# A Cyclotomic Contruction for Leech&#39;s Lattices

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236 A  CYCLOTOMIC CONSTRUCTION  FOR LEECH S LATTICE MAURICE CRAIG §1.  Introduction.  The 24 classes of even-valued, unimodular, positive quadraticforms in  24  variables have been determined by Niemeier . One class is distinguishedfrom the rest by the property that its forms have arithmetic minimum 4, rather than 2.The 24-dimensional lattice A corresponding to this form-class can therefore beidentified with one found earlier by Leech , which has been studied extensively inconnection with sporadic simple groups  (e.g.  ).We construct here a 24-variable unimodular form—with minimum 4—by imple-menting a variant of the Minkowski embedding for number fields developed in .Specifically, A is realized as the lattice representing an ideal divisor of 39 in the fieldof 39th roots of unity. It will be seen that even unimodular forms are readilylocated by this method. (Thus the 52nd and 72nd roots of unity likewise yield 24-variable forms with determinant 1, both having minimum 2.) The general problemof finding minima, however, seems to require special arguments. §2.  The  unimodular  form.  Put 0 = exp  (2ni/39)  and  K =  Q(0). We have idealfactorizations(3) = CPiMi^PiO 2 . (13) = (CQ C ) 12 where  ty u  ^B 2 > & are prime ideals with norms 3 3 , 3 3 , 13 respectively, and  c  denotescomplex conjugation  (9 C  = 0 1 ). Leta^ ..., a 24  be an integral basis for  91  = ^31^^and suppose £ =  Y, z i a i>  z i  e  Z. Then with G = Aut  (K/Q),  the expressionE |^| 2  =  p(x),  say, G is an even-valued, positive, integral quadratic form.For, as the sum of squared norms, it is positive if  £  # 0. Since G is abelian, wehave 2l 8(1+c)  = X)£ (1+e)g > where the right member is G-invariant. Hence  p(z)  isrational, and also an integer of  K.  Finally, by combining for each  h  e  G  the termscorresponding to  g = h, g  =  ch  in the above sum, we see that  p(z)  is even.We complete the construction by showing that  p(z) = 39q(z),  where  q(z)  isintegral and unimodular (and, of course, even). In fact,  t, e  91  implies that £ 1+c  iscontained in 9I 1+C , which is G-invariant. It follows that £( 1+c >«  6  9I 1+C  for all  g e G, hence by summation ^(z) eZn 5I 1+C  = (39). The proof given for Lemma  1  of shows, however, that Det  p(z) =  |Discr. 9I|. We haveDiscr. 91 = (Norm 5I) 2 Discr.  K = (3 6 -13) 2 (3 12 -13 22 ) =  39 24 . So  q(z)  is as required.While showing the existence of the even unimodular formg(z), the argument above [MATHEMATIKA  25  (1978),  236-241]  A CYCLOTOMIC CONSTRUCTION  FOR  LEECH'S LATTICE  337 is  not  sufficiently explicit  to  yield much further information regarding  it.  Thus,  for determining  the  minimum of  q(z),  we  shall approach  the  construction more carefully.We first form  the  expression £|£*| 2  with  £ an  arbitrary integer, written  in  terms  of a particular integral basis  for  K  to  produce  a  quadratic form  x T H\. This will correspond   to a  24-dimensional lattice  A o  representing  the  fullinteger ring  of  K.  We  shall then take successive account  of  the conditions  £ e \$^2 and  £ e Q, so as to  arrive  gradually  at the  sublattice  A  C  A o  which represents 91  and corresponds  to the  form  q. For handling  the  former condition above,  it is  convenient  to  decompose  A o relative  to the  12-dimensional section  of  itself that represents  the  subfield Q(0 3 ).  A subsequent digression  can be  avoided  by  beginning with  the  necessary detailed studyof the 12-variable form r(y) corresponding  to  this section. §3.  The  auxiliary form.  Let e ;  e Z 12  be the  column vector with  j-th  entry  <5 y  for 1  <  i, j  < 12.  Each  y =  J^y^t  is  uniquely expressible  in the  form  ±  ^ Yfii  with0  <  s  < 6.  Here  £'  denotes summation from  0 to  12,  e 0  = -  £>,-,  and s = 2'  ^i- Next,  let LEMMA.  The  vectors realizing  the  first seven distinct successive minima  of  r(y)  are given  by  the following list,  the  suffixes within each expression being chosen withoutrepetition from  0, 1, 2, ..., 12: r(s)  =12, ± y = e ; 22  e, + e, 26  e (  - e, 30  e s  + e,  + e k 38  e, + e, - e* 40  e,  + e, +  e k  + e, +  e m . Proof.  Assume  r(y) < 40. We  have  r(y) = 13/  —  J 2 , where  / =  J^'Y 2 .  Hence 13/  =? 40 + 36, or / < 5.  Moreover,  if  s  ^ 4  then  / < 4, and if  s  < 3  then  / < 3. The list comprises  all  vectors compatible with these restraints. §4.  The  reduction.  