0 in the seond free ftor represent the sme element of A, then A A=K is nilpotent (of nilpoteny lss 2). In n erlier pper [9], we introdued group of pth

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1 Tiling with squres nd squre-tilele surfes. Rihrd Kenyon Astrt We introdue the `squre tiling group' nd use it to give neessry onditions on plnr polygon to e tilele with squres. We dene squre tilings on Riemnn surfes, nd ompute the Euliden strutures on torus whih re squre-tilele. We give neessry onditions on surfes of higher genus to e tilele. A higher-dimensionl version of the squre-tiling group yields neessry onditions on n n-dimensionl retiliner polyhedron to e tilele y oxes, eh hving k rtionl reltions mong its side lengths. We dene the retiliner sissors ongruene prolem for R n relted to ox tilings nd onstrut omplete invrint. AMS Clssition: 52C20, 52C22. Key Words: Tiling, qudrti dierentil. 1 Introdution Wht retngles n e tiled 1 y squres? This well-known prolem is in prt the motivtion for this pper. A retngle of side lengths nd n e tiled y squres if nd only if = is rtionl. The proof of this ft, originlly due to Dehn [4], (see lso [17]) does not generlize to polygons other thn retngles; however, it turns out tht there is lot more to e sid out this more generl prolem. Squre tilings re not new; prt of the interest lies in their onnetion with resistor networks ([5, 17]). Reently other interesting results were otined y Shrmm [13, 14] who otined squre tilings from ertin kinds of extreml length prolems nd uniformiztion of multiply-onneted plnr domins, nd y Cnnon, Floyd nd Prry [2] who used squre tilings to dene \nite Riemnn mpping" prolem. One other motivtion for this work is of n lgeri nture: s we shll see, squre tilings re intimtely onneted with the group H = R R, (the free produt of R with itself), nd ertin nilpotent quotient group of H. In Theorem 2 we show tht if A is ny elin group, K the norml sugroup of A A generted y ommuttors [; 0 ] where in the rst free ftor nd CNRS URM 128, Eole Normle Superieure de Lyon, 46, llee d'itlie, Lyon, Frne. 1 Our tilings re not required to e edge-to-edge tilings. 1

2 0 in the seond free ftor represent the sme element of A, then A A=K is nilpotent (of nilpoteny lss 2). In n erlier pper [9], we introdued group of pths in the plne, nd suggested tht it hd some pplitions to tiling prolems. This group is ontinuous generliztion of the (disrete) tiling groups of Conwy nd Lgris [3]. Conwy nd Lgris nd lter Thurston [16], nd others [10, 11], re le to solve ertin tiling prolems (of lttie polygons) using omintoril group theory, in prtiulr y dening wht is now known s the `tiling group' of set of tiles. Briey, given nite set T of polyominoes or other lttie polygons, one denes disrete group G whose reltors re the oundries of tiles in T, in suh wy tht polygon tiled with trnsltes of the desired tiles n e lifted to surfe in the Cyley grph of the group G. By studying the geometry of the Cyley grph one n sometimes (i.e. for ertin sets of tiles T ) give n lgorithm for the tiling prolem: given simple polygon, n it e tiled with trnsltes of tiles in T? In the present se we wish to do the sme onstrution. Here the mjor dierene is tht the `tiling group' will not e disrete group, ut n innite dimensionl topologil group. However it is rther tme s suh ojets go, nd we will e le to rry the nlysis quite fr. In prtiulr, to eh retiliner polygon P in the plne we ssoite qudrti form J(P ) on Q-vetor spe, nd show (Theorem 6) tht if P is tilele then J(P ) is symmetri nd positive semidenite. These two onditions re suient to determine preisely whih Euliden tori re squre tilele (Theorem 9); the squre-tilele tori represent dense one-dimensionl suset of the (2-dimensionl) spe of Euliden strutures on torus. We give notion of squre tiling of Riemnn surfes using holomorphi qudrti dierentils. For holomorphi qudrti dierentil! we dene n invrint J(!) whih gives neessry onditions on! to e squre-tilele. We onjeture tht these onditions re lso suient. We nd n optiml ound of 2g for the dimension of the Q-vetor spe generted y the edge lengths of squres in ny tiling of surfe of genus g (Corollry 12). In setion 6 we disuss tilings of retiliner polyhedr in R n y oxes, where the oxes hve vrious restritions on their shpes. As n exmple of our results, we generlize the retngle-tiled-y-squres result ove: if ox is tiled with oxes, eh tile hving side lengths generting rtionl vetor spe of dimension k, then the sme will e true of the lrge ox (Corollry 15). We dene the retiliner sissors ongruene prolem: two retiliner polyhedr in R n re retiliner sissors ongruent if one n ut the rst long oordinte (n? 1)-plnes nd rerrnge the resulting piees y trnsltion to form the seond polygon. We show (Theorem 16) tht the invrint J 2 N n Q R we onstrut is omplete retiliner sissors ongruene invrint. 2

