Counting intersections of spirls on torus 1 The problem Consider unit squre with opposite sides identified. For emple, if we leve the centre of the squre trveling long line of slope 2 (s shown in the first figure below), we shortl rrive t point A, which is 3 4 of the w long the top side. We identif this point with tht is, consider it to be the sme point s point B, which is 3 4 of the w long the bottom side. Thus (s shown in the second figure) upon pssing through the top side t A we emerge from the bottom side t B. A A D C B Similrl, on pssing through point C, we emerge t point D (third figure), becuse the left nd right sides re identified. Continuing in this mnner, we eventull return to the centre of the squre (lst figure). The collection of points formed b such trvel is not wht we usull cll line ; we will cll it spirl. (This one, t lest, wrps round the squre few times.) Add to the lst figure bove second spirl, lso pssing through the centre of the squre, but of slope 1 3. Note tht, since opposite sides of the squre re identified, the four corners re ll the sme point. Thus, when pssing through corner on constnt bering, we emerge from the opposite corner. As the ltest figure shows, these two spirls hve 7 intersection points. We will show tht the number of intersection points in such figures is (with few qulifictions) given b the determinnt formed b the spirls slopes: 2 1 1 3 = 7. The proof uses onl some elementr nltic geometr nd liner lgebr, nd one theorem of number theor (stted s proposition 6, on pge 4). Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 1
2 Formultion on the plne The squre with opposite sides identified is less fmilir environment thn the Crtesin plne; we would prefer to reformulte the problem on the plne so we cn deplo our knowledge of tht surfce. We will do so b tiling the plne with copies of the squre. This tiling induces n equivlence reltion on the plne: points occuping the sme position in different squres re in certin sense the sme. For emple, when counting intersections we will wnt to tret ll the centres of these squres s just one intersection. Reclling tht the squre is unit squre, we see tht in this sense, n point is the sme s the point one unit to the right, nd the point one unit up, nd the point two units up nd three left, nd in generl, n integer mount left/ right nd n integer mount up/down. Thus we hve the following definition. Definition 1 Let be the binr reltion on R 2 given b 1 1 2 2 1 2 2 Z nd 1 2 2 Z. When 1 1 2 2, we s tht 1 1 nd 2 2 re equivlent. It is es to verif tht thus defined is indeed n equivlence reltion. It might now be nturl to define spirl in the plne s consisting of such-nd-such points in one of the squres, nd ll points equivlent to those. But there is more convenient definition, bsed on the observtion tht the line segments in ech squre in the tiling join up with the line segments in the neighbouring squres, s seen in the figure bove. In the emple on the previous pge, the jump from point A to point B is jump down b one unit; but in the tiling, jumping down one unit from point A moves us to point A of the net squre down. Thus the line segment in one squre tht strts t B will Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 2
join up with the line segment ending t A in the net squre down, forming lrger line segment. The sme thing hppens t ll the points where spirl crosses from one squre to nother, so spirl s line segments join up on the whole plne to form set of prllel lines. Accordingl we will define spirl in the plne s consisting, not of certin points in one of the squres (nd ll equivlent points), but of line in the plne (nd ll equivlent points). For the purposes of this note, nmel counting intersection points, we cn restrict our ttention to certin kinds of line. First, we will consider onl lines tht pss through the origin: if two spirls under considertion intersect t ll, we cn tke one of their intersection points s the origin; if the don t intersect we re not ver interested in them. Second, we will consider onl lines with rtionl slope. Since line through the origin psses through the point (m, n) if nd onl if its slope is n/m, hving rtionl slope is the sme s pssing through some lttice point other thn the origin. (A lttice point is point with integer coordintes.) On the tiled plne, the lttice points re ll equivlent to ech other, so in terms of the squre with opposite sides identified, hving rtionl slope is the sme s returning to one