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Proposition 3.30. Suppose that \( G, H \), and \( K \) are matrix Lie groups and \( \Phi : H \rightarrow K \) and \( \Psi : G \rightarrow H \) are Lie group homomorphisms. Let \( \Lambda : G \rightarrow K \) be the composition of \( \Phi \) and \( \Psi \) and let \( \phi ,\psi \), and \( \lambda \) be the Lie algebra m... | Proof. For any \( X \in \mathfrak{g} \) ,\n\n\[ \Lambda \left( {e}^{tX}\right) = \Phi \left( {\Psi \left( {e}^{tX}\right) }\right) = \Phi \left( {e}^{{t\psi }\left( X\right) }\right) = {e}^{{t\phi }\left( {\psi \left( X\right) }\right) }.\]\n\nThus, \( \lambda \left( X\right) = \phi \left( {\psi \left( X\right) }\right... | Yes |
Proposition 3.31. If \( \Phi : G \rightarrow H \) is a Lie group homomorphism and \( \phi : \mathfrak{g} \rightarrow \mathfrak{h} \) is the associated Lie algebra homomorphism, then the kernel of \( \Phi \) is a closed, normal subgroup of \( G \) and the Lie algebra of the kernel is given by \[ \operatorname{Lie}\left(... | Proof. The usual algebraic argument shows that \( \ker \left( \Phi \right) \) is normal subgroup of \( G \) . Since, also, \( \Phi \) is continuous, \( \ker \left( \Phi \right) \) is closed. If \( X \in \ker \left( \phi \right) \), then \[ \Phi \left( {e}^{tX}\right) = {e}^{{t\phi }\left( X\right) } = I \] for all \( t... | Yes |
Proposition 3.33. Let \( G \) be a matrix Lie group, with Lie algebra \( \mathfrak{g} \). Let \( \mathrm{{GL}}\left( \mathfrak{g}\right) \) denote the group of all invertible linear transformations of \( \mathfrak{g} \). Then the map \( A \rightarrow \) \( {\operatorname{Ad}}_{A} \) is a homomorphism of \( G \) into \(... | Proof. Easy. Note that Point 1 of Theorem 3.20 guarantees that \( {\operatorname{Ad}}_{A}\left( X\right) \) is actually in \( \mathfrak{g} \) for all \( X \in \mathfrak{g} \). | No |
Proposition 3.34. Let \( G \) be a matrix Lie group, let \( \mathfrak{g} \) be its Lie algebra, and let \( \mathrm{{Ad}} : G \rightarrow \mathrm{{GL}}\left( \mathfrak{g}\right) \) be as in Proposition 3.33. Let \( \mathrm{{ad}} : \mathfrak{g} \rightarrow \mathfrak{{gl}}\left( \mathfrak{g}\right) \) be the associated Li... | Proof. By Point 3 of Theorem 3.28, ad can be computed as follows:\n\n\[ \n{\left. {\operatorname{ad}}_{X} = \frac{d}{dt}{\operatorname{Ad}}_{{e}^{tX}}\right| }_{t = 0}. \n\]\n\nThus,\n\n\[ \n{\operatorname{ad}}_{X}\left( Y\right) = {\left. \frac{d}{dt}{e}^{tX}Y{e}^{-{tX}}\right| }_{t = 0} = \left\lbrack {X, Y}\right\rb... | Yes |
For any \( X \) in \( {M}_{n}\left( \mathbb{C}\right) \), let \( {\operatorname{ad}}_{X} : {M}_{n}\left( \mathbb{C}\right) \rightarrow {M}_{n}\left( \mathbb{C}\right) \) be given by \( {\operatorname{ad}}_{X}Y = \left\lbrack {X, Y}\right\rbrack \). Then for any \( Y \) in \( {M}_{n}\left( \mathbb{C}\right) \), we have\... | This result can also be proved by direct calculation-see Exercise 14. | No |
Proposition 3.37. Let \( \mathfrak{g} \) be a finite-dimensional real Lie algebra and \( {\mathfrak{g}}_{\mathbb{C}} \) its complexification. Then the bracket operation on \( \mathfrak{g} \) has a unique extension to \( {\mathfrak{g}}_{\mathbb{C}} \) that makes \( {\mathfrak{g}}_{\mathbb{C}} \) into a complex Lie algeb... | Proof. The uniqueness of the extension is obvious, since if the bracket operation on \( {\mathfrak{g}}_{\mathbb{C}} \) is to be bilinear, then it must be given by\n\n\[ \left\lbrack {{X}_{1} + i{X}_{2},{Y}_{1} + i{Y}_{2}}\right\rbrack = \left( {\left\lbrack {{X}_{1},{Y}_{1}}\right\rbrack - \left\lbrack {{X}_{2},{Y}_{2}... | Yes |
Proposition 3.39. Let \( \mathfrak{g} \) be a real Lie algebra, \( {\mathfrak{g}}_{\mathbb{C}} \) its complexification, and \( \mathfrak{h} \) an arbitrary complex Lie algebra. Then every real Lie algebra homomorphism of \( \mathfrak{g} \) into \( \mathfrak{h} \) extends uniquely to a complex Lie algebra homomorphism o... | Proof. The unique extension is given by \( \pi \left( {X + {iY}}\right) = \pi \left( X\right) + {i\pi }\left( Y\right) \) for all \( X, Y \in \) \( \mathfrak{g} \) . It is easy to check that this map is, indeed, a homomorphism of complex Lie algebras. | Yes |
There does not exist a matrix \( X \in \operatorname{sl}\left( {2;\mathbb{C}}\right) \) with\n\n\[ \n{e}^{X} = \left( \begin{array}{rr} - 1 & 1 \\ 0 & - 1 \end{array}\right)\n\]\n\neven though the matrix on the right-hand side of (3.18) is in \( \mathrm{{SL}}\left( {2;\mathbb{C}}\right) \) . | Proof. If \( X \in \operatorname{sl}\left( {2;\mathbb{C}}\right) \) has distinct eigenvalues, then \( X \) is diagonalizable and \( {e}^{X} \) will also be diagonalizable, unlike the matrix on the right-hand side of (3.18). If \( X \in \operatorname{sl}\left( {2;\mathbb{C}}\right) \) has a repeated eigenvalue, this eig... | Yes |
Theorem 3.42. For \( 0 < \varepsilon < \log 2 \), let \( {U}_{\varepsilon } = \left\{ {X \in {M}_{n}\left( \mathbb{C}\right) \mid \parallel X\parallel < \varepsilon }\right\} \) and let \( {V}_{\varepsilon } = \exp \left( {U}_{\varepsilon }\right) \) . Suppose \( G \subset \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) i... | The condition \( \varepsilon < \log 2 \) guarantees (Theorem 2.8) that for all \( X \in {U}_{\varepsilon },\log \left( {e}^{X}\right) \) is defined and equal to \( X \) . Note that if \( X = \log A \) is in \( \mathfrak{g} \), then \( A = {e}^{X} \) is in \( G \) . Thus, the content of the theorem is that for some \( \... | Yes |
Lemma 3.43. Suppose \( {B}_{m} \) are elements of \( G \) and that \( {B}_{m} \rightarrow I \) . Let \( {Y}_{m} = \log {B}_{m} \) , which is defined for all sufficiently large \( m \) . Suppose that \( {Y}_{m} \) is nonzero for all \( m \) and that \( {Y}_{m}/\begin{Vmatrix}{Y}_{m}\end{Vmatrix} \rightarrow Y \in {M}_{n... | Proof. For any \( t \in \mathbb{R} \), we have \( \left( {t/\begin{Vmatrix}{Y}_{m}\end{Vmatrix}}\right) {Y}_{m} \rightarrow {tY} \) . Note that since \( {B}_{m} \rightarrow I \), we have \( \begin{Vmatrix}{Y}_{m}\end{Vmatrix} \rightarrow 0 \) . Thus, we can find integers \( {k}_{m} \) such that \( {k}_{m}\begin{Vmatrix... | Yes |
Corollary 3.44. If \( G \) is a matrix Lie group with Lie algebra \( \mathfrak{g} \), there exists a neighborhood \( U \) of \( 0 \) in \( \mathfrak{g} \) and a neighborhood \( V \) of \( I \) in \( G \) such that the exponential map takes \( U \) homeomorphically onto \( V \) . | Proof. Let \( \varepsilon \) be such that Theorem 3.42 holds and set \( U = {U}_{\varepsilon } \cap \mathfrak{g} \) and \( V = \) \( {V}_{\varepsilon } \cap G \) . The theorem implies that exp takes \( U \) onto \( V \) . Furthermore, exp is a homeomorphism of \( U \) onto \( V \), since there is a continuous inverse m... | Yes |
Corollary 3.45. Let \( G \) be a matrix Lie group with Lie algebra \( \mathfrak{g} \) and let \( k \) be the dimension of \( \mathfrak{g} \) as a real vector space. Then \( G \) is a smooth embedded submanifold of \( {M}_{n}\left( \mathbb{C}\right) \) of dimension \( k \) and hence a Lie group. | Proof. Let \( \varepsilon \in \left( {0,\log 2}\right) \) be such that Theorem 3.42 holds. Then for any \( {A}_{0} \in G \) , consider the neighborhood \( {A}_{0}{V}_{\varepsilon } \) of \( {A}_{0} \) in \( {M}_{n}\left( \mathbb{C}\right) \) . Note that \( A \in {A}_{0}{V}_{\varepsilon } \) if and only if \( {A}_{0}^{-... | Yes |
