Balanced configurations of 2n + 1 plane vectors

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1 Balanced configurations of n + 1 plane vectors RessayreNicolas (ressayre@math.univ-montp.fr) Université Montpellier II, Département de Mathématiques, UMR 5030 Case courrier 51- Place Eugène Bataillon, Montpellier Cedex 5 (France) Abstract. A configuration {v 1, v,, v m} (where m is a positive integer) of vectors of R is said to be uniform and balanced if for any index i {1,, m} the set with multiplicities D i = {det(v i, v j) : j i} is symmetric around the origin and does not contain it. Solving a conjecture of E. Cattani and A. Dickenstein, we prove that the linear group GL (R) acts transitively on the set of uniform and balanced configurations of m vectors. Keywords: Balanced configuration, polygon linearly regular, linear algebra 1. Introduction A configuration {v 1, v,, v m } of vectors of R is said to be balanced if for any index i {1,, m} the set with multiplicities {det(v i, v j ) : j i} is symmetric around the origin. It is said to be uniform if every pair of vectors is linearly independent. E. Cattani, A. Dickenstein and B. Sturmfels introduced the notion of balanced configuration in (CDS99; CD0) for its relationship with multivariable hypergeometric functions in the sense of Gel fand, Kapranov and Zelevinsky (see (GKZ89; GKZ90)). These functions include, as particular example, the classical Gauss hypergeometric functions, as well as the multivariable generalizations of Appell, Horn, and Lauricella (see (EMOT81)). Balanced planar configurations with at most six vectors have been classified in (CDS99). In a previous version of (CD0), E. Cattani and A. Dickenstein classified, with the help of a computer calculation, the balanced planar configurations of seven vectors. Moreover, they conjectured that any uniform balanced planar configuration is GL (R)- equivalent to a regular (n + 1)-gon (where n is a positive integer). In this note, we prove this conjecture for all n. In (CD0), our method is adapted to the case n = 3. In Section, we precisely state our result. The following section is the proof. To obtain a sketch of proof, one can read the begining of Paragraphs 3.1 to 3.6. c 004 Kluwer Academic Publishers. Printed in the Netherlands. balanced_kluwer.tex; 8/04/004; 11:30; p.1

2 . Statement of the result We start with giving a precise definition of balanced configurations: Definition. A planar configuration {v 1,, v m } is said to be balanced if for all i = 1,, m and for all x in R the cardinality of the set {j i : det(v i, v j ) = x} equals those of the set {j i : det(v i, v j ) = x}. Remark. Assume {v 1,, v m } is balanced and m is even. Then, the set {det(v 1, v j ) : j =,, m} is symmetric around 0 and its cardinality (with multiplicities) is odd; so it contains 0. Then, {v 1,, v m } is not uniform. In this note, we are only interested in uniform and balanced configurations. So, from now on, we assume that m = n + 1 for some positive integer n. Let us identify R with the field C of complex numbers. To avoid any confusion with index-numbers, we denote by 1 the complex number i. Denote by U m the set of m th -roots of 1. Set ω = e 1π m. Then, U m = {w k : k = 0, n} is uniform and balanced. Indeed, for all integers k and a, we have det(ω k, ω k+a ) = det(ω k, ω k a ). (1) One can notice that the group GL (R) acts naturally on the set of balanced (resp. uniform balanced) configurations of m vectors. Indeed, if g GL (R) then det(g.v i, g.v j ) = det(g) det(v i, v j ). The aim of this note is to prove the THEOREM 1. For any odd integer m, GL (R) acts transitively on the set of uniform balanced configurations of m vectors. In other words, modulo GL (R), U m is the only uniform balanced configuration of m vectors. 3. The proof 3.1 The set {0,, n} is denoted by I. We denote by P (I) the set of pairs of elements of I. Let us fix a uniform planar configuration C = {v 0,, v n }. balanced_kluwer.tex; 8/04/004; 11:30; p.

