Coincidences of Hypercubic Lattices in 4 dimensions

Transcription

1 Concdences of Hypercubc Lattces n 4 dmensons P. Zener Insttute for Theoretca Physcs & CMS, TU Wen, Wedner Hauptsraße 8 0, 040 Venna, Austra May 7, 006 Abstract We consder the CSLs of 4 dmensona hypercubc attces. In partcuar, we derve the concdence ndex Σ and cacuate the number of dfferent CSLs as we as the number of nequvaent CSLs for a gven Σ. The hypercubc face centered case s deat wth n deta and t s sketched how to derve the correspondng resuts for the prmtve hypercubc attce. Introducton Concdence ste attces (CSL) for three dmensona attces have been studed ntensvey snce they are an mportant too to characterze and anayze the structure of gran boundares n crystas ([, ] and references theren). For quascrystas these concepts have to be adapted. Snce a ot of quasperodc structures can be obtaned by the we known cut and proecton scheme [3, 4] from a perodc structure n superspace, t s natura to aso nvestgate CSLs n hgher dmenson. An mportant exampe are the four dmensona hypercubc attces, whch sha be dscussed here. Four dmensona attces are partcuary nterestng snce they are the frst ones that aow 5 fod, 8 fod, 0 fod and fod symmetres whch are actuay observed n quascrystas. In partcuar the four dmensona hypercubc attces aow 8 fod symmetres, from whch we can obtan e.g. the promnent Ammann Beenker tng [4]. Snce rotatons n four dmensona space can be parameterzed by quaternons, one has a strong too to nvestgate the CSLs of the hypercubc attces. In partcuar, one knows a concdence rotatons [5] and thus a CSLs can be characterzed. But one can go much further and ths w be done n the present paper. We frst cacuate the concdence ndex Σ and then we try to fnd a CSLs for a gven ndex Σ. It turns out that one can cacuate the tota number of dfferent CSLs for a gven Σ and furthermore, we can derve even the number of nequvaent CSLs and for each CSL we can cacuate the number of equvaent CSLs. These cacuatons are factated by the fact that the four dmensona rotatons are cosey reated to the ther three dmensona counterparts. In partcuar we expot the fact that SO(4) SU() SU()/C,.e. we can make use of the resuts of the three dmensona cubc case that have been pubshed recenty [6]. Thus the resuts of the hypercubc case are qute smar to the three dmensona resuts, athough proofs are a bt more engthy and the resutng formuas are a bt more compex. However, there s one bg dfference between three and four dmensons: Whereas a mportant quanttes ke Σ, number of CSLs etc. are equa for a three knds of cubc attces, ths s no onger true for four dmensons. For a gven concdence rotaton R, the concdence ndces for the prmtve and the face centered hypercubc attce are n genera not the same, whch s not surprsng snce the pont groups are dfferent, too. Thus we must dea wth both cases separatey. However, one can derve the resuts of the prmtve attce from the correspondng resuts of the face centered attce. Thus we concentrate on the atter and sketch how these resuts can then be used for the prmtve hypercubc attce. Now et us reca some basc facts and fx the notaton. Let L R n be an n-dmensona attce and R a rotaton. Then L(R) = L RL s caed a concdence ste attce (CSL) f t s a subattce of fnte ndex of L, the correspondng rotaton s caed a concdence rotaton [5]. The concdence ndex Σ(R) s defned as the ndex of L(R) n L. By ndex we mean the group theoretca ndex of L(R) n L, where we vew L(R) and L as addtve groups. Any rotaton n 4 dmensons can be parameterzed by two quaternons p = (k,, m, n) and q = (a, b, c, d) n the foowng way [7, 8, 9]: R(p, q) = M(p, q) () pq p q pu q pu q pu 3 q M(p, q) = p u q pu u q pu u q pu 3 u q p u q pu u q pu u q pu 3 u q () p u 3 q pu u 3 q pu u 3 q pu 3 u 3 q

2 ak + b + cm + dn a + bk + cn dm am bn + ck + d an + bm c + dk = a bk + cn dm ak + b cm dn an + bm + c dk am + bn + ck + d am bn ck + d an + bm + c + dk ak b + cm dn a bk + cn + dm (3) an + bm c dk am + bn ck + d a + bk + cn + dm ak b cm + dn Here, u are the unt quaternons u 0 = (, 0, 0, 0), u = (0,, 0, 0), u = (0, 0,, 0) and u 3 = (0, 0, 0, ) and p = k + + m + n s the norm of p. Furthermore we have made use of the nner product p q := ak + b + cm + dn. If we dentfy the quaternons wth the eements of Z 4 (or R 4 ) n the obvous way, then the acton of M(p, q) on a vector x Z 4 can be wrtten as M(p, q)x = px q. Here q = (a, b, c, d) denotes the conugate of q. By an ntegra quaternon we mean a quaternon wth ntegra coeffcents. If the greatest common dvsor of a coeffcents s, we ca the quaternon prmtve. In the foowng, a quaternons w be ether prmtve or normazed to unty. It w aways be cear from the context whch conventon has been chosen. Obvousy