Corollary 3.46. Suppose \( G \subset \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) is a matrix Lie group with Lie algebra \( \mathfrak{g} \) . Then a matrix \( X \) is in \( \mathfrak{g} \) if and only if there exists a smooth curve \( \gamma \) in \( {M}_{n}\left( \mathbb{C}\right) \) with \( \gamma \left( t\right) \in... | Proof. If \( X \) is in \( \mathfrak{g} \), then we may take \( \gamma \left( t\right) = {e}^{tX} \) and then \( \gamma \left( 0\right) = I \) and \( {d\gamma }/{\left. dt\right| }_{t = 0} = X \) . In the other direction, suppose that \( \gamma \left( t\right) \) is a smooth curve in \( G \) with \( \gamma \left( 0\rig... | Yes |
Lemma 3.48. Suppose \( A : \left\lbrack {a, b}\right\rbrack \rightarrow \mathrm{{Gl}}\left( {n;\mathbb{C}}\right) \) is a continuous map. Then for all \( \varepsilon > 0 \) there exists \( \delta > 0 \) such that if \( s, t \in \left\lbrack {a, b}\right\rbrack \) satisfy \( \left| {s - t}\right| < \delta \), then\n\n\[... | Proof. We note that\n\n\[ \begin{Vmatrix}{A\left( s\right) A{\left( t\right) }^{-1} - I}\end{Vmatrix} = \begin{Vmatrix}{\left( {A\left( s\right) - A\left( t\right) }\right) A{\left( t\right) }^{-1}}\end{Vmatrix} \]\n\n\[ \leq \parallel A\left( s\right) - A\left( t\right) \parallel \begin{Vmatrix}{A{\left( t\right) }^{-... | Yes |
Corollary 3.49. Let \( G \) and \( H \) be matrix Lie groups with Lie algebras \( \mathfrak{g} \) and \( \mathfrak{h} \), respectively, and assume \( G \) is connected. Suppose \( {\Phi }_{1} \) and \( {\Phi }_{2} \) are Lie group homomorphisms of \( G \) into \( H \) and that \( {\phi }_{1} \) and \( {\phi }_{2} \) be... | Proof. Let \( g \) be any element of \( G \) . Since \( G \) is connected, Corollary 3.47 tells us that every element of \( G \) can be written as \( {e}^{{X}_{1}}{e}^{{X}_{2}}\cdots {e}^{{X}_{m}} \), with \( {X}_{j} \in \mathfrak{g} \) . Thus,\n\n\[ \n{\Phi }_{1}\left( {{e}^{{X}_{1}}\cdots {e}^{{X}_{m}}}\right) = {e}^... | Yes |
Corollary 3.50. Every continuous homomorphism between two matrix Lie groups is smooth. | Proof. For all \( g \in G \), we write nearby elements \( h \in G \) as \( h = g{e}^{X} \), with \( X \in \mathfrak{g} \), so that\n\n\[ \Phi \left( h\right) = \Phi \left( g\right) \Phi \left( {e}^{X}\right) = \Phi \left( g\right) {e}^{\phi \left( X\right) }.\]\n\nThis relation says that in the exponential coordinates ... | Yes |
Corollary 3.51. If \( G \) is a connected matrix Lie group and the Lie algebra \( \mathfrak{g} \) of \( G \) is commutative, then \( G \) is commutative. | Proof. Since \( \mathfrak{g} \) is commutative, any two elements of \( G \), when written as in Corollary 3.47, will commute. | No |
Corollary 3.52. If \( G \subset {M}_{n}\left( \mathbb{C}\right) \) is a matrix Lie group, the identity component \( {G}_{0} \) of \( G \) is a closed subgroup of \( \mathrm{{GL}}\left( {N;\mathbb{C}}\right) \) and thus a matrix Lie group. Furthermore, the Lie algebra of \( {G}_{0} \) is the same as the Lie algebra of \... | Proof. Suppose that \( \left\langle {A}_{m}\right\rangle \) is a sequence in \( {G}_{0} \) converging to some \( A \in \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) . Then certainly \( A \in G \), since \( G \) is closed. Furthermore, \( {A}_{m}{A}^{-1} \) lies in \( G \) for all \( m \) and \( {A}_{m}{A}^{-1} \rightarr... | Yes |
Proposition 4.6. Let \( \mathfrak{g} \) be a real Lie algebra and \( {\mathfrak{g}}_{\mathbb{C}} \) its complexification. Then every finite-dimensional complex representation \( \pi \) of \( \mathfrak{g} \) has a unique extension to a complex-linear representation of \( {\mathfrak{g}}_{\mathbb{C}} \), also denoted \( \... | Of course, the extension of \( \pi \) to \( {\mathfrak{g}}_{\mathbb{C}} \) is given by \( \pi \left( {X + {iY}}\right) = \pi \left( X\right) + {i\pi }\left( Y\right) \) for all \( X, Y \in \mathfrak{g} \) . Proof. The existence and uniqueness of the extension follow from Proposition 3.39. The claim about irreducibility... | Yes |
Proposition 4.8. Suppose \( G \) is a matrix Lie group with Lie algebra \( \mathfrak{g} \) . Suppose \( V \) is a finite-dimensional inner product space, \( \Pi \) is a representation of \( G \) acting on \( V \) , and \( \pi \) is the associated representation of \( \mathfrak{g} \) . If \( \Pi \) is unitary, then \( \... | Proof. The proof is similar to the computation of the Lie algebra of the unitary group \( \mathrm{U}\left( n\right) \) . If \( \Pi \) is unitary, then for all \( X \in \mathfrak{g} \) we have\n\n\[ \n{\left( {e}^{{t\pi }\left( X\right) }\right) }^{ * } = \Pi {\left( {e}^{tX}\right) }^{ * } = \Pi {\left( {e}^{tX}\right)... | Yes |
Example 4.10. Let \( {V}_{m} \) denote the space of homogeneous polynomials of degree \( m \) in two complex variables. For each \( U \in \mathrm{{SU}}\left( 2\right) \), define a linear transformation \( {\Pi }_{m}\left( U\right) \) on the space \( {V}_{m} \) by the formula \[ \left\lbrack {{\Pi }_{m}\left( U\right) f... | To see that \( {\Pi }_{m} \) is actually a representation, compute that \[ {\Pi }_{m}\left( {U}_{1}\right) \left\lbrack {{\Pi }_{m}\left( {U}_{2}\right) f}\right\rbrack \left( z\right) = \left\lbrack {{\Pi }_{m}\left( {U}_{2}\right) f}\right\rbrack \left( {{U}_{1}^{-1}z}\right) = f\left( {{U}_{2}^{-1}{U}_{1}^{-1}z}\rig... | Yes |
Proposition 4.11. For each \( m \geq 0 \), the representation \( {\pi }_{m} \) is irreducible. | Proof. It suffices to show that every nonzero invariant subspace of \( {V}_{m} \) is equal to \( {V}_{m} \) . So, let \( W \) be such a space and let \( w \) be a nonzero element of \( W \) . Then \( w \) can be written in the form\n\n\[ w = {a}_{0}{z}_{1}^{m} + {a}_{1}{z}_{1}^{m - 1}{z}_{2} + {a}_{2}{z}_{1}^{m - 2}{z}... | Yes |
Theorem 4.14. If \( U \) and \( V \) are any finite-dimensional real or complex vector spaces, then a tensor product \( \left( {W,\phi }\right) \) exists. Furthermore, \( \left( {W,\phi }\right) \) is unique up to canonical isomorphism. That is, if \( \left( {{W}_{1},{\phi }_{1}}\right) \) and \( \left( {{W}_{2},{\phi ... | Proof. Exercise 7. | No |
Proposition 4.16. Let \( U \) and \( V \) be finite-dimensional real or complex vector spaces. Let \( A : U \rightarrow U \) and \( B : V \rightarrow V \) be linear operators. Then there exists a unique linear operator from \( U \otimes V \) to \( U \otimes V \), denoted \( A \otimes B \), such that\n\n\[ \left( {A \ot... | Proof. Define a map \( \psi \) from \( U \times V \) into \( U \otimes V \) by\n\n\[ \psi \left( {u, v}\right) = \left( {Au}\right) \otimes \left( {Bv}\right) . \]\n\nSince \( A \) and \( B \) are linear and since \( \otimes \) is bilinear, \( \psi \) is a bilinear map of \( U \times V \) into \( U \otimes V \) . By th... | Yes |