3 2 The squre tiling group A retiliner polygon is losed region in R 2 ounded y simple losed polygonl urve whose sides re prllel to the xes. If P is retiliner polygon, tiling of P y squres is nite olletion T of squres with pirwise disjoint interiors, eh ontined in P, whose union overs P. We shll e onerned with the question of tileility of P y squres. Let H e the group H = R R, the free produt of R with itself. We estlish the following nottion for n element of H. We denote the elements of the rst ftor R y symols h(t) for t 2 R, nd those of the seond ftor y symols v(t) for t 2 R. Thus word in H is either 0 (the identity) or nite ontention of symols h(t i ) nd v(t j ). A word is in redued form if it is either 0 or the h nd v symols ourring in the word lternte, nd none of the t i re zero. Any word n e put into redued form y pplying repetedly the rules h(t 1 )h(t 2 ) = h(t 1 + t 2 ), v(t 1 )v(t 2 ) = v(t 1 + t 2 ), nd removing ourrenes of h(0); v(0). Note tht h(t)?1 = h(?t) nd likewise for v. It is now esy to model elements of H s retiliner polygonl pths in the plne strting t the origin. Suh pth is n ordered sequene of direted horizontl nd vertil segments: to eh horizontl segment from (x; y) to (x + t; y) ssoite the symol h(t), nd to eh vertil segment from (x; y) to (x; y +t) ssoite the symol v(t). Then to pth from the origin is ssoited unique (unredued) word in H. Note tht mny pths desrie the sme element of H: if pth ontins two onseutive horizontl or vertil segments then the word ssoited to it is not in redued form in H. In prtiulr if the pth ktrks long itself, tht is, hs two onseutive horizontl or vertil segments of opposite orienttion, then this pth orresponds to the sme element of H s the pth in whih the ktrk is removed. Thus H is in ijetion with the set of nonktrking retiliner polygonl pths strting t the origin. We n give this set produt struture mking it into group isomorphi to H. We refer to this pth group y the sme nme H. With this produt struture it is sugroup of the group of pths dened in [9]. The eliniztion of H is the group R 2. Viewed s pth group, the eliniztion is the homomorphism sending pth to its endpoint in R 2. The importne of the group H in regrds to our tiling prolem is now esily understood: given polygon P with the origin on its oundry, the urve strting t the origin nd winding one round the oundry of P n e thought of s n element of H. Given tiling of P y squres, we n write the urve s produt of lssos, tht is, produt of pths of the form xh(t)v(t)h(?t)v(?t)x?1 where x is n ritrry element of H. Eh lsso is in the norml sugroup K of H generted y ommuttors [h(t); v(t)] = h(t)v(t)h(?t)v(?t) for t 2 R. Thus if we let G def = H=K we hve: Proposition 1 If polygon P is tilele y squres, then the oundry word is trivil in the group G. 3

4 This group G is lled the squre tiling group. Understnding G nd the mp H! G is essentil in understnding squre tilings. Theorem 2 The group G is 2-step nilpotent, i.e. [G; G] is in the enter of G. Proof of Theorem 2. Let G 3 = [G; [G; G]] denote the group generted y [x; [y; z]] for x; y; z 2 G. We show tht G 3 = feg, tht is, for ll x; y; z 2 G we hve [x; [y; z]] = e. The proof involves just plying gmes with ommuttors. In prtiulr, the sme result holds if one reples R with ny elin group. It is well known (see [12]) tht the group G 3 = [G; [G; G]] is generted y [x 1 ; [x 2 ; x 3 ]] where x 1 ; x 2 ; x 3 re genertors of G. This follows esily from the identities [xy; z] = x[y; z]x?1 [x; z] (1) [x; yz] = [x; y]y[x; z]y?1 (2) Thus we need only hek tht [x 1 ; [x 2 ; x 3 ]] = e for x i eh in one of the rel ftors. Sine for ; 2 R, h() ommutes with h(), we hve for ny x tht [x; [h(); h()]] = e, nd similrly [x; [v(); v()]] = e. Thus we need only onsider elements [h(); [h(); v()]] nd [h(); [v(); h()]] (the other two ses eing symmetri). In ddition sine [v(); h()] = [h(); v()]?1, we hve tht [h(); [v(); h()]] is onjugte of [h(?); [h(); v()]]. Thus we need only hek term of the type [h(); [h(); v()]]. The proof is muh simpler to understnd using pitures. We refer the reder to gures 1 nd 2. First let us show tht [h(); [h(); v()]] = e (Figure 1). We hve nd so Also, so tht e = h(?)[h( + ); v( + )]h() = h(?)[h( + ); v()]v()[h( + ); v()]v(?)h() = [h(); v()]v()[v(); h(?)]v(?) = h()v()h(?)v()h(?)v(?)h()v(?); v(?)h(?)v() = h(?)v()h(?)v(?)h(): (3) e = [h( + ); v( + )] = [h( + ); v()]v()[h( + ); v()]v(?) = [h(); v()]v()[h(); v()]v(?) = h()v()h(? + )v()h(?)v(?)v(?); v(?)h(?)v() = h(? + )v()h(?)v(?): (4) 4

5 e= = e = = Figure 1: Proof tht [h(); [h(); v()]] = e. Comining (3) nd (4) gives h(?)v()h(?)v(?)h() = h(? + )v()h(?)v(?); Now nel the rst terms h(?) from either side; the left hnd side followed y the inverse of the righthnd side is e = h()v()h(?)v(?)h(?)v()h()v(?) = h(?)[h(); [h(); v()]]h(): This proves the rst step. Interhnging h nd v yields e = [v(); [v(); h()]] whih n e rewritten [h(); v()] = [v(?); h()] (5) We will use this form elow. The proof of the generl se [h(); [h(); v()]] = e is similr ut more omplited: we refer to Figure 2. We strt with e = [h( + + ); v( + + )] whih implies [v(?); h( + )] = [h(?); v( + )]: Comined with the ft tht this gives [v(?); h()] = [h(?); v()]; h()[v(?); h()]h(?) = v()[h(?); v()]v(?) (6) 5

