Suren Poghosyan and Daniel
نویسنده
چکیده
CLUSTER EXPANSION WITH APPLICATIONS 3 In order to guess the correct form of a, one should consider the left side of the equation above with a(y) ≡ 0. The integral may depend on x; a typical situation is that x is characterized by a length l(x), which is a positive number, so that the left side is roughly proportional to l(x). This suggests to try a(x) = cl(x), and one can then optimize on the value of c. We also consider an alternate criterion that involves u rather than ζ. It is inspired by the recent work of Procacci [22]. Let u(x, y) = { u(x, y) if Reu(x, y) 6= ∞, 1 if Reu(x, y) = ∞. (2.5) Assumption 2’. There exists a nonnegative function a on X such that for almost all x ∈ X, ∫ d|μ|(y) |u(x, y)| e 6 a(x). For positive u we can take b(x) ≡ 0; and since 1 − e−u 6 u, Assumption 2 is always better than Assumption 2’. We actually conjecture that, together with Assumption 1, a sufficient condition is ∫ d|μ|(y) min ( |ζ(x, y)|, |u(x, y)| ) e 6 a(x). (2.6) That is, it should be possible to combine the best of both assumptions. In this respect Assumption 2 is optimal in the case of positive potentials, and Assumption 2’ is optimal in the case of hard core plus negative potentials. We denote by Gn the set of all graphs with n vertices (unoriented, no loops) and Cn ⊂ Gn the set of connected graphs with n vertices. We introduce the following combinatorial function on finite sequences (x1, . . . , xn) of elements of X: φ(x1, . . . , xn) = { 1 if n = 1, 1 n! ∑ G∈Cn ∏ {i,j}∈G ζ(xi, xj) if n > 2. (2.7) The product is over edges of G. Theorem 2.1 (Cluster expansions). Suppose that Assumptions 1 and 2, or 1 and 2’, hold true. We also suppose that ∫ d|μ|(y)| e < ∞. Then we have Z = exp { ∑
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CLUSTER EXPANSION WITH APPLICATIONS 3 In order to guess the correct form of a, one should consider the left side of the equation above with a(y) ≡ 0. The integral may depend on x; a typical situation is that x is characterized by a length l(x), which is a positive number, so that the left side is roughly proportional to l(x). This suggests to try a(x) = cl(x), and one can then optimize on the v...
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