Bourbaki

习题 17, § 9

标签 0AKG,永久不变。引用为 [Bourbaki, Tag 0AKG]。

李群与李代数, III, 400

Bourbaki marks this one as harder than the rest.

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(a) 让cc表示实数可解李代数,其中基(a,b,c)(a, b, c)使得[a,b]=c,[a,c]=−b,[b,c]=0[a, b] = c, [a, c] = -b, [b, c] = 0。让dd表示实数可解李代数,其中基(a,b,c,d)(a, b, c, d)使得[a,b]=c,[a,c]=−b,[b,c]=d,[a,d]=[b,d]=[c,d]=0[a, b] = c, [a, c] = -b, [b, c] = d, [a, d] = [b, d] = [c, d] = 0。令gg为实数可解李代数。如果gg包含非零元素x,y,zx, y, z使得[x,y]=z[x, y] = z和[x,z]=−y[x, z] = -y,则gg包含子代数同构于cc或dd。 (通过ai=[ai−1,ci−1],bi=[bi−1,ci−1],ci=[ai,bi]a_i = [a_{i-1}, c_{i-1}], b_i = [b_{i-1}, c_{i-1}], c_i = [a_i, b_i]归纳写出a1=y,b1=z,c1=[y,z]a_1 = y, b_1 = z, c_1 = [y, z];定义ai,bi,cia_i, b_i, c_i。考虑最小的整数kk,使得ck=0c_k = 0。)

(b) 设gg和g∗g^*为gg的实可解李代数和ϕ\phi同态到g∗g^*。如果 g∗g^* 包含子代数同构于 cc 或 dd,则 gg 具有相同的性质。

(c) 设hh为复数可解李代数。考虑一个 Jordan–Hölder 级数作为hh 的伴随表示; 级数定义一维表示与 hh 的商,因此线性形式对 hh。这些仅依赖于hh的线性形式,称为hh的根。如果h′h'是实数可解李代数,则h′h'的根是限制到h′⊗RCh' \otimes_{\mathbf{R}} \mathbf{C}的根h′h'。

Then let GG be a finite-dimensional simply connected solvable real Lie group and gg its Lie algebra. Show that the following conditions are equivalent: (α\alpha) exp⁡G\exp_G is injective; (β\beta) exp⁡G\exp_G is surjective; (γ\gamma) exp⁡G\exp_G is bijective; (δ\delta) exp⁡G\exp_G is an isomorphism of the analytic manifold L(G)L(G) onto the analytic manifold GG; (ε\varepsilon) L(G)L(G) contains no subalgebra isomorphic to cc or dd; (ζ\zeta) there exists no quotient algebra of L(G)L(G) admitting a subalgebra isomorphic to cc; (η\eta) every root of gg is of the form ϕ+iϕ′\phi + i\phi' where ϕ,ϕ′\phi, \phi' are in g∗g^* and ϕ′\phi' is proportional to ϕ\phi; (θ\theta) for all x∈gx \in g, the only pure imaginary eigenvalue (in C\mathbf{C}) of ad x\mathrm{ad}\, x is 0.\footnote{For more details, cf. M. Saito, Sur certains groupes de Lie résolubles, Sci. Papers of the College of General Education, Univ. of Tokyo, 7 (1957), pp. 1–11 and 157–168.}

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