### Download A Biordered Set Representation of Regular Semigroups by Yu B.J., Xu M. PDF

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1. Structure Constants. Let g be a finite-dimensional real or complex Lie algebra, and let X1 , · · · , Xn be a basis for g (as a vector space). Then for each i, j, [Xi , Xj ] can be written uniquely in the form n [Xi , Xj ] = cijk Xk . k=1 The constants cijk are called the structure constants of g (with respect to the chosen basis). Clearly, the structure constants determine the bracket operation on g. In some of the literature, the structure constants play an important role, although we will not have occasion to use them in this course.

To see this, consider A ∈ E. Then look at the set of matrices of the form AB, with B ∈ V . This will be a neighborhood of A. But 46 3. LIE ALGEBRAS AND THE EXPONENTIAL MAPPING every such B can be written as B = eX and A can be written as A = eX1 eX2 · · · eXn , so AB = eX1 eX2 · · · eXn eX . Now we claim that E is closed (in G). Suppose A ∈ G, and there is a sequence An ∈ E with An → A. Then AA−1 → I. Thus we can choose some n0 such n X −1 = e and A = An0 eX . But by assumption, An0 = ∈ V . Then AA that AA−1 n0 n0 eX1 eX2 · · · eXn , so A = eX1 eX2 · · · eXn eX .

1 2. eX is invertible, and eX = e−X . 3. e(α+β)X = eαX eβX for all real or complex numbers α, β. 4. If XY = Y X, then eX+Y = eX eY = eY eX . −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.

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