# Holzapfel R.-P.'s Arithmetic and Geometry around Hypergeomet PDF

By Holzapfel R.-P.

This quantity contains lecture notes, survey and study articles originating from the CIMPA summer season tuition mathematics and Geometry round Hypergeometric services held at Galatasaray college, Istanbul in the course of June 13-25, 2005. quite a lot of themes regarding hypergeometric capabilities is roofed, hence giving a wide standpoint of the state-of-the-art within the box.

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Additional resources for Arithmetic and Geometry around Hypergeomet

Example text

The local exponents can be read oﬀ immediately, λ, λ at 0, μ, μ at 1 and −λ − μ − deg(q), −λ − μ − deg(p) at ∞. The sum of the local exponents must be 1, hence − deg(p) − deg(q) = 1. Clearly this is a contradiction. 13. Suppose that A, B ∈ GL(2, C) have disjoint sets of eigenvalues and suppose that AB −1 has eigenvalue 1. Then, letting X 2 +a1 X +a2 and X 2 +b1 X +b2 be the characteristic polynomials of A, B, we have up to common conjugation, A= 0 −a2 , 1 −a1 B= 0 −b2 . 1 −b1 Proof. Choose v ∈ ker(A − B) and w = Av = Bv.

Without loss of generality we can assume that a ≤ b. We consider the following cases. Case i) 0 < a < c < b < 1. We take a = a, b = b, c = c. Then, λ = 1 − c, μ = a + b − c, ν = b − a and the inequalities are satisﬁed. Moreover, λ + μ + ν = 1 + 2b − 2c > 1. Case ii) 0 < a ≤ b < c < 1. We take a = a, b = b, c = c. When c ≥ a + b we get λ = 1 − c, μ = c − a − b, ν = b − a and the inequalities hold. Moreover, λ + μ + ν = 1 − 2a < 1. When c ≤ a + b we get λ = 1 − c, μ = a + b − c, ν = b − a and the inequalities hold.

We call these three cases the euclidean, spherical and hyperbolic case respectively. Proof. Let v ∈ ker(A − B) and w = Av. Suppose ﬁrst that v, w form a basis of C2 . Of course, with respect to this basis A and B have the form given in the previous lemma. In particular we see that A, B cannot have the same characteristic equation, since this would imply that A = B. Gauss’ Hypergeometric Function 35 We have to ﬁnd a hermitean form F such that F (gv, gv) = F (v, v), F (gv, gw) = F (v, w), F (gw, gv) F (w, v), F (gw, gw) = F (w, w), = for every g ∈ G.