplus {\rho }_{2}}\right) \otimes \left( {{\rho }_{1} \oplus {\rho }_{2}}\right) \) . Se... | Yes |
Corollary 11.2.6. Suppose \( G \subset \mathbf{GL}\left( {n,\mathbb{C}}\right) \) is an algebraic subgroup and the action of \( G \) on \( {\mathbb{C}}^{n} \) is completely reducible. Then \( G \) is a reductive group. | Proof. The right multiplication representation of \( G \) on \( {M}_{n}\left( \mathbb{C}\right) \) is completely reducible. Hence by Theorem 11.2.5, the \( k \) -fold tensor product of this representation on \( {M}_{n}{\left( \mathbb{C}\right) }^{\otimes k} \) is completely reducible for all integers \( k \) . Thus the... | Yes |
Theorem 11.2.7. Let \( G \) be a linear algebraic group. Then \( G \) is reductive if and only if \( {\operatorname{Rad}}_{u}\left( G\right) = \{ 1\} \) . | Proof. We have \( G \subset \mathbf{{GL}}\left( V\right) \) as an algebraic subgroup for some finite-dimensional vector space \( V \) . If \( G \) is reductive, there is a decomposition \( V = {\bigoplus }_{i}{V}_{i} \), where \( {V}_{i} \) is an irreducible \( G \) -module. Since \( {\operatorname{Rad}}_{u}\left( G\ri... | Yes |
Corollary 11.2.8. Let \( G \) be a linear algebraic group. Set \( U = {\operatorname{Rad}}_{u}\left( G\right) \) . Then \( G/U \) is reductive. | Proof. Suppose \( N/U \subset G/U \) is a unipotent normal subgroup, where \( N \) is a normal subgroup of \( G \) . The elements of \( N \) must be unipotent, by the preservation of the Jordan decomposition under homomorphisms. Hence \( N = U \) and \( N/U = \{ 1\} \) . Thus \( G/U \) is reductive. | Yes |
Lemma 11.2.10. Let \( G \) be a linear algebraic group and let \( N \) be a Zariski-connected normal subgroup of \( G \) . If \( G/N \) is Zariski connected, then so is \( G \) . If \( N \) and \( G/N \) are connected in the Lie group topology, then \( G \) is connected in the Lie group topology. | Proof. Fix either of the topologies in the statement of the lemma. Let \( {G}^{ \circ } \) be the identity component of \( G \) . Then \( N \subset {G}^{ \circ } \) . Let \( \pi \) be the natural projection of \( G \) onto \( G/N \) . Then \( \pi \left( {G}^{ \circ }\right) \) is an open subgroup of \( G \) . Hence \( ... | Yes |
Theorem 11.2.11. Let \( G \) be a Zariski-connected reductive linear algebraic group with finite center. Then \( G \) is generated by its unipotent elements. | Proof. Let \( {G}^{\prime } \) be the subgroup generated by the unipotent elements of \( G \) . Then \( {G}^{\prime } \) is a normal subgroup. If we show that \( {G}^{\prime } \) has a nonempty Zariski interior then it will be open and closed in the Zariski topology and hence equal to \( G \) . We may assume \( G \subs... | Yes |
Proposition 11.2.12. The Lie algebra of the adjoint group \( G \) is \( \operatorname{Der}\left( \mathfrak{g}\right) \), and the adjoint representation \( \operatorname{ad} : \mathfrak{g} \rightarrow \operatorname{Der}\left( \mathfrak{g}\right) \) is a Lie algebra isomorphism. | Proof. Let \( g\left( t\right) = \exp \left( {tD}\right) \), for \( t \in \mathbb{C} \), be a one-parameter subgroup of \( \operatorname{Aut}\left( \mathfrak{g}\right) \), where \( D \in \operatorname{End}\left( \mathfrak{g}\right) \) . Let \( X, Y \in \mathfrak{g} \) . Differentiating the equation \( g\left( t\right) ... | Yes |
Lemma 11.2.13. Let \( {G}^{\lambda } \) be the subgroup of \( \mathbf{{SL}}\left( {V}^{\lambda }\right) \) generated by \( {G}_{1}^{\lambda },\ldots ,{G}_{l}^{\lambda } \) .\n\n1. \( {G}^{\lambda } \) is a connected algebraic subgroup of \( \mathbf{{SL}}\left( {V}^{\lambda }\right) \) .\n\n2. Assume that \( \mathfrak{g... | Proof. To prove (1), define the set \( M = {G}_{{j}_{1}}\cdots {G}_{{j}_{p}} \), where \( p = l \cdot l \) ! and the sequence \( {j}_{1},\ldots ,{j}_{p} \) is the concatenation of the \( l \) ! sequences \( \gamma \left( 1\right) ,\ldots ,\gamma \left( l\right) \) as \( \gamma \) runs over all permutations of \( 1,\ldo... | Yes |
Theorem 11.2.14. Assume that \( \mathfrak{g} \) is simple. Then the group \( \widetilde{G} \) is algebraically simply connected and its Lie algebra is isomorphic to \( \mathfrak{g} \) . | Proof. From the definition of \( \Gamma \) we see that\n\n\[ \operatorname{Lie}\left( \widetilde{G}\right) = \left\{ {\left( {{X}_{1},\ldots ,{X}_{l}, X}\right) : \mathrm{d}{\varphi }_{j}\left( {X}_{j}\right) = X\text{ for }j = 1,\ldots, l}\right\} \]\n\n\[ \subset \operatorname{Lie}\left( {G}_{1}\right) \oplus \cdots ... | Yes |
For every \( x \in M \), the stabilizer \( {G}_{x} \) of \( x \) is an algebraic subgroup of \( G \) and the orbit \( G \cdot x \) is a smooth quasiprojective subset of \( M \) . | Proof. By the regularity of the map \( g \mapsto g \cdot x \), we know that \( {G}_{x} \) is closed and \( G \cdot x \) contains an open subset of its closure (cf. Theorem A.2.8 and the remarks at the end of Section A.4.3). By homogeneity the orbit is thus open in its closure; hence it is quasiprojective. The set of si... | Yes |
Corollary 11.3.2. There exists a point \( x \in M \) such that \( G \cdot x \) is closed in \( M \) . | Proof. Let \( y \in M \) and let \( Y \) be the closure of \( G \cdot y \) . Then \( G \cdot y \) is open in \( Y \), by the argument in the proof of Theorem 11.3.1, and hence \( Z = Y - G \cdot y \) is closed in \( Y \) Thus \( Z \) is quasiprojective. Furthermore, \( \dim Z < \dim Y \) by Theorem A.1.19, and \( Z \) ... | No |
Theorem 11.3.3. Let \( H \) be a closed subgroup of a linear algebraic group \( G \). 1. There exist a regular action of \( G \) on \( {\mathbb{P}}^{n} \) and a point \( {x}_{0} \in \mathbb{P}\left( V\right) \) such that \( H \) is the stabilizer of \( {x}_{0} \). The map \( g \mapsto g \cdot {x}_{0} \) is a bijection ... | Proof. The first assertion in (1) follows from Theorem 11.1.13. The independence of choices and the proofs of (2) and (3) follow by arguments similar to the proof of Theorem 11.1.15, taking into account the validity of Theorem A.2.9 for projective algebraic sets (cf. the remarks at the end of Section A.4.3). | No |