Proposition 4.18. Let \( G \) and \( H \) be matrix Lie groups with Lie algebras \( \mathfrak{g} \) and \( \mathfrak{h} \), respectively. Let \( {\Pi }_{1} \) and \( {\Pi }_{2} \) be representations of \( G \) and \( H \), respectively, and consider the representation \( {\Pi }_{1} \otimes {\Pi }_{2} \) of \( G \times ... | Proof. Suppose that \( u\left( t\right) \) is a smooth curve in \( U \) and \( v\left( t\right) \) is a smooth curve in \( V \) . Then, by repeating the proof of the product rule for scalar-valued functions (or by calculating everything in a basis), we have\n\n\[ \frac{d}{dt}\left( {u\left( t\right) \otimes v\left( t\r... | Yes |
Proposition 4.22. If \( \Pi \) is a representation of a matrix Lie group \( G \), then (1) \( {\Pi }^{ * } \) is irreducible if and only if \( \Pi \) is irreducible and (2) \( {\left( {\Pi }^{ * }\right) }^{ * } \) is isomorphic to \( \Pi \) . Similar statements apply to Lie algebra representations. | Proof. See Exercise 6. | No |
Example 4.25. Let \( \Pi : \mathbb{R} \rightarrow \mathrm{{GL}}\left( {2;\mathbb{C}}\right) \) be given by\n\n\[ \Pi \left( x\right) = \left( \begin{array}{ll} 1 & x \\ 0 & 1 \end{array}\right) \]\n\nThen \( \Pi \) is not completely reducible. | Proof. Direct calculation shows that \( \Pi \) is, in fact, a representation of \( \mathbb{R} \) . If \( \left\{ {{e}_{1},{e}_{2}}\right\} \) is the standard basis for \( {\mathbb{C}}^{2} \), then clearly the span of \( {e}_{1} \) is an invariant subspace. We now claim that \( \left\langle {e}_{1}\right\rangle \) is th... | Yes |
Proposition 4.26. If \( V \) is a completely reducible representation of a group or Lie algebra, then the following properties hold.\n\n1. For every invariant subspace \( U \) of \( V \), there is another invariant subspace \( W \) such that \( V \) is the direct sum of \( U \) and \( W \) .\n\n2. Every invariant subsp... | Proof. For Point 1, suppose that \( V \) decomposes as\n\n\[ V = {U}_{1} \oplus {U}_{2} \oplus \cdots \oplus {U}_{k} \]\n\nwhere the \( {U}_{j} \) ’s are irreducible invariant subspaces, and that \( U \) is any invariant subspace of \( V \) . If \( U \) is all of \( V \), then we can take \( W = \{ 0\} \) and we are do... | Yes |
Theorem 4.28. If \( G \) is a compact matrix Lie group, every finite-dimensional representation of \( G \) is completely reducible. | Proof of Theorem 4.28. Choose an arbitrary inner product \( \langle \cdot , \cdot \rangle \) on \( V \), and then define a map \( \langle \cdot , \cdot {\rangle }_{G} : V \times V \rightarrow \mathbb{C} \) by the formula\n\n\[ \langle v, w{\rangle }_{G} = {\int }_{G}\langle \Pi \left( A\right) v,\Pi \left( A\right) w\r... | Yes |
Corollary 4.30. Let \( \Pi \) be an irreducible complex representation of a matrix Lie group \( G \) . If \( A \) is in the center of \( G \), then \( \Pi \left( A\right) = {\lambda I} \), for some \( \lambda \in \mathbb{C} \) . Similarly, if \( \pi \) is an irreducible complex representation of a Lie algebra \( \mathf... | Proof. We prove the group case; the proof of the Lie algebra case is similar. If \( A \) is in the center of \( G \), then for all \( B \in G \) ,\n\n\[ \Pi \left( A\right) \Pi \left( B\right) = \Pi \left( {AB}\right) = \Pi \left( {BA}\right) = \Pi \left( B\right) \Pi \left( A\right) . \]\n\nHowever, this says exactly ... | Yes |
Corollary 4.31. An irreducible complex representation of a commutative group or Lie algebra is one dimensional. | Proof. Again, we prove only the group case. If \( G \) is commutative, the center of \( G \) is all of \( G \), so by the previous corollary \( \Pi \left( A\right) \) is a multiple of the identity for each \( A \in G \) . However, this means that every subspace of \( V \) is invariant! Thus, the only way that \( V \) c... | Yes |
Lemma 4.33. Let \( u \) be an eigenvector of \( \pi \left( H\right) \) with eigenvalue \( \alpha \in \mathbb{C} \). Then we have\n\n\[ \pi \left( H\right) \pi \left( X\right) u = \left( {\alpha + 2}\right) \pi \left( X\right) u. \]\n\nThus, either \( \pi \left( X\right) u = 0 \) or \( \pi \left( X\right) u \) is an eig... | Proof. We know that \( \left\lbrack {\pi \left( H\right) ,\pi \left( X\right) }\right\rbrack = \pi \left( \left\lbrack {H, X}\right\rbrack \right) = {2\pi }\left( X\right) \). Thus,\n\n\[ \pi \left( H\right) \pi \left( X\right) u = \pi \left( X\right) \pi \left( H\right) u + {2\pi }\left( X\right) u \]\n\n\[ = \pi \lef... | Yes |
Theorem 4.34. If \( \left( {\pi, V}\right) \) is a finite-dimensional representation of \( \operatorname{sl}\left( {2;\mathbb{C}}\right) \), not necessarily irreducible, the following results hold.\n\n1. Every eigenvalue of \( \pi \left( H\right) \) is an integer. Furthermore, if \( v \) is an eigenvector for \( \pi \l... | Proof. For Point 1, suppose \( v \) is an eigenvector of \( \pi \left( H\right) \) with eigenvalue \( \lambda \) . Then there is some \( N \geq 0 \) such that \( \pi {\left( X\right) }^{N}v \neq 0 \) but \( \pi {\left( X\right) }^{N + 1}v = 0 \), where \( \pi {\left( X\right) }^{N}v \) is an eigenvector of \( \pi \left... | Yes |
Proposition 4.35. Let \( {\sigma }_{m} = {\pi }_{m} \circ {\phi }^{-1} \) be an irreducible complex representation of the Lie algebra \( \operatorname{so}\left( 3\right) \left( {m \geq 0}\right) \) . If \( m \) is even, there is a representation \( {\sum }_{m} \) of the group \( \mathrm{{SO}}\left( 3\right) \) such tha... | Proof. Suppose, first, that \( m \) is odd and suppose that such a \( {\sum }_{m} \) existed. Computing as in Sect. 2.2, we see that\n\n\[ \n{e}^{{2\pi }{F}_{1}} = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & \cos {2\pi } & - \sin {2\pi } \\ 0 & \sin {2\pi } & \cos {2\pi } \end{matrix}\right) = I. \n\]\n\nMeanwhile, \( {\sig... | No |
Lemma 4.38. If \( X \) is a nilpotent matrix and \( {e}^{tX} = I \) for some nonzero \( t \), then \( X = 0 \) . | Proof. Since \( X \) is nilpotent, the power series for \( {e}^{tX} \) terminates after a finite number of terms. Thus, each entry of \( {e}^{tX} \) depends polynomially on \( t \) ; that is, there exist polynomials \( {p}_{jk}\left( t\right) \) such that \( {\left( {e}^{tX}\right) }_{jk} = {p}_{jk}\left( t\right) \) .... | Yes |
Theorem 5.1. Suppose \( X \) and \( Y \) are \( n \times n \) complex matrices, and that \( X \) and \( Y \) commute with their commutator:\n\n\[ \left\lbrack {X,\left\lbrack {X, Y}\right\rbrack }\right\rbrack = \left\lbrack {Y,\left\lbrack {X, Y}\right\rbrack }\right\rbrack = 0. \]\n\nThen we have\n\n\[ {e}^{X}{e}^{Y}... | Proof. Consider \( X \) and \( Y \) in \( {M}_{n}\left( \mathbb{C}\right) \) satisfying (5.5). We will prove that\n\n\[ {e}^{tX}{e}^{tY} = \exp \left( {{tX} + {tY} + \frac{{t}^{2}}{2}\left\lbrack {X, Y}\right\rbrack }\right) ,\]\n\nwhich reduces to the desired result in the case \( t = 1 \) . Since, by assumption, \( \... | Yes |
Theorem 5.2. Let \( H \) denote the Heisenberg group and \( \mathfrak{h} \) its Lie algebra. Let \( G \) be a matrix Lie group with Lie algebra \( \mathfrak{g} \) and let \( \phi : \mathfrak{h} \rightarrow \mathfrak{g} \) be a Lie algebra homomorphism. Then there exists a unique Lie group homomorphism \( \Phi : H \righ... | Proof. Recall (Exercise 18 in Chapter 3) that the Heisenberg group has the special property that its exponential map is one-to-one and onto. Let \ | No |