R(p, q) s a ratona matrx f p and q are ntegra quaternons such that pq s an nteger. In ths case we ca the par (p, q) admssbe [5]. On the other hand any ratona orthogona matrx R can be parameterzed by an admssbe par of ntegra quaternons. Furthermore reca p = p p. The CSLs and ther Σ vaues In 4 dmensons there are ony two dfferent hypercubc attces, namey the prmtve and the centered hypercubc attces. They are equvaent to L P = Z 4 and L F = D 4, respectvey. D 4 Z 4 s of ndex and conssts of a nteger vectors n wth n even. It s known that R s a concdence rotaton of Z 4 or D 4, f and ony f a ts entres are ratona [5],.e. R = R(p, q) for some admssbe par of prmtve quaternons (p, q). In order to anayze the CSLs t s often convenent to fnd some approprate subattces of the CSLs. To ths end we defne the denomnator den(r) = gcd{k N krl L(R)}, (4) where gcd denotes the greatest common dvsor. Snce den(r) L s a subattce of L(R) t foows that den(r) Σ(R) den(r) n, (5) where n = 4 s the dmenson of L. In case of the prmtve cubc attce ths defnton concdes wth [5] whereas for the centered attce we fnd den P (R) = gcd{k N kr nteger matrx}, (6) den F (R) = den P (R), (7) where = 0, s the maxma power such that dvdes den P (R). In partcuar we fnd for any admssbe par p, q den F (R(p, q)) = pq, (8) where = 0,, s the maxma power such that dvdes pq. It foows from Eq. (5) that R s a symmetry operaton of L f and ony f den(r) =. Thus R(p, q) s a symmetry operaton of the centered attce D 4 f and ony f p =, 4 and q =, 4 or p = q =. Ths gves the we known 576 pure symmetry rotatons of D 4. Note that not a of them are nteger matrces, whch refects the fact that the symmetry group of D 4 s arger than those of Z 4. In fact ony 9 rotatons are nteger matrces, namey the pure symmetry rotatons of Z 4. These are the rotatons correspondng to the pars (p, q) such that p = q = or p = q = wth p q even or p = q = 4 wth p q dvsbe by 4. We consder the face centered attce frst, formuatng and provng a resut that was frst stated (wthout proof) n [5], Eq. (3.). Theorem. Let p, q be an admssbe par of prmtve nteger quaternons and et Σ(p) = p, where = 0,, s the maxma power such that dvdes p. Then, for the fcc attce, the rotaton R(p, q) has concdence ndex Σ F (p, q) := Σ F (R(p, q)) = cm(σ(p), Σ(q)). (9) Proof: Let us wrte p = α γ, q = β γ, where γ = gcd( p, q ). Further et p () = pu and q () = ū q. Then βp () and α q () are nteger vectors wth nteger pre-mages. Thus they are n L P (R) and hence certany βp () and α q () are n L F (R). Thus f and k the four vectors βp (), βp (), α q (k), α q () span a subattce of L F (R). Now ( det βp (), βp (), α q (k), α q ()) = α β ( p () q (k ) p () q ( ) p () q ( ) p () q (k ), (0) Uness p and q are both odd, even βp () and α q () are eements of L F (R). In any case β(p () +p () ) L F (R) and α( q (k) + q () ) L F (R).

3 where k, are chosen such that (k,, k, ) s an even permutaton of (0,,, 3). Hence we concude that Σ F (R) dvdes 8α β c, where c s the greatest common dvsor of c k = p () q (k) p () q () p () q () p () q (k) ). Usng the expanson we see that c dvdes 3 p a = p () a p () () =0 3 c k p () a = p ( a q (k) p () q () a q () p () q (k) ) () =0 for any nteger quaternon a. We now choose a such that a q (k) = 0, n partcuar f k = 0 we choose a = q u 0 +q 0 u and a = q () q u 0 q 0 u. Hence c must dvde p (qm +q n ) p() q. Now q s prmtve so that the greatest common dvsor of a combnatons qm + qn s at most. Thus c dvdes p p () q. Smary one proofs that c dvdes p p () q (k) for arbtrary k, and hence c must dvde 8 p because q and p are both prmtve. In the same way one shows that c dvdes 8 q. Thus c dvdes 8γ and Σ F (R) dvdes 64α β γ. But Σ F (R) dvdes den F (R) 4, whch s odd. So Σ F (R) dvdes cm(σ(p), Σ(q)). It remans to show the converse statement, cm(σ(p), Σ(q)) Σ F (R). To ths end, we count the vectors y L F (R) contaned n the hypercube H(βp () ) spanned by βp (). If there are n F of them then Σ F (R) = 8β 4 p 4 /n F = 8α 4 q 4 /n F. Now L F (R) s a subattce of L P (R), so that Σ F (R) s a mutpe of 8β 4 p 4 /n P f n P denotes the number of vectors y L P (R) contaned n the hypercube H(βp () ). Now n P = 6n P, where n P s the number of the vectors y L P (R) contaned n the smaer hypercube H(βp () ) spanned by βp (). Equvaenty we can count ther pre-mages x = R y yng nsde the hypercube H(αu q). In the foowng we dentfy H(αu q) wth the factor group L P /L q, where L q denotes the Z span of the vectors αu q. Observe that any vector x of H(αu q) can be expressed as x = q such that 0 x u q < α q. Now x s n L P (R) f ts mage s an ntegra vector. Snce 3 x u q u q, (3) =0 R(p, q)x = pq 3 p () x u q (4) =0 p () Rx = p q x u q = α β x u q (5) a coeffcents x u