Let \( \dim V = n \). Let \( 0 \neq u \in \mathop{\bigwedge }\limits^{p}V \). Then \( \dim V\left( u\right) \leq p \) and \( \operatorname{Rank}\left( {T}_{u}\right) \geq n - p \). Furthermore, \( \operatorname{Rank}\left( {T}_{u}\right) = n - p \) if and only if \( u \) is decomposable. | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) be a basis for \( V\left( u\right) \). We complete it to a basis for \( V \), and we write\n\n\[ u = \mathop{\sum }\limits_{J}{c}_{J}{v}_{J} \]\n\nwhere \( {v}_{J} = {v}_{{j}_{1}} \land \cdots \land {v}_{{j}_{p}} \) for \( J \) a \( p \)-tuple with \( {j}_{1} <... | Yes |
Proposition 11.3.5. \( {\operatorname{Grass}}_{p}\left( V\right) \) is an irreducible projective algebraic set. | Proof. We use the notation of Lemma 11.3.4. If \( u \) is a \( p \) -vector, then \( \operatorname{Rank}\left( {T}_{u}\right) = \) \( n - \dim V\left( u\right) \geq n - p \) . Hence the \( p \) -vectors \( u \neq 0 \) with \( \dim V\left( u\right) = p \) are those for which all minors of size \( n - p + 1 \) in \( {T}_... | Yes |
Corollary 11.3.8. Let \( P = G \star 1 = \left\{ {{g\theta }{\left( g\right) }^{-1} : g \in G}\right\} \) be the orbit of the identity element under the \( \theta \) -twisted conjugation action. Then \( P \) is a closed irreducible subset of \( G \) isomorphic to \( G/K \) as a \( G \) -space (relative to the \( \theta... | Proof. Since \( g \star 1 = 1 \) if and only if \( g \in K \), the map \( \psi : G/K \rightarrow P \) with \( \psi \left( {gK}\right) = \) \( g \star 1 \) is bijective. This map is regular by Theorem 11.3.3, and its differential at the identity coset sends \( X \) to \( X - \theta \left( X\right) \), for \( X \in \math... | Yes |
Proposition 11.3.9. Let \( \sigma \) be a regular automorphism of the classical group \( G \). 1. If \( G = \mathbf{{SL}}\left( {n,\mathbb{C}}\right) \) then there exists \( s \in G \) such that \( \sigma \) is either \( \sigma \left( g\right) = {sg}{s}^{-1} \) or \( \sigma \left( g\right) = s{\left( {g}^{t}\right) }^{... | Proof. Let \( \pi \) be the defining representation of \( G \) on \( {\mathbb{C}}^{m} \) (where \( m = n \) in cases (1) and (3), and \( m = {2n} \) in case (2)). The representation \( {\pi }^{\sigma }\left( g\right) = \pi \left( {\sigma \left( g\right) }\right) \) also acts irreducibly on \( {\mathbb{C}}^{m} \). The W... | Yes |
Theorem 11.3.10. Let \( \theta \) be an involution of the classical group \( G \) . Assume that \( \operatorname{Lie}\left( G\right) \) is simple. Then \( \theta \) is given as follows, up to conjugation by an element of \( G \) . 1. If \( G = \mathbf{{SL}}\left( {n,\mathbb{C}}\right) \), then there are three possibili... | Proof. We use Proposition 11.3.9. Suppose \( G = \mathbf{{SL}}\left( {n,\mathbb{C}}\right) \) and \( \theta \left( x\right) = J{\left( {x}^{t}\right) }^{-1}{J}^{-1} \) . Since \[ x = {\theta }^{2}\left( x\right) = J{\left( {J}^{t}\right) }^{-1}x{J}^{t}{J}^{-1}\;\text{ for all }x \in G, \] we have \( {J}^{t}{J}^{-1} \) ... | Yes |
Proposition 11.4.1. Assume that \( G \) is a connected linear algebraic group. Then \( \mathcal{D}\left( G\right) \) is closed and connected. | Proof. Set \( C = \left\{ {{xy}{x}^{-1}{y}^{-1} : x, y \in G}\right\} \) . Then \( C = {C}^{-1},1 \in C \), and by definition\n\n\[ \mathcal{D}\left( G\right) = \mathop{\bigcup }\limits_{{n \geq 1}}{C}^{n} \]\n\nwhere \( {C}^{n} \) is all products of \( n \) commutators. Because \( {C}^{n} \) is the image of \( G \time... | Yes |
Theorem 11.4.4. Let \( G \) be a connected solvable linear algebraic group and let \( N = \) \( {\operatorname{Rad}}_{u}\left( G\right) \) . Then there exists a torus \( T \subset G \) such that \( G = T \cdot N \) . Furthermore, \( G \cong \) \( T \ltimes N \) (semidirect product) as a group, and \( G \cong T \times N... | Proof. If \( G = N \) there is nothing to prove, so we may assume that \( N \) is a proper subgroup of \( G \) . We have \( \mathcal{D}\left( G\right) \subset N \) by Corollary 11.4.3, so \( S = G/N \) is a connected commutative algebraic group. From Corollary 11.2.8, \( S \) is reductive, and hence \( S \) is a torus,... | Yes |
Corollary 11.4.6. Suppose \( G \) is a connected solvable linear algebraic group. Let \( A \subset G \) be a torus.\n\n1. There exists \( s \in G \) such that \( {sA}{s}^{-1} \subset T \) . In particular, if \( A \) is a maximal torus in \( G \), then \( {sA}{s}^{-1} = T \) . Thus all maximal tori in \( G \) are conjug... | Proof. Take \( g \in A \) such that the subgroup generated by \( g \) is Zariski dense in \( A \) (Lemma 2.1.4). There exists \( s \in G \) such that \( {sg}{s}^{-1} \in T \) . This implies that \( {sA}{s}^{-1} \subset T \) , so (1) holds. Since \( {\operatorname{Cent}}_{G}\left( s\right) = {\operatorname{Cent}}_{G}\le... | Yes |
Theorem 11.4.7. Let \( G \) be a connected linear algebraic group. Then \( G \) contains a Borel subgroup \( B \), and all other Borel subgroups of \( G \) are conjugate to \( B \). The homogeneous space \( G/B \) is a projective variety. Furthermore, if \( S \) is any connected solvable subgroup of \( G \) such that \... | To prove this theorem we shall use the following geometric generalization of the Lie-Kolchin theorem. | No |
Theorem 11.4.8 (Borel Fixed Point). Let \( S \) be a connected solvable group that acts algebraically on a projective variety \( X \). Then there exists a point \( {x}_{0} \in \bar{X} \) such that \( s \cdot {x}_{0} = {x}_{0} \) for all \( s \in S \). | Proof. We proceed by induction on \( \dim S \), as in the proof of the Lie-Kolchin theorem. The theorem is true when \( \dim S = 0 \), since \( S = \{ 1\} \) in this case by connectedness. We may assume that the theorem is true for the derived group \( \mathcal{D}\left( S\right) \). Thus we know that\n\n\[ Y = \{ x \in... | Yes |