6 = = + = = Figure 2: (this is the upper sequene in Figure 2). In similr fshion, writing e = [h(++); v(++)] in dierent order gives h()[h(); v()]h(?) = v(?)[v(?); h(?)]v(): (7) The ommuttor on the lefthnd side n e rewritten using (5) so tht (7) eomes h()[v(?); h()]h(?) = v(?)[v(?); h(?)]v() (8) (this is the lower sequene in Figure 2). Comining (6) nd (8) gives v()[h(?); v()]v(?) = v(?)[v(?); h(?)]v(); nd the LHS followed y the inverse of the RHS gives e = v()[h(?); v()]v(?2)[h(?); v(?)]v() = = v(?? )[v(2 + ); [h(?); v()]]v( + ): Sine ; ; were ritrry, this implies [v( 0 ); [v( 0 ); h( 0 )]] = e. Interhnging h nd v gives the desired result. Let L e the elin group (under ddition) of step funtions f : R 2! Z whih is generted y the hrteristi funtions R of (retiliner) retngles. Here we onsider two step funtions to e equivlent (i.e. they represent the sme element of L) if they dier on set of mesure 0. Rell tht H = R R. Let H 2 = [H; H], the ommuttor sugroup (generted y [x; y] where x; y 2 H). Similrly H 3 = [H; [H; H]], the sugroup of H 2 generted y [x; [y; z]], where x; y; z 2 H, nd H 2;2 = [[H; H]; [H; H]]. Both H 2 =H 3 nd H 2 =H 2;2 re elin groups. 6

7 Theorem 3 We hve H 2 =H 2;2 = L, H2 =H 3 = R Q R, nd [G; G] = R ^Q R. The elin groups R Q R nd R ^Q R re respetively the tensor produt nd the exterior produt of R, onsidered s Q-vetor spe, with itself. They re thus vetor spes over Q of unountle dimension onsisting of elements of the form P k 1 ix i y i ; respetively P k 0 1 ix i ^ y i ; where the i 2 Q nd x i ; y i 2 R, the symols x y nd x ^ y re iliner over Q, nd x ^ y is ntisymmetri: x ^ y =?y ^ x. Proof of Theorem 3. Let 2 H 2. The pth is losed loop in the plne, sine the imge of in H=H 2 = R 2 is trivil. Dene homomorphism : H 2! L s follows: () is the step funtion whose vlue t point x is the winding numer of round x. (Ignore wht hppens to points on the urve.) The mp is well dened, sine the pth orresponding to n element of H 2 is well dened up to ktrking, whih doesn't hnge (). It is lso ler tht is homomorphism. The mp is lso surjetive, sine ny (retiliner) retngle in the plne is the imge of lsso in H 2 (nd the imges of retngles generte L). Wht is the kernel of? Firstly, if = [[x; y]; [z; w]] then the winding numer of round ny point is zero. So H 2;2 ker(): On the other hnd, suppose 2 H 2 nd () = 0. Write s produt of onjugtes of elementry ommuttors, = ky 1 x i [h(s i ); v(t i )]x?1 i : (9) The imge of eh term x i [h(s i ); v(t i )]x?1 i is 1 or?1 times the hrteristi funtion of retngle R i in the plne. Tke the set of the retngles fr i g for ll i, nd sudivide eh retngle R i long the horizontl nd vertil lines extending the oundry edges of eh other retngle R j in the set fr i g. Eh R i is rened in this wy into nite set of retngles S i. Eh time we rene n R i y utting it into two retngles y horizontl or vertil line, we n rewrite the orresponding term in (9): x[h(s + s 0 ); v(t)]x?1 = (xh(s)[h(s 0 ); v(t)]h(?s)x?1 )(x[h(s); v(t)]x?1 ) for vertil line nd x[h(s); v(t + t 0 )]x?1 = (x[h(s); v(t)]x?1 )(xv(t)[h(s); v(t 0 )]v(?t)x?1 ) for horizontl line. Sine the sum of the hrteristi funtions of the R i is zero y hypothesis, eh retngle in ny of the S i reours in s mny renements S j with positive orienttion s with negtive orienttion. Using the ft tht the group H 2 =H 2;2 is elin nd the ft tht if xy?1 2 H 2 then (x[h(s); v(t)]x?1 )(y[h(s); v(t)]y?1 )?1 2 H 2;2 ; we see tht we n rerrnge the sudivided terms in the produt so tht everything nels modulo H 2;2. Thus must e in H 2;2. This proves the rst prt of the theorem. 7

8 Dene mp : L! R Q R s follows. If R is retngle with horizontl nd vertil edge lengths nd respetively, then ( R ) =. Extend y linerity to ll of L. To show tht this is well dened, if we hve two representtions P P of f 2 L s sum of hrteristi funtions of retngles, f = i f i = j g j, P then there is ommon renement s efore, tht is, there is sum f = k h k where the set of retngles of the h k renes the sets of retngles of the f i nd s j. Hene we need only show tht rening single retngle doesn't hnge. But this is ler y the ilinerity of : if retngle R is ( + 0 ) then rening it into two retngles R 0 whih is nd R 00 whih is 0 gives ( R 0 + R 00) = ( R ) = ( + 0 ) = + 0 = ( R 0) + ( R 00): Thus the mp is well dened; it is esily seen to e surjetive. To ompute the kernel, if () = 0 then gin y renement rgument we see tht is sum of terms of the form R? R 0, where R 0 is trnslte of R. Using the isomorphism etween L nd H 2;2, suh term is element [x; [h(s); v(t)]] in H 3 =H 2;2 (if R is n s t retngle). On the other hnd, s everything in H 3 =H 2;2 is in the kernel of, we hve ker = H 3 =H 2;2. Sine L=ker( ) = R Q R, the seond sttement of the theorem follows. Rell tht K is the norml sugroup of H generted y terms of the form [h(t); v(t)]. We hve K H 2, nd (K=H 2;2 ) is equl to the suspe V of R Q R generted y terms of the form t t. Thus indues n isomorphism of [G; G] = H 2 =K with R Q R=V = R ^Q R. This ompletes the proof of the theorem. Exmple 1. Let ; > 0 nd let R e n retngle. The pth round the oundry of R strting t the lower left orner is [h(); v()]. If R is squretilele, then [h(); v()] = e in [G; G]. Therefore ^ = 0. This implies tht = 2 Q. So R is tilele only if = 2 Q. Sine = 2 Q implies tht R is tilele, we hve Corollry 4 ([4]) An retngle is squre-tilele if nd only if = 2 Q. Exmple 2. When ll the lengths of polygon P re given in terms of sis of R over Q, it is esy to lulte these invrints. For the polygon in gure 3, deompose it into retngles s shown; the invrint in R ^Q R is (1 + 2 p 2) ^ (2 p 2? 1) + 1 ^ 1 + p 2 ^ (1 + p 2) = 3(1 ^ p2) 6= 0; so the polygon is not squre tilele. For polygon P with oundry urve, the ondition () = 0 is neessry for the existene of tiling ut not suient, s we shll see in the next setion. 3 The R R invrint. Dene J : H 2! R Q R y J =. If P is retiliner polygon, let (P ) e the pth orresponding to the oundry of P. Dene J(P ) = J((P )) 2 RQR. 8

