Proof of Atiyah's conjecture for two special types of configurations

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1 Electronic Journal of Linear Algebra Volume 9 Volume 9 (2002) Article Proof of Atiyah's conjecture for two special types of configurations Dragomir Z. Djokovic dragomir@herod.uwaterloo.ca Follow this and additional works at: Recommended Citation Djokovic, Dragomir Z.. (2002), "Proof of Atiyah's conjecture for two special types of configurations", Electronic Journal of Linear Algebra, Volume 9. DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

2 PROOF OF ATIYAH S CONJECTURE FOR TWO SPECIAL TYPES OF CONFIGURATIONS DRAGOMIR Ž. D OKOVIĆ Abstract. To an ordered N-tuple (x 1,...,x N ) of distinct points in the three-dimensional Euclidean space Atiyah has associated an ordered N-tuple of complex homogeneouspolynomials (p 1,...,p N )intwovariablesx, y of degree N 1, each p i determined only up to a scalar factor. He hasconjectured that these polynomialsare linearly independent. In thisnote it isshown that Atiyah sconjecture istrue for two special configurationsof N points. For one of these configurations, it is shown that a stronger conjecture of Atiyah and Sutcliffe is also valid. Key words. Atiyah sconjecture, Hopf map, Configuration of N pointsin the three-dimensional Euclidean space, Complex projective line. AMS subject classifications. 51M04, 51M16, 70G25 1. Two conjectures. Let (x 1,...,x N ) be an ordered N-tuple of distinct points in the three-dimensional Euclidean space. Each ordered pair (x i,x j )withi j determines a point x j x i x j x i on the unit sphere S 2.IdentifyS 2 with the complex projective line by using a stereographic projection. Hence one obtains a point (u ij,v ij ) on this projective line and a complex nonzero linear form l ij = u ij x + v ij y in two variables x and y. Define homogeneous polynomials p i of degree N 1by p i = l ij (x, y), i =1,...,N. (1.1) j i Conjecture 1.1. (Atiyah [2]) The polynomials p 1,...,p N are linearly independent. Atiyah [1], [2] has observed that his conjecture is true if the points x 1,...,x N are collinear. He has also verified the conjecture for N =3. ThecaseN = 4 has been verified by Eastwood and Norbury [4]. For additional information on the conjecture (further conjectures, generalizations, and numerical evidence) see [2], [3]. In order to state the second conjecture, one has to be more explicit. Identify the three-dimensional Euclidean space with R C and denote the origin by O. Following Eastwood and Norbury [4], we make use of the Hopf map h : C 2 \{O} (R C)\{O} defined by h(z,w) =(( z 2 w 2 )/2,z w). Received by the editorson 26 May Accepted for publication on 17 July Handling editor: Shmuel Friedland. Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, N2L 3G1, Canada (djokovic@uwaterloo.ca). The author wassupported in part by the NSERC Grant A

3 Atiyah sconjecture 133 This map is surjective and its fibers are the circles {(zu,wu) :u S 1 },wheres 1 is the unit circle. If h(z,w) =(a, v), we say that (z,w) is a lift of (a, v). For instance, we can take λ 1/2 (λ, v), λ = a + a 2 + v 2, as the lift of (a, v). Assume that our points are x i =(a i,z i ). For the sake of simplicity assume that if i<jand z i = z j then a i <a j. As the lift of the vector x j x i, i<j,wechoose 1 λij (λ ij, z j z i ), where λ ij = a j a i + (a j a i ) 2 + z j z i 2. According to the recipe in [2], [3], [4], we always use the lift ( w, z) for the vector x i x j if (z,w) has been chosen as the lift of x j x i. Hence we introduce the linear forms l ij (x, y) =λ ij x +( z j z i )y, i < j; l ij (x, y) =(z j z i )x + λ ji y, i > j. Define P to be the N N coefficient matrix of the binary forms p i (x, y) defined by (1.1) using the above l ij s. The second conjecture that we are interested in can now be formulated as follows. Conjecture 1.2. (Atiyah and Sutcliffe [3, Conjecture 2]; see also [4]) If r ij = x j x i,then det(p ) i<j(2λ ij r ij ). As 2λ ij r ij = λ 2 ij + z j z i 2, this conjecture can be rewritten as det(p ) i<j ( λ 2 ij + z j z i 2). (1.2) Obviously, this conjecture is stronger than Conjecture Two special cases of Atiyah s conjecture. We shall prove Atiyah s conjecture in the following two cases: (A) N 1ofthepointsx 1,...,x N are collinear. (B) N 2ofthepointsx 1,...,x N are on a line L and the line segment joining the remaining two points has its midpoint on L and is perpendicular to L.

