Recall that a locally compact group G is called unimodular if the left Haar measure on G is equal to the right one. It is proved in this paper that G is unimodular iff it is approximable by finite quasigroups (Latin squares).

In this paper, we construct odd unimodular lattices in dimensions n = 36, 37 having minimum norm 3 and 4s = n − 16, where s is the minimum norm of the shadow. We also construct odd unimodular lattices in dimensions n = 41, 43, 44 having minimum norm 4 and 4s = n− 24.

We explain how one can construct a class of discrete hypergroups which are non-unimodular. They arise as double coset hypergroups induced by the transitive action of a non-unimodular group of permutations on an innnite set. A concrete example is given in terms of the aane group of a homogeneous tree.

All indecomposable unimodular hermitian lattices in dimensions 14 and 15 over the ring of integers in Q( √ −3) are determined. Precisely one lattice in dimension 14 and two lattices in dimension 15 have minimal norm 3. In 1978 W. Feit [10] classified the unimodular hermitian lattices of dimensions up to 12 over the ring Z[ω] of Eisenstein integers, where ω is a primitive third root of unity. Th...

The highest possible minimal norm of a unimodular lattice is determined in dimensions n ≤ 33. There are precisely five odd 32-dimensional lattices with the highest possible minimal norm (compared with more than 8.1020 in dimension 33). Unimodular lattices with no roots exist if and only if n ≥ 23, n 6= 25.

Let M be a finite set of vectors in Rn of cardinality m and H(M) = {{x ∈ Rn : cTx = 0} : c ∈ M} the central hyperplane arrangement represented by M. An independent subset of M of cardinality n is called a Camion basis, if it determines a simplex region in the arrangementH(M). In this paper, we first present a new characterization of Camion bases, in the case whereM is the column set of the node...

Let $a, b, k,in K$ and $u, v in U(K)$. We show for any idempotent $ein K$, $(a 0|b 0)$ is e-clean iff $(a 0|u(vb + ka) 0)$ is e-clean and if $(a 0|b 0)$ is 0-clean, $(ua 0|u(vb + ka) 0)$ is too.