Theorem 11.4.9. Let \( G \) be a connected linear algebraic group and \( B \) a fixed Borel subgroup of \( G \). Then\n\n\[ G = \mathop{\bigcup }\limits_{{x \in G}}{xB}{x}^{-1}. \]\n\nThus every element of \( G \) is contained in a Borel subgroup. | Proof. Let \( Y = \mathop{\bigcup }\limits_{{x \in G}}{xB}{x}^{-1} \). We first show that \( Y \) is closed in \( G \) (all topological assertions in the proof will refer to the Zariski topology). To see this we define\n\n\[ Z = \{ \left( {x, y}\right) : x \in G/B, y \in G,\text{ and }y \cdot x = x\} .\n\nHere \( y \cd... | Yes |
Theorem 11.4.10. Let \( G \) be a connected linear algebraic group. Suppose \( A \subset G \) is a torus. Then \( {\operatorname{Cent}}_{G}\left( A\right) \) is connected. Furthermore, if \( x \in {\operatorname{Cent}}_{G}\left( A\right) \) is semisimple, then there exists a torus \( S \subset {\operatorname{Cent}}_{G}... | Proof. Let \( x \in {\operatorname{Cent}}_{G}\left( A\right) \) . By Theorem 11.4.9 there exists a Borel subgroup \( B \) containing \( x \) . Let \( Y \subset G/B \) be the fixed-point set for the action of \( x \) on \( G/B \) . Then \( Y \) is nonempty (since \( B \in Y \) ) and is closed in the projective variety \... | Yes |
Corollary 11.4.13. The natural inclusion map \( {\operatorname{Norm}}_{U}\left( T\right) /T \rightarrow {\operatorname{Norm}}_{G}\left( H\right) /H \) is an isomorphism. | Proof. This follows from Theorem 11.4.11 and Proposition 7.3.2. | No |
Lemma 11.4.15. Define a map \( \Psi : \left( {G/H}\right) \times H \rightarrow G \) by \( \Psi \left( {{gH}, h}\right) = {gh}{g}^{-1} \) . If \( g \in G \) and \( h \in {H}^{\prime \prime } \), then\n\n\[{\Psi }^{-1}\left( {{gh}{g}^{-1}}\right) = \left\{ {\left( {{gwH},{w}^{-1}{hw}}\right) : w \in {W}_{G}}\right\} ,\]\... | Proof. Let \( h \in {H}^{\prime \prime } \) . Suppose \( {g}_{1} \in G \) and \( {h}_{1} \in H \) satisfy \( {g}_{1}{h}_{1}{g}_{1}^{-1} = {gh}{g}^{-1} \) . Set \( w = \) \( {g}^{-1}{g}_{1} \) . Then \( w{h}_{1} = {hw} \) . Given any \( {h}_{2} \in H \), we have\n\n\[{hw}{h}_{2}{w}^{-1} = w{h}_{1}{h}_{2}{w}^{-1} = w{h}_... | Yes |
Lemma 11.4.16. An element \( b \) is in \( B \cap {G}^{\prime } \) if and only if \( {b}^{\alpha } \neq 1 \) for all \( \alpha \in \Phi \) . Thus \( B \cap {G}^{\prime } = {H}^{\prime }{N}^{ + } \) is open and Zariski dense in \( B \) . | Proof. Write \( b = {hn} \) with \( h \in H \) and \( n \in {N}^{ + } \) . By Theorem 11.4.5 (3), \( b \) is \( {N}^{ + } \) - conjugate to \( {b}^{\prime } = {hu} \), where \( u \in {N}^{ + } \) and \( {hu} = {uh} \) . Hence \( {b}^{\alpha } = {\left( {b}^{\prime }\right) }^{\alpha } \) .\n\nBy definition, \( b \in {G... | Yes |
Theorem 11.4.18. The set \( {G}^{\prime } \) of regular semisimple elements is Zariski dense and open in \( G \) . | Proof. If \( f \in \mathcal{O}\left\lbrack G\right\rbrack \) and \( f\left( {G}^{\prime }\right) = 0 \), then by (11.4.16) we have \( f\left( {g{H}^{\prime }{N}^{ + }{g}^{-1}}\right) = 0 \) and hence \( f\left( {{gB}{g}^{-1}}\right) = 0 \) for all \( g \in G \) . Thus Theorem 11.4.9 implies that \( f = 0 \), proving th... | Yes |
Theorem 11.5.1. A connected linear algebraic group is reductive if and only if it has a compact real form. | Proof. A connected linear algebraic group \( G \) is also connected in the Lie group topology, by Theorem 11.2.9. If it has a compact real form, then it is reductive, by Theorem 3.3.15. Conversely, if \( G \) is reductive and has Lie algebra \( \mathfrak{g} \), then \( \mathfrak{g} = \) \( \mathfrak{z} \oplus \left\lbr... | Yes |
Lemma 11.5.4. Let \( G \) and \( {\tau }_{0} \) be as above and assume that \( G \) has finite center. Set \( \mathfrak{g} = \operatorname{Lie}\left( G\right) \) . Let \( U = \left\{ {g \in G : {\tau }_{0}\left( g\right) = g}\right\} \) and identify \( \mathfrak{u} = \operatorname{Lie}\left( U\right) \) with the space ... | Proof. We apply the unitary trick (Section 3.3.4) to the representation \( \left( {{\left. \operatorname{Ad}\right| }_{U},\mathfrak{g}}\right) \) . Thus there is a positive definite Hermitian inner product \( \left( {\cdot , \cdot }\right) \) on \( \mathfrak{g} \) such that\n\n\[ \left( {\operatorname{Ad}\left( u\right... | Yes |
Corollary 11.5.5. If \( G \) is reductive with finite center and if \( {U}_{1} \) and \( {U}_{2} \) are compact real forms of \( G \), then there exists \( g \in G \) such that \( g{U}_{1}{g}^{-1} = {U}_{2} \) . | Proof. Let \( {\tau }_{1} \) and \( {\tau }_{2} \) be the conjugations of \( G \) corresponding to \( {U}_{1} \) and \( {U}_{2} \) . Then Theorem 11.5.3 implies that there exists \( g \in G \) such that the automorphism \( {\tau }_{3}\left( x\right) = \) \( g{\tau }_{1}\left( x\right) {g}^{-1} \), for \( x \in G \), sa... | Yes |
Lemma 11.5.6. Let \( A \in {M}_{n}\left( \mathbb{C}\right) \) be such that \( {A}^{ * } = A \) . Let \( f \) be a polynomial on \( {M}_{n}\left( \mathbb{C}\right) \) such that \( f\left( {\exp \left( {mA}\right) }\right) = 0 \) for all positive integers \( m \) . Then \( f\left( {\exp \left( {tA}\right) }\right) = 0 \)... | Proof. There is a basis \( \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \) of \( {\mathbb{C}}^{n} \) and \( {\lambda }_{i} \in \mathbb{R} \) such that \( A{e}_{i} = {\lambda }_{i}{e}_{i} \) . Rewriting \( f \) in terms of matrix entries with respect to this basis, we see that the lemma reduces to the following assertion:\... | Yes |