q must be dvsbe by β. In order to determne the vectors that satsfy ths condton we frst observe that there exsts a vector x such that x q = snce q s prmtve. Regardng x as an eement of the abean group L P /L q we see that t has order α q. Among a vectors x wth x q = 0 there exsts one of order α q / or α q, dependng on whether q s dvsbe by 4 or not. Hence x and x generate a subgroup of order α q 4 or α q 4 / of L P /L q. Condton (5) s satsfed by α q 4 /β or α q 4 /(β ) of them, respectvey. Thus L P /L q contans at most α 4 q 4 /β vectors satsfyng condton (5) and hence n P s a dvsor of α4 q 4 /β. Let L β P denote the subgroup of L P /L q that s formed by the vectors satsfyng cond. (5) and assume x L β P n the foowng. We can rewrte Eq. (4) as R(p, q)x = pq 3 u x pu q = =0 3 =0 x pu q u, (6) αβγ.e. pq must dvde x pu q By assumpton, β dvdes x pu q. On the other hand, snce den(r) = pq /, there exsts an eement x of order αγ/ or hgher. Thus at most L β P /(αγ) vectors x L P/L q satsfy condton (4) and hence n P dvdes + α 3 q 4 /(β γ) = + α 3 q. From ths we nfer that Σ F (R) s a mutpe of α q / + and hence a mutpe of Σ(q). Anaogousy we prove that Σ F (R) s a mutpe of Σ(p). Thus cm(σ(p), Σ(q)) Σ F (R) and the cam foows. Consder the vectors q u 0 q 0 u and u q + q u 0 q 0 u. Ther orders are mutpes of α q /gcd(α q, q + q m ) and an approprate combnaton thereof gves the desred vector x. 3

4 From ths resut we can easy nfer the concdence ndex Σ P (R) for the prmtve attce. Snce L F s a subattce of ndex of L P, Σ P (R) must dvde Σ F (R) and Σ F (R) must dvde Σ P (R).[5] Snce Σ F (R) s odd we have Σ P (R) = Σ F (R) or Σ P (R) = Σ F (R). Due to Eq. (5) the ndex Σ P (R) s odd f den(r) s odd and even f den(r) s even. Hence we have proved Theorem. Let p, q be an admssbe par of prmtve nteger quaternons and et Σ(q) = q, where = 0,, s the maxma power such that dvdes q. Then, for the prmtve attce, the rotaton R(p, q) has concdence ndex Ths was frst stated, wthout proof, n [5]. 3 Equvaent CSLs Σ P (p, q) := Σ P (R(p, q)) = cm[σ(p), Σ(q), den(r(p, q))]. (7) Dfferent concdence rotatons may generate the same CSL or rotated copes of each other. It s natura to group these rotatons and CSLs n appropratey chosen equvaence casses. The natura way s to ca two concdence rotatons equvaent f they are n the same doube coset of the symmetry group of the attce [6, 0, ]. To be precse, et G P and G F denote the symmetry groups of the prmtve and the face centered hypercubc attce. Then we ca two concdence rotatons R, R P equvaent (F equvaent) f there exst two rotatons Q, Q G P (Q, Q G F ) such that R = QR Q. Accordngy, we ca two CSLs P equvaent (F equvaent) f the correspondng concdence rotatons are P equvaent (F equvaent). In partcuar, R and RQ, Q G P,F gve rse to the same CSL. Hence two concdence rotatons are equvaent f they beong to the same doube coset G P RG P or G F RG F. These doube cosets can be cacuated f one knows the subgroups H (R) := G RG R, = P, F. In order to determne these groups we make use of the fact that SU() SU() s a doube cover of the 4 dmensona rotaton group SO(4), whch s refected n the parameterzaton Eq. (). Athough the correspondng doube cover of G F and G P s not a drect product but a subdrect product, we can make use of ths speca property and reduce the 4 dmensona case to the 3 dmensona one. In order to do ths we reca that the 3 dmensona rotatons can be parameterzed by quaternons as we [7, 8, 9]. The group G of order G = 48 generated by the quaternons (±, 0, 0, 0), (±, ±, 0, 0), (±, ±, ±, ±) and permutatons thereof s a doube cover of the cubc symmetry group O of order O = 4. Based on the noton of equvaence of 3 dmensona concdence rotatons we ntroduce the foowng equvaence noton for quaternons: Two quaternons q and q are equvaent (q q ) f there exst quaternons s, s G such that q = sqs. Ther equvaence casses are known [6] and the dfferent types are summarzed n Tabe. Here H(q) := G qgq. Furthermore the number of nequvaent CSLs for a gven Σ s known [, 6]. These numbers are summarzed n Tabe for a speca quaternons q. The number of nequvaent CSLs for a genera q can be obtaned by consderng the tota number of CSLs [6]. Let G G be the group generated by the quaternons (±, 0, 0, 0), (±, ±, ±, ±) and permutatons thereof. Now the group G F = G G ( (,, 0, 0), (,, 0, 0))G G s a doube cover of G F. We ca two pars of quaternons (p, q) and (p, q ) F equvaent f the correspondng rotatons R(p, q) and R(p, q ) are F equvaent. If (p, q) and (p, q ) are F equvaent then p p and q q, but the converse s not true n genera. Let us anayze the converse stuaton. Let p p and q q,.e. there exst r, r, s, s G such that p = rpr and q = sqs. If both pars (p, q) and (p, q ) are