9 Figure 3: The ft tht the imge of J(P ) in R ^Q R is zero is the sme s sying tht J(P ) is symmetri iliner form. Proposition 5 If P is squre-tilele, then eh edge length of eh squre is in the Q-vetor spe generted y the lengths of the edges of P. Proof. Let V e the Q-vetor spe generted y the edges of P. Then J(P ) 2 V Q V, sine there is sudivision of P into retngles hving sides P in V. If the squres in squre-tiling of P hve lengths i then we hve i i = J(P ) 2 V Q V. We lim tht this implies i 2 V for eh i: for if not, let W = V V 0 e the vetor spe generted y the i, nd let 0 i e the omponent of P P i in V 0. Then the imge of i i under the projetion W W! V 0 V 0 is 0 i 0 i, whih is nonzero unless 0 i = 0 for ll i. The invrint J gives us more informtion out whih polygons re squretilele. If P is squre-tilele P polygon tiled with squres of side i for 1 n i n, then J(P ) = 1 i i where i > 0. Let e 1 ; : : : ; e k e sis for the Q-vetor spe generted y the i. Then onsidering J(P ) s qudrti P form in the vriles e i, it is positive denite sine it n e written 2 i. In the vetor spe generted y the edges of P (whih ontins the e i ), this shows tht J(P ) is t lest positive semi-denite. (We revite J(P ) positive semi-denite s J(P ) 0.) To sum up, we hve proved Theorem 6 If P is squre-tilele, then J(P ) is symmetri positive semidenite iliner form on the Q-vetor spe generted y the edge lengths of P. Exmple. For the polygon in Figure 4 we hve J(P ) = p 2 p 2? 1 1; 9

10 Figure 4: whih is symmetri ut on the Q-vetor spe generted y e 1 = 1; e 2 = p 2, J(P ) hs signture (1;?1) nd so is not positive semi-denite. Thus P nnot e squre-tiled. 4 The squre-tilele tori Let M e losed (i.e. without oundry) Euliden 2-mnifold (y whih we men the universl over of M is R 2 with its Euliden metri.) A squre in Euliden 2-mnifold M is suset whih is the projetion of (not neessrily retiliner) squre S in the universl over R 2 of M, the projetion eing injetive on the interior of S. A squre tiling of Euliden 2-mnifold M is lolly nite overing of M y squres hving pirwise disjoint interiors. We shll e interested here only in the se M is Euliden torus. (Of the other ses, only the Klein ottle is non-trivil, ut the rgument is similr nd left to the reder.) For disrete 2-dimensionl lttie in R 2, dene T to e the torus R 2 =. 4.1 Neessry onditions In ny squre tiling of torus, the squres re ll oriented in the sme wy. Thus it mkes sense to tlk out squre tilings in given diretion on the torus. 10

11 Let us x the lttie generted y two vetors (x; y) nd (x 0 ; y 0 ) with xy 0? x 0 y > 0. Suppose the torus is tiled with retiliner squres in these oordintes. Let e ny retiliner pth from (0; 0) to (x; y), nd e ny retiliner pth from (0; 0) to (x 0 ; y 0 ). Let ; e the elements of H = R R orresponding to nd. Dene =?1?1 2 H. We lim tht J() is n invrint of, tht is, it does not depend on the hoies of the pths nd. To see this, note tht if 2 is nother pth from (0; 0) to (x; y), then?1 2 2 H 2, so tht 2?1 2?1?1 2 H 3, nd so?1 = 2?1 2 modulo H 3. Hene they represent the sme element J() 2 H 2 =H 3. A similr rgument shows invrine under the hoie of. Dene J() = J(). To ompute J(), let 2 = h(x)v(y) nd 2 = h(x 0 )v(y 0 ) (see Figure 5). Then we nd (x+x,y+y ) (x,y ) (x,y) (0,0) Figure 5: J() = (x? x 0 ) y + x (y 0? y) = x y 0? x 0 y: If the torus T is tiled with squres, the lift of the tiling to the universl over R 2 gives periodi squre-tiling of the plne, periodi with lttie. Suppose tht (0; 0) is on the oundry of squre. We n hoose pths nd lying on the oundries of the squres so tht outlines fundmentl region for the lttie. Indeed, tke ny two simple losed pths on the torus whih re disjoint exept t the origin nd re homotopi to the sis (x; y); (x 0 ; y 0 ) for the fundmentl group. Homotope these pths xing (0; 0) to simple losed pths 11