Lemma 11.5.7. The map \( \Psi : {\operatorname{Herm}}_{n} \rightarrow {\Omega }_{n} \) given by \( \Psi \left( X\right) = \exp X \) is a diffeomorphism of \( {\operatorname{Herm}}_{n} \) onto \( {\Omega }_{n} \) . | Proof. Suppose \( A \in {\Omega }_{n} \) and \( \exp X = \exp Y = A \) with \( X, Y \in {\operatorname{Herm}}_{n} \) . Then\n\n\[ \exp \left( {mX}\right) = \exp \left( {mY}\right) = {A}^{m}\;\text{ for }m = 1,2,\ldots . \]\n\nThus Lemma 11.5.6 implies that \( \exp \left( {tX}\right) = \exp \left( {tY}\right) \) for all... | Yes |
Lemma 11.5.8. There exists a regular homomorphism \( \Psi : G \rightarrow \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) such that \( \Psi \) is an isomorphism of \( G \) onto its image and such that \( \Psi \left( {\tau \left( g\right) }\right) = {\left( \Psi {\left( g\right) }^{ * }\right) }^{-1} \) . | Proof. We may assume that \( G \subset \mathbf{{GL}}\left( V\right) \) as a Zariski-closed subgroup, where \( V \) is an \( n \) -dimensional complex vector space. The unitary trick (Section 3.3.4) implies that there exists an inner product \( \left( {\cdot , \cdot }\right) \) on \( V \) such that \( \left( {{uv}, w}\r... | Yes |
Theorem 11.5.9. The map \( \Phi : U \times \mathfrak{u} \rightarrow G \) defined by \( \Phi \left( {u, X}\right) = u\exp \left( {\mathrm{i}X}\right) \), for \( u \in U \) and \( X \in \mathfrak{u} \), is a diffeomorphism onto \( G \) . In particular, \( U \) is connected. | Proof. By Lemma 11.5.8 we may assume that \( G \subset \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) and \( \tau \left( g\right) = {\left( {g}^{ * }\right) }^{-1} \) . If \( g \in G \) then \( {g}^{ * }g \) is positive definite. Since \( {\left( {g}^{ * }g\right) }^{m} \in G \) for all \( m \in \mathbb{Z} \), Lemma 11.5... | Yes |
Theorem 11.5.10. Let \( G \) be a connected reductive linear algebraic group. Let \( \tau \) be a conjugation on \( G \) corresponding to a compact real form \( U \) . Let \( \theta \) be an involutive automorphism of \( G \) such that \( {\tau \theta } = {\theta \tau } \) . Set \( K = \{ g \in G : \theta \left( g\righ... | Proof. Since \( {\theta \tau } = {\tau \theta } \), the restriction of \( \tau \) to \( K \) is a conjugation of \( K \) . Hence \( {K}_{0} \) is a real form of \( K \) that is compact, since \( U \) is compact. Thus \( K \) is reductive by Theorem 11.5.1. It remains to show that \( {K}_{0} \) is Zariski dense in \( K ... | Yes |
Lemma 11.6.1 (Gauss Decomposition). Set \( \Omega = {V}^{ - }L{V}^{ + } \) . Let \( {\Delta }_{i}\left( g\right) \) denote the upper left-hand corner minor of \( g \) of size \( {m}_{1} + \cdots + {m}_{i} \) . Then\n\n\[ \Omega = \left\{ {g \in \mathbf{{GL}}\left( {n,\mathbb{C}}\right) : {\Delta }_{i}\left( g\right) \n... | Proof. This follows by induction on \( r \) using the proof of Lemma B.2.6, with the scalar matrix entries \( {x}_{ij} \) in that proof replaced by \( {m}_{i} \times {m}_{j} \) matrices. | No |
Theorem 11.6.2. If \( g \in G \cap \Omega \), then \( {\gamma }_{0}\left( g\right) ,{\gamma }_{ + }\left( g\right) \), and \( {\gamma }_{ - }\left( g\right) \) are in \( G \) . | Proof. The key to the theorem is the following assertion:\n\n\( \left( \star \right) \;{N}^{ - }M{N}^{ + } \) contains a neighborhood of \( I \) in \( G \) relative to the Zariski topology.\n\nLet us show how \( \left( \star \right) \) implies the theorem. Suppose \( f \in \mathcal{O}\left\lbrack {\mathbf{{GL}}\left( {... | Yes |
Lemma 11.6.4. Suppose that \( A \) is \( \sigma \) -split. There exists a regular homomorphism \( \varphi : G \rightarrow \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) such that \( \varphi : G \rightarrow \varphi \left( G\right) \) is a regular isomorphism and such that\n\n1. \( \varphi \left( {\sigma \left( g\right) }\... | Proof. We use the notation in the proof of Theorem 1.7.5. Let \( V \) and \( \rho \) be as in that proof. Since \( \left( {\rho, V}\right) \) is a regular representation of \( G \), Proposition 2.1.3 furnishes a weight-space decomposition\n\n\[ V = {\bigoplus }_{\chi \in F}V\left( \chi \right) \]\n\nwith \( F \subset \... | Yes |
Proposition 12.1.1. There is an isotypic decomposition \( E = {\bigoplus }_{\omega \in \widehat{G}}{E}_{\left( \omega \right) } \) . Furthermore, for each \( \omega \in \widehat{G} \) the map \( T \otimes v \mapsto T\left( v\right) \) for \( T \in {\operatorname{Hom}}_{G}\left( {\omega ,\rho }\right) \) and \( v \in {V... | Proof. This follows from Proposition 4.1.15 with \( \mathcal{A} = \mathcal{A}\left\lbrack G\right\rbrack \) . | No |
For \( \lambda \in {P}_{+ + }\left( G\right) \), the isotypic subspace of type \( {\pi }^{\lambda } \) in \( \mathcal{O}\left\lbrack X\right\rbrack \) is the span of \( {\rho }_{X}\left( G\right) \mathcal{O}{\left\lbrack X\right\rbrack }^{{N}^{ + }}\left( \lambda \right) \). This subspace is isomorphic to \( {V}^{\lamb... | Proof. We define a bijection between covariants and highest-weight vectors for the representation \( {\rho }_{X} \) as follows: Given \( T : {V}^{\lambda } \rightarrow \mathcal{O}\left\lbrack X\right\rbrack \), a covariant of type \( {\pi }^{\lambda } \), we set \( \psi \left( T\right) = T{v}_{\lambda } \). Then \( \ps... | Yes |
Theorem 12.1.3. Let \( \left( {\pi ,{V}_{\pi }}\right) \) be a regular representation of \( G \) . Then there is a vector-space isomorphism\n\n\[ \n{\operatorname{Hom}}_{G}\left( {\pi ,{\operatorname{Ind}}_{K}^{G}\left( \mu \right) }\right) \cong {\operatorname{Hom}}_{K}\left( {{\operatorname{Res}}_{K}^{G}\left( \pi \r... | Proof. The proof of Theorem 4.4.1 applies without change, because the maps defined in that proof are regular when \( G \) and \( K \) are linear algebraic groups. When \( G \) and \( K \) are reductive, the reciprocity statement about multiplicities follows from (12.2) and (12.4). | Yes |