3 We now explain the first step of the proof in the case when for all distinct i, k and l in I det(v i, v k ) det(v i, v l ). Then, one easily check that C is balanced if and only if i, k I l I det(v i, v k + v l ) = 0. () The default of Condition () is that the quantifiers does not respect the symetry in k and l of the equation det(v i, v k + v l ) = 0. Then following lemma show that Condition () is equivalent to the better following one: k, l I i I det(v i, v k + v l ) = 0. LEMMA 1. Let us recall that C is uniform. Then, C is balanced if and only if there exists a map φ : P (I) I such that 1. {k, l} P (I) det(v φ({k,l}), v k + v l ) = 0, and. {k, l}, {k, l } P (I) (φ({k, l}) = φ({k, l }) l = l ). Proof. If E is a finite set, we denote its cardinality by E. Assume that there exists φ as in the lemma. If there exists a pair {i, l} such that φ({i, l}) = i, then det(v i, v l ) = det(v i, v i ) = 0; this contradicts the fact that C is uniform. Therefore, for all i I, φ 1 ({i}) is formed by pairs of elements of I {i}. Moreover, by Assertion these pairs are pairwise disjoint. Therefore, φ 1 ({i}) n. But, P (I) = mn and I = m. It follows that for all i I, φ 1 ({i}) = n. One easily deduces that φ 1 ({i}) is a partition of I {i}. It follows that C is balanced. Conversely, let us assume that C is balanced. Then, for all i I, there exists a (a priori non unique) part P i (I) of P (I) such that: 3 I {i} is the disjoint union of the elements of P i (I), and {k, l} P i (I) det(v i, v k + v l ) = 0. Let {k, l} P (I). Since C is uniform, the set of vectors v R such that det(v, v k + v l ) = 0 is the line generated by v k + v l and there exists at most one i I such that det(v i, v k + v l ) = 0. This means that for any i j the set P i(i) Pj (I) is empty. Moreover, P i (I) = n, for all i I. Therefore, the cardinality of i I Pi (I) equals nm, which is the cardinality of P (I). It follows that P(I) i = P (I). i I balanced_kluwer.tex; 8/04/004; 11:30; p.3

4 4 Therefore, for all {k, l} P (I), there exists a unique φ({k, l}) I such that {k, l} P φ({k,l}) (I). Then, φ satisfies Condition 1. Condition follows from the fact that the elements of P i are pairwise disjoint. Notation. From now on, we identify I from Z/mZ, by k k + mz. Since m is odd, is invertible in the ring Z/mZ. For example, with this identification, we have v 1 = v n+1 and v 1 = v n. 3. We assume that C is uniform and balanced. The second step of the proof is to show that C has the same combinatorics as U m. Precisely, we show that by relabeling the vectors, we can assume that C satisfies the equations det(v k, v k+a + v k a ) = 0 similar to Equalities (1). Definition. Each v k has a unique polar form v k = ρ k e α k 1 with ρ k in ]0; + [ and α k in [0; π[. The set C is said to be labeled by increasing arguments if there exists k I such that α k < α k+1 < < α n < α 0 < α 1 < < α k 1. LEMMA. If C is labeled by increasing arguments then the unique map φ satisfying Lemma 1 is φ({k, l}) = k+l for all {k, l} P (I). Proof. By relabeling the vectors, it is sufficient to prove that φ({0, k}) = k, for all k = n,, 1, 1,, n. By symmetry, we may assume that k = 1,, n. Let us fix such a k. Notice that the set of i I such that det(v 0, v i ) is positive (that is, such that α i α 0 < π) is of cardinality n. Therefore, since C is labelled by increasing arguments α n α 0 < π < α n+1 α 0. In the same way, we have: α n+k α k < π < α n+k+1 α k (see Figure ). With our convention, we have α 1 = α n+1 and α k 1 = α n+k. Since v φ({0,k}) belongs to R(v 0 + v k ), we have: φ({0, k}) { 1,, 3,, k 1 }. (3) This ends the proof when k = 1. We now proceed by induction on k. Let us assume that φ({i, i + j}) = i + j, for all j = 1,, k 1. By Lemma 1, φ({0, k}) φ({0, j}), for all j = 1,, k 1. Therefore, φ({0, k}) { 1,,, k 1 }. (4) balanced_kluwer.tex; 8/04/004; 11:30; p.4

