By Jenca G.

We end up that if E1 and E2 are a-complete impact algebras such that E1 is an element of E2 and E2 is an element of E1, then E1 and E2 are isomorphic.

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Av(n)) denotes the (signed) volume in Up corresponding to u$. 17. For x = [A] a simplex A = (a 0 ,ai,a 2 , a3) in £ 3 and g = l , the components x$ for each pair $ = (f/o D U\) of a plane and a line incident with A can be read off the following figure: EXAMPLE Chapter 4. Translational scissors congruences 35 Finally let us state the following Lusztig exact sequences. 18. Let V be a vector space of dimension Then for q= 1, 2 , . . , n there is an exact sequence n over F. o -»vq{v) A 0 vq{un~l) A • • • 4 0 vq{uq+1) -»• jyn-1 ljq+l ^ ® AqF(Uq)-> AqF(U) ^ 0 where d : Vq(Ul) —»• Vg(U%~1) is induced by sending a strict flag (C/n D • • O Ui-q-i) in Ul to (Ui D • • • D C/i_ 9 _i) in U1'1 = U0.

A more general question is if this induces an isomorphism on the level of Ho(0(n), —). This is known to be true for q = n and q = n — 2 for all n (see [ D u p o n t , 1982]). 4) and the corresponding result in E4 (cf. 16). We refer to [ D u p o n t - S a h , 1990] for the details. Another proof has been given by [ C a t h e l i n e a u , 1998]. 11). 1 # ; ( 0 ( n ) , A^(R n )) = 0 if i + j < n, j > 0. Note t h a t this does not directly give the desired vanishing for n = 3 and j = 1 since the coefficients are untwisted (so t h a t in this case the homology vanishes trivially by "center kills").

Finally for U C X a p-dimensional subspace let f/1 denote the subspace perpendicular to U. 5 for the polytope module. We refer to [ D u p o n t , 1982] for the proof. 10. Let X = 5 " , n > 0. Then i) H , ( C * ( X ) / a ( X ) n - 1 ) = 0 for q ± n. ii) There is a filtration of I(X)-modules 0 C 0{X) = F_i C F 0 C • • • C Fp C • • • C Fn = P t ( X ) and natural Fp/Fp^ isomorphisms 9* 0 S t ( C / P ) ® 0 ( C / p l ) , p = 0 , 1 , 2 , . . , n. UP p where U runs through all p-dimensional iii) In particular subspaces of X.