Theorem 12.1.4. As a G-module under left translation, \n\n\[ \n\mathcal{R}\left( {G/K}\right) \cong {\bigoplus }_{\omega \in \widehat{G}}{V}_{\omega } \otimes {\left( {V}_{\omega }^{ * }\right) }^{K}, \n\] \n\nwhere \( g \in G \) acts by \( {\pi }_{\omega }\left( g\right) \otimes 1 \) in each summand and \( {\left( {V}... | Proof. By part (2) of Theorem 4.2.7, a function \( f \in \mathcal{O}\left\lbrack G\right\rbrack \) is fixed under right translations by \( K \) if and only if its components in the decomposition 4.18 are in \( {\phi }_{\omega }\left( {\left( {V}_{\omega }^{ * }\right) \otimes {V}_{\omega }^{K}}\right) \) for all \( \om... | No |
Theorem 12.1.5. The space \( \mathcal{R}\left( {{N}^{ - } \smallsetminus G}\right) \) decomposes under \( G \) as\n\n\[ \mathcal{R}\left( {{N}^{ - } \smallsetminus G}\right) = {\bigoplus }_{\lambda \in {P}_{+ + }\left( G\right) }{\varphi }_{\lambda }\left( {{v}_{\lambda }^{ * } \otimes {V}^{\lambda }}\right) \cong {\bi... | Proof. By Theorem 3.2.13 the space of \( {N}^{ - } \) -fixed vectors in \( {\left( {V}^{\lambda }\right) }^{ * } \) is spanned by \( {v}_{\lambda }^{ * } \) . We use the map \( f \mapsto \check{f} \), where \( \check{f}\left( x\right) = f\left( {x}^{-1}\right) \), to change \( \mathcal{R}\left( {{N}^{ - } \smallsetminu... | Yes |
Theorem 12.1.6. Let \( \lambda \in {P}_{+ + }\left( G\right) \) . Let \( {\mathcal{R}}_{\lambda } \subset \mathcal{O}\left\lbrack G\right\rbrack \) be the subspace of functions such that \[ f\left( {\bar{n}{hg}}\right) = {h}^{\lambda }f\left( g\right) \;\text{ for }\bar{n} \in {N}^{ - }, h \in H,\text{ and }g \in G. \]... | Proof. The space \( \mathcal{R}\left( {{N}^{ - } \smallsetminus G}\right) \) is invariant under left translations by the subgroup \( H \) (since \( H \) normalizes \( {N}^{ - } \) ). The vector \( {v}_{\lambda }^{ * } \) has weight \( - \lambda \) . Hence by (12.6) we see that the subspace \( {\varphi }_{\lambda }\left... | Yes |
Corollary 12.1.7. Let \( {\lambda }_{1},\ldots ,{\lambda }_{r} \) be generators for the additive semigroup \( {P}_{+ + }\left( G\right) \) . Set \( {f}_{i}\left( g\right) = {f}_{{\lambda }_{i}}\left( g\right) \) . Let \( \lambda \in {P}_{+ + }\left( G\right) \) and write \( \lambda = {m}_{1}{\lambda }_{1} + \cdots + {m... | Proof. Set \( {V}_{i} = {V}^{{\lambda }_{i}},{v}_{i} = {v}_{{\lambda }_{i}} \), and \( {v}_{i}^{ * } = {v}_{{\lambda }_{i}}^{ * } \) . Then \( {V}^{\lambda } \) can be realized as the \( G \) -cyclic space generated by the highest-weight vector\n\n\[ \n{v}_{\lambda } = {v}_{1}^{\otimes {m}_{1}} \otimes \cdots \otimes {... | Yes |
Theorem 12.2.1. Let \( X \) be an irreducible affine \( G \) -space. Suppose \( B \cdot {x}_{0} \) is open in \( X \) for some point \( {x}_{0} \in X \) (equivalently, \( \dim \mathfrak{b} = \dim X + \dim {\mathfrak{b}}_{{x}_{0}} \) ). Then\n\n1. \( X \) is multiplicity-free as a \( G \) -space, and\n\n2. if \( \lambda... | Proof. (1): It suffices by Theorem 12.1.2 to show that \( \dim \mathcal{O}{\left\lbrack X\right\rbrack }^{N}\left( \lambda \right) \leq 1 \) for all \( \lambda \in {P}_{+ + }\left( G\right) \) . Suppose \( B \cdot {x}_{0} \) is open in \( X \) (and hence dense in \( X \), by the irreducibility of \( X \) ). Then \( f \... | Yes |
Proposition 12.2.4. The pair \( \left( {G, K}\right) \) is spherical if and only if the space \( \mathcal{R}\left( {G/K}\right) \) is multiplicity-free as a \( G \) -module. | Proof. This follows immediately from Theorem 12.1.4. | No |
Theorem 12.2.5. Suppose \( G \) is a connected reductive group. If there exists a connected solvable algebraic subgroup \( S \) of \( G \) such that \( \operatorname{Lie}\left( S\right) + \operatorname{Lie}\left( K\right) = \mathfrak{g} \), then \( \left( {G, K}\right) \) is spherical. | Proof. The Lie algebra assumption implies that the map \( S \times K \rightarrow G \) given by multiplication has surjective differential at \( \left( {1,1}\right) \) . Hence \( {SK} \) is Zariski dense in \( G \) by Theorem A.3.4. By Theorem 11.4.7 there exists \( {x}_{0} \in G \) such that \( {x}_{0}S{x}_{0}^{-1} \su... | Yes |
Theorem 12.2.6. Assume that there exists \( {x}_{0} \in X \) such that \( \sigma \left( B\right) {x}_{0} \) is open in \( X \) . Let \( {H}_{0} = \left\{ {h \in H : h \cdot {x}_{0} = {x}_{0}}\right\} \) . Let \( \mathcal{E}\left( X\right) \) be the set of all irreducible polynomials \( f \in \mathcal{P}\left( X\right) ... | Proof. Let \( \left\{ {{f}_{1},\ldots ,{f}_{k}}\right\} \) be any finite subset of \( \mathcal{E}\left( X\right) \) . Since \( B \cdot {x}_{0} \) is open, each \( {f}_{i} \) is uniquely determined by its weight \( {\lambda }_{i} \) and the normalization \( f\left( {x}_{0}\right) = 1 \) . Also \( {f}_{i} \) must be a ho... | Yes |
Lemma 12.2.8. The B orbit of \( {x}_{0} \) is open in \( {M}_{n, n - 1} \) and consists of all \( x \) such that\n\n\[ \n{\Delta }_{i}\left( x\right) \neq 0,\;{\Gamma }_{i}\left( x\right) \neq 0\;\text{ for }i = 1,\ldots, n - 1.\n\] | Proof. Let \( g = \left( {\left\lbrack \begin{array}{ll} a & 0 \\ 0 & 1 \end{array}\right\rbrack, b}\right) \), where \( a, b \in {B}_{n - 1} \) . For \( x = \left\lbrack \begin{array}{l} y \\ z \end{array}\right\rbrack \in {M}_{n, n - 1} \) we have\n\n\[ \n{g}^{-1} \cdot x = \left\lbrack \begin{matrix} {a}^{t} & 0 \\ ... | Yes |
Theorem 12.2.9. Let \( X = {M}_{n, n - 1}\left( \mathbb{C}\right) \) and let \( {x}_{0} \) be given by (12.12). The space \( \mathcal{P}\left( X\right) \) is multiplicity-free under the action of \( G = \mathbf{{GL}}\left( {n - 1,\mathbb{C}}\right) \times \mathbf{{GL}}\left( {n - 1,\mathbb{C}}\right) \) . For \( \mathb... | Proof. The action of \( G \) is multiplicity-free by Lemma 12.2.8. To find which representations of \( G \) occur, we observe that \( {\Delta }_{1},\ldots ,{\Delta }_{n - 1} \) and \( {\Gamma }_{1},\ldots ,{\Gamma }_{n - 1} \) are \( B \) - eigenfunctions, by (12.14) and (12.15). They are also irreducible polynomials b... | Yes |