admssbe, then (rr, ss ) must be admssbe, too. If (r, s) s admssbe, then so s (r, s ), and (p, q) and (p, q ) are F equvaent. If (r, s) s not admssbe, then (p, q) and (p, q ) are F equvaent ony f there exst admssbe pars (r, s ) and (r, s ) such that rpr = r pr and sqs = s qs. Ths s possbe f and ony f H(p) or H(q) contans one of the quaternons (±, ±, 0, 0) or a permutaton thereof. The atter statement s equvaent to the statement that p or q s equvaent to one of the foowng quaternons: (, 0, 0, 0), (0,,, ), (m, n, 0, 0) or (m, n, n, 0). We can summarze these consderatons as foows: If (p, q) and (p, q ) are F equvaent then p p and q q. Conversey p p and q q mpes that (p, q) and (p, q ) are F equvaent f p or q s equvaent to one of the foowng quaternons: (, 0, 0, 0), (0,,, ), (m, n, 0, 0) or (m, n, n, 0). Assume now that p (m, n, n, n) and q (m, n, n, n ). Then we may not concude that the admssbe par (p, q) s F equvaent to ((m, n, n, n), (m, n, n, n )). However, we may concude that (p, q) s F equvaent ether to ((m, n, n, n), (m, n, n, n )) or ((m, n, n, n), (m, n, n, n )). Note that the atter pars are not F equvaent. Nevertheess, they are of the same type. Havng ths n mnd we can use Tabe to cacuate a types of possbe F equvaence casses. Instead of cacuatng the groups H F (R(p, q)) drecty we compute ther correspondng doube covers H F (p, q). It turns out that they are smpy gven by H F (p, q) = (H(p) H(q)) G. The resuts are sted n Tabe 3. In order to save space we have omtted some pars. These can be easy obtaned by nterchangng the roe of p and q and adaptng the correspondng subgroup H F. In addton, we have used the defnton H := H G. 4

5 The fact that G F s a speca subgroup of G G enabes us to derve the number of dfferent and nequvaent CSLs from the 3 dmensona case. Frst, we consder the tota number of dfferent CSLs f F (Σ). Reca that the tota number of dfferent CSLs f(σ) for a gven Σ n the 3 dmensona case s gven by [6, 5] f() = (8) f() = 0 (9) f(p r ) = (p + )p r f p s an odd prme, r (0) f(mn) = f(m)f(n) f m, n are coprme. () The mutpcatvty of f(σ) s due to the unqueness of the (eft) prme factorzaton of the nteger quaternons [8]. The same reasonng hods true n four dmensons, too, so we ony need to cacuate f F (p r ). To ths end we note that there are precsey f(σ(p))f(σ(q)) dfferent CSLs for gven Σ(p) and Σ(q). Summng up a admssbe combnatons of (Σ(p), Σ(q)) that gve a fxed p r we obtan [5] f F (p r ) = p + p pr (p r+ + p r ). () In a smar way we can cacuate the number of nequvaent CSLs of a certan type, say (p, q) ((0,,, ), (0,,, )) or ((m, n, n, n), (m, n, n, n )). These resuts are summarzed n Tabes 4 and 5. Let us dscuss some of them. Consder pars of type ((, 0, 0, 0), (m, n, n, n)) frst. Then Σ(p, q) = Σ F mpes Σ(q) = Σ F. Hence the number of nequvaent CSLs s equa to the number of nequvaent quaternons q = (m, n, n, n), whch can be read off drecty from Tabe. Thus there are precsey k nequvaent CSLs f p = mod 6 for a the k dfferent prme factors of Σ F. Note that a prme factor 3 cannot exst, snce Σ F must be a square as (p, q) must be an admssbe par. Consder now pars of type ((m, n, n, n), (m, n, n, n )). Such pars can ony exst f Σ F (p, q) = 3 t pr where a prme factors p, q = mod 6 and t = 0,. Ths mpes that Σ(p) = 3 t pr 3 t pr + qs qs+ qs + and Σ(q) = wth r = max(r, r ), s = max(s, s ). For a fxed combnaton of {r, s }, there are k nequvaent quaternons p, where k s the number of dfferent prme factors p, q 3 contaned n Σ(p). If we use the notaton ν(a) = 0, for a = 0, a, respectvey, we can wrte m = / ν(r ) ν(s +). In order to get the number of nequvaent admssbe pars (p, q) we have to take the sum over a possbe combnatons of r, r, s, s. Note that r runs through 0,...,r f r = r and vce versa. Hence the number of nequvaent admssbe pars reads (r,r,s,s ) = r s r =0 ν(r s =0 ν(s ν(r )+ν(r ) r )+ν(r) + ν(s +)+ν(s +) (3) r =0 ν(r)+ν(r s +)+ν(s+) + ) + ν(r)+ν(r) (4) s =0 ν(s+)+ν(s +) + ν(s+)+ν(s +) (5) = (8r ) [4(s + )] = 4 k (r ) (s + ), (6) where k s the number of dfferent prme factors p, q 3. If Σ F (p, q) contans at east one odd prme power q s+, we have fnshed. Otherwse we have to take nto account that the sum above ncudes k pars of the form ((m, n, n, n), (, 0, 0, 0)) or ((m, n, n, n), (0,,, )). Hence a term k must be subtracted from the sum above. Thus there exst n F = 4 k t δ k (7) nequvaent admssbe pars ((m, n, n, n), (m, n, n, n )) for a fxed Σ F (p, q) = 3 t pt wth p = mod 6 and t = 0,. Here δ = f a t are even and δ = 0 otherwse. Next we consder the case ((m, n, n, n), (m, n, 0, 0)), whch s an exampe where p and q are of dfferent type. Frst observe that Σ(q) may ony contan prme factors p = mod 4, whereas Σ(p) may ony contan prme factors p = mod 6 and p = 3, for the atter ony the powers 3 0 and 3 are aowed. Snce the par must be admssbe, the factor p = 3 s rued out and the concdence ndex takes the form Σ F (p, q) = pr where 5 p r qs