5 5 α k α n π α n+1 π α 0 α n+k α n+k+1 Figure 1. The positions of α 0, α k, α n, α n+1, α n+k and α n+k+1 In the same way, φ({0, k}) φ({k j, k}), for all j = 1,, k 1. Therefore, φ({0, k}) { k + 1,,, k 1 }. (5) The lemma follows from (3), (4) and (5). We will say that C is labeled in respect with Lemma if the function φ : P (I) I = Z/mZ, {k, l} k+l satisfies Conditions 1 (and ) of Lemma 1. Lemma shows that if C is labeled by increasing arguments, then it is labeled in respect with Lemma. 3.3 In this paragraph, we assume that C is labeled in respect with Lemma ; and we give a method to construct C starting from v n, v 0 and v n. Moreover, we have det(v 0, v n + v n ) = 0. This leave us at most one degree of liberty to construct all uniform balanced configurations labeled in respect with Lemma modulo the action of GL (R). Let us start with LEMMA 3. With above notation, for all k I we have: det(v k, v k+1 ) = det(v 0, v 1 ) and det(v k, v k+ 1 ) = det(v 0, v n ). Proof. For all k I, since φ({k, k+}) = k+1, we have det(v k, v k+1 ) = det(v k+1, v k+ ). The first assertion follows immediately. For all k, we also have det(v k, v k+n ) = det(v k+n, v k+n ). Since n is prime with m = n + 1, this implies the second assertion. v 1 Set A 1 := det(v 1 = v n )., v 1 ) and A n := det(v 0, v 1 ) (let us recall that balanced_kluwer.tex; 8/04/004; 11:30; p.5

6 6 LEMMA 4. For all k Z/mZ, we have: av k 1 v k = A 1 v k 1 A n v k. Proof. Since C is uniform there exist a and b in R such that v k det(v k a = A 1 + bv k, v k. But, by Lemma 3, we have: det(v k ) = A 1 and det(v k, v k 1 A n and b = 1. =, v k 1 ) = A n, ) = A n. One easily deduces that 3.4 Any uniform balanced configuration is GL (R)-equivalent to one with v 1 = (0, 1) and v 0 = (1, 0). Then, since det(v 0, v 1 +v 1 ) = 0, v 1 = (t, 1) for one t in R. Moreover, by Lemma 4, we have: v k = tv k 1 v k. Conversely, we define a sequence (w k (t)) k 1 of vector of R with a parameter t R as follows. We start with ( ) 0 w 1 = 1 and define w k (t) by induction: ( ) 1 w 0 = 0 w 1 (t) = ( ) t, 1 k w k (t) = tw k 1 (t) w k (t). (6) Now, C is any uniform balanced configuration labeled in respect with Lemma. Formally, the above discussion gives LEMMA 5. There exits a unique g C GL (R) such that g C.v k = w k, for k = 1, 0 and a unique t C R such that g C.v 1 = w 1 (t C ). Moreover, for all k 1, we have: w k (t C ) = g C.v k. Proof. The existence and the uniqueness of g C are obvious; those of t C is a direct consequence of the equality det(v 0, v 1 + v 1 ) = 0. The configuration C := {g C.v k } k I is uniform and balanced and labeled in respect with Lemma, since C be. Then, C satisfies Lemma 4 with A 1 = det(g C.v 1, g C.v 1 ) = t and A n = det(g C.v 0, g C.v 1 ) = 1. The lemma follows. 3.5 To obtain all the GL (R)-orbits of uniform balanced configurations, it remains to find the t C which occur in Lemma 5. But, this balanced_kluwer.tex; 8/04/004; 11:30; p.6

