Bourbaki

Exercise 6, § 3

Lie Groups and Lie Algebras, III, 374

Bourbaki marks this one as harder than the rest.

Let F be the vector space of Hermitian complex matrices with 2 rows and 2 columns. For all s∈SL(2,C)s \in \mathbf{SL}(2, \mathbf{C}), we define an automorphism σ(s)\sigma(s) of F by σ(s)(m)=sms∗\sigma(s)(m) = sms^*. We identify the element (t+xy+izy−izt−x)\begin{pmatrix} t + x & y + iz \\ y - iz & t - x \end{pmatrix} of F with the element (t,x,y,z)(t, x, y, z) of R4\mathbf{R}^4 and write

q(t,x,y,z)=t2−x2−y2−z2. q(t, x, y, z) = t^2 - x^2 - y^2 - z^2.
Let C be the set of (t,x,y,z)∈R4(t, x, y, z) \in \mathbf{R}^4 such that t2−x2−y2−z2>0, t>0t^2 - x^2 - y^2 - z^2 > 0,\ t > 0. Let SO+(q)\mathbf{SO}^+(q) be the set of g∈SO(q)g \in \mathbf{SO}(q) such that g(C)=Cg(C) = C; it is an open normal subgroup of SO(q)\mathbf{SO}(q) of index 2 in SO(q)\mathbf{SO}(q) (cf. Exercise 4). Show that σ\sigma is a morphism of the underlying real Lie group of SL(2,C)\mathbf{SL}(2, \mathbf{C}) onto the real Lie group SO+(q)\mathbf{SO}^+(q) with kernel {I,−I}\{I, -I\}. (For the surjectivity, use Exercise 5.) Deduce that the real Lie group SO+(q)\mathbf{SO}^+(q) is isomorphic to the underlying real Lie group of SL(2,C)/{I,−I}\mathbf{SL}(2, \mathbf{C})/\{I, -I\}, that is (Exercise 3) the underlying real Lie group of SO(3,C)\mathbf{SO}(3, \mathbf{C}).

Solution

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Transcribed from the printed volume. Not translated.