6 p = mod 4, p mod 6, p = mod 6, p mod 4, q = mod 4, q = mod 6. Hence Σ(q) = pr qs qs, Σ(p) = p r where s = max(s, s ). Agan we have to sum over a possbe combnatons s, s and fnay obtan the number n F3 of F nequvaent admssbe pars n F3 = k+k 4 k3 s, (8) f k and k. Here k, k, k 3 are the number of dfferent prme factors p, p, q. If k = 0, k 0 ths expresson ncudes the pars of type ((m, n, n, n), (, 0, 0, 0)), so that a term k+k3 must be subtracted. Thus n F3 = k (4 k3 s k3 ). (9) A smar expresson s obtaned for k = 0. Fnay, f k = k = 0, we get n F3 = 4 k3 s k3. (30) At ast, et us consder pars where at east one quaternon s competey genera. As an exampe, we use ((m, n, n, n), q). In ths case, the approach s sghty dfferent from the prevous cases, snce we ack a nce formua for the three dmensona case. But we can proceed as foows: We frst cacuate the number of dfferent admssbe pars ((m, n, n, n), q), where q s a genera or a speca quaternon. We then subtract the number of a speca combnatons ((m, n, n, n), q) and fnay dvde by the number of equvaent pars. We frst note that Σ((m, n, n, n), q) must be of the form Σ = 3 r ps qt, where p = (mod 6) and q (mod 6) and r 0 and at east qt one s. We have to sum over a pars wth Σ(m, n, n, n) = 3 r ps, Σ(q) = 3r ps such that r, r = r (mod ), s = max(s, s ). For fxed Σ(m, n, n, n) and Σ(q) we have the foowng stuaton: There are k = / δ 0,s nequvaent quaternons of type (m, n, n, n) (f at east one s > 0, k s the number of dfferent prme factors > 3) and there are 48 (4 3 r ) δ0,r (p +)p s (q +)q t dfferent (n genera not nequvaent) quaternons q. Note that the product ranges ony over those for whch s > 0. If we use Gauss symbo [x] n order to denote the argest nteger n x we may rewrte ths as 48 [4 3 r ] [(p + )p s ] (q + )q t and take the product over a. Hence for fxed Σ(m, n, n, n) and Σ(q) we have / 8 48 / δ 0,s 48 [4 3 r ] [(p + )p s ] (q + )q t (3) dfferent (n genera not nequvaent admssbe pars). Note that we have added a factor / takng nto account that ony haf of the pars are admssbe. Summng over a possbe combnatons of Σ(m, n, n, n) and Σ(q) we get 5m F = 4 5 [4 3 r ] [s /] [s /] δ 0,s (p + )p s + [(p + )p s ] (q + )q t (3) = =0 = 4 5 [4 3 r ] ( ) (s + )(p + )p s + ps (q + )q t (33) p dfferent admssbe pars f there s at east one s s odd. Otherwse we must excude the term wth = s / for a n the frst sum,.e. ( ( ) ) m F = 4 [4 3 r ] (s + )(p + )p s + ps δ F (p + )p s (q + )q t, (34) p where δ F = 0, accordng to whether there exsts an odd s or not. From ths expresson we subtract a admssbe pars wth speca q, dvde by the number of equvaent pars and obtan the foowng expresson for the number of nequvaent admssbe quaternons of type ((m, n, n, n), q): n F5 = 96 ( m F ) 4 g F n F. (35) Smar expresson are obtaned for n F35 and n F45. And fnay we can compute n F55 by recang the tota number of dfferent quaternons f F gven n Eq. (): f F = =0 5 g F n F. (36),=0 6

7 Fnay et us have a short ook on the prmtve hypercubc attce. Smar resuts can be proved for ths case. The best way to obtan them s to derve them drecty from the prevous resuts. We ust have to keep n mnd that the symmetry group G P s a subgroup of ndex 3 of G F. In partcuar, the coset decomposton for the correspondng groups of quaternons reads G F = G P ( (, 0, 0, 0), (,,, )) G P ( (, 0, 0, 0), (,,, )) G P. (37) If we appy ths decomposton to the doube cosets G F (p, q)g F, we get the doube cosets of G P, whch are ust the P equvaence casses of admssbe pars, see Tab. 6. The correspondng groups H P (p, q) can now be nferred from the correspondng groups H F (p, q). In partcuar, we have H P (p, q) H F (p, q) G P, whch smpfes the determnaton of H P (p, q) consderaby. The resuts are shown n Tab. 7. Combnng these resuts wth the numbers n F of F nequvaent admssbe pars, we get the number of P nequvaent admssbe pars, whch are sted n Tab Concusons and Outook We have cacuated the concdence ndex Σ for both knds of four dmensona hypercubc attces. Moreover, we have determned a CSLs and ther equvaence casses as we as the tota number of dfferent and nequvaent CSLs for fxed Σ. Here, equvaence aways means equvaence up to proper rotatons. But of course there exst refectons that eave the hypercubc attces nvarant and one can be nterested n extendng the noton of equvaence to the fu symmetry group. We brefy sketch how one can ncude the mproper rotatons. Frst note that the speca refecton m : q (q 0, q, q, q 3 ) ust corresponds to quaternon conugaton. Now any symmetry operaton s a