Theorem 5.3 (Baker-Campbell-Hausdorff). For all \( n \times n \) complex matrices \( X \) and \( Y \) with \( \parallel X\parallel \) and \( \parallel Y\parallel \) sufficiently small, we have\n\n\[ \log \left( {{e}^{X}{e}^{Y}}\right) = X + {\int }_{0}^{1}g\left( {{e}^{{\mathrm{{ad}}}_{X}}{e}^{t{\mathrm{{ad}}}_{Y}}}\ri... | The proof of this theorem is given in Sect. 5.5 of this chapter. Note that \( {e}^{{\mathrm{{ad}}}_{X}}{e}^{t{\mathrm{{ad}}}_{Y}} \) and, hence, also \( g\left( {{e}^{{\operatorname{ad}}_{X}}{e}^{t{\operatorname{ad}}_{Y}}}\right) \) are linear operators on the space \( {M}_{n}\left( \mathbb{C}\right) \) of all \( n \ti... | No |
Theorem 5.4 (Derivative of Exponential). For all \( X, Y \in {M}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \n{\left. \frac{d}{dt}{e}^{X + {tY}}\right| }_{t = 0} = {e}^{X}\left\{ {\frac{I - {e}^{-{\mathrm{{ad}}}_{X}}}{{\mathrm{{ad}}}_{X}}\left( Y\right) }\right\} \n\]\n\n\[ \n= {e}^{X}\left\{ {Y - \frac{\left\lbrack... | Our proof follows [Tuy]. | No |
Lemma 5.5. If \( Z \) is a linear operator on a finite-dimensional vector space, then\n\n\[ \mathop{\lim }\limits_{{m \rightarrow \infty }}\frac{1}{m}\mathop{\sum }\limits_{{k = 0}}^{{m - 1}}{\left( {e}^{-Z/m}\right) }^{k} = \frac{1 - {e}^{-Z}}{Z}. \] | Proof. If we formally applied the formula for the sum of a finite geometric series to \( {e}^{-Z/m} \), we would get\n\n\[ \frac{1}{m}\mathop{\sum }\limits_{{k = 0}}^{{m - 1}}{\left( {e}^{-Z/m}\right) }^{k} = \frac{1}{m}\frac{1 - {e}^{-Z}}{1 - {e}^{-Z/m}}\underbrace{m \rightarrow \infty }\frac{1 - {e}^{-Z}}{Z}. \]\n\nT... | Yes |
Corollary 5.7. Suppose \( G \) and \( H \) are simply connected matrix Lie groups with Lie algebras \( \mathfrak{g} \) and \( \mathfrak{h} \), respectively. If \( \mathfrak{g} \) is isomorphic to \( \mathfrak{h} \), then \( G \) is isomorphic to \( H \) . | Proof. Let \( \phi : \mathfrak{g} \rightarrow \mathfrak{h} \) be a Lie algebra isomorphism. By Theorem 5.6, there exists an associated Lie group homomorphism \( \Phi : G \rightarrow H \) . Since \( \psi \mathrel{\text{:=}} {\phi }^{-1} \) is also a Lie algebra homomorphism, there is a corresponding Lie group homomorphi... | Yes |
Proposition 5.9. Let \( G \) and \( H \) be matrix Lie groups with Lie algebras \( \mathfrak{g} \) and \( \mathfrak{h} \) , respectively, and let \( \phi : \mathfrak{g} \rightarrow \mathfrak{h} \) be a Lie algebra homomorphism. Define \( {U}_{\varepsilon } \subset G \) by\n\n\[ \n{U}_{\varepsilon } = \{ A \in G \mid \p... | Proof. Choose \( \varepsilon \) small enough that Theorem 3.42 applies and small enough that for all \( A, B \in {U}_{\varepsilon } \), the BCH formula applies to \( X \mathrel{\text{:=}} \log A \) and \( Y \mathrel{\text{:=}} \log B \) and also to \( \phi \left( X\right) \) and \( \phi \left( Y\right) \) . Then if \( ... | Yes |
Theorem 5.10. Let \( G \) and \( H \) be matrix Lie groups, with \( G \) simply connected. If \( \left( {U, f}\right) \) is a local homomorphism of \( G \) into \( H \), there exists a unique (global) Lie group homomorphism \( \Phi : G \rightarrow H \) such that \( \Phi \) agrees with \( f \) on \( U \) . | Proof. Step 1: Define \( \Phi \) along a path. Since \( G \) is simply connected and thus connected, for any \( A \in G \), there exists a path \( A\left( t\right) \in G \) with \( A\left( 0\right) = I \) and \( A\left( 1\right) = A \) . Let us call a partition \( 0 = {t}_{0} < {t}_{1} < {t}_{2}\cdots < {t}_{m} = 1 \) ... | No |
Theorem 5.11. Suppose that \( G \) is a simply connected matrix Lie group and that the Lie algebra \( \mathfrak{g} \) of \( G \) decomposes as a Lie algebra direct sum \( \mathfrak{g} = {\mathfrak{h}}_{1} \oplus {\mathfrak{h}}_{2} \), for two subalgebras \( {\mathfrak{h}}_{1} \) and \( {\mathfrak{h}}_{2} \) of \( \math... | Proof. Consider the Lie algebra homomorphism \( \phi : \mathfrak{g} \rightarrow \mathfrak{g} \) that sends \( X + Y \) to \( X \), where \( X \in {\mathfrak{h}}_{1} \) and \( Y \in {\mathfrak{h}}_{2} \) . Since \( G \) is simply connected, there is a Lie group homomorphism \( \Phi : G \rightarrow G \) associated to \( ... | Yes |
Proposition 5.13. If \( G \) is a connected matrix Lie group and \( \left( {{H}_{1},{\Phi }_{1}}\right) \) and \( \left( {{H}_{2},{\Phi }_{2}}\right) \) are universal covers of \( G \), then there exists a Lie group isomorphism \( \Psi : {H}_{1} \rightarrow {H}_{2} \) such that \( {\Phi }_{2} \circ \Psi = {\Phi }_{1} \... | Proof. See Exercise 9. | No |
The universal cover of \( \mathrm{{SO}}\left( 3\right) \) is \( \mathrm{{SU}}\left( 2\right) \). | Proof. The group SU(2) is simply connected by Proposition 1.15. Proposition 1.19 and Example 3.29 then provide the desired covering map. | Yes |
Lemma 5.17. Suppose \( \psi : \mathfrak{{sl}}\left( {2;\mathbb{R}}\right) \rightarrow \mathfrak{{gl}}\left( {n;\mathbb{C}}\right) \) is a Lie algebra homomorphism. Then there exists a Lie group homomorphism \( \Phi : \mathrm{{SL}}\left( {2;\mathbb{R}}\right) \rightarrow \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) such... | Proof. Let \( {\psi }_{\mathbb{C}} : \mathrm{{sl}}\left( {2;\mathbb{C}}\right) \rightarrow \mathrm{{gl}}\left( {n;\mathbb{C}}\right) \) be the complex-linear extension of \( \psi \) to \( \mathrm{{sl}}\left( {2;\mathbb{C}}\right) \cong \mathrm{{sl}}{\left( 2;\mathbb{R}\right) }_{\mathbb{C}} \), which is a Lie algebra h... | Yes |
Theorem 5.23. Let \( H \) be a connected Lie subgroup of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) with Lie algebra \( \mathfrak{h} \) . Then \( H \) can be given the structure of a smooth manifold in such a way that the group operations on \( H \) are smooth and the inclusion map of \( H \) into \( \mathrm{{GL}}... | Proof. For any \( A \in H \) and any \( \varepsilon > 0 \), define\n\n\[ \n{U}_{A,\varepsilon } = \left\{ {A{e}^{X} \mid X \in \mathfrak{h}\text{ and }\parallel X\parallel < \varepsilon }\right\} .\n\]\n\nNow define a topology on \( H \) as follows: A set \( U \subset H \) is open if for each \( A \in U \) there exists... | No |
Proposition 5.24. Suppose \( G \subset \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) is a matrix Lie group with Lie algebra \( \mathfrak{g} \) and that \( \mathfrak{h} \) is a maximal commutative subalgebra of \( \mathfrak{g} \), meaning that \( \mathfrak{h} \) is commutative and \( \mathfrak{h} \) is not contained in a... | Proof. Since \( \mathfrak{h} \) is commutative, \( H \) is also commutative, since every element of \( H \) is a product of exponentials of elements of \( \mathfrak{h} \) . It easily follows that the closure \( \bar{H} \) of \( H \) in \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) is also commutative. We now claim th... | Yes |
