Let o be a complete discrete valuation ring with finite residue field k of odd characteristic. Let G be a general or special linear group or a unitary group defined over o and let g denote its Lie algebra. For every positive integer ℓ, let K ℓ be the ℓ-th principal congruence subgroup of G(o). A continuous irreducible representation of G(o) is called regular of level ℓ if it is trivial on K ℓ+1 and its restriction to K ℓ /K ℓ+1 ≃ g(k) consists of characters with G(k)-stabiliser of minimal dimension. In this paper we construct the regular characters of G(o), compute their degrees and show that the latter satisfy Ennola duality. We give explicit uniform formulae for the regular part of the representation zeta functions of these groups.

Abstract. In FOCS 2002, Even et al. showed that any set of n discs in the plane can be Conflict-Free colored with a total of at most O(log n) colors. That is, it can be colored with O(log n) colors such that for any (covered) point p there is some disc whose color is distinct from all other colors of discs containing p. They also showed that this bound is asymptotically tight. In this paper we prove the following stronger results:(i) Any set of n discs in the plane can be colored with a total of at most O(k log n) colors such that (a) for any point p that is covered by at least k discs, there are at least k distinct discs each of which is colored by a color distinct from all other discs containing p and (b) for any point p covered by at most k discs, all discs covering p are colored distinctively. We call such a coloring a k-Strong Conflict-Free coloring. We extend this result to pseudo-discs and arbitrary regions with linear union-complexity. (ii) More generally, for families of n simple closed Jordan regions with union-complexity bounded by O(n 1+α ), we prove that there exists a k-Strong Conflict-Free coloring with at most O(kn α ) colors. (iii) We prove that any set of n axis-parallel rectangles can be k-Strong Conflict-Free colored with at most O(k log 2 n) colors. (iv) We provide a general framework for k-Strong Conflict-Free coloring arbitrary hypergraphs. This framework relates the notion of k-Strong Conflict-Free coloring and the recently studied notion of k-colorful coloring. All of our proofs are constructive. That is, there exist polynomial time algorithms for computing such colorings.

For an integer n ≥ 2, let X n be the Cayley graph on the symmetric group S n generated by the set of transpositions {(1 2), (1 3), . . . , (1 n)}. It is shown that the spectrum of X n contains all integers from −(n − 1) to n − 1 (except 0 if n = 2 or n = 3).

Abstract:A polychromatic k-coloring of a plane graph G is an assignment of k colors to the vertices of G such that every face of G has all k colors on its boundary. For a given plane graph G, one seeks the maximum number k such that G admits a polychromatic k-coloring. In this paper, it is proven that every connected plane graph of order at least three, and maximum degree three, other than K 4 or a subdivision of K 4 on five vertices, admits a 3-coloring in the regular sense (i.e., no monochromatic edges) that is also a polychromatic 3-coloring. Our proof is constructive and implies a polynomial-time algorithm. ᭧

It is shown that a Cayley multigraph over a group G with generating multiset S is integral (i.e., all of its eigenvalues are integers) if S lies in the integral cone over the boolean algebra generated by the normal subgroups of G. The converse holds in the case when G is abelian. This in particular gives an alternative, character theoretic proof of a theorem of Bridges and Mena (1982). We extend this result to provide a necessary and sufficient condition for a Cayley multigraph over a Hamiltonian group to be integral, in terms of character sums and the structure of the generating set. Proof. To prove (b), observe thatSince Λ 1 and Λ 2 are integral, so is their difference, hence X ∩ Y is B-integral.If ½ ∈ B, then the complete graph K n is B-integral since A(K n ) has eigenvalue n−1 with eigenvector ½ and eigenvalue −1 with eigenspace ½ ⊥ . Thus (a) follows from (b) by taking K n and X playing the role of X and Y , respectively.By observing that A(X ∪ Y ) = A(X) + A(Y ) − A(X ∩ Y ), a proof similar to the above proof of (b) shows that (c) holds.