product of ths refecton and a rotaton, and t s suffcent to consder ths refecton n deta. Snce mr(p, q) = R(q, p)m, t foows that the admssbe pars (p, q) and (q, p) are equvaent. Thus we have two stuatons: If (p, q) and (q, p) are not equvaent under proper rotatons, than ther equvaence casses merge to form a snge equvaence cass. If (p, q) and (q, p) are aready equvaent under proper rotatons, than the equvaence cass stays the same and the correspondng symmetry group H(p, q) contans a symmetry operaton whch s a conugate of m. Thus we know a equvaence casses and ther symmetry groups H(p, q). It s then straghtforward to cacuate the number of nequvaent CSLs. Acknowedgements The author s very gratefu to Mchae Baake for nterestng dscussons on the present subect and to the Facuty of Mathematcs, Unversty Beefed, for ts hosptaty. Fnanca support by the Austran Academy of Scences (APART-program) s gratefuy acknowedged. References [] Bomann, W.: Crysta Defects and Crystane Interfaces. Sprnger, Bern, 970. [] Bomann, W.: Crysta attces, nterfaces, matrces. pubshed by the author, Geneva, 98. [3] Duneau, M.; Katz, A.: Quasperodc patterns. Phys. Rev. Lett. 54 (985) [4] Baake, M.: A Gude to Mathematca Quascrystas In: Quascrystas, (Eds. J.-B. Suck, M. Schreber, P. Häußer), p. 7 48, Sprnger, Bern, 00. [5] Baake, M.: Souton of the concdence probem n dmensons d 4. In: The Mathematcs of Long-Range Aperodc Order (Ed. R. V. Moody), p. 9 44, Kuwer, Dordrecht, 997. [6] Zener, P.: Symmetres of concdence ste attces of cubc attces. Z. Krstaogr. 0 (005) [7] Koecher, M.; Remmert, R.: Hamton s Quaternons. In: Numbers (Eds. H.-D. Ebbnghaus et. a.), p Sprnger, 99. [8] Hurwtz, A.: Voresungen über de Zahentheore der Quaternonen. Sprnger, Bern 99. [9] du Va, P Homographes, Quaternons and rotatons. Carendon Press, Oxford, 964. [0] Grmmer, H.: Dsorentatons and concdence rotatons for cubc attces. Acta Cryst. A 30 (974) [] Grmmer, H: Concdence ste attces: New resuts and comments on papers by Fortnow and Worgard-de Fouquet. Scrpta Met. 0 (976)

8 q H(q) H(q) GqG (, 0, 0, 0) G (0,,, ) (3,,, ) H = [(, 0, 0, 0), (,,, ), (0,,, 0)] 4 48 = 9 (m, n, n, n) H = [(, 0, 0, 0), (,,, )] = 384 (m, n, 0, 0) H 3 = [(, 0, 0, 0), (,, 0, 0)] = 88 (m, n, n, 0) H 4 = [(, 0, 0, 0), (0,,, 0)] 4 48 = 576 otherwse H 5 = [(, 0, 0, 0)] 4 48 = 5 Tabe : Equvaence casses of quaternons: Any prmtve quaternon s equvaent to one of the quaternons n the frst coumn. The second coumn gves a set of generators of H(q). The thrd coumn gves the order of H(q) and the forth coumn states the number of equvaent q, whch s 48 tmes the number of equvaent 3 dmensona CSLs. q nequv. CSLs condton (, 0, 0, 0) Σ = (0,,, ) Σ = 3 (m, n, n, n) k at most once and k s the number of dfferent prme factors p = mod 6 p = mod 6 for a prme factors p 3 of Σ > 3, the factor p = 3 occurs of Σ (m, n, 0, 0) k p = mod 4 for a prme factors p of Σ and k s the number of dfferent prme factors of Σ. (m, n, n, 0) k p = or 3 mod 8 for a prme factors p of Σ, where k s the number of dfferent prme factors of Σ > 3. Tabe : Number of nequvaent cubc CSLs/concdence rotatons for a fxed vaue Σ. The ast coumn gves the condton under whch these vaues hod. If ths condton s not satsfed, the correspondng number of nequvaent CSLs s 0 for the partcuar type of q. p q H F (p, q) H F (p, q) G F R(p, q)g F (, 0, 0, 0) (, 0, 0, 0) G F 5 576g F00 = 576 (, 0, 0, 0) (m, n, n, n) G H g F0 = (, 0, 0, 0) (m, n, 0, 0) G H 3 ( (,, 0, 0), (,, 0, 0))G H g F03 = (, 0, 0, 0) (m, n, n, 0) G H 4 ( (,, 0, 0), (0,,, 0))G H g F04 = 576 (, 0, 0, 0) genera G H g F05 = (0,,, ) (0,,, ) H H ( (0,,, 0), (0,,, 0))H H 7 576g F = (0,,, ) (m, n, n, n) H H g F = (0,,, ) (m, n, 0, 0) H H 3 ( (0,,, 0), (,, 0, 0))H H g F3 = (0,,, ) (m, n, n, 0) H H 4 ( (0,,, 0), (0,,, 0))H H g F4 = (0,,, ) genera H H 5 576g F5 = (m, n, n, n) (m, n, n, n ) H H g F = (m, n, n, n) (m, n, 0, 0) H H g F3 = (m, n, n, n) (m, n, n, 0) H H 4 576g F4 = (m, n, n, n) genera H H 5 576g F5 = (m, n, 0, 0) (m, n, 0, 0) H 3 H 3 ( (,, 0, 0), (,, 0, 0))H 3 H g F33 = (m, n, 0, 0) (m, n, n, 0) H 3 H 4 ( (,, 0, 0), (0,,, 0))H 3 H g F34 = (m, n, 0, 0) genera H 3 H g F35 = (m, n, n, 0) (m, n, n, 0) H 4 H 4 ( (0,,, 0), (0,,, 0))H 4 H g F44 = (m, n, n, 0) genera H 4 H g F45 = genera genera H 5 H g F55 = Tabe 3: F Equvaence casses of admssbe pars. For each admssbe par the correspondng group H F (p, q) and ts order s sted. The ast coumn gves the number of equvaent concdence rotatons R(p, q). By dvdng these numbers by 576 we obtan the number of equvaent CSLs. In order to save space we have omtted some pars. These can be easy obtaned by nterchangng the roe of p and q and by adaptng the subgroup H F correspondngy. 8

