# A-Simplicial Objects and A-Topological Groups by Smirnov V. A.

By Smirnov V. A.

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1 A1 . . A1i−1 0  .. ..  . .  j−1 j−1 A . . A 0 i−1  1 0 ... 0 1 C(A)ij := det   j+1 j+1 A . . A 0 i−1  1  . .  .. An1 ... Ani−1 0 Let C(A) be the matrix of the A1i+1 .. Aj−1 i+1 0 Aj+1 i+1 .. Ani+1 ... ... ...  A1n ..  .    Aj−1 n  0    Aj+1 n  ..  .  Ann Prove that C(A)A = AC(A) = det(A) · I (Cramer’s rule)! This can be done by remembering the the expansion formula for the determinant during multiplying it out. Prove that d(det)(A)X = Trace(C(A)X)! There are two ways to do this.

The unique solution is ∞ N (s) = 1 (p+1)! sp+1 , and so p=0 δ(exp)(X) = M (X) = N (1) = ∞ 1 (p+1)! ad(X)p . 28. Corollary. TX exp is bijective if and only if no eigenvalue of ad(X) : g → g √ is of the form −1 2kπ for k ∈ Z \ {0}. √ z Proof. The zeros of g(z) = e z−1 are exactly z = 2kπ −1 for k ∈ Z \ {0}. TX exp is 0. But the eigenvalues of g(ad(X)) are the images under g of the eigenvalues of ad(X). 29. Theorem. The Baker-Campbell-Hausdorff formula. Let G be a Lie group with Lie algebra g. For complex z near 1 we consider the (−1)n n function f (z) := log(z) n≥0 n+1 (z − 1) .

K −1 ⊂ V . K. , Vi open and dense for i ∈ N implies Vi dense). The set ϕ(ai )ϕ(K) is compact, thus closed. ϕ(K), there is some i such that ϕ(ai )ϕ(K) has non empty interior, so ϕ(K) has non empty interior. Choose b ∈ G such that ϕ(b) is an interior point of ϕ(K) in H. Then eH = ϕ(b)ϕ(b−1 ) is an interior point of ϕ(K)ϕ(K −1 ) ⊂ ϕ(V ). So if U is open in G and a ∈ U , then eH is an interior point of ϕ(a−1 U ), so ϕ(a) is in the interior of ϕ(U ). Thus ϕ(U ) is open in H, and ϕ is a homeomorphism.