Theorem 12.2.12. (SFT for \( {\mathbf{{GL}}}_{n} \) ) Assume \( n < \min \left( {k, m}\right) \). 1. The set of all \( \left( {n + 1}\right) \times \left( {n + 1}\right) \) minors is a minimal generating set for \( {\mathcal{J}}_{k, m, n} \). 2. As a module for \( {\mathbf{{GL}}}_{k} \times {\mathbf{{GL}}}_{m} \), the ... | Proof. Set \( G = {\mathbf{{GL}}}_{k} \times {\mathbf{{GL}}}_{m} \) and take \( {N}^{ + } = {N}_{k}^{ - } \times {N}_{m}^{ + } \). Since \( {\mathcal{J}}_{k, m, n} \) is a \( G \) -invariant ideal in \( \mathcal{P}\left( {M}_{k, m}\right) \), Proposition 12.2.10 implies that \[ {\mathcal{J}}_{k, m, n} = \operatorname{S... | Yes |
Lemma 12.2.13. \( \operatorname{Span}\left\{ {{\pi }_{m, k}\left( G\right) {\Delta }_{p}}\right\} \) is the space spanned by the set of all \( p \times p \) minors. It is isomorphic to \( {\left( {F}_{k}^{{\lambda }_{p}}\right) }^{ * } \otimes {F}_{m}^{{\lambda }_{p}} \) as a G-module. | Proof. We embed \( {\mathfrak{S}}_{k} \) into \( {\mathbf{{GL}}}_{k} \) as the permutation matrices as usual. The space spanned by the set of all \( p \times p \) minors is then\n\n\[ \operatorname{Span}\left\{ {{\pi }_{k, m}\left( {s, t}\right) {\Delta }_{p} : s \in {\mathfrak{S}}_{k}, t \in {\mathfrak{S}}_{m}}\right\... | Yes |
Theorem 12.2.14. (SFT for \( \mathbf{O}\left( n\right) \) ) Assume \( n < k \) .\n\n1. The restrictions to \( S{M}_{k} \) of the \( \left( {n + 1}\right) \times \left( {n + 1}\right) \) minors is a minimal generating set for the ideal \( {\mathrm{{SJ}}}_{k, n} \).\n\n2. As a module for \( {\mathbf{{GL}}}_{k} \), the sy... | Proof. We follow the same general line of argument as in Theorem 12.2.12. Let \( G = {\mathbf{{GL}}}_{k} \) and \( {N}^{ + } = {N}_{k}^{ + } \) . Since \( {\mathcal{{SI}}}_{k, m, n} \) is a \( G \) -invariant ideal in \( \mathcal{P}\left( {S{M}_{k}}\right) \), Proposition 12.2.10 implies that\n\n\[ \n{\mathcal{{SJ}}}_{... | Yes |
Theorem 12.2.15. (SFT for \( \operatorname{Sp}\left( {\mathbb{C}}^{n}\right) \) ) Assume \( n < k \) . 1. The set \( \left\{ {\pi \left( s\right) \mathop{\operatorname{Pf}}\limits_{{\left( {n/2}\right) + 1}} : s \in {\mathfrak{S}}_{k}}\right\} \) is a minimal generating set for \( {\mathcal{A}}_{k, n} \) . 2. As a modu... | Proof. We follow the same general line of argument as in Theorem 12.2.14. Let \( G = {\mathbf{{GL}}}_{k} \) and \( {N}^{ + } = {N}_{k}^{ + } \) . Since \( {\mathcal{{AJ}}}_{k, m, n} \) is a \( G \) -invariant ideal in \( \mathcal{P}\left( {A{M}_{k}}\right) \), Proposition 12.2.10 implies that \[ {\mathcal{{AJ}}}_{k, n}... | Yes |
Lemma 12.3.3. Let \( X, Y \in {V}_{0} \) . There exists \( {k}_{0} \in {K}_{0}^{ \circ } \) such that \( \left\lbrack {\operatorname{Ad}\left( {k}_{0}\right) X, Y}\right\rbrack = 0 \) . | Proof. Let \( f\left( k\right) = \langle \operatorname{Ad}\left( k\right) X, Y\rangle \) for \( k \in {K}_{0}^{ \circ } \) . Since \( {K}_{0}^{ \circ } \) is compact and \( f \) is real-valued, \( f \) has a critical point, say \( {k}_{0} \) . Set \( Z = \operatorname{Ad}\left( {k}_{0}\right) X \) . If \( T \in {\mathf... | Yes |
Corollary 12.3.4. There is a polar decomposition \( {V}_{0} = \operatorname{Ad}\left( {K}_{0}^{ \circ }\right) {\mathfrak{a}}_{0} \) . Furthermore, if \( {\mathfrak{a}}_{1} \) is another maximal abelian subspace in \( {V}_{0} \), then there exists \( k \in {K}_{0}^{ \circ } \) such that \( \operatorname{Ad}\left( k\rig... | Proof. This is clear from Lemma 12.3.3 and statement \( \left( \star \right) \) . | No |
Lemma 12.3.7. Let \( H \) be a \( \theta \) -stable and \( \tau \) -stable maximal algebraic torus in \( G \) with Lie algebra \( \mathfrak{h} \) and root system \( \Phi \) . Set \( \mathfrak{a} = \{ X \in \mathfrak{h} : \theta \left( X\right) = - X\} \) and\n\n\[ \n{\Phi }_{0} = \{ \alpha \in \Phi : \langle \alpha, X\... | Assume that \( H \) is maximally \( \theta \) anisotropic. Let \( \mathfrak{l} \) be the centralizer of \( \mathfrak{a} \) in \( \mathfrak{g} \) and let \( \mathfrak{m} \) be the centralizer of \( \mathfrak{a} \) in \( \mathfrak{k} \) . Then \( \mathfrak{l} = \mathfrak{a} \oplus \mathfrak{m} \) . Proof. The equivalence... | Yes |
Lemma 12.3.8. The group \( L = A{M}^{ \circ } \) and \( M = T{M}^{ \circ } \) . Hence \( M \) is connected if and only if \( T \) is connected. | Proof. Let \( x \in L \) . Then the semisimple and unipotent components \( {x}_{s} \) and \( {x}_{u} \) are in \( L \) , since they commute with \( A \) . We can write \( {x}_{u} = \exp Y \), where \( Y \) is a nilpotent element of \( \mathfrak{l} \) . Since \( \mathfrak{l} = \mathfrak{a} \oplus \mathfrak{m},\left\lbra... | Yes |
Lemma 12.3.9. The maximal torus \( H = A{T}^{ \circ } \) and \( A \cap T = \left\{ {a \in A : {a}^{2} = 1}\right\} \cong \) \( {\left( \mathbb{Z}/2\mathbb{Z}\right) }^{m} \), where \( m = \operatorname{rank}\left( A\right) \) . Thus \( C \cong {\left( \mathbb{Z}/2\mathbb{Z}\right) }^{r} \) for some \( r \) with \( 0 \l... | Proof. There is a decomposition \( \mathfrak{h} = \mathfrak{t} \oplus \mathfrak{a} \), where \( \theta = 1 \) on \( \mathfrak{t} \) and \( \theta = - 1 \) on \( \mathfrak{a} \) . Clearly we have \( \operatorname{Lie}\left( A\right) = \mathfrak{a} \) and \( \operatorname{Lie}\left( T\right) = \mathfrak{t} \) . Since\n\n... | Yes |
Lemma 12.3.11. There are vector-space direct sum decompositions\n\n\[ \mathfrak{g} = {\mathfrak{n}}^{-} \oplus \mathfrak{m} \oplus \mathfrak{a} \oplus {\mathfrak{n}}^{+} = \mathfrak{k} \oplus \mathfrak{a} \oplus {\mathfrak{n}}^{+} \]\n\nHence \( {N}^{-}{MA}{N}^{+} \) and \( {KA}{N}^{+} \) are open Zariski-dense subsets... | Proof. Since \( \Phi \) is the disjoint union \( {\Phi }_{0} \cup {\Phi }_{1}^{+} \cup \left( {-{\Phi }_{1}^{+}}\right) \), we can use the rootspace decomposition of \( \mathfrak{g} \) to write \( X \in \mathfrak{g} \) as\n\n\[ X = \mathop{\sum }\limits_{{\beta \in {\Phi }_{1}^{+}}}{X}_{-\beta } + \left\{ {{H}_{0} + \m... | Yes |
