Bourbaki

Exercise 26, § 1

Lie Groups and Lie Algebras, I, 81

Bourbaki marks this one as harder than the rest.

(a) Let K be a field of characteristic ≠2\neq 2, E an n-dimensional vector space over K and Φ\Phi a non-degenerate symmetric bilinear form on E. Suppose that E has an orthogonal basis with respect to Φ\Phi and let R denote the (diagonal) matrix of Φ\Phi with respect to this basis. We know (Algebra, Chapter IX, § 4, Exercise 11) that every matrix (relative to the basis in question ) UU of the orthogonal group G(Φ)\mathbf{G}(\Phi) such that det⁡(I+U)≠0\det(I + U) \neq 0 can be written

U=(I−R−1S)−1(I+R−1S), U = (I - R^{-1}S)^{-1}(I + R^{-1}S),

where S=R(U−I)(U+I)−1S = R(U - I)(U + I)^{-1} is an alternating matrix such that det⁡(I−R−1S)≠0\det(I - R^{-1}S) \neq 0; conversely, for every matrix SS satisfying these conditions, U=(I−R−1S)−1(I+R−1S)U = (I - R^{-1}S)^{-1}(I + R^{-1}S) is a matrix of G(Φ)\mathbf{G}(\Phi) such that det⁡(I+U)≠0\det(I + U) \neq 0. Let Sij,Sij′,Sij′′S_{ij}, S'_{ij}, S''_{ij} be 3n(n−1)/23n(n-1)/2 indeterminates (1≤i<j≤n)(1 \leq i < j \leq n) and let L be a field containing the ring of formal power series K[[S_{ij}, S'_{ij}, S''_{ij}]]. If we denote by S, S', S'' the matrices

∑i<jSij(Eij−Eji),∑i<jSij′(Eij−Eji),∑i<jSij′′(Eij−Eji), \sum_{i<j} S_{ij}(E_{ij} - E_{ji}), \quad \sum_{i<j} S'_{ij}(E_{ij} - E_{ji}), \quad \sum_{i<j} S''_{ij}(E_{ij} - E_{ji}),

then det(I - R^{-1}S) \neq 0 and the analogues for S' and S'' hold; if we write

U=(I−R−1S)−1(I+R−1S),U′=(I−R−1S′)−1(I+R−1S′), U = (I - R^{-1}S)^{-1}(I + R^{-1}S), \quad U' = (I - R^{-1}S')^{-1}(I + R^{-1}S'),

then also det(I + UU') \neq 0. We write

F(S,S′)=R(UU′−I)(UU′+I)−1=(fij(S,S′)); F(S, S') = R(UU' - I)(UU' + I)^{-1} = (f_{ij}(S, S'));

the f_{ij}(S, S') belong to the ring of formal power series K[[S_{ij}, S'_{ij}]] and, by considering F(S, S') as a formal power series with coefficients in the algebra of matrices of order n over K, we can write:

F(S,S′)=S+S′+o2(S,S′) F(S, S') = S + S' + o_2(S, S')

where in the elements of the matrix o_2(S, S') the terms are all of degree \geq 1 in the S_{ij} and of degree \geq 1 in the S'_{ij}; then similarly:

U=I+2R−1S+o2′(S) U = I + 2R^{-1}S + o'_2(S)

the elements of o'_2(S) being formal power series of order \geq 2. Show that F(S, S') is a formal group law over K of dimension n(n - 1)/2; this formal group is called the formal orthogonal group (corresponding to \Phi) and is denoted by G(\Phi) by an abuse of notation.

(b) If we write M(S) = (m_{ij}(S)) = (I - R^{-1}S)^{-1}(I + R^{-1}S), M is a formal homomorphism of G(\Phi) into GL(n, K). Let \mathfrak{o}(\Phi) be the Lie algebra of G(\Phi) and let (T_{ij})_{i < j} denote the basis of this algebra such that T^{(e)}_{ij} = D^{(e)}_{ij} (cf. Exercise 24 (d)); if an element D = \sum_{i < j} c_{ij} T_{ij} of \mathfrak{o}(\Phi) is identified with the matrix D=∑i<jcij(Eij−Eji)D = \sum_{i < j} c_{ij}(E_{ij} - E_{ji}), show that (in the notation of Exercise 24 (d) with the identification made in Exercise 25 of the Lie algebra of GL(n, K) with gl(n, K)):

M∗(D)=2R−1D. M^*(D) = 2R^{-1}D.

Deduce that M* is an isomorphism of \mathfrak{o}(\Phi) onto the subalgebra of gl(n, K) consisting of the matrices X such that ^tXR + RX = 0 (for every ordered pair (a, b) of elements of E, identified with matrices with one column, consider \Phi(a + M(S).a, b + M(S).b) = ^ta.(I + ^tM(S))R(I + M(S)).b as a formal power series reduced to its constant term and use the fact that the subalgebra of X \in gl(n, K) such that ^tXR + RX = 0 is of dimension n(n - 1)/2).

(c) Define similarly the formal symplectic group in 2n variables over an arbitrary (commutative) field and show that its Lie algebra is identified with the subalgebra of gl(2n,K)\mathfrak{gl}(2n, \mathbf{K}) consisting of the matrices XX such that

tXA+AX=0,whereA=(0In−In0). {}^t X A + A X = 0, \quad \text{where} \quad A = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}.
Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.