9 p q nequvaent CSLs condton (, 0, 0, 0) (, 0, 0, 0) n F00 = Σ F = (, 0, 0, 0) (m, n, n, n) n F0 = k tors, a prme factors p = mod 6, Σ F k s the number of dfferent prme fac- s a square (, 0, 0, 0) (m, n, 0, 0) n F03 = k tors, a prme factors p = mod 4, Σ F k s the number of dfferent prme fac- s a square (, 0, 0, 0) (m, n, n, 0) n F04 = k tors, a prme factors p = or 3 mod 8, k s the number of dfferent prme fac- Σ F s a square (, 0, 0, 0) genera n F05 Σ F s a square (0,,, ) (0,,, ) n F = Σ F = 3 (0,,, ) (m, n, n, n) n F = k prme factors of a, a prme factors p = Σ F = 3a, k s the number of dfferent mod 6 (0,,, ) (m, n, n, 0) n F4 = k prme factors of Σ F, a prme factors Σ F = 3a, k s the number of dfferent p = or 3 mod 8 (0,,, ) genera n F5 Σ F = 3a Σ F = 3 r a, r = 0,, k s the number (m, n, n, n) (m, n, n, n ) n F = 4 k of dfferent prme factors of a, whch s t δ k not dvsbe by 3, a prme factors p = mod 6, δ = f a s a square and δ = 0 otherwse Σ F = pr p r qt, p = (mod 4) (mod 6), p = (mod 6) (mod 4), q = (m, n, n, n) (m, n, 0, 0) n F3 = k+k 4 k3 (mod 4) = (mod 6), k t, k, k 3 denote the number of dfferent prme factors of δ k+k3 δ k+k3 type p, p and q, respectvey. δ = 0 uness a t are even and k = 0, where δ =. An anaogous defnton appes for δ wth k = 0 repaced by k = 0. (m, n, n, n) genera n F5 Tabe 4: Number of F nequvaent CSLs (Part ). The ast coumn gves the condton under whch these vaues hod. If ths condton s not satsfed, the correspondng number of nequvaent CSLs s 0 for the partcuar type of (p, q). In order to save space we have omtted some pars. These can be easy obtaned by nterchangng p and q and readng of the correspondng vaue n F = n F. 9

10 p q nequvaent CSLs condton = 3 s pr (m, n, n, n) (m, n, n, 0) n F4 = k+k 4 k3 t /(n F0 + n F + n F04 + n F4 ) Σ F p r qt, p = (mod 6) or 3 (mod 8), p = or 3 (mod 8) (mod 6), q = (mod 6) = or 3 (mod 8), there must be at east one prme factor = (mod 6) and one = or 3 (mod 8). k, k denote the number of dfferent prme factors of type p and p, respectvey. k 3 s the number of prme factors of type q f s = 0 and the number of prme factors of type q pus f s >. (m, n, 0, 0) (m, n, 0, 0) n F33 = 4 k t δ k of Σ F, a prme factors p = mod 4, δ = f Σ F s a square and δ = 0 otherwse k s the number of dfferent prme factors (m, n, 0, 0) (m, n, n, 0) n F4 = k+k 4 k3 t /(n F0 + n F + n F04 + n F4 ) (m, n, 0, 0) genera n F35 Σ F = pr p r qt, p = (mod 4) or 3 (mod 8), p = 3 (mod 8), q = (mod 8), there must be at east one prme factor = (mod 4) and one = or 3 (mod 8). k, k, k 3 denote the number of dfferent prme factors of type p, p, and q, respectvey. (m, n, n, 0) (m, n, n, 0) n F44 = 4 k t δ k of Σ F, a prme factors p =, 3 mod 8, δ = f Σ F = a, 3a and δ = 0 k s the number of dfferent prme factors (m, n, n, 0) genera n F45 otherwse genera genera n F55 Tabe 5: Number of F nequvaent CSLs (Part ) 0

11 g = (p, q) doube coset decomposton of G F gg F ((, 0, 0, 0), (, 0, 0, 0)) G P s G P ((, 0, 0, 0), (m, n, n, n)) G P gg P G P gs G P G P gs G P ((, 0, 0, 0), (m, n, 0, 0)) G P gg P G P gs G P ((, 0, 0, 0), (m, n, n, 0)) G P gg P G P gs G P ((, 0, 0, 0), (m, n, p, q)) G P gg P G P gs G P G P gs G P ((0,,, ), (0,,, )) G P gg P G P gs G P ((0,,, ), (m, n, n, n)) G P gg P G P gs G P G P gs G P ((0,,, ), (m, n, 0, 0)) G P gg P G P gs G P ((0,,, ), (m, n, n, 0)) G P gg P G P gs G P ((0,,, ), (m, n, p, q)) G P gg P G P gs G P G P gs G P ((m, n, n, n), (m, n, n, n )) G P gg P G P gs G P G P gs G P ((m, n, n, n), (m, n, 0, 0)) G P gg P G P gs G P G P gs G P ((m, n, n, n), (m, n, n, 0)) G P gg P G P gs G P G P gs G P ((m, n, n, n), (m, n, p, q )) G P gg P G P gs G P G P gs G P ((m, n, 0, 0), (m, n, 0, 0)) G P gg P G P gs G P G P s gg P G P s gs G P G P s gs G P ((m, n, 0, 0), (m, n, n, 0)) G P gg P G P gs G P G P gs G P G P s gg P G P s gs G P ((m, n, 0, 0), (m, n, p, q )) G P gg P G P gs G P G P gs G P G P s gg P G P s gg P G P s gs G P G P s gs G P G P s gs G P G P s gs G P ((m, n, n, 0), (m, n, n, 0)) G P gg P G P gs G P G P s gg P G P s gs G P G P s gs G P ((m, n, n, 0), (m, n, p, q G )) P gg P G P gs G P G P gs G P G P s gg P G P s gg P G P s gs G P G P s gs G P G P s gs G P G P s gs G P ((m, n, p, q), (m, n, p, q G )) P gg P G P gs G P G P gs G P G P s gg P G P s gg P G P s gs G P G P s gs G P G P s gs G P G P s gs G P Tabe 6: Spttng of F equvaence casses nto P equvaence casses. The ast coumn gves the decomposton of the F equvaence cass G F (p, q)g F nto doube cosets of G P. Here, s = (u 0, (,,, )), s = (u 0, (,,, )). In order to save space we have omtted some pars. These can be easy obtaned by nterchangng p and q and adaptng the decomposton correspondngy,.e. we have to nterchange the correspondng quaternons of the pars gs, s g,... as we.