Theorem 12.3.12. \( K \) is a spherical subgroup of \( G \) . If \( \lambda \) is the \( B \) -highest weight of an irreducible \( K \) -spherical representation of \( G \), then\n\n\[ \n{t}^{\lambda } = 1\;\text{ for all }t \in T \n\] | Proof. Lemma 12.3.11 and Theorem 12.2.5 imply that \( K \) is spherical. Since \( T = \) \( K \cap B \), condition (12.43) is satisfied by the highest weight of a \( K \) -spherical representation by Theorem 12.2.1. | No |
Theorem 12.3.13. Let \( \left( {{\pi }^{\lambda },{V}^{\lambda }}\right) \) be an irreducible regular representation of \( G \) with highest weight \( \lambda \) (relative to \( B \) ). The following are equivalent:\n\n1. \( {V}^{\lambda } \) contains a nonzero vector fixed by \( K \) .\n\n2. \( {V}^{\lambda } \) conta... | Proof. We have already shown (in Theorem 12.3.12) that (1) implies (3) . We observe that condition (2) is equivalent to\n\n\( \left( {\star \star }\right) \;{\pi }^{\lambda }\left( M\right) {v}_{\lambda } = {v}_{\lambda } \), where \( {v}_{\lambda } \) is a nonzero \( B \) eigenvector in \( {V}^{\lambda } \) .\n\nThis ... | Yes |
Lemma 12.3.14. Let \( \lambda \in \mathcal{X}\left( H\right) \) . Suppose \( {t}^{\lambda } = 1 \) for all \( t \in T \) . Then \( {a}^{\lambda } > 0 \) for all \( a \in {A}_{0} \) . | Proof. From Proposition 12.3.5 we know that \( A \) is \( \sigma \) -split. Hence by Lemma 11.6.4 there is an embedding \( \varphi : G \rightarrow \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) such that \( \varphi : A \cong {D}_{p} \), with \( {A}_{0} \) corresponding to the real matrices in \( {D}_{p} \) (as done for t... | Yes |
Corollary 12.3.15. The space of regular functions on \( G/K \) is isomorphic to \( { \oplus }_{\lambda }{V}^{\lambda } \) as a \( G \) -module, where \( \lambda \) runs over all \( \theta \) -admissible dominant weights of \( H \) . | Proof. This follows by Theorems 12.3.12 and 12.3.13. | Yes |
Theorem 12.4.1 (Kostant-Rallis). The map \( h \otimes f \mapsto {hf} \) (pointwise multiplication) from \( \mathcal{H} \otimes \mathcal{P}{\left( V\right) }^{K} \) to \( \mathcal{P}\left( V\right) \) is a linear bijection. Furthermore, \( \mathcal{H} \) is equivalent to \( \mathcal{O}\left\lbrack {K/M}\right\rbrack = {... | The last statement of the theorem follows by Frobenius reciprocity (Theorem 12.1.3). The proof of the rest of the theorem will be given later. | No |
Theorem 12.4.2 (Kostant). Let \( G \) be a connected, semisimple, linear algebraic group. Let \( T \) be a maximal torus in \( G \) . Let \( \mathfrak{g} \) be the Lie algebra of \( G \) and let \( \mu \left( g\right) f\left( X\right) = f\left( {\operatorname{Ad}{\left( g\right) }^{-1}X}\right) \) for \( g \in G, f \in... | Proof. Take \( {G}_{1} = G \times G \) in place of \( G \) in Theorem 12.4.1 and let \( \theta \left( {g, h}\right) = \left( {h, g}\right) \) for \( \left( {g, h}\right) \in {G}_{1} \) . Then \( {G}_{1} \) is semisimple and \( K = {G}_{1}^{\theta } \cong G \) (embedded diagonally). Let \( \mathfrak{g} \) be the Lie alg... | No |
Theorem 12.4.3. If \( \lambda \in \sum \) then \( {s}_{\lambda } \in W\left( \mathfrak{a}\right) \) . Furthermore, | For the adjoint representation (as in Theorem 12.4.2), this was proved in Section 11.4.6. For the case of general symmetric spaces we refer the reader to Helgason [66, Chapter VIII, §2] for a proof. | No |
Theorem 12.4.5. The map res : \( \mathcal{P}{\left( V\right) }^{K} \rightarrow \mathcal{P}{\left( \mathfrak{a}\right) }^{W\left( \mathfrak{a}\right) } \) defined by \( \operatorname{res}\left( f\right) = {\left. f\right| }_{\mathfrak{a}} \) for \( f \in \mathcal{P}{\left( V\right) }^{K} \) is an algebra isomorphism. | We have already shown that the restriction map is injective. For a proof of surjectivity see Helgason [67, Chapter II, §5.2]. We will verify this in the next section by direct calculations when \( K \) is a classical group. | No |
Lemma 12.4.6. Let \( \mathcal{J} \) be a homogeneous ideal in \( \mathcal{P} \) . Then the following are equivalent:\n\n1. \( \mathcal{V}\left( \mathcal{I}\right) = \{ 0\} \) .\n\n2. There exists \( j > 0 \) such that \( \mathcal{J} \supset {\mathcal{P}}^{j} \) . | Proof. We first prove that (2) implies (1). If \( {\mathcal{P}}^{j} \subset \mathcal{I} \) then \( {x}_{i}^{j} \in \mathcal{I} \) for \( i = 1,\ldots, n \) . This implies that if \( x \in \mathcal{V}\left( \mathcal{J}\right) \) then \( {x}_{i}^{j} = 0 \) for \( i = 1,\ldots, n \) ; thus \( x = 0 \) . We now prove that ... | Yes |
Lemma 12.4.7. The set of all \( f \in V\left( d\right) \) such that \( \mathcal{V}\left( {\mathcal{I}\left( f\right) }\right) = \{ 0\} \) is Zariski open in \( V\left( d\right) \) . It is nonempty if \( m \geq n \) . | Proof. Let \( W \) be a vector space over \( \mathbb{C} \) with basis \( {w}_{1},\ldots ,{w}_{n} \) . We grade \( W \) by setting\n\n\[ \n{W}^{j} = \mathop{\sum }\limits_{{{d}_{i} = j}}\mathbb{C}{w}_{i} \n\]\n\nWe put a grading on \( \mathcal{P} \otimes W \) by setting \( {\left( \mathcal{P} \otimes W\right) }^{k} = \m... | Yes |
Lemma 12.4.8. Let \( \mathcal{J} \) be an ideal in \( \mathcal{P} \) and set \( \mathcal{J} = \operatorname{Span}\left\{ {{f}_{\text{top }} : f \in \mathcal{J}}\right\} \) . Then \( \mathcal{J} \) is a homogeneous ideal in \( \mathcal{P} \) and \( \operatorname{Gr}\left( {\mathcal{P}/\mathcal{I}}\right) \) is isomorphi... | Proof. If \( u \in \mathcal{P} \) then \( u = \sum {u}_{i} \) with \( {u}_{i} \) homogeneous of degree \( i \) . Thus to show that \( {u\varphi } \in \mathcal{J} \) for all \( \varphi \in \mathcal{J} \), we may assume that \( u \) and \( \varphi \) are homogeneous. In this case take any \( f \in \mathcal{I} \) such tha... | Yes |
