Arithmetic and Geometry Around Hypergeometric Functions: by Rolf-Peter Holzapfel, Muhammed Uludag, M. Yoshida

By Rolf-Peter Holzapfel, Muhammed Uludag, M. Yoshida

This quantity includes lecture notes, survey and study articles originating from the CIMPA summer time tuition mathematics and Geometry round Hypergeometric capabilities held at Galatasaray collage, Istanbul, June 13-25, 2005. It covers a variety of themes with regards to hypergeometric capabilities, hence giving a extensive point of view of the cutting-edge within the box.

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Extra info for Arithmetic and Geometry Around Hypergeometric Functions: Lecture Notes of a CIMPA Summer School held at Galatasaray University, Istanbul, 2005

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Publish or Perish, Houston, 1994. [12] V. V. Nikulin, Involutions of integral quadratic forms and their applications to real algebraic geometry, Math. USSR Izvestiya 22 (1984), 99–172. [13] V. V. Nikulin, Discrete reflection groups in Lobachevsky spaces and algebraic surfaces, Proc. M. Berkeley 1986, 654–671. [14] M. Sebastiani and R. Thom, Un r´esultat sur la monodromie, Invent. Math. 13 (1971), 90–96. [15] W. org/publications/books/gt3m. [16] W. Thurston, Three-dimensional geometry and topology, Princeton Mathematical Series, 35.

N. The indicial equation at ∞ is given by X(X + 1) · · · (X + n − 1) − a1 X(X + 1) · · · (X + n − 2) + · · · +(−1)n−1 an−1 X + (−1)n an = 0. Proof. Exercise. 2 (Cauchy). Suppose P ∈ C is a regular point of (5). Then there exist n C-linear independent Taylor series solutions f1 , . . , fn in z − P with positive radius of convergence. Moreover, any Taylor series solution of (5) is a C-linear combination of f1 , . . , fn . 3. Any analytic solution of (5) near a regular point can be continued analytically along any path in C not meeting any singularity.

Notice that our equation has local exponents 0, 1/2 in z = 1. Hence the new equation has local exponents 0, 1 in t = 2, with regular solutions, and t = 2 turns out to be a regular point. At t = 0 we get the local exponents 0, 2(1/2 − a − b) and in t = 1, ∞, the 32 Frits Beukers points above z = ∞, we have the local exponents a, b and a, b. Thus our equation in t has again three singular points and Riemann scheme 0 1 0 a 1 − 2a − 2b b ∞ a b By the method sketched above, one easily sees that (1 − t)a F (2a, a + b, 2a + 2b|t) is a solution of this equation.

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