Theorem 5.25. If \( \mathfrak{g} \) is any finite-dimensional, real Lie algebra, there exists a connected Lie subgroup \( G \) of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) whose Lie algebra is isomorphic to \( \mathfrak{g} \) . | Proof. By Ado’s theorem, we may identify \( \mathfrak{g} \) with a real subalgebra of \( \mathfrak{{gl}}\left( {n;\mathbb{C}}\right) \) . Then by Theorem 5.20, there is a connected Lie subgroup of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) with Lie algebra \( \mathfrak{g} \) . | Yes |
Proposition 6.2. Every representation of \( \mathbf{sl}\left( {3;\mathbb{C}}\right) \) has at least one weight. | Proof. Since we are working over the complex numbers, \( \pi \left( {H}_{1}\right) \) has at least one eigenvalue \( {m}_{1} \in \mathbb{C} \) . Let \( W \subset V \) be the eigenspace for \( \pi \left( {H}_{1}\right) \) with eigenvalue \( {m}_{1} \) . Since \( \left\lbrack {{H}_{1},{H}_{2}}\right\rbrack = 0,\pi \left(... | Yes |
Proposition 6.3. If \( \left( {\pi, V}\right) \) is a representation of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \) and \( \mu = \left( {{m}_{1},{m}_{2}}\right) \) is a weight of \( V \), then both \( {m}_{1} \) and \( {m}_{2} \) are integers. | Proof. Apply Point 1 of Theorem 4.34 to the restriction of \( \pi \) to \( \left\langle {{H}_{1},{X}_{1},{Y}_{1}}\right\rangle \) and to the restriction of \( \pi \) to \( \left\langle {{H}_{2},{X}_{2},{Y}_{2}}\right\rangle \) . | Yes |
Lemma 6.5. Let \( \alpha = \left( {{a}_{1},{a}_{2}}\right) \) be a root and let \( {Z}_{\alpha } \in \mathrm{{sl}}\left( {3;\mathbb{C}}\right) \) be a corresponding root vector. Let \( \pi \) be a representation of \( \operatorname{sl}\left( {3;\mathbb{C}}\right) \), let \( \mu = \left( {{m}_{1},{m}_{2}}\right) \) be a... | Proof. By the definition of a root, we have the commutation relation \( \left\lbrack {{H}_{1},{Z}_{\alpha }}\right\rbrack = \) \( {a}_{1}{Z}_{\alpha } \) . Thus,\n\n\[ \pi \left( {H}_{1}\right) \pi \left( {Z}_{\alpha }\right) v = \left( {\pi \left( {Z}_{\alpha }\right) \pi \left( {H}_{1}\right) + {a}_{1}\pi \left( {Z}_... | Yes |
Theorem 6.8. For every pair \( \left( {{m}_{1},{m}_{2}}\right) \) of non-negative integers, there exists an irreducible representation \( \Pi \) of \( \mathrm{{SU}}\left( 3\right) \) such that the associated representation \( \pi \) of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \) has highest weight \( \left( {{m}_{1... | ## 6.4 Proof of the Theorem\n\nThe proof consists of a series of propositions.\n\nProposition 6.9. In eve | No |
In every irreducible representation \( \left( {\pi, V}\right) \) of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \), the operators \( \pi \left( {H}_{1}\right) \) and \( \pi \left( {H}_{2}\right) \) can be simultaneously diagonalized; that is, \( V \) is the direct sum of its weight spaces. | Let \( W \) be the sum of the weight spaces in \( V \) . Equivalently, \( W \) is the space of all vectors \( w \in V \) such that \( w \) can be written as a linear combination of simultaneous eigenvectors for \( \pi \left( {H}_{1}\right) \) and \( \pi \left( {H}_{2}\right) \) . Since (Proposition 6.2) \( \pi \) alway... | Yes |
Lemma 6.12 (Reordering Lemma). Suppose that \( \mathfrak{g} \) is any Lie algebra and that \( \pi \) is a representation of \( \mathfrak{g} \) . Suppose that \( {X}_{1},\ldots ,{X}_{m} \) is an ordered basis for \( \mathfrak{g} \) as a vector space. Then any expression of the form\n\n\[ \pi \left( {X}_{{j}_{1}}\right) ... | Proof. The idea is to use the commutation relations of \( \mathfrak{g} \) to re-order the factors into the desired order, at the expense of generating terms with one fewer factors, which then be handled by the same method. To be more formal, we use induction on \( N \) . If \( N = 1 \), there is nothing to do: Any expr... | Yes |
Proposition 6.13. Every irreducible representation of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \) is a highest weight cyclic representation, with a unique highest weight \( \mu \) . | Proof. We have already shown that every irreducible representation \( \pi \) is the direct sum of its weight spaces. Since the representation is finite dimensional, there can be only finitely many weights, so there must be a maximal weight \( \mu \), that is, such that there is no weight strictly higher than \( \mu \) ... | Yes |
Proposition 6.14. Suppose \( \left( {\pi, V}\right) \) is a completely reducible representation of \( \mathfrak{{sl}}\left( {3;\mathbb{C}}\right) \) that is also highest weight cyclic. Then \( \pi \) is irreducible. | Proof. Let \( \left( {\pi, V}\right) \) be a highest weight cyclic representation with highest weight \( \mu \) and let \( v \) be a weight vector with weight \( \mu \) . By assumption, \( V \) decomposes as a direct sum of irreducible representations\n\n\[ V \cong {\bigoplus }_{j}{V}_{j} \]\n\n(6.9)\n\nBy Proposition ... | Yes |
Proposition 6.15. Two irreducible representations of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \) with the same highest weight are isomorphic. | Proof. Suppose \( \left( {\pi, V}\right) \) and \( \left( {\sigma, W}\right) \) are irreducible representations with the same highest weight \( \mu \) and let \( v \) and \( w \) be the highest weight vectors for \( V \) and \( W \) , respectively. Consider the representation \( V \oplus W \) and let \( U \) be smalles... | No |
Proposition 6.16. If \( \pi \) is an irreducible representation of \( \operatorname{sl}\left( {3;\mathbb{C}}\right) \) with highest weight \( \mu = \left( {{m}_{1},{m}_{2}}\right) \), then \( {m}_{1} \) and \( {m}_{2} \) non-negative integers. | Proof. By Proposition 6.3, \( {m}_{1} \) and \( {m}_{2} \) are integers. If \( v \) is a weight vector with weight \( \mu \), then \( \pi \left( {X}_{1}\right) v \) and \( \pi \left( {X}_{2}\right) v \) must be zero, or \( \mu \) would not be the highest weight for \( \pi \) . Thus, if we then apply Point 1 of Theorem ... | Yes |
If \( {m}_{1} \) and \( {m}_{2} \) are non-negative integers, then there exists an irreducible representation of \( \mathbf{{sl}}\left( {3;\mathbb{C}}\right) \) with highest weight \( \mu = \left( {{m}_{1},{m}_{2}}\right) \). | Proof. Since the trivial representation is an irreducible representation with highest weight \( \left( {0,0}\right) \), we need only construct representations with at least one of \( {m}_{1} \) and \( {m}_{2} \) positive.\n\nFirst, we construct two irreducible representations, with highest weights \( \left( {1,0}\right... | Yes |
Proposition 6.20. The group \( Z \) consists precisely of the diagonal matrices inside \( \mathrm{{SU}}\left( 3\right) \), namely the diagonal matrices with diagonal entries \( \left( {{e}^{i\theta },{e}^{i\phi },{e}^{-i\left( {\theta + \phi }\right) }}\right) \) with \( \theta ,\phi \in \mathbb{R} \) . The group \( N ... | Proof. Suppose \( A \) is in \( Z \), which means that \( A \) commutes with all elements of \( \mathfrak{h} \) , including \( {H}_{1} \), which has eigenvectors \( {e}_{1},{e}_{2} \), and \( {e}_{3} \), with corresponding eigenvalues \( 1, - 1 \), and 0 . Since \( A \) commutes with \( {H}_{1} \), it must preserve eac... | Yes |