12 whih lie long the tile oundries, nd re non-rossing. Let nd e lifts of these two pths strting t the origin. Then is retiliner pth with winding numer 0 or 1 round lmost every point in the plne. The region enlosed y is tiled y squres. Thus J() is symmetri nd 0. We hve proved: Lemm 7 If T is squre tiled, then x y 0? x 0 y is symmetri nd 0. The onditions of Lemm 7 re very strong: Lemm 8 If ; ; ; d 2 R nd ^ = ^ d then either oth = nd =d 2 Q [ f1g or there is mtrix A = ( ij ) 2 SL 2 Q suh tht d = A If in ddition? d 0 then either 11 ; ; 21 or 11 ; ; 21. Proof. Lemm 13 elow shows tht ; ; ; d ll lie in Q-vetor spe of dimension t most 2. If ^ = 0 then either = 0 or = is rtionl. If ^ 6= 0 then nd generte 2-dimensionl Q-vetor spev. Also, ; d 2 Vimplies the existene of mtrix A with oeients in Q nd = A. Tht d det(a) = 1 follows from the ft tht A preserves the volume form ^ on V. The lst sttement is n esy lultion. Theorem 9 The torus T is squre-tilele if nd only if J() is symmetri nd 0. Proof. We lredy know the `only if' prt from Lemm 7. Let T = T. Suppose J() is symmetri nd 0. Firstly, if ll four oordintes x; y; x 0 ; y 0 re in the sme 1-dimensionl Q-vetor spe then it is trivil to tile T with squres. Seondly, if x ^ y 0 = 0 (then x 0 ^ y = 0) ut we re not in the rst se ove then x y 0? x 0 y 0 implies xy 0 0 nd x 0 y 0. Figure 6 shows fundmentl domin in the se x; y 0 ; y > 0 > x 0 ; oth retngles tiling the fundmentl domin re rtionl, so the torus hs rtionl tiling. This fundmentl domin n e extended to the se x; y 0 ; y 0 x 0. The other hoies of signs re similr. p q Lstly suppose x ^ y 0 6= 0 nd x 0 ^ y 6= 0. Let A = 2 SL r s 2 Q e the mtrix given y Lemm 8, so tht x 0 p = y r q s : x y 0 : (10) Assume p; s 0 q; r. (For the other se p; s 0 q; r, reet the lttie in the x-xis, nd hnge (x 0 ;?y 0 ) to (?x 0 ; y 0 ). This hs the eet of multiplying 12

13 (x,y ) x y x (x,y) y (0,0) Figure 6: the mtrix A y?i. The new torus is tilele if nd only if the old one ws, nd now we hve p; s 0 q; r). If p < 1 or s < 1 we still hve ps 1 sine ps? qr = 1 nd qr 0. Let = m=n e rtionl so tht p 1 nd s= 1; Then onsider the torus T 0 with lttie generted y (x; y)=m nd (x 0 ; y 0 )=n; its mtrix is m A 0 = n p q n r m s : For this mtrix the digonl entries re 1. We onstrut elow tiling of T 0. The originl torus T is mn-sheeted over of T 0, so T is lso squre tilele. If y < 0, hnge sign of oth (x; y) nd (x 0 ; y 0 ). This doesn't hnge the lttie or the mtrix A, so we n ssume y 0. So we hve p 1 nd s 1, q 0; r 0 nd y 0. If r = 0 then ps = 1 implies p = s = 1. Then y = y 0 nd fundmentl region for T is (x? x 0 ) y retngle (with verties (x 0 ; 0); (x; 0); (x; y); (x 0 ; y)). Sine x? x 0 = x? (x + qy 0 ) =?qy 0 =?qy, the retngle hs rtionl edge rtio?q, nd so T is squre-tilele. If r 6= 0, dene Here ; ; re rtionl nd 0. = 1? p ; =?1 r r ; = 1? s : r 13

14 There re now two ses. In se y 0? y 0, Figure 7 shows fundmentl domin for the torus T with hving sis (x; y) = (y 0 + (y 0? y); y); (x 0 ; y 0 ) = (?y + (y 0? y); y 0 ): Sine ; nd re rtionl this torus n e tiled y squres (the three retngles shown hve edge rtios ; nd ). In se y 0? y < 0, Figure 7 shows fundmentl domin tiled with 3 retngles of edge rtios ; ;. y (y y) (y y ) y y y y y y y y y y (0,0) (0,0) Figure 7: y This ompletes the proof. If we ignore the diretion nd just onsider the question of whih ltties give squre-tilele tori in some diretion, we rst notie tht: if T is squretilele, then for ll 2 C? f0g, so is T, where is the lttie fx j x 2 g. In other words we n ssume our lttie is generted y the vetors 1 nd z 2 C, where z is in the upper hlf plne. Let T z e the torus with this lttie. Theorem 10 The torus T z is squre tilele if nd only if z lies on either irle in R 2 of rtionl enter nd rtionl rdius, not rossing the rel xis, or on horizontl line t rtionl distne from the x-xis. Proof. Let us ompute those z orresponding to tori T with generted y (x; y) nd (x 0 ; y 0 ), stisfying x ^ y 0 = x 0 ^ y nd x y 0? x 0 y 0. 14

