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By Nicolas Monod

Recent examine has time and again ended in connections among very important stress questions and bounded cohomology. despite the fact that, the latter has remained frequently intractable. This monograph introduces the functorial research of the continual bounded cohomology for topological teams, with coefficients in Banach modules. The strong options of this extra common thought have effectively solved a few of the unique difficulties in bounded cohomology. As functions, one obtains, particularly, tension effects for activities at the circle, for representations on advanced hyperbolic areas and on Teichmüller areas. a different attempt has been made to supply certain proofs or references in particularly a few generality.

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Geometrically, the index of an angle ∠BAC is the integer area of the smallest triangle AB C whose edges AB and AC generate the sublattices of lines containing the corresponding edges. For instance, the angle ∠BAC in Fig. 2 has index equal to 2: Let us conclude this subsection with a general remark. 9 We write l , ld, lS, and lα (starting with the letter “l”) to indicate that all these notions are well defined for any lattice, and not just for the integer lattice. 2 Empty Triangles: Their Integer and Euclidean Areas Let us completely study the case of the following “smallest” triangles.

6 Consider an arbitrary angle with integer vertex. Let the sail for this angle be a broken line with sequence of vertices (Ai ). Define: a2k = l Ak Ak+1 , a2k−1 = lsin(∠Ak−1 Ak Ak+1 ) for all admissible indices. The lattice length sine sequence (LLS sequence for short) for the sail is the sequence (an ). The LLS sequence can be either finite or infinite on one or both sides. 7 Notice that LLS sequences contain only positive integer elements. Consider a finite to the left LLS sequence. Regular continued fractions define a one-to-one correspondence between such LLS sequences and real numbers not smaller than 1; namely, the sequence (a0 , a1 , .

6, we get that lα(∠Ai Ai+1 Ai+2 ) = l (Bi Bi+1 ) and lα(∠Bi Bi+1 Bi+2 ) = l (Ai+1 Ai+2 ). For the example of α = 7/5 (see Figs. 4), we get lα(∠A0 A1 A2 ) = l (B0 B1 ) = a1 = 2; lα(∠B0 B1 B2 ) = l (A1 A2 ) = a2 = 2. 4 Exercises 39 Fig. 1 Prove that every sail is homeomorphic to R1 . 2 Describe the edge–angle duality for the case 0 < α < 1. 3 Construct the sails of continued fractions for rational numbers 3/2, 10/7, 18/13, and 13/18. 4 Find geometrically all the convergents to the rational numbers 8/5, 9/4, and 17/8.

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