Lemma 12.4.9. Assume that \( \mathcal{A} \subset \mathbb{C}\left\lbrack {{y}_{1},\ldots ,{y}_{n}}\right\rbrack \) is a finite-dimensional graded subspace such that the map \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \otimes \mathcal{A} \otimes \mathbb{C}\left\lbrack {{u}_{1},\ldots ,{u}_{n}}\right\... | Proof. We first note that \( \mathcal{J} \supset \mathop{\sum }\limits_{i}\mathcal{P}{u}_{i} \) . Indeed, \( {\left( {u}_{i} - {c}_{i}\right) }_{\text{top }} = {u}_{i} \) . Hence \( \mathcal{J} \supset {\mathcal{J}}_{0} \) . Let\n\n\[{T}_{c} : \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \otimes \mathc... | Yes |
Lemma 12.4.10 (Nakayama). Let \( \mathcal{R} \) be a commutative ring with unit 1. Let \( \mathcal{S} \) be a subring of \( \mathcal{R} \) with unit 1 and let \( \mathcal{J} \subset \mathcal{S} \) be a proper ideal of \( \mathcal{S} \) . If \( \mathcal{R} \) is finitely generated as an \( \mathcal{S} \) -module, then \... | Proof. By the finite generation hypothesis, there exist elements \( {m}_{1},\ldots ,{m}_{d} \) in \( \mathcal{R} \) such that \( \mathcal{R} = \mathop{\sum }\limits_{i}\mathcal{S}{m}_{i} \) . We assume that \( \mathcal{R} = \mathcal{R} \) and derive a contradiction. Under this assumption there exist \( {a}_{ij} \in \ma... | Yes |
Proposition 12.4.12. Let \( h \in {\mathfrak{a}}_{0} \) satisfy (12.68). Then \( {\mathcal{J}}_{h} \) is a prime ideal and \( \mathcal{V}\left( {\mathcal{J}}_{h}\right) = \operatorname{Ad}\left( K\right) h = \operatorname{Ad}\left( {K}^{ \circ }\right) h \), where \( {K}^{ \circ } \) is the identity component of \( K \... | We first show that \( {\mathcal{J}}_{h} \) is a radical ideal. To prove this, we claim that it suffices to show the following:\n\n\( \left( \star \right) \)\n\n\[ \text{If}f \in \mathcal{P}\left( V\right) /{\mathcal{I}}_{h}\text{and}{f}^{2} = 0\text{, then}f = 0\text{.} \]\n\nIndeed, assume that \( \left( \star \right)... | No |
Let’s find the prime ideals of \( \overline{\mathbb{C}}\left\lbrack x\right\rbrack \) . | As \( \mathbb{C}\left\lbrack x\right\rbrack \) is an integral domain, \( 0 \) is prime. Also, \( \left( {x - a}\right) \) is prime, for any \( a \in \mathbb{C} \) : it is even a maximal ideal, as the quotient by this ideal is a field:\n\n\[ 0 \rightarrow \left( {x - a}\right) \rightarrow \mathbb{C}\left\lbrack x\right\... | Yes |
Example 2 (the affine line over \( k = \bar{k} \) ): \( {\mathbb{A}}_{k}^{1} \mathrel{\text{:=}} \operatorname{Spec}k\left\lbrack x\right\rbrack \) where \( k \) is an algebraically closed field. This is called the affine line over \( k \). | All of our discussion in the previous example carries over without change. We will use the same picture, which is after all intended to just be a metaphor. | No |
Show that for the last type of prime, of the form \( \left( {{x}^{2} + {ax} + b}\right) \), the quotient is always isomorphic to \( \mathbb{C} \). | So we have the points that we would normally expect to see on the real line, corresponding to real numbers; the generic point 0 ; and new points which we may interpret as conjugate pairs of complex numbers (the roots of the quadratic). This last type of point should be seen as more akin to the real numbers than to the ... | No |
Example 6 (the affine line over \( {\mathbb{F}}_{\mathrm{p}} \) ): \( {\mathbb{A}}_{{\mathbb{F}}_{\mathrm{p}}}^{1} = \operatorname{Spec}{\mathbb{F}}_{\mathrm{p}}\left\lbrack x\right\rbrack \) . | As in the previous examples, \( {\mathbb{F}}_{\mathrm{p}}\left\lbrack \mathrm{x}\right\rbrack \) is a Euclidean domain, so the prime ideals are of the form (0) or \( \left( {f\left( x\right) }\right) \) where \( f\left( x\right) \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) is an irreducible polynomial, which can ... | Yes |
What are the prime ideals of \( \mathbb{C}\left\lbrack {x, y, z}\right\rbrack \) ? | Analogously to before, \( \left( {x - a, y - b, z - c}\right) \) is a prime ideal. This is a maximal ideal, because its residue ring is a field \( \left( \mathbb{C}\right) \) ; we think of these as \ | No |
Consider \( \pi : \operatorname{Spec}k\left\lbrack \epsilon \right\rbrack /\left( {\epsilon }^{2}\right) \rightarrow \operatorname{Spec}k\left\lbrack x\right\rbrack = {\mathbb{A}}_{k}^{1} \) given by \( x \mapsto \epsilon \) . | Then the scheme-theoretic image of \( \pi \) is given by \( \operatorname{Spec}k\left\lbrack x\right\rbrack /\left( {x}^{2}\right) \) (the polynomials pulling back to 0 are precisely multiples of \( {x}^{2} \) ). Thus the image of the fuzzy point still has some fuzz. | Yes |
Consider \( \pi : \operatorname{Spec}k\left\lbrack \epsilon \right\rbrack /\left( {\epsilon }^{2}\right) \rightarrow \operatorname{Spec}k\left\lbrack x\right\rbrack = {\mathbb{A}}_{k}^{1} \) given by \( x \mapsto 0 \) . | Then the scheme-theoretic image is given by \( \operatorname{Spec}k\left\lbrack x\right\rbrack /\left( x\right) \) : the image is reduced. In this picture, the fuzz is \ | No |
Consider \( \pi : \operatorname{Spec}k\left\lbrack {t,{t}^{-1}}\right\rbrack = {\mathbb{A}}^{1} - \{ 0\} \rightarrow {\mathbb{A}}^{1} = \operatorname{Spec}k\left\lbrack u\right\rbrack \) given by \( u \mapsto t \) . Any function \( g\left( u\right) \) which pulls back to 0 as a function of \( t \) must be the zero-func... | The set-theoretic image, on the other hand, is the distinguished open set \( {\mathbb{A}}^{1} - \{ 0\} \) . Thus in not-too-pathological cases, the underlying set of the scheme-theoretic image is not the set-theoretic image. But the situation isn't terrible: the underlying set of the scheme-theoretic image must be clos... | No |
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