15 First of ll if y 0 =x nd x 0 =y re nite nd rtionl, y 0 =x = p nd x 0 =y = q then z = x0 + iy 0 x + iy nd setting t = x=y this eomes = qy + ipx x + iy ; z = pi + q + p t + i : Also x y 0? x 0 y = p(x x)? q(y y), so tht either y=x is rtionl or p 0 q. In the rst se z is rtionl nd in the seond s t vries in R? f0g, z(t) desries irle of enter (p? q)i=2 nd rdius jq + pj=2, exept for the points?qi when t = 0 nd ip when t = 1. These points re overed y the se x or y = 0, elow. This irle stises the onlusion of the theorem. Seondly suppose one of x or y is zero. Assume y = 0; then y 0 2 Qx, xy 0 > 0 nd x 0 n e ny vlue. Thus z = x0 +iy 0 x is horizontl line in the upper hlf plne of rtionl distne y 0 =x from the x-xis. When x = 0 then x 0 2 Qy, x 0 y 0 nd y 0 n e ny vlue. Then x0 +iy 0 iy desries horizontl line of rtionl height?x 0 =y > 0. Lstly, ssume tht oth x=y 0 nd x 0 =y re not rtionl. Let B e the mtrix B = 1 1 q ; s?r 1 x giving (x 0 ; y 0 0 x ) in terms of (x; y): y 0 = B (here the q; r; s re s in y (10); rell tht ps 1 so tht s 6= 0). We hve z = x0 + iy 0 x + iy nd letting t = x=y we hve x + qy + i(?rx + y) = ; s(x + iy) z = 1 s + q? irt s(t + i) = 1 s? ir s + q? r s(t + i) : (11) Relling tht qs 0 nd rs 0, the eqution (11) s t rnges over R? f0g denes irle (minus 2 points) entered t 1 s? i r+q 2s with rdius jq?rj 2s. Sine r s 0 nd q s 0 the irle is ontined in the losure of the upper hlf plne. The two missing points t t = 0 nd t = 1 re rtionl. Furthermore for ny irle of rtionl enter nd rdius whih is ontined in the losure of the upper hlf-plne, either the enter is pure imginry, in whih se the rst se of this proof gives tiling, or there is triple s; q; r s ove to whih it orresponds. Exmple. Tke the irle with enter 1 + i nd rdius 1. It rosses the y- 2 xis t the point (1? p 3)i. Thus the retngulr torus of length 1 nd height 2 1? p 3 is squre-tilele. Working kwrds through these two proofs, one n 2 onstrut tiling (Figure 8). In ontrst, it is not hrd to show tht for exmple the equilterl torus, for whih z = 1 + i p 3, is not tilele in ny diretion

16 1 1 3/2 (0,0) Figure 8: Tiling of retngulr torus with height z. 5 Squre-tilele surfes When we del with Riemnn surfes whih don't hve Euliden metris we need to generlize gin our notion of squre tiling. For kground in qudrti dierentils, the reder n onsult [6, 1]. Let e Riemnn surfe nd! holomorphi qudrti dierentil on. Rell tht this mens in lol oordinte! n e written! = (z)dz 2, where (z) is holomorphi. If = (z) is nother lol oordinte with! = ()d 2 then the oordinte hnge trnsformtion is given y ()(d=dz) 2 = (z). The set of zeros Z of! (points where (z) = 0) hs nite rdinlity nd is independent of hoie of lol oordinte. The order of eh zero is lso welldened. If z is lol oordinte for! = (z)dz 2, the metri dened y ds = j(z)j 1=2 jdzj is t wy from the zeros of. Thus to holomorphi dierentil is ssoited nonil t metri on? Z. Furthermore, t zero of! of order m, the metri extends to one-metri of ngle (m + 2). Lstly, the metri ds dmits prllel line eld, whose lines re the lines on whih! is rel nd positive. It is lled the horizontl line eld. Another wy of sying this is tht the liner holonomy group of the metri is f1g. If ll the zeros of! hve even order, the holonomy group is trivil nd there is prllel vetor eld on? Z. In this se! hs well-dened squre root p!, whih is holomorphi one-form on. Let M e Riemnn surfe with holomorphi qudrti dierentil!, nd ssoited t metri ds (hving onil singulrities of ngles multiples of 16

17 greter thn 2). A squre on suh surfe is the losure of n open suset U isometri (in the metri ds ) to n open squre (0; ) (0; ) in the Euliden plne, nd hving side prllel to the horizontl line eld. We llow squre to hve some oundry identitions. A squre tiling is s efore lolly nite overing of M y squres with pirwise disjoint interiors. Note tht the ondition tht the squres e oriented prllel to the horizontl line eld is not serious, sine for ngle # the qudrti dierentil e i#! indues the sme t metri ut hs horizontl line eld rotted y #=2 with respet to the horizontl line eld for!. 5.1 The J invrint Suppose the zeros of! hve even order. For eh 1-yle g 2 H 1 (; Z) let [g] = g x + ig y 2 C denote its period, dened y [g] = Z g p!: We dene J-invrint for (;!) s follows: let 1 ; 1 ; : : : ; g ; g e genertors for the fundmentl group of, with Q [ i ; i ] = 1. Dene J(!) = P ( ix iy? ix iy ), where [ i ] = ( ix ; iy ) nd [ i ] = ( ix ; iy ). If the zeros of! do not ll hve even order, let 0 e rnhed over of, rnhed of degree 2 over the zeros of odd order. The dierentil! lifts to dierentil! 0 on 0 whih hs zeros of even order. Dene J(!) = 1 n J(!0 ), where n is the numer of sheets of the over. Theorem 11 If the pir (;!) is squre-tilele then J(!) is symmetri nd 0. Corollry 12 If (;!) is squre-tiled then the Q-vetor spe generted y the lengths of the sides of the squres is of dimension 2g, where g is the genus of. Furthermore, there is squre-tiling of genus-g surfe whih relizes the ound 2g. Proof of Theorem 11. Suppose tht! hs some zeros of odd order. If (;!) is squre-tilele then so is ( 0 ;! 0 ): squre-tiling lifts to squre tiling of the doule over. Thus we need only onsider the se when the zeros of! re ll of even order. Tke ny tiling of y retngles fr j g. All the retngles hve the sme orienttion with respet to the prllel vetor eld. If j nd j re the horizontl nd vertil side lengths (wrt the metri ds) of retngle R j, we lim tht J(!) = P j j j. Thus if is squre-tilele, J(!) = P j j nd we re done. The rst step is to ut prt long horizontl nd vertil line segments nd ly it t on the plne to mke (possily nonsimple) retiliner polygon P. As in the se of the torus, J(P ) does not depend on the utting: for ny hoie of uts, one n sudivide the retngultion R j so tht the uts re 17

