Number of Substitutions Omitting at Least one Letter in a by Miler G. A.

By Miler G. A.

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19) is i {([X1 , Y1 ] − [X2 , Y2 ]) + i ([X2 , Y1 ] + [X1 , Y2 ])} = (− [X2 , Y1 ] − [X1 , Y2 ]) + i ([X1 , Y1 ] − [X2 , Y2 ]) , and indeed these are equal. It remains to check the Jacobi identity. Of course, the Jacobi identity holds if X, Y, and Z are in g. ) linear in X for fixed Y and Z. It follows that the Jacobi identity holds if X is in gC , and Y, Z in g. The same argument then shows that we can extend to Y in gC , and then to Z in gC . Thus the Jacobi identity holds in gC . 36. The Lie algebras gl(n; C), sl(n; C), so(n; C), and sp(n; C) are complex Lie algebras, as is the Lie algebra of the complex Heisenberg group.

18. This result is extremely important because it implies that if G is connected and simply connected, then there is a natural one-to-one correspondence between the representations of G and the representations of its Lie algebra g (as explained in Chapter 5). In practice, it is much easier to determine the representations of the Lie algebra than to determine directly the representations of the corresponding group. This result (relating Lie algebra homomorphisms and Lie group homomorphisms) is deep.

Suppose not. Then we can find a sequence gn ∈ G with gn → I such that no gn is in exp (U ). Since Φ is locally invertible, we can write gn (for large n) uniquely as gn = exp (Xn ) exp (Yn ), with Xn ∈ g and Yn ∈ D. Since gn → I and Φ−1 is continuous, Xn and Yn tend to zero. Thus (for large n), Xn ∈ U . So we must have (for large n) Yn = 0, otherwise gn would be in exp (U ). Let gn = exp (Yn ) = exp (−Xn ) gn . Note that gn ∈ G and gn → I. Since the unit ball in D is compact, we can choose a subsequence of {Yn } (still called {Yn }) so that Yn / Yn converges to some Y ∈ D, with Y = 1.

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