A Bernstein-Chernoff deviation inequality, and geometric by Artstein-Avidan S.

By Artstein-Avidan S.

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Proof. Set (3) where U denotes the orthogonal complement of R. It is clear that = U. By 1'0 we denote an element of R such that [jro, "0] = 1'0 It is clear (by axiom III) that jU m(ro) = m(Liro, roJ) > 0. 54 THE GEOMETRY OF CLASSICAL DOMAINS First of all we will show that [R, UJ = o. Indeed, w([ro, uD = 0 by orthogonality. On the other hand, [ro, u] E R because R is an ideal of the algebra G. As a result, [ro, u] O. R + U is the centralizer of the element 1"0 and, therefore, a sub algebra of the algebra G.

Here we will present a brief survey of the contents of the present chapter. , Lie algebra containing an additional structure associated with the fact that the corresponding group acts analytically on a complex manifold. , j-algebras that are simultaneously solvable and split over the field of real numbers as Lie algebras. As was shown (see the appendix), there is a transitive group whose Lie algebra is solvable and split over R defined in each bounded homogeneous domain. Moreover, it is known that all maximal solvable and split (over R) subalgebras in any Lie algebra are conjugate (see Borel [4] and E.

19) U sing the fact that the transformation is analytic with respect to u, it is easy to use (19) to show that an involution must have the form (16). In order to complete the proof, note that z -+- c/>(z) is an analytic automorphism of the Siegel domain of genus 1 with cone V and the unique stationary point Zo; its square is the identity transformation, and, therefore, this transformation is an involution of this domain. The lemma is proved. We will now show that the domain Sl is not symmetric.

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