18 long the oundries of the retngles; thus there is P n indued tiling of P y the sme set of retngles, nd we see tht J(P ) = j j. It remins to show tht J(P ) = J(!). For this x se point p nd hoose 2g losed pths sed t p, disjoint exept t p, homotopi to the genertors 1 ; 1 ; : : : ; g ; g of the fundmentl group, suh tht for eh zero of! there is pth pssing through it. We n hoose these pths to onsist of horizontl nd vertil segments (ny pth n e pproximted ritrrily losely in the C 0 topology y suh pth). The only diulty is ner the vertex p: we must relx the requirement tht the pths e disjoint nd only require tht they re disjoint exept in smll neighorhood of p, where they re required to e nonrossing. When we ut long these pths (we denote the pths orresponding to genertors i ; i y the sme nmes i ; i ) the resulting spe X is topologilly disk, metrilly t (sine there re no one singulrities in its interior). When we develop X on the plne, the imge is self-overlpping polygon P 0 ounded y non-simple polygonl urve. (Alterntively, let P 0 e the imge of X under the essentilly unique lol isometry mpping X to R 2.) The oundry of P 0 is the pth 1 1?1 1?1 1 : : : g g?1 g?1 g. Sine the one ngles t the singulrities re multiples of 2, the supths i i?1 i?1 i for eh i re losed loops, regrdless of whether or P not i or i pssed through singulrity. So we nd y linerity J(P 0 ) = i J( i i?1 i?1 i ). Furthermore J( i i?1 i?1 i ) only depends on the vlues of pths i nd i in H=H 2 = R 2, tht is, on the endpoints P ( ix ; iy ) nd ( ix ; iy ) of the pths i ; i. From Figure 5 we see tht J(P 0 ) = i ix iy? ix ij. This ompletes the proof. Proof P of Corollry 12. We sw tht P if (;!) ws tiled y squres then g 1 ix iy? ix iy = J(!) = J(P ) = j j j, where f j g is the set of side lengths of the squres in the tiling. We sw in Proposition 5 tht this implies j 2 V, where V is the vetor spe generted y the ix ; iy ; ix ; iy. On the other hnd, the spe V hs dimension t most 2g, y Lemm 13, elow. To onstrut squre tiling using squres of sides generting 2g-dimensionl Q-vetor spe, tke g squre-tiled tori, eh tiled with squres of sides generting 2-dimensionl Q-vetor spe (s in Figure 7, for exmple). For eh i sle the ith torus y rel numer i so tht the sides of squres in dierent tori re rtionlly independent. Tke suiently smll nd slie eh torus long horizontl segment of length ontined in one of the sides of the squres in its tiling. Then the ith torus hs oundry loop onsisting of two segments `i; `0i eh of length. Glue this set of tori together long their oundries, gluing `0i to `i+1modg in n orienttion-preserving fshion for ll i. This mkes losed genus g surfe with squre-tiling hving the desired property. Lemm 13 P If x i ; y i re elements of vetor spe V over eld K for 1 k i k, nd 1 x i ^ y i = 0 then the suspe W generted y x 1 ; y 1 ; : : : ; x k ; y k hs dimension t most k. 18

19 Proof. The proof is y indution. If k = 1 the result is ler. Suppose it is true up to k? 1. P k?1 Suppose the y i re not ll independent: reindex so tht y k = i=1 q iy i, where q i 2 K. Then 0 = k?1 X i=1 (x i + q i x k ) ^ y i ; nd so y indution the set of vetors fx i + q i x k ; y i g for 1 i k? 1 genertes vetor spe W 0 of dimension k?1. Sine y k 2 W 0 ; we hve W = W 0 Kx k whih hs dimension t most k y indution. On the other hnd if the y i re ll independent, let w = y 1 ^ y 2 ^ : : : ^ y k. Then x i ^ w ontins the term x i ^ y i ; repling this term in the produt y? P j6=i x j ^ y j, we see tht everything nels. Hene x i ^ w = 0 whih implies tht x i is in the spn of fy 1 ; : : : ; y k g. Sine this is true for ny i we re done. To lose this setion we would like to onjeture tht, s in the se of the torus, the onditions in Theorem 11 re suient for surfe to e squretilele. However the rther rutish methods of Theorem 9 re less esily pplied here sine there re mny more ses to onsider. 6 Tiling with oxes. A ox in R n is losed set of the form [x 1 ; x ] : : : [x n ; x n + n ], where i > 0 is the ith edge length. A retiliner polyhedron in R n is polyhedron (homeomorphi to the ll B n ) eh of whose fes is prllel to oordinte (n? 1)-plne. We shll e interested in the prolem of tiling retiliner polyhedr with restrited fmilies of oxes. A ox is lled k-rtionl if the set of its edge lengths genertes vetor spe over Q of dimension k. For exmple, n n-ue is 1-rtionl. For retiliner polyhedron P 2 R n we dene n invrint J(P ) 2 N n Q R in similr mnner s in setion 3: sudivide P into oxes B i of edge lengths i1 ; i2 ; : : : ; in ; then J(P ) = X i i1 i2 : : : in : The rgument of Theorem 3 generlizes esily to this se to show tht J does not depend on the sudivision hosen. For ny suset S f1; 2; : : : ; ng dene the symol ( 1 2 : : : n ) S where i 2 R, to e Q-liner in eh oordinte nd ntisymmetri in ny permuttion of oordintes 1; : : : ; n whih is the identity outside S. For exmple if S = f1; 2; : : : ; ig then ( 1 2 : : : n ) S = ( 1 ^ 2 ^ : : : ^ i ) i+1 : : : n : 19

