By Kondratev A.S.
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Extra info for 2-Local subgroups of finite groups
Moreover, A (0) = B (0) = I. Thus by standard uniqueness results for ordinary differential equations, A (t) = B (t) for all t. 2. Let H denote the Heisenberg group, and h its Lie algebra. Let G be a matrix Lie group with Lie algebra g, and let φ : h → g be a Lie algebra homomorphism. Then there exists a unique Lie group homomorphism φ : H → G such that e φ eX = eφ(X) for all X ∈ h. Proof. Recall that the Heisenberg group has the very special property that its exponential mapping is one-to-one and onto.
1 5. If C is invertible, then eCXC = CeX C −1. 6. eX ≤ e X . It is not true in general that eX+Y = eX eY , although by 4) it is true if X and Y commute. This is a crucial point, which we will consider in detail later. ) Proof. Point 1) is obvious. Points 2) and 3) are special cases of point 4). To verify point 4), we simply multiply power series term by term. ) Thus eX eY = I +X+ X2 + ··· 2! I +Y + Y2 +··· 2! 2. 5) eX eY = ∞ m m=0 k=0 ∞ 1 X k Y m−k = k! (m − k)! m=0 m! m k=0 m! X k Y m−k . (m − k)!
For each A ∈ G, show that AdA is a Lie algebra automorphism of g. Ad and ad. Let X and Y be matrices. Show by induction that n n (adX) (Y ) = k=0 n X k Y (−X)n−k . k Now show by direct computation that eadX (Y ) = Ad(eX )Y = eX Y e−X . You may assume that it is legal to multiply power series term-by-term. ) 52 3. LIE ALGEBRAS AND THE EXPONENTIAL MAPPING 14. 15. 16. 17. 18. 19. 20. Hint: Recall that Pascal’s Triangle gives a relationship between things of the form n+1 and things of the form nk .