Theorem 6.22. Suppose that \( \left( {\Pi, V}\right) \) is a finite-dimensional representation of \( \mathrm{{SU}}\left( 3\right) \) with associated representation \( \left( {\pi, V}\right) \) of \( \mathrm{{sl}}\left( {3;\mathbb{C}}\right) \) . If \( \lambda \in \mathfrak{h} \) is a weight for \( V \) then \( w \cdot ... | Proof. Suppose that \( \lambda \) is a weight for \( V \) with weight vector \( v \) . Then for all \( U \in N \) and \( H \in \mathfrak{h} \), we have\n\n\[ \pi \left( H\right) \Pi \left( U\right) v = \Pi \left( U\right) \left( {\Pi {\left( U\right) }^{-1}\pi \left( H\right) \Pi \left( U\right) }\right) v \]\n\n\[ = \... | Yes |
Theorem 6.24. Let \( \mu \) be a dominant integral element and let \( {V}_{\mu } \) be the irreducible representation with highest weight \( \mu \) . If \( \lambda \) is a weight of \( {V}_{\mu } \), then \( \lambda \) satisfies the following two conditions: (1) \( \mu - \lambda \) can be expressed as an integer combin... | Proof. According to the proof of Proposition 6.11, \( {V}_{\mu } \) is spanned by vectors of the form in (6.8). These vectors are weight vectors with weights of the form \( \lambda \mathrel{\text{:=}} \) \( \mu - {\alpha }_{{j}_{1}} - \cdots - {\alpha }_{{j}_{N}} \) . Thus, every weight of \( {V}_{\mu } \) satisfies th... | No |
Theorem 6.27. The dimension of the irreducible representation with highest weight \( \left( {{m}_{1},{m}_{2}}\right) \) is\n\n\[ \frac{1}{2}\left( {{m}_{1} + 1}\right) \left( {{m}_{2} + 1}\right) \left( {{m}_{1} + {m}_{2} + 2}\right) . \] | The reader is invited to verify this formula by direct computation in the representations depicted in Figure 6.4. | No |
The following Lie algebras are semisimple: \[ \operatorname{sl}\left( {n;\mathbb{C}}\right) ,\;n \geq 2 \] \[ \operatorname{so}\left( {n;\mathbb{C}}\right) ,\;n \geq 3 \] \[ \operatorname{sp}\left( {n;\mathbb{C}}\right) ,\;n \geq 1. \] | Proof. It is easy to see that the listed Lie algebras are reductive, with the corresponding compact groups \( K \) being \( \mathrm{{SU}}\left( n\right) ,\mathrm{{SO}}\left( n\right) ,\mathrm{{Sp}}\left( n\right) ,\mathrm{U}\left( n\right) \), and \( \mathrm{{SO}}\left( 2\right) \), respectively. (Compare (3.17) in Sec... | Yes |
Proposition 7.5. Suppose \( \mathfrak{g} = {\mathfrak{k}}_{\mathbb{C}} \) is a reductive Lie algebra. Choose an inner product on \( \mathfrak{g} \) as in Proposition 7.4. Then if \( \mathfrak{h} \) is an ideal in \( \mathfrak{g} \), the orthogonal complement of \( \mathfrak{h} \) is also an ideal, and \( \mathfrak{g} \... | Proof. An ideal in \( \mathfrak{g} \) is nothing but an invariant subspace for the adjoint action of \( \mathfrak{g} \) on itself. If \( \mathfrak{h} \) is a complex subspace of \( \mathfrak{g} \) and \( \mathfrak{h} \) is invariant under the adjoint action of \( \mathfrak{k} \), it will also be invariant under adjoint... | Yes |
Proposition 7.6. Every reductive Lie algebra \( \mathfrak{g} \) over \( \mathbb{C} \) decomposes as a Lie algebra direct sum \( \mathfrak{g} = {\mathfrak{g}}_{1} \oplus \mathfrak{z} \), where \( \mathfrak{z} \) is the center of \( \mathfrak{g} \) and where \( {\mathfrak{g}}_{1} \) is semisimple. | Proof. Since \( \mathfrak{z} \) is an ideal in \( \mathfrak{g} \), Proposition 7.5 shows that \( {\mathfrak{g}}_{1} \mathrel{\text{:=}} {\mathfrak{z}}^{ \bot } \) is also an ideal in \( \mathfrak{g} \) and that \( \mathfrak{g} \) decomposes as a Lie algebra direct sum \( \mathfrak{g} = {\mathfrak{g}}_{1} \oplus \mathfr... | No |
Proposition 7.7. If \( K \) is a simply connected compact matrix Lie group with Lie algebra \( \mathfrak{k} \), then \( \mathfrak{g} \mathrel{\text{:=}} {\mathfrak{k}}_{\mathbb{C}} \) is semisimple. | Proof. As in the proof of Proposition 7.6, \( \varepsilon \) decomposes as a Lie algebra direct sum \( \mathfrak{k} = {\mathfrak{k}}_{1} \oplus {\mathfrak{z}}^{\prime } \), where \( {\mathfrak{z}}^{\prime } \) is the center of \( \mathfrak{k} \) and where \( {\mathfrak{g}}_{1} \mathrel{\text{:=}} {\left( {\mathfrak{k}}... | Yes |
Theorem 7.8. Suppose that \( \mathfrak{g} \) is semisimple in the sense of Definition 7.1. Then \( \mathfrak{g} \) decomposes as a Lie algebra direct sum\n\n\[ \mathfrak{g} = {\bigoplus }_{j = 1}^{m}{\mathfrak{g}}_{j} \]\n\n(7.4)\n\nwhere each \( {\mathfrak{g}}_{j} \subset \mathfrak{g} \) is a simple Lie algebra. | Proof. If \( \mathfrak{g} \) has a nontrivial ideal \( \mathfrak{h} \), then by Proposition 7.5, \( \mathfrak{g} \) decomposes as the Lie algebra direct sum \( \mathfrak{g} = \mathfrak{h} \oplus {\mathfrak{h}}^{ \bot } \), where \( {\mathfrak{h}}^{ \bot } \) is also an ideal in \( \mathfrak{g} \) . Suppose that, say, \... | Yes |
Proposition 7.11. Let \( \mathfrak{g} = {\mathfrak{k}}_{\mathbb{C}} \) be a complex semisimple Lie algebra and let \( \mathfrak{t} \) be any maximal commutative subalgebra of \( \mathfrak{k} \) . Define \( \mathfrak{h} \subset \mathfrak{g} \) by\n\n\[ \mathfrak{h} = {\mathfrak{t}}_{\mathbb{C}} = \mathfrak{t} + i\mathfr... | Proof of Proposition 7.11. It is clear that \( \mathfrak{h} \) is a commutative subalgebra of \( \mathfrak{g} \) . We must first show that \( \mathfrak{h} \) is maximal commutative. So, suppose that \( X \in \mathfrak{g} \) commutes with every element of \( \mathfrak{h} \), which certainly means that it commutes with e... | Yes |
Proposition 7.14. Each root \( \alpha \) belongs to \( i\mathfrak{t} \subset \mathfrak{h} \) . | Proof. As we have already noted, each \( {\operatorname{ad}}_{H}, H \in \mathfrak{t} \), is a skew self-adjoint operator on \( \mathfrak{h} \), which means that \( {\operatorname{ad}}_{H} \) has pure imaginary eigenvalues. It follows that if \( \alpha \) is a root, \( \langle \alpha, H\rangle \) must be pure imaginary ... | Yes |
Proposition 7.17. For any \( \alpha \) and \( \beta \) in \( \mathfrak{h} \), we have\n\n\[ \left\lbrack {{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }}\right\rbrack \subset {\mathfrak{g}}_{\alpha + \beta } \]\n\nIn particular, if \( X \) is in \( {\mathfrak{g}}_{\alpha } \) and \( Y \) is in \( {\mathfrak{g}}_{-\al... | Proof. It follows from the Jacobi identity that \( {\operatorname{ad}}_{H} \) is a derivation:\n\n\[ \left\lbrack {H,\left\lbrack {X, Y}\right\rbrack }\right\rbrack = \left\lbrack {\left\lbrack {H, X}\right\rbrack, Y}\right\rbrack + \left\lbrack {X,\left\lbrack {H, Y}\right\rbrack }\right\rbrack . \]\n\nThus, if \( X \... | Yes |
Proposition 7.18. 1. If \( \alpha \in \mathfrak{h} \) is a root, so is \( - \alpha \) . Specifically, if \( X \) is in \( {\mathfrak{g}}_{\alpha } \), then \( {X}^{ * } \) is in \( {\mathfrak{g}}_{-\alpha } \), where \( {X}^{ * } \) is defined by (7.2) in Proposition 7.4. | Proof. For Point 1, if \( X = {X}_{1} + i{X}_{2} \) with \( {X}_{1},{X}_{2} \in \mathfrak{k} \), let\n\n\[ \bar{X} = {X}_{1} - i{X}_{2} \]\n\n(7.5)\n\nSince \( \mathfrak{k} \) is closed under brackets, if \( H \in \mathfrak{t} \subset \mathfrak{k} \) and \( X \in \mathfrak{g} \), we have\n\n\[ \overline{\left\lbrack H,... | Yes |
