Graph Theory

Elementary number theory, group theory, and Ramanujan graphs by Giuliana Davidoff, Peter Sarnak, Alain Valette

By Giuliana Davidoff, Peter Sarnak, Alain Valette

This article is a self-contained examine of expander graphs, in particular, their specific building. Expander graphs are hugely attached yet sparse, and whereas being of curiosity inside of combinatorics and graph thought, they could even be utilized to desktop technological know-how and engineering. just a wisdom of effortless algebra, research and combinatorics is needed as the authors give you the precious heritage from graph concept, quantity concept, staff idea and illustration concept. hence the textual content can be utilized as a quick advent to those topics and their synthesis in glossy arithmetic.

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X → x1 , with 3 ≤ ≤ k. Indeed each such circuit contributes 1 to the sum, n − 2 for each of graphs X ’s. Now there are n(n − 1) . . 6. Large Girth and Large Chromatic Number 35 such circuits of length . (( 2 )− +1) k = n m =3 (by the first step) n 2 =3 = − m− n 2 n(n − 1) . . (( n2 )− +1) −1 +1) . The term in parentheses is a o(1), as n → ∞. This gives the estimate k A(n, k) ≤ (1 + o(1)) =3 ≤ since m = [n 1+ε (1 + o(1)) k · ] and ε < k n m n 2 ( ) 2m k n−1 = (1 + o(1)) =3 2m n−1 = o(n), 1 .

5 1. What do the results of this section become for bipartite graphs? 2. 3. 6. Large Girth and Large Chromatic Number A combinatorial problem that has attracted much attention is to construct graphs with large chromatic number and large girth. Note that adding edges increases (or at least does not decrease) the chromatic number but that it does decrease the girth. Given this tension, it is by no means obvious that such graphs exist. A method, known as the probabilistic method, and due to Erd¨os [24], has proven to be very powerful in demonstrating the existence of such combinatorial objects.

P−1 , write kq = p kqp + u k , with u k ∈ 2 {1, . . , p − 1}; note that u k is nothing but the remainder in the Euclidean division of kq by p. If u k < 2p , then u k is the minimal residue of qk, so that u k = ri for exactly one i; if u k > 2p , then u k − p is the minimal residue of kq, so that u k − p = −r j for a unique j. Set R = that R = k:u k < 2p u k and R = µp − λ ri and R = i=1 j=1 that is, µ + uk . k = R + R = µp + uk ≡ p−1 , 2 q p 2 −1 8 (by Claim 1), uk − k:u k < 2p k=1 k=1 hence, p−1 2 ( p−1)/2 ( p−1)/2 from k = 1 to k = r j , so k:u k > 2p Since r1 , .

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