20 For eh S dene f S : N n Q R! J n S R y f S ( 1 : : : n ) = ( 1 2 : : : n ) S : The f S re liner mps for ll S. Dene J S (P ) = f S (J(P )). Note tht if J S (P ) = 0 then J S 0(P ) = 0 for ll S 0 ontining S. Theorem 14 A retiliner polyhedron P n e tiled y k-rtionl oxes only if J S (P ) = 0 for ll S of rdinlity k + 1. Corollry 15 A ox B n e tiled y k-rtionl oxes if nd only if B is k- rtionl. Proof of Theorem 14. If s1 ; : : : ; sk+1 generte Q-vetor spe of dimension less thn k + 1 then s1 ^ : : : ^ sk+1 = 0. Thus for eh k-rtionl ox in R n nd eh S of rdinlity k + 1 we hve ( 1 : : : n ) S = 0. If P is retiliner polyhedron P tiled y k-rtionl oxes B i nd rd(s) = k + 1 we hve J S (P ) = i J S(B i ) = 0. This invrint J n e used to give similr informtion out tilings with oxes hving other shpe restritions, for exmple oxes hving squre fe, ue fe, two squre fes, nd so on. 6.1 Retiliner sissors ongruenes Two retiliner polyhedr P 1 ; P 2 re sid to e retiliner sissors ongruent if one n ut P 1 long plnes prllel to the oordinte (n? 1)-plnes, nd trnslte the resulting piees round to form P 2. This is very restritive form of sissors ongruene; the usul sissors ongruene llows utting long ritrry diretions, nd rottions of the piees. Sissors ongruene is n old nd well-studied prolem (it is in ft Hilert's 3rd prolem): in the plne, the re is omplete invrint; in higher dimensions there is in ddition the Dehn invrint [7]. Sissors ongruene llowing only trnsltions of the piees is lso wellstudied prolem; in this se there is nother invrint, the Hdwiger invrint, generlizing the Dehn invrint, whih together with the volume is omplete invrint; see [8]. Retiliner sissors ongruene s dened ove is distint from these prolems, ut n e pprohed in similr mnner. Theorem 16 The invrint J is omplete retiliner sissors ongruene invrint. In other words, two retiliner polyhedr P 1 nd P 2 re retiliner sissors ongruent if nd only if J(P 1 ) = J(P 2 ). Proof. This follows immeditely from the seond prt of Theorem 3; it is ler tht J(P ) is invrint under retiliner sissors ongruene; on the other hnd if J(P 1 ) = J(P 2 ) then there re renements of P 1 nd P 2 so tht eh ox in the renement of P 1 ours with the sme multipliity in the renement of P 2. 20

21 7 open prolems 1. Theorem 11 ove desries suset of the otngent undle to the Teihmuller spe T g of losed surfe of genus g. It would e nie to understnd the projetion of this suset to T g itself. In prtiulr, given two t metris with onil singulrities on losed surfe of genus g, n one tell if they re onformlly equivlent? 2. Sht nd Voevodsky [15] showed tht the set of Riemnn surfes whih hve edge-to-edge tilings y equilterl tringles re preisely those dened y omplex lgeri urve with lgeri oeients. Is there n lgeri desription of the set of squre-tilele surfes? Referenes [1] W. Aiko; The rel-nlyti theory of Teihmuller spe, Springer Leture notes no. 820, [2] J. W. Cnnon, W. J. Floyd, W. R. Prry; Squring retngles: the nite Riemnn mpping theorem. preprint. [3] J.H.Conwy, J. C. Lgris; Tilings with polyominoes nd omintoril group theory, J. Comin. Theory Ser. A. 53 (1990): [4] M. Dehn; Zerlegung von Rehteke in Rehteken, Mth. Annlen 57 (1903): [5] D. G. Med, S. K. Stein; More on retngles tiled y retngles, Am. Mth. Monthly 100 no. 7 (1993): [6] F. Grdiner; Teihmuller Theory nd Qudrti Dierentils, Wiley Intersiene, [7] H. Hopf, Dierentil Geometry in the Lrge, Springer Leture Notes in Mth no [8] B. Jessen, A. Thorup; The lger of polytopes in ne spes, Mth. Snd.43, (1978): [9] R. Kenyon; The group of pths in R 2. preprint. [10] C. Kenyon, R. Kenyon; Tiling polygon with retngles, Pro. 33rd FOCS, (1992): [11] J. C. Lgris, D. S. Romno; A polyomino tiling prolem of Thurston nd its ongurtionl entropy. J. Com. Theory Ser. A 63, (1993):

22 [12] W. Mgnus, A, Krrss, D. Solitr; Comintoril Group Theory: presenttions of groups in terms of genertors nd reltions. 2nd ed. Dover pulitions, New York. [13] O. Shrmm; Squre tilings with presried omintoris. Isr. J. Mth 84 (1993), [14] O. Shrmm; Trnsoundry extreml length, preprint. [15] G. B. Sht, V. A. Voevodsky; Equilterl tringultions of Riemnn surfes nd urves over lgeri numers, Soviet Mth. Dokl. 39 (1989) no. 1, [16] W. P. Thurston; Conwy's Tiling Groups, Am. Mth. Monthly, 97 (1990): [17] R. L. Brooks, C. A. B. Smith, A. H. Stone, W. T. Tutte; Disseting retngle into squres, Duke Mth J. 7 (1940)