Theorem 7.19. For each root \( \alpha \), we can find linearly independent elements \( {X}_{\alpha } \) in \( {\mathfrak{g}}_{\alpha },{Y}_{\alpha } \) in \( {\mathfrak{g}}_{-\alpha } \), and \( {H}_{\alpha } \) in \( \mathfrak{h} \) such that \( {H}_{\alpha } \) is a multiple of \( \alpha \) and such that\n\n\[ \left\... | If \( {X}_{\alpha },{Y}_{\alpha },{H}_{\alpha } \) are as in the theorem, then on the one hand, \( \left\lbrack {{H}_{\alpha },{X}_{\alpha }}\right\rbrack = 2{X}_{\alpha } \) , while on the other hand, \( \left\lbrack {{H}_{\alpha },{X}_{\alpha }}\right\rbrack = \left\langle {\alpha ,{H}_{\alpha }}\right\rangle {X}_{\a... | Yes |
Corollary 7.20. For each root \( \alpha \), let \( {X}_{\alpha },{Y}_{\alpha } \), and \( {H}_{\alpha } \) be as in Theorem 7.19, with \( {Y}_{\alpha } = {X}_{\alpha }^{ * } \) . Then the elements\n\n\[ \n{E}_{1}^{\alpha } \mathrel{\text{:=}} \frac{i}{2}{H}_{\alpha };\;{E}_{2}^{\alpha } \mathrel{\text{:=}} \frac{i}{2}\... | Proof. Since \( \alpha \) belongs to \( i\mathrm{t} \) and \( {H}_{\alpha } \) is, by (7.9), a real multiple of \( \alpha \), the element \( \left( {i/2}\right) {H}_{\alpha } \) will belong to \( \mathfrak{t} \subset \mathfrak{k} \) . Meanwhile, we may check that \( {\left( {E}_{2}^{\alpha }\right) }^{ * } = - {E}_{2}^... | Yes |
Lemma 7.22. Suppose that \( X \) is in \( {\mathfrak{g}}_{\alpha } \), that \( Y \) is in \( {\mathfrak{g}}_{-\alpha } \), and that \( H \) is in \( \mathfrak{h} \). Then \( \left\lbrack {X, Y}\right\rbrack \) is in \( \mathfrak{h} \) and\n\n\[ \langle \left\lbrack {X, Y}\right\rbrack, H\rangle = \langle \alpha, H\rang... | Proof. That \( \left\lbrack {X, Y}\right\rbrack \) is in \( \mathfrak{h} \) follows from Proposition 7.17. Then, using Proposition 7.4 , we compute that \n\n\[ \langle \left\lbrack {X, Y}\right\rbrack, H\rangle = \left\langle {{\operatorname{ad}}_{X}\left( Y\right), H}\right\rangle = \left\langle {Y,{\operatorname{ad}}... | Yes |
Lemma 7.24. If \( \alpha \) and \( {c\alpha } \) are both roots with \( \left| c\right| > 1 \), then \( c = \pm 2 \) . | Proof. Let \( {\mathfrak{s}}^{\alpha } \) be as in (7.13). If \( \beta = {c\alpha } \) is also a root and \( X \) is a nonzero element of \( {\mathfrak{g}}_{\beta } \), then by (7.8), we have\n\n\[ \left\lbrack {{H}_{\alpha }, X}\right\rbrack = \left\langle {\beta ,{H}_{\alpha }}\right\rangle X = \bar{c}\left\langle {\... | Yes |
Theorem 7.26. The action of \( W \) on it preserves \( R \) . That is to say, if \( \alpha \) is a root, then \( w \cdot \alpha \) is a root for all \( w \in W \) . | Proof. For each \( \alpha \in R \), consider the invertible linear operator \( {S}_{\alpha } \) on \( \mathfrak{g} \) given by\n\n\[ {S}_{\alpha } = {e}^{{\operatorname{ad}}_{{X}_{\alpha }}}{e}^{-{\operatorname{ad}}_{{Y}_{\alpha }}}{e}^{{\operatorname{ad}}_{{X}_{\alpha }}}.\]\n\nNow, if \( H \in \mathfrak{h} \) satisfi... | Yes |
Corollary 7.27. The Weyl group is finite. | Proof. Since the roots span \( \mathfrak{h} \), each \( w \in W \) is determined by its action on \( R \) . Since, also, \( w \) maps \( R \) onto \( R \), we see that \( W \) may be thought of as a subgroup of the permutation group on the roots. | Yes |
Proposition 7.29. For all roots \( \alpha \) and \( \beta \), we have that\n\n\[ \left\langle {\beta ,{H}_{\alpha }}\right\rangle = 2\frac{\langle \alpha ,\beta \rangle }{\langle \alpha ,\alpha \rangle } \]\n\n(7.18)\nis an integer. | Proof. If \( {\mathfrak{s}}^{\alpha } = \left\langle {{X}_{\alpha },{Y}_{\alpha },{H}_{\alpha }}\right\rangle \) is as in Theorem 7.19 and \( X \) is a root vector associated to the root \( \beta \), then \( \left\lbrack {{H}_{\alpha }, X}\right\rbrack = \left\langle {\beta ,{H}_{\alpha }}\right\rangle X \) . Thus, \( ... | Yes |
Proposition 7.31. Suppose \( \mathfrak{g} \) is a real Lie algebra and that the complexification \( {\mathfrak{g}}_{\mathbb{C}} \) of \( \mathfrak{g} \) is simple. Then \( \mathfrak{g} \) is also simple. | Proof. Since \( {\mathfrak{g}}_{\mathbb{C}} \) is simple, the dimension of \( {\mathfrak{g}}_{\mathbb{C}} \) over \( \mathbb{C} \) is at least 2, so that the dimension of \( \mathfrak{g} \) over \( \mathbb{R} \) is also at least 2 . If \( \mathfrak{g} \) had a nontrivial ideal \( \mathfrak{h} \), then the complexificat... | No |
Lemma 7.34. Suppose that \( K \) is a compact matrix Lie group whose Lie algebra \( \mathfrak{k} \) is noncommutative. Then \( \mathfrak{k} \) does not admit a complex structure. | Proof. Suppose, toward a contradiction, that \( \mathfrak{k} \) does admit a complex structure \( J \) . By Proposition 7.4, there exists a real inner product on \( \mathfrak{k} \) with respect to which \( {\operatorname{ad}}_{H} \) is skew symmetric for all \( H \in \mathfrak{k} \) . If we choose \( H \) not in the ce... | Yes |
Theorem 7.35. Let \( \mathfrak{g} \cong {\mathfrak{k}}_{\mathbb{C}} \) be a complex semisimple Lie algebra, let \( \mathfrak{t} \) be a maximal commutative subalgebra of \( \mathfrak{k} \), and let \( \mathfrak{h} = {\mathfrak{t}}_{\mathbb{C}} \) be the associated Cartan subalgebra of \( \mathfrak{g} \) . Let \( R \sub... | Proof of Theorem 7.35, Part 1. Assume first that \( \mathfrak{g} \) is not simple, so that, by Theorem 7.32, \( \mathfrak{k} \) is not simple either. Thus, \( \mathfrak{k} \) has a nontrivial ideal \( {\mathfrak{k}}_{1} \) . Now, ideals in \( \mathfrak{k} \) are precisely invariant subspaces for the adjoint action of \... | Yes |
Proposition 8.3. Suppose \( \left( {E, R}\right) \) and \( \left( {F, S}\right) \) are root systems. Consider the vector space \( E \oplus F \), with the natural inner product determined by the inner products on \( E \) and \( F \) . Then \( R \cup S \) is a root system in \( E \oplus F \), called the direct sum of \( ... | Proof. If \( R \) spans \( E \) and \( S \) spans \( F \), then \( R \cup S \) spans \( E \oplus F \), so Condition 1 is satisfied. Condition 2 holds because \( R \) and \( S \) are root systems in \( E \) and \( F \) , respectively. For Condition 3, if \( \alpha \) and \( \beta \) are both in \( R \) or both in \( S \... | Yes |
Proposition 8.6. Suppose \( \alpha \) and \( \beta \) are roots, \( \alpha \) is not a multiple of \( \beta \), and \( \langle \alpha ,\alpha \rangle \geq \) \( \langle \beta ,\beta \rangle \) . Then one of the following holds:\n\n1. \( \langle \alpha ,\beta \rangle = 0 \)\n\n2. \( \langle \alpha ,\alpha \rangle = \lan... | Proof. Suppose that \( \alpha \) and \( \beta \) are roots and let \( {m}_{1} = 2\langle \alpha ,\beta \rangle /\langle \alpha ,\alpha \rangle \) and \( {m}_{2} = \) \( 2\langle \beta ,\alpha \rangle /\langle \beta ,\beta \rangle \), so that \( {m}_{1} \) and \( {m}_{2} \) are integers. Assume \( \langle \alpha ,\alpha... | Yes |
Corollary 8.7. Suppose \( \alpha \) and \( \beta \) are roots. If the angle between \( \alpha \) and \( \beta \) is strictly obtuse (i.e., strictly between \( \pi /2 \) and \( \pi \) ), then \( \alpha + \beta \) is a root. If the angle between \( \alpha \) and \( \beta \) is strictly acute (i.e., strictly between 0 and... | Proof. The proof is by examining each of the three obtuse angles and each of the three acute angles allowed by Proposition 8.6. Consider first the acute case and adjust the labeling so that \( \langle \alpha ,\alpha \rangle \geq \langle \beta ,\beta \rangle \) . An examination of Cases 2,3, and 4 in the proposition (se... | Yes |