s strting point. So the restriction to lines with rtionl slope ssures us tht the spirl will not pss through the squre infinitel mn times (which would mke infinitel mn intersection points). Actull, rtionl slope is slightl imprecise; our definitions should llow verticl line through the origin (tht is, the -is), which certinl returns to its strting point, even though its slope is 1/0, which is not rtionl number. Definition 2 gcd(, b) = 1. (, b) is coprime pir if nd onl if nd b re integers nd Recll tht gcd(, 0) = for n, so in prticulr, gcd(0, 0) = 0; thus (0, 0) is not coprime pir. But (1, 0) nd (0, 1) re. Definition 3 If (, b) is coprime pir, then L(, b) is the line given b 2 L(, b) + b = 0. L(, b) is the line through the origin with slope /b (ecept tht b m be zero, in which cse the line is verticl), nd /b is not onl rtionl but in lowest terms. Now we cn define spirls s plnned: spirl consists of points tht re equivlent to the points on certin line. Definition 4 If (, b) is coprime pir, then the spirl S(, b) is the subset of the plne given b 2 S(, b) 9u, v: nd 2 L(, b). Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 3
(The ngled brckets round the eistentil sttement show eplicitl the scope of the dumm vribles u nd v.) The intersections of two spirls on the plne re, then, the elements of the set S(, b) \ S(c, d). However, we don t wish to count tht set, since it contins infinitel mn copies of ech intersection point (considering equivlent points on the plne s copies of the sme point on the squre with opposite sides identified). Leving side for now the question of how we will count ech intersection point onl once, we cn t lest stte the question: Given coprime pirs (, b) nd (c, d), how mn nonequivlent points re there in S(, b) \ S(c, d)? 3 Preliminries We will need two fmilir results, which we stte formll here for clrit. Their proofs re omitted, since the re well-known nd cn be found in introductions to the relevnt subjects. From nltic geometr we know tht the line L(, b), being the locus of the eqution +b = 0, hs the direction vector b ; in other words, L(, b) consists of the multiples of b. Proposition 5 If (, b) is coprime pir, then 2 L(, b) 9s: = s. From number theor we know tht, for n two integers, their integer liner combintions re precisel the multiples of their gretest common divisor. Proposition 6 For n integers nd b, nd n rel number, 9m, n: m 2 Z nd n 2 Z nd = m + bn gcd(, b) divides. (This proposition would usull be stted onl for integers ; the generliztion to rel is trivil.) We will lso need the following corollr of proposition 6. Proposition 7 If (, b) is coprime pir, then for n rel number, 2 Z nd b 2 Z = 2 Z. Proof Suppose (, b) is coprime pir, so tht gcd(, b) = 1. Then certinl gcd(, b) divides 1, so b proposition 6, there eist integers m nd n such tht m + bn = 1. Thus = 1 = (m + bn) = ()m + (b)n, which is n integer if nd b re. Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 4
4 Solution Definition 4 chrcterizes the points in spirl in nturl w, s the points equivlent to certin line. Eploiting proposition 6 ields simpler, if perhps less intuitive, chrcteriztion of these points. Proposition 8 If (, b) is coprime pir, then 2 S(, b) + b 2 Z. Proof If (, b) is coprime pir, then 2 S(, b) {definition 4} 9u, v: nd 2 L(, b) {definition 1} 9u, v: u 2 Z nd v 2 Z nd u v 2 L(, b) {reinde the quntifiction; let m, n := u, v} 9m, n: m 2 Z nd n 2 Z nd m n 2 L(, b) {definition 3} 9m, n: m 2 Z nd n 2 Z nd ( m) + b( n) = 0 {lgebr} 9m, n: m 2 Z nd n 2 Z nd + b = m + bn {proposition 6, with := + b} gcd(, b) divides + b {b hpothesis, (, b) is coprime pir; definition 2} 1 divides + b {ll nd onl the integers re divisible b 1} + b 2 Z As formulted bove, the problem is to count the nonequivlent points in S(, b) \ S(c, d). A tpicl w to count nonequivlent objects is to restrict our ttention to complete (or complete enough) sstem of equivlence clss representtives in other words, set of points such tht ever point we re interested in is equivlent to ectl one of the points in the set. (Hving t lest one equivlent point in the set ensures we count everthing of interest; hving t most one ensures we don t count nthing twice.) The obvious choice is the unit squre [0, 1) 2 = { : 0 < 1 nd 0 < 1}, Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 5
