Ideal multigrades with trigonometric coefficients

Transcription

1 Ideal multgrades wth trgoometrc coeffcets Zarathustra Brady December 13, The problem A (, k) multgrade s defed as a par of dstct sets of tegers such that (a 1,..., a ; b 1,..., b ) a j = =1 for all postve tegers j k. A commo otato for such sets s =1 b j a 1,..., a k = b1,..., b. A soluto s sad to be deal whe = k + 1, sce t s easy to see that whe = k the oly solutos are trval. The Tarry-Escott-Prouhet Problem s to prove (or dsprove) that there s a deal multgrade of degree k for every k. Approach Let X be the varety of deal multgrades of degree k ( other words, the multgrades for whch, the umber of varables each set, s k + 1). We detfy a hyperplae of trval multgrades wth P k. Let C be ay curve cotaed that hyperplae. The we have the clusos C P k X P 1, ad f we let R be the rg Q[x 0,..., x k, y 0,..., x k ], I be the deal of C R, H the deal of the hyperplae of trval solutos, ad J be the deal of X, we have the reverse clusos R I H J. 1

2 We wat to study the deformatos of C. Accordg to [1], the space of frst order deformatos ca be detfed wth the global sectos of the ormal sheaf to C. For the deformatos that rema P k, the ormal sheaf s N C/P k = Hom(I/H, R/I), whle for deformatos that rema X the ormal sheaf s N C/X = Hom(I/J, R/I). Thus, f dm H 0 (N C/X ) > dm H 0 (N C/P k), the the curve C has a otrval frst order deformato. Hopefully, ths frst order deformato could the be exteded to a full deformato, fally gvg us a otrval ratoal curve cotaed X. 3 Prevous Results If we restrct to the case where C s a ratoal curve defed by a homogeous degree d map (s, t) (f 0 (s, t),..., f k (s, t); f 0 (s, t),..., f k (s, t)) (.e., x = y = f (s, t)), the as log as o two f are detcal C has a otrval frst order deformato wth the varety X f ad oly f deg gcd a<b,a,b (f a (s, t) f b (s, t)) ( 3) d. Ths s the result of our prevous paper. It turs out that t s more useful to subtract both sdes from d ( 1)( ), so we get deg lcm (s, t) f j (s, t) d. j(f We ca terpret ths result geometrcally as follows. Get a pece of O P 1(d) graph paper, ad draw graphs of f 0,..., f k o t. Mark every x- coordate at whch two fuctos f, f j are equal, ad the cout them wth multplcty ad compare ths sum wth d. If t s smaller tha or equal to d, the the curve C has otrval frst order deformatos. To cout wth multplcty, we ca use the followg rules of thumb: If there s more tha oe pot over a sgle x coordate where two graphs come together, we take the maxmum of the multplctes over all such pots to get the multplcty of ths x coordate. If all of the curves tersect trasversally, the the multplcty at a pot wth several curves passg through t s the umber of curves passg through t mus oe.

3 For more complcated stuatos, we ca refer back to the algebrac defto above. Ths paper wll be cocered wth the case d = 1, whch case the rules of thumb gve above are eough to properly calculate the multplcty. 4 Lear curves wth otrval deformatos Sce we are lookg for ratoal solutos to the multgrade problem, we restrct to the case f R[s, t], ad we ca assume wthout loss of geeralty that o tersectos occur at fty. Now we ca rephrase the problem terms of coutg x coordates of tersectos of oparallel, overtcal les the real plae. I clam that f ay three les pass through a pot, the all les must pass through that pot. To see ths, assume the cotrary, ad perform a projectve trasformato o the x axs to make the x coordate of ths pot occur betwee two other x-coordates where tersectos occur. Takg the two closest such x coordates o ether sde of ths x coordate, we see that the mddle le ca ot tersect ay other le at ether of these two x coordates. Now f we cout tersectos wth the mddle le wth multplcty, we get 1, ad there are at least two addtoal x coordates ot volvg the mddle le, cotradcto. Thus we have two cases: the case where all les go through a sgle pot, ad the case where all tersectos have multplcty oe. If all les go through a sgle pot, we get the curve C gve by (s, t) (s + c 0 t, s + c 1 t,..., s + c k t; s + c 0 t,..., s + c k t) descrbed the prevous paper. Now we tur to the case where all tersectos have multplcty oe. Call a x coordate specal f two les tersect above t. Call a pot o a le above ths x coordate full f t s the tersecto of two les, ad call t empty otherwse. We see that there must be at least oe empty pot by lookg at a mmal tragle formed by our les, so there are ecessarly more tha 1 specal x coordates. Thus there are specal x coordates, ad there must be exactly empty pots. Betwee two cosecutve specal x coordates, all of the les have a fxed orderg, so there s a le wth a smallest heght, ad ts heght s strctly smallest at oe of the two specal x coordates. Thus, the tersecto of ths le wth that x coordate gves us a empty pot below all of the full pots wth ths x coordate. Smlarly, for ay two adjacet specal x coordates, at least oe of them has a empty pot above all of ts full pots. Sce there are oly empty pots, they must all be ether above 3

4 or below all the full pots wth the same x coordate. Ths formato together wth the party of gves us the full cdece structure of our les, whch tur gves us the equatos of our les up to lear equvalece. Istead of solvg for the equatos of these les, we ca costruct them as follows: take a le 3 dmesoal space that does t tersect the x axs, ad rotate t aroud the x axs by the agles π for = 0,..., k. The project all of these les oto the xy plae. Ths gves us the explct equatos ( ) π f (s, t) = cos s + s ( ) π s ( ) π t. These equatos have coeffcets Q[cos ( ) π ], whch has degree φ()/ over the ratoals. = 6 s the last for whch ths s defed over the ratoals, ad for = 6 we ca rewrte our sx les as s, t, s+t, s, t, s t by a lear chage of varables. Cocdetally, t s kow [referece?] that we have ±s, ±t, ±(s + t) = 5 ±u, ±v, ±(u + v) Smlarly, for = 8 we get that f ad oly f s + t + st = u + v + uv. ±a ± b( + 1), ±a( + 1) ± b 7 = ±c ± d( + 1), ±c( + 1) ± d I geeral, we have Theorem 1. f ad oly f a + b = c + d. f 0 (s, t),..., f k (s, t) k = f 0 (u, v),..., f k (u, v) f ad oly f f 0 (s, t),..., f k (s, t) = f 0 (u, v),..., f k (u, v) Proof. Set ζ = e π. After a complex lear chage of varables, we ca wrte f = ζ s + ζ t. Now for j = 1,..., k, we have f j = ( ) j ζ j s j + ζ (j ) s j 1 t + + ζ j t j, 1 whch s 0 f j s odd ad ( j j/) s j/ t j/ f j s eve. Thus f j s completely determed by st = ( f )/ for j k. 4

5 Refereces [1] Joe Harrs ad Ia Morrso. Modul of curves, volume 187 of Graduate Texts Mathematcs. Sprger-Verlag, New York,