12 Tabe 7: P Equvaence casses of admssbe pars. The groups HP(p, q) are gven n terms of ther generators sted n the thrd coumn, where we aways have to add the trva generators ((, 0, 0, 0), (, 0, 0, 0)) and ((, 0, 0, 0), (, 0, 0, 0)). The ast coumn gves the number of P equvaent concdence rotatons R(p, q). By dvdng these numbers by 9 we obtan the number of P equvaent CSLs. In order to save space we have omtted some pars. These can be easy obtaned by nterchangng p and q and adaptng the generators correspondngy. p q non trva generators of H P (p, q) H P (p, q) G P R(p, q)g P (, 0, 0, 0) (, 0, 0, 0) G P (, 0, 0, 0) (,,, ) (u, u 0 ), (u, u 0 ), (u 0, u ), (u 0, u ), ( (,,, ), (,,, )) 9 9 (, 0, 0, 0) (m, n, n, n) (u, u 0 ), (u, u 0 ), ( (,,, ), (,,, )) (, 0, 0, 0) (m, n, 0, 0) (u, u 0 ), (u, u 0 ), ( (,, 0, 0), (,, 0, 0)) (, 0, 0, 0) ( m n, m+n, m n, m+n ) (u, u 0 ), (u, u 0 ), (u 0, u ) 3 9 (, 0, 0, 0) (m, n, n, 0) (u, u 0 ), (u, u 0 ), ( (0,,, 0), (0,,, 0)) 3 9 (, 0, 0, 0) ( m n, m + n, m, m ) (u, u 0 ), (u, u 0 ) (, 0, 0, 0) genera (u, u 0 ), (u, u 0 ) (0,,, ) (0,,, ) ( (,,, ), (,,, )), ( (0,,, 0), (0,,, 0)) (0,,, ) (3,,, ) ( (,,, ), (,,, )) 3 9 (0,,, ) (m, n, n, n) ( (,,, ), (,,, )) 3 9 (0,,, ) (m, n, 0, 0) ( (0, 0,, ), (,, 0, 0)) (0,,, ) ( m n, m+n, m n, m+n ) (u 0, u ) (0,,, ) (m, n, n, 0) ( (0,,, 0), (0,,, 0)) (0,,, ) ( m n, m + n, m, m ) (0,,, ) genera (m, n, n, n) (m, n, n, n ) ( (,,, ), (,,, )) 3 9 (m, n, n, n) (m, n, 0, 0) (u 0, u ) (m, n, n, n) (m, n, n, 0) (m, n, n, n) genera (m, n, 0, 0) (m, n, 0, 0) ( (,, 0, 0), (,, 0, 0)), ( (,, 0, 0), (,, 0, 0)) 3 9 (m, n, 0, 0) (m, 0, n, 0) (u, u 0 ), (u 0, u ) (m, n, 0, 0) ( m n, m +n, m n, m +n ) (u, u 0 ), (u 0, u ) (m, n, 0, 0) ( m n, m +n, m +n, m n ) (u, u 0 ), (u 0, u ) (m, n, 0, 0) ( m n, m n, m +n, m +n ) (u, u 0 ), (u 0, u 3 ) (m, n, 0, 0) (m, n, n, 0) (u, u 0 ) (m, n, 0, 0) (m, 0, n, n ) ( (,, 0, 0), (0, 0,, )) (m, n, 0, 0) ( m n, m + n, m, m ) (u, u 0 ) (m, n, 0, 0) ( m n, m, m n, m ) (u, u 0 ) (m, n, 0, 0) ( m n, m, m + n, m ) (u, u 0 ) (m, n, 0, 0) genera (u, u 0 ) (m, n, n, 0) (m, n, n, 0) ( (0,,, 0), (0,,, 0)) (m, n, n, 0) (m, 0, n, n ) (m, n, n, 0) ( m n, m + n, m, m ) (m, n, n, 0) ( m n, m, m n, m ) (m, n, n, 0) ( m n, m, m + n, m ) (m, n, n, 0) genera genera genera

13 (p, q) # nequv. pars # nequv. pars Σ P = Σ F Σ P = Σ F ((, 0, 0, 0), (, 0, 0, 0)) (, 0, 0, 0), (,,, ) ((, 0, 0, 0), (m, n, n, n)) n F0 n F0 ((, 0, 0, 0), (m, n, 0, 0)) n F03 ((, 0, 0, 0), (m n, m + n, m n, m + n)) n F03 ((, 0, 0, 0), (m, n, n, 0)) n F04 ((, 0, 0, 0), (m n, m + n, m, m)) n F04 ((, 0, 0, 0), (m, n, p, q)) n F05 n F05 ((0,,, ), (0,,, )) ((0,,, ), (3,,, )) ((0,,, ), (m, n, n, n)) n F n F ((0,,, ), (m, n, n, 0)) n F4 ((0,,, ), (m n, m + n, m, m)) n F4 ((0,,, ), (m, n, p, q)) n F5 n F5 ((m, n, n, n), (m, n, n, n )) n F n F ((m, n, n, n), (m, n, 0, 0)) n F3 n F3 ((m, n, n, n), (m, n, n, 0)) n F4 n F4 ((m, n, n, n), (m, n, p, q )) n F5 n F5 ((m, n, 0, 0), (m, n, 0, 0)) n F33 ((m, n, 0, 0), (m, 0, n, 0)) n F33 ((m, n, 0, 0), (m n, m + n, m n, m + n )) n F33 ((m, n, 0, 0), (m n, m + n, m + n, m n )) n F33 ((m, n, 0, 0), (m n, m n, m + n, m + n )) n F33 ((m, n, 0, 0), (m, n, n, 0)) n F34 ((m, n, 0, 0), (m, 0, n, n )) n F34 ((m, n, 0, 0), (m n, m + n, m, m )) n F34 ((m, n, 0, 0), ( m n, m, m n, m )) n F34 ((m, n, 0, 0), (m n, m, m + n, m )) n F34 ((m, n, 0, 0), (m, n, p, q )) 3n F35 6n F35 ((m, n, n, 0), (m, n, n, 0)) n F44 ((m, n, n, 0), (m, 0, n, n )) n F44 ((m, n, n, 0), (m n, m + n,, m )) n F44 ((m, n, n, 0), ( m n, m, m n, m )) n F44 ((m, n, n, 0), (m n, m, m + n, m )) n F44 ((m, n, n, 0), (m, n, p, q )) 3n F45 6n F45 ((m, n, p, q), (m, n, p, q )) 3n F55 6n F55 Tabe 8: Number of P nequvaent admssbe pars. The second coumn gves the number of nequvaent pars for odd vaues of Σ whereas the thrd coumn gves the same nformaton for even vaues of Σ. In order to save space we have omtted some pars. These can be easy obtaned by nterchangng p and q. 3