but we will use nother, described b the net proposition. Proposition 9 If (, b) is coprime pir, then ever point in S(, b) is equivlent to ectl one point of the form t b with 0 t < 1, nd ever point of tht form is in S(, b). Proof First we show tht such t eists for n point in S(, b). Note tht 2 S(, b) {definition 4} 9u, v: nd 2 L(, b) {proposition 5} 9u, v: nd 9s: u v = s {logic} 9s: s {definition 1} 9s: + bs 2 Z nd s 2 Z Choose such n s, nd let t = s bsc. Certinl, then, 0 t < 1. Moreover, + bt = ( + bs) bbsc 2 Z nd t = ( s) + bsc 2 Z, whence t. Net we show tht such t is unique. Suppose 0 t 1 < 1 nd 0 t 2 < 1. Then t 1 nd t2 = { is n equivlence reltion} t 1 t2 {definition 1} bt 1 + bt 2 2 Z nd t 1 t 2 2 Z { 2 Z 2 Z} bt 1 bt 2 2 Z nd t 1 t 2 2 Z {lgebr} b(t 1 t 2 ) 2 Z nd (t 1 t 2 ) 2 Z = {proposition 7, with := t 1 t 2 } t 1 t 2 2 Z { 1 < t 1 t 2 < 1} t 1 t 2 = 0 t 1 = t 2 {lgebr} Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 6
Finll, we show tht ll such points re in S(, b). Indeed, t b 2 L(, b) b proposition 5; since is refleive, t b t. Thus t 2 S(, b) b definition 4. Now we re in position to describe the intersections of two spirls. Proposition 10 d bc. Then: Suppose (, b) nd (c, d) re coprime pirs. Let = b c d = If = 0, then S(, b) nd S(c, d) coincide. If 6= 0, then ever point in S(, b) \ S(c, d) is equivlent to ectl one point of the form n, where n 2 Z nd 0 n < 1. Proof In the first cse, when = 0, we know from liner lgebr tht the rows of the mtri b c d re linerl dependent. Since neither row is zero (s noted fter definition 2), ech row is multiple of the other. Thus, s we know from nltic geometr, L(, b) nd L(c, d) coincide, whence S(, b) nd S(c, d) lso coincide. So now suppose 6= 0, nd consider the points in S(, b) \ S(c, d). In fct, rther thn considering ll of S(, b), consider just the representtive line segment described in proposition 9; tht is, consider the points of the form t b, where 0 t < 1, which re lso in S(c, d). t b 2 S(c, d) {definition of sclr multipliction} tb t 2 S(c, d) {proposition 8} tbc + td 2 Z {definition of } t 2 Z {logic} 9n: n 2 Z nd t = n { 6= 0} 9n: n 2 Z nd t = n So the intersection points of S(, b) nd S(c, d) tht lie on the represent- Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 7
tive line segment described b proposition 9 re those for which 9t: = t nd 0 t < 1 nd t 2 S(c, d) {b the clcultion bove} 9t: = t nd 0 t < 1 nd 9n: n 2 Z nd t = n {logic} 9n: = n nd 0 n < 1 nd n 2 Z tht is, the points described in the proposition. A corollr to this proposition is tht if 6= 0 then there re intersection points, which ws wht we set out to prove. Or nerl so. In proposition 10 the relevnt determinnt is formed b plcing the spirls norml vectors in rows, while the initil sttement of the result on pge 1 plced the slopes in columns (without specifing how to hndle signs). It is es to show tht these determinnts differ t most in sign. The initil sttement lso mde no mention of the need to tke the bsolute vlue of the determinnt, nor of the need to epress the slopes in lowest terms, nor of the mening of zero determinnt. 5 Alterntive pproches The determinnt ppers in the proof of proposition 10 in somewht enigmtic w. Two lterntive pproches, in which the determinnt ppers more nturll, begin with the observtion tht 2 S(, b) \ S(c, d) {proposition 8} + b 2 Z nd c + d 2 Z {mtri lgebr} b c d 2 Z 2 Tht is, the intersection points of S(, b) nd S(c, d) re precisel the preimges of lttice points under the liner trnsformtion whose mtri is b c d. In the first lterntive pproch, we consider which lttice points hve preimges on the representtive line segment described in proposition 9; if 6= 0, these re the points such tht = 1 d b mn c nd = t for suitble m, n, t. Equting these two epressions for nd ppling some mtri lgebr ields m d c + n = 0 d c + t. Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 8
Since 6= 0, the vectors d c nd form bsis for R 2 ; identifing the corresponding coefficients ields proposition 10. In the second lterntive pproch, we consider which lttice points hve preimges in the unit squre [0, 1) 2. The imge of tht squre is prllelogrm with re nd whose vertices lie on lttice points. B Pick s theorem, = I + 1 2 B 1 where B is the number of lttice points on the edges of the prllelogrm nd I is the number of lttice points in its interior. From coprimeness it cn be shown tht the onl points on the edges re the vertices; thus B = 4. The intersection points we wish to count correspond to the lttice points in the interior, plus one for the origin, tht is, I + 1. Steven Tschuk 2007 Mrch 15 http://www.motlp.org/mth/spirls.pdf 9