4 134 D.Ž. D oković Let L and M be two perpendicular lines in the three-dimensional Euclidean space intersecting at the origin, O. LetN = m + n and assume that the points x 1,...,x m are on L and x m+1,...,x N are on M but not on L. Sety j = x m+j for j =1,...,n. Without any loss of generality, we may assume that L = R {0} and M = {0} R. Write x i =(a i, 0) for i =1,...,m and y j =(0,b j )forj =1,...,n. We may also assume that a 1 <a 2 < <a m and b 1 <b 2 < <b n. The lifts of the nonzero vectors x j x i, i, j {1,...,N} are given in Table 2.1, wherewehaveset λ ij = a i + a 2 i + b2 j. Vectors Index restrictions Lifts Linear forms x r x i 1 i<r m (2(a r a i )) 1/2 (1, 0) 2(a r a i )x x i x r 1 i<r m (2(a r a i )) 1/2 (0, 1) 2(a r a i )y y s y j 1 j<s n (b s b j ) 1/2 (1, 1) (b s b j )(y + x) y j y s 1 j<s n (b s b j ) 1/2 ( 1, 1) (b s b j )(y x) x i y j 1 i m, 1 j n λ 1/2 ij (λ ij, b j ) λ ij x b j y y j x i 1 i m, 1 j n λ 1/2 ij (b j,λ ij ) b j x + λ ij y Table 2.1 The lifts of the vectors x j x i. The associated polynomials p i (up to scalar factors) are given by p i (x, y) =x m i y i 1 n j=1 p m+j (x, y) =(y + x) n j (y x) j 1 (b j x + λ ij y), 1 i m; (2.1) m (λ ij x b j y), 1 j n. (2.2) Theorem 2.1. Conjecture 1.1 is valid under the hypothesis (A). Proof. Inthiscasewehaven = 1. Without any loss of generality we may assume that b 1 = 1. After dehomogenizing the polynomials p i (or p i ) by setting x =1, we obtain the polynomials: y i 1 (1 λ i y), 1 i m; (y + λ i ),

5 Atiyah sconjecture 135 where λ i = λ i1 > 0. The coefficient matrix of these polynomials is 1 λ λ λ 3 0 0, λ m E m E m 1 E m 2 E m 3 E 1 1 where E k is the k-th elementary symmetric function of λ 1,...,λ m. Its determinant, 1+λ m E 1 + λ m 1 λ m E λ 1 λ 2 λ m E m, is positive. Theorem 2.2. Conjecture 1.1 is valid under the hypothesis (B). Proof. Inthiscasen =2andb 1 + b 2 = 0. Without any loss of generality we may assume that b 1 = 1. After dehomogenizing the polynomials p i (or p i ) by setting x = 1, we obtain the polynomials: y i 1 (1 λ 2 i y 2 ), 1 i m; (y +1) (y + λ i ), (y 1) (y λ i ), where λ i = λ i1 > 0. The coefficient matrix of these polynomials is 1 0 λ λ , λ 2 m Ẽ m+1 Ẽ m Ẽ m 1 Ẽ 2 Ẽ 1 1 ( 1) m+1 Ẽ m+1 ( 1) m Ẽ m ( 1) m 1 Ẽ m 1 Ẽ 2 Ẽ1 1 where Ẽk is the k-th elementary symmetric function of 1,λ 1,...,λ m. Its determinant is 2pq where p =1+λ 2 mẽ2 + λ 2 m 2 λ2 mẽ4 +, q = Ẽ1 + λ 2 m 1Ẽ3 + λ 2 m 3 λ2 m 1Ẽ5 +, and thus it is positive. 3. Atiyah and Sutcliffe conjecture is valid in case (A). In the general setup of the previous section, the Conjecture 1.2 asserts that det(p ) 2 (n 2) i,j ( λ 2 ij + b 2 ) j. (3.1)

6 136 D.Ž. D oković where P is the coefficient matrix (of order N = m + n) of the polynomials (2.1) and (2.2). In case (A) this inequality takes the form 1+λ m E 1 + λ m 1 λ m E λ 1 λ 2 λ m E m (1 + λ 2 i ), (3.2) where, as in the proof of Theorem 2.1, we assume that b 1 = 1 ande k denotes the k-th elementary symmetric function of λ 1,...,λ m.thuswehave λ i = a i + 1+a 2 i > 0 and λ 1 <λ 2 < <λ m. (3.3) Let E (2) k denote the k-th elementary symmetric function of λ 2 1,...,λ2 m. In view of (3.3), we have λ m k+1 λ m k+2 λ m E k E (2) k, 0 k m. The inequality (3.2) is a consequence of the inequalities just written since m (1 + λ 2 i )= E (2) k. Hence we have the following result. Theorem 3.1. Conjecture 1.2 is valid in case (A). In case (B) the inequality (3.1) takes the form: ( ) 1+λ 2 mẽ2 + λ 2 m 2 λ2 mẽ4 + )(Ẽ1 + λ 2 m 1Ẽ3 + λ 2 m 3 λ2 m 1Ẽ5 + ( ) 1+λ 2 2 i, where Ẽk are as in the proof of Theorem 2.2. It is easy to verify that this inequality holds for m = 1, but we were not able to prove it in general. If we set all λ i = λ>0, then the above inequality specializes to (1 + λ 2 ) m + ( ) m (λ 4k+3 λ 4k+2 ) 2k +1 k 0 (1 + λ 2 ) m k 0 ( ) m 2k +1 k=0 (λ 4k+2 λ 4k+1 ) (1 + λ 2 ) 2m.

7 Atiyah sconjecture 137 Since k 0 ( ) m (λ 4k+3 λ 4k+2 )= 1 2k +1 2 (λ 1) [ (1 + λ 2 ) m (1 λ 2 ) m], it is easy to verify that the specialized inequality is valid. REFERENCES [1] M. Atiyah. The geometry of classical particles. In Surveys in Differential Geometry, Vol.VII, S.-T. Yau, ed., International Press, 2000, pp [2] M. Atiyah. Configurationsof points. Philosophical Transactions of the Royal Society of London, Series A, 359: , [3] M. Atiyah and P. Sutcliffe. The geometry of point particles. arxiv:hep-th/ v2, 15 October [4] M. Eastwood and P. Norbury. A proof of Atiyah sconjecture on configurationsof four pointsin Euclidean three-space. Geometry and Topology, 5: , 2001.