Proposition 8.8. Every rank-two root system is isomorphic to one of the systems in Figure 8.3. | Proof. It is harmless to assume \( E = {\mathbb{R}}^{2} \) ; thus, let \( R \subset {\mathbb{R}}^{2} \) be a root system. Let \( \theta \) be the smallest angle occurring between any two vectors in \( R \) . Since the elements of \( R \) span \( {\mathbb{R}}^{2} \), we can find two linearly independent vectors \( \alph... | Yes |
Proposition 8.9. If \( R \) is a rank-two root system with minimum angle \( \theta = {2\pi }/n \) , \( n = 4,6,8,{12} \), then the Weyl group of \( R \) is the symmetry group of a regular \( n/2 \) -gon. | Proof. The group \( W \) will contain at least \( n/2 \) reflections, one for each pair \( \pm \alpha \) of roots. If \( \alpha \) and \( \beta \) are roots with some angle \( \phi \) between them, then the composition of the reflections \( {s}_{\alpha } \) and \( {s}_{\beta } \) will be a rotation by angle \( \pm {2\p... | Yes |
Proposition 8.11. If \( R \) is a root system, then \( {R}^{ \vee } \) is also a root system and the Weyl group for \( {R}^{ \vee } \) is the same as the Weyl group for \( R \) . Furthermore, \( {\left( {R}^{ \vee }\right) }^{ \vee } = R \) . | Proof. We compute that\n\n\[ \left\langle {{H}_{\alpha },{H}_{\alpha }}\right\rangle = 4\frac{\langle \alpha ,\alpha \rangle }{\langle \alpha ,\alpha {\rangle }^{2}} = \frac{4}{\langle \alpha ,\alpha \rangle } \]\n\nand, therefore,\n\n\[ 2\frac{{H}_{\alpha }}{\left\langle {H}_{\alpha },{H}_{\alpha }\right\rangle } = 2\... | Yes |
Proposition 8.13. If \( \alpha \) and \( \beta \) are distinct elements of a base \( \Delta \) for \( R \), then \( \langle \alpha ,\beta \rangle \leq 0 \) . | Proof. Since \( \alpha \neq \beta \), if we had \( \langle \alpha ,\beta \rangle > 0 \), then the angle between \( \alpha \) and \( \beta \) would be strictly between 0 and \( \pi /2 \) . Then, by Corollary \( {8.7},\alpha - \beta \) would be an element of \( R \) . Since the elements of \( \Delta \) form a basis for \... | Yes |
Proposition 8.14. There exists a hyperplane \( V \) through the origin in \( E \) that does not contain any element of \( R \) . | Proof. For each \( \alpha \in R \), let \( {V}_{\alpha } \) denote the hyperplane\n\n\[ \n{V}_{\alpha } = \{ H \in E \mid \langle \alpha, H\rangle = 0\} .\n\]\n\nSince there are only finitely many of these hyperplanes, there exists \( H \in E \) not in any \( {V}_{\alpha } \) . (See Exercise 2.) Let \( V \) be the hype... | No |
Theorem 8.17. For any base \( \Delta \) for \( R \), there exists a hyperplane \( V \) and a side of \( V \) such that \( \Delta \) arises as in Theorem 8.16. | Proof. If \( \Delta = \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right\} \) is a base for \( R \), then \( \Delta \) is a basis for \( E \) in the vector space sense. Then, by elementary linear algebra, for any sequence of numbers \( {c}_{1},\ldots ,{c}_{r} \) there exists a unique \( \gamma \in E \) with \( \left\l... | Yes |
Lemma 8.19. Let \( \Delta \) be a base, \( {R}^{ + } \) the associated set of positive roots, and \( \alpha \) an element of \( \Delta \) . Then \( \alpha \) cannot be expressed as a linear combination of elements of \( {R}^{ + } \smallsetminus \{ \alpha \} \) with non-negative real coefficients. | Proof. Let \( {\alpha }_{1} = \alpha \) and let \( {\alpha }_{2},\ldots ,{\alpha }_{r} \) be the remaining elements of \( \Delta \) . Suppose \( {\alpha }_{1} \) is a linear combination elements \( \beta \neq {\alpha }_{1} \) in \( {R}^{ + } \) with non-negative coefficients. Each such \( \beta \) can then be expanded ... | Yes |
Proposition 8.21. For each open Weyl chamber \( C \), there exists a unique base \( {\Delta }_{C} \) for \( R \) such that \( C \) is the open fundamental Weyl chamber associated to \( {\Delta }_{C} \) . The positive roots with respect to \( {\Delta }_{C} \) are precisely those elements \( \alpha \) of \( R \) such tha... | Proof. Let \( H \) be any element of \( C \) and let \( V \) be the hyperplane orthogonal to \( H \) . Since \( H \) is contained in an open chamber, \( H \) is not orthogonal to any root, and, thus, \( V \) does not contain any root. Thus, by Theorem 8.16, there exists a base \( \Delta = \left\{ {{\alpha }_{1},\ldots ... | Yes |
Proposition 8.24. If \( \Delta \) is a base, then \( W \) is generated by the reflections \( {s}_{\alpha } \) with \( \alpha \in \Delta \) . | To prove Proposition 8.24, we must show that \( {W}^{\prime } = W \) . Let \( {s}_{\alpha } \) be the reflection associated to an arbitrary root \( \alpha \) . By Proposition 8.22, \( \alpha \) belongs to \( {\Delta }_{D} \) for some chamber \( D \) . If \( w \in {W}^{\prime } \) is chosen so that \( w \cdot D = C \), ... | Yes |
Lemma 8.26. Let \( \Delta \) be a base for \( R \) and let \( C \) be the associated fundamental Weyl chamber. Let \( w \) be an element of \( W \) with \( w \neq I \) and let \( w = {s}_{{\alpha }_{1}}{s}_{{\alpha }_{2}}\cdots {s}_{{\alpha }_{k}} \), with \( {\alpha }_{j} \in \Delta \), be a minimal expression for \( ... | Proof. Since \( w \neq I \), we must have \( k \geq 1 \) . If \( k = 1 \), then \( w = {s}_{{\alpha }_{1}} \), so that \( w \cdot C = {s}_{{\alpha }_{1}} \cdot C \) is on the opposite side of \( {V}_{{\alpha }_{1}} \) from \( C \) . Assume, inductively, that the result holds for \( u \in W \) where the minimal number o... | Yes |
Proposition 8.27. The Weyl group acts freely on the set of open Weyl chambers. If \( H \) belongs to an open chamber \( C \) and \( w \cdot H = H \) for some \( w \in W \), then \( w = I \) . | Proof. Let \( C \) be any chamber and let \( \Delta \) be the associated base (Proposition 8.21). If \( w \in W \) and \( W \neq I \), Lemma 8.26 tells us that \( w \cdot C \) and \( C \) lie on opposite sides of a hyperplane, so that \( w \cdot C \) cannot equal \( C \) .\n\nMeanwhile, if \( H \) belongs to an open ch... | Yes |
Proposition 8.28. For any two bases \( {\Delta }_{1} \) and \( {\Delta }_{2} \) for \( R \), there exists a unique \( w \in W \) such that \( w \cdot {\Delta }_{1} = {\Delta }_{2} \) . | Proof. By Proposition 8.21, there is a bijective correspondence between bases and open Weyl chambers, and this correspondence is easily seen to respect the action of the Weyl group. Since, by Propositions 8.23 and 8.27, \( W \) acts freely and transitively on the chambers, the same is true for bases. | Yes |
Proposition 8.29. Let \( C \) be a Weyl chamber and \( H \) an element of \( E \) . Then there exists exactly one point in the \( W \) -orbit of \( H \) that lies in the closure \( \bar{C} \) of \( C \) . | Proof. If \( U \) is any neighborhood of \( H \), then by the argument in Exercise 2, the hyperplanes \( {V}_{\alpha },\alpha \in R \), cannot fill up \( U \), which means \( U \) contains points in some open Weyl chamber. It follows that \( H \) belongs to \( \bar{D} \) for some chamber \( D \) . By Proposition 8.23, ... | Yes |
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