7 lemma shows that w n+1 (t C ) = w 0. We will deduce that the parameter t C must be choosen among at most n values: LEMMA 6. Denote by (x, y ) the coordinate forms of R. Then, for all k 0, the function x (w k (t)) (resp. y (w k (t))) is a polynomial function of degree k (resp. k 1) and of same parity as k (resp. k 1). In particular, the equation w n+1 (t C ) = w 0 has at most n solutions. Proof. One easily check the first assertion by induction on k. One can notice that y (w n+1 (0)) 0. Then, the equation y (w n+1 (t)) = 0 has j solutions: t j < < t 1 < 0 < t 1 < < t j with j n. Since x (w n+1 (t)) is an odd function, at most one element of a pair ±t k is a solution of the equation x (w n+1 (t)) = 1. This ends the proof of the lemma To end the proof, we show that the equation w n+1 (t C ) = w 0 has indeed n solutions corresponding to various labeling of the regular d-gons where d runs over the divisors of n + 1. LEMMA 7. We have: 1 {t R : w n+1 (t) = w 0 } = { sin(kπ/m) : k = 1, n}. Proof. We are going to construct n balanced configurations labeled in respect with Lemma. Each one will give a solution of the equation w n+1 (t) = w 0. Consider U m = {ω k : k = 0, n}. Let us fix k {1, n}. Set ζ = ω k and denote by d the order of ζ in the group U m. Then, C k := {ζ l : l = 0,, d 1} is a uniform balanced configuration. Moreover, C k is labeled in respect with Lemma. The integer d divides m and is odd. Let e be the positive integer such that d = e + 1. Denote by g k the unique element of GL (R) such that g k.ζ 0 = w 0 and g k.ζ e = w 1. Let t k be the unique real number such that g k.ζ e = w 1 (t k ). One easily checks that t k = (sin(kπ/m)) 1. But, by Lemma 5, we have for all l 1, w l (t k ) = g k.ζ l. Since d divides m = n + 1, we deduce that w n+1 (t k ) = w Proof. [of Theorem 1] We may assume that C = {v 0,, v n } is labeled by increasing arguments. We define g C GL (R) and t C R as balanced_kluwer.tex; 8/04/004; 11:30; p.7

8 8 in Lemma 5. Then, by Lemma 7, there exists a unique k C {1, n} such that t C = (sin(k C π/m)) 1. Set ζ = ω k C. Let g kc GL (R) defined as in the proof of Lemma 7. Then, by Lemma 5, for all k = 0,, n, we have w k (t C ) = g C.v k = g kc ζ k. Since C is uniform, it follows that the order of ζ in U m is m. Theorem 1 is proved. References E. Cattani & A. Dickenstein Planar configurations of lattice vectors and GKZ-rational toric fourfolds in P 6, Preprint (00), arxiv:math.co/ E. Cattani, A. Dickenstein & B. Strumfels Rational hypergeometric functions, Preprint (1999), arxiv:math.ag/ A. Erdélyi, W. Magnus, F. Oberhettinger & F. G. Tricomi Higher transcendental functions. Vol. I, Robert E. Krieger Publishing Co. Inc., Melbourne, Fla., 1981, Based on notes left by Harry Bateman, With a preface by Mina Rees, With a foreword by E. C. Watson, Reprint of the 1953 original. I. Gelfand, M. Kapranov & A. Zelevinsky Hypergeometric functions and toral manifolds, Functional Analysis and its Applications (1989), no. 3, p , Generalized Euler integrals and A-hypergeometric functions, Advances in Mathematics (1990), no. 84, p balanced_kluwer.tex; 8/04/004; 11:30; p.8