Let  x  denote  a  primitive  13th  root  of  unity.  The  three-termperiods A t  =  T  +  T 3  + T 9 , V = T 4  + T 10  + T 12 , ^ 2  =  T 7  +  X 8  +  X 11 , A 2C  =  X 2  +  X 5  +  X 6 , are  the  roots  of the  cyclic quartic polynomialin particular, X^iXfXi = 3. (1)  238 MAURICE CRAIG Hence Q(/li)  is the  decompositon field  for ^J x  (and its  conjugate ideals).  Now  Q(T) is  the  fixed field  of  b  e  G,  where  9 b  = 0 14 . It  follows that Sey& 2  implies  { +  ? e  M 2  n  Q(T).  (2) We shall identify  ^S 1;  ^2,  <p tc ,  \$ 2C  with the ideals which intersect  Q(t) in  the principalideals  (A x ), (A 2 ), (V), (A 2e )  respectively.We claim next that  for  S,  e  K, (That  is to say, the  identity  2S,  = (^ + £*) +  ((*  -  £ b )  affords  an orthogonal decomposition  of  K  relative  to its  subspace  Q(T).  Analogous results hold  for the other subfields.)  In  fact, and  writing  t]  =  £ g , we  have 2(ri 1+c  +  »7 (1+c)i> )  = (i; +  r, b ) 1+c  + (r, -  /? 6 ) 1+c   n e\ 2e   \z~  e\ 2g - The results become explicit when £  is  expressed  in  terms  of  an integral basis  for  K. One such basis  is  6, ...,  0 24 . However,  the  relations 6 3J (6 39  -  1)1 (6 13  - 1) = 0 =  0(0 39  -  l)/(0 3  - 1) (3) permit  6 3 , 0 6 ,  ...,0 24  and  9 13 ,  hence  the  entire basis,  to be  expressed  in  terms  of elements  6'  with  (i,  39) = 1. The 24  primitive 39th roots  of  unity thus constitute  a second Z-basis.Noting that  G =  <a,  b},  with  6 = 9' 2 ,  we  order  the new  basis  as  9, 9~ 2 , 9 A , 6~ 8 ,...,  0 14 , 0 28 , ... . A  typical integer  of  K  then  has the  formwhere  (i) is  written  in  place  of  a'  1  to  avoid second-order superscripts.  The  argumentin Lemma  3 of  can now be  adapted  to  show  the  expression  for £  l£l 2g  is  x r /ix,with  x =  (x lf  ...,  x 24 ) r  and  with 2R  -R] R  2R\' R  being  the  symmetric matrix such that  y T Ry  = r(y). (In the  notation  of  ,  H  is the the direct product  V(12, - 1, ..., - 1) ® V(2, - 1).)  Moreover,DetH  =  3 12 -13 22  =  Discr.  K. Since  £ +  £*  = £  ( x fi (>)  +  Xi+izd {i)b )  for(/  being  the 12 x 12  identity matrix),  we  have ~  = 2x T  A CYCLOTOMIC CONSTRUCTION FOR LEECH'S LATTICE 239 We are thus led to the decomposition 2ff =  (I,I) T R(I,I) +  3(1,  -I) T R(I,  -I) or 2x T Hx  =  u T Ru + 3\ T Rv  = r(u) + 3r(v), (4)where u = £ (x ;  + Ar,- +12 )e £  and v = £ (x ;  - x,+  12 )e,-,  so that if x 6 Z 24 ,u = v (mod 2). (5)Equation 3(0 for; = 9 yields  9 + 6 b  = - 6 21 .  Hence with  x = d 27  (conformingwith the earlier use of the symbol  T ,  we have £ + £ 6  =  —  £ «;T (l) . The condition£ 6  ^Pi^8 2  i s now eas Y  to  incorporate. From (2), f + £ 6  must lie in the principal ideal(A^j) of  Q(T).  However,The elements  T   (1  + T 4  + T 10  + z 12 ) therefore generate (A^)  as a  Z-module, andthe implied condition in (2) takes the form u = Mw, w e Z 12 , where M is the matrixwhose i-th column equals e -  4 e i+10  + e i+12 , 1 «S  i  < 12,the subscripts to be read as least non-negative residues modulo 13. From equation (1)it can be seen that  MN  = 3/,  N  being the matrix derived from  X 1C X 2C  as M wasderived from /M 2 > tnat  is, with e ;  + e i+1  + e i+3  + e (+9  for its i-th column.Since  M T RM = 3R(N.B. Re {  = e 0  + 13(1 -  ^ 0> -)e ; ),  (4) yields 2\ T Hx =  3(r(w) + r(v)), (6)and by (5) we have the mutually equivalent conditionsv  =  Mw (mod 2), w =  Nv  (mod  2).  (7) §5.  7%e  minimum.  The expression 22 l£| 2g  = x r Hx assumes the value 4 x 39 for £, — 8(1  —  6 3 )(l  —  0 13 )  e 9t (all conjugates of £ are associates), with  JC,-  =  —  x i+12  = (5 U  —  ^ 3i . By (6), the smaller value 2 x 39 would entailr(w) + r(v) = 52. (8)Now if  v  = 0, by (5) we obtain u = 2u' (u' e Z 12 ) and (4) shows that2x T Hx = 4r(u')  =  0 (mod 8).(We may alternatively observe that since  <f*  = ^,  (9£) b  =  61;  cannot both be true,and since 2l^l 2g  = Sl<^l 2g '  tne case  v = 0 does not require separate attention.)Similarly, the possibility u = 0 can be dismissed.When r(w), r(v) are both positive, we see from our list of small values of  r  that theleft side in (8) can only be one of 12 +40, 22 + 30, 26 + 26, 30 + 22 or 40 + 12.For the case r(w) = 12, we have w = e ; with0 < i < 12. Soby(7)(i),v = e ;  + e i+4  + e i+10  + e i+12  (mod 2). (9)This is clearly incompatible with the form which v must have for r(v) = 40.If r(w) = 22 or 26, then v will be congruent (mod 2) to a sum of two vectors such

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