1 Generalization of Pigeon Hole Bound Toni Ernvall Department of Mathematics, University of Turku Abstract—H.-F. Lu, J. Lahtonen, R. Vehkalahti, and C. Hol- Then we have det(AA†)=det(BB†). lanti introducedso called Pigeon Hole Boundfor decayfunction Proof: If k > n then det(AA†) = 0 = det(BB†). If 2 of MIMO-MAC codes. Here we give a generalization for it. k =n then det(A) is 1 c e e e 20 I. DECAY FUNCTION (cid:12)(cid:12) c12 (cid:12)(cid:12) (cid:12)(cid:12) c21 (cid:12)(cid:12) (cid:12)(cid:12) e12 (cid:12)(cid:12) (cid:12)(cid:12) e12 (cid:12)(cid:12) WeconsiderheredecayfunctionofMIMO-MACcodes.Ev- (cid:12) . (cid:12) (cid:12) . (cid:12) (cid:12) . (cid:12) (cid:12) . (cid:12) n ery user has some specific lattice L ⊆ M ,j = 1,...,U (cid:12)(cid:12) .. (cid:12)(cid:12)=(cid:12)(cid:12) .. (cid:12)(cid:12)=(cid:12)(cid:12) .. (cid:12)(cid:12)=···=(cid:12)(cid:12) .. (cid:12)(cid:12) a j n×k (cid:12) c (cid:12) (cid:12) c (cid:12) (cid:12) c (cid:12) (cid:12) e (cid:12) J withk ≥Un.Weassumethateachuser’slatticeisoffullrank (cid:12) k−1 (cid:12) (cid:12) k−1 (cid:12) (cid:12) k−1 (cid:12) (cid:12) k−1 (cid:12) 7 r =2kn so the lattice Lj hasan integralbasis Bj,1,...,Bj,r. (cid:12)(cid:12)(cid:12) ck (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ck (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ck (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ck (cid:12)(cid:12)(cid:12) 2 Nowthecodeassociatedwithjthuserisarestrictionoflattice i.e. d(cid:12)et(B) a(cid:12)nd h(cid:12)ence de(cid:12)t(AA(cid:12)†)=de(cid:12)t(BB†). (cid:12) (cid:12) Lj such that Assume k <n. Let v ,...,v ∈Cn be such that v ∈ ] 1 n−k 1 A r L(c ,c ,...,c ,)⊥\{0},v ∈L(v ,c ,c ,...,c )⊥\{0}, 1 2 k 2 1 1 2 k R Lj(Nj)={XbiBj,i|bi ∈Z,−Nj ≤bi ≤Nj} ...,vn−k ∈L(v1,v2,...vn−k−1,c1,c2,...,ck)⊥\{0}.Now . i=1 (as in the case n=k) we have h where N is a given positive number. j c e at Using these definitions the U-user mimo-mac code is .1 .1 m (L (N ),L (N ),...,L (N )). .. .. 1 1 2 2 U U [ For this we define ck−1 ek−1 1 D(N1,...,NU)= min det(MM†) det( vck )=det( vck ) v Xj∈Lj(Nj)\{0} 1 1 4 where M =M(X1,...,XU). For a special case N1 =···= ... ... 74 NU =N we write vn−k vn−k 5 D(N)=D(N1 =N,...,NU =N). and hence . 01 In the special case k =Un we have (cid:12) c1.c∗1 ... c1.c∗k c1.v1∗ ... c1v.n∗−k (cid:12) 2 D(N1,...,NU)=D(N1,...,NU)2 (cid:12)(cid:12) .. .. .. .. (cid:12)(cid:12) v:1 and especially D(N)=D(N)2. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) vck1cc∗1∗1 ...... vck1cc∗k∗k vck1vv1∗1∗ ...... vck1vvn∗n∗−−kk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Xi (cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12) ar SIIo.cAalNledUPPPigEeRonBOHUoNleDBUoSuInNdGfoPIrGdEeOcaNyHfuOnLcEtiPoRnIoNfCMIPILMEO- (cid:12)(cid:12)(cid:12) vn−kc∗1 ... vn−kc∗k vn−kv1∗ ... vn−kvn∗−k (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) is equal than MACcodeswasintroducedin[1].Here wegiveageneraliza- actiliosLno−efmeoimr ∈iat.2L.1(c:i+L1e,tcci1+,2c,2.,....,.c,kc)kc,1feo1r,ie2=,.1.,..,.e.k,−k1−∈1C.nW,arintde (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) vcek11...eee∗1∗1∗1 ......... vcek11...ccc∗k∗k∗k vcek11...vvv1∗1∗1∗ ......... vcek11vvv...nn∗∗n∗−−−kkk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12). c (cid:12) .. .. .. .. (cid:12) 2 (cid:12) . . . . (cid:12) . (cid:12) (cid:12) A= .. (cid:12)(cid:12) vn−kc∗1 ... vn−kc∗k vn−kv1∗ ... vn−kvn∗−k (cid:12)(cid:12) ckc−1 A(cid:12)nd since the way we chose v1,...,vn−k this means tha(cid:12)t k c c∗ ... c c∗ 0 ... 0 and (cid:12) 1. 1 1. k . . (cid:12) e1 (cid:12)(cid:12) .. .. .. .. (cid:12)(cid:12) B = ecek...−k21 . (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ck00...c∗1 ......... ck00...c∗k v100...v1∗ ......... vn−k00...vn∗−k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12) (cid:12) 2 is equal than and especially K e e∗ ... e c∗ 0 ... 0 D(N)≤ (cid:12) 1. 1 1. k . . (cid:12) Nα (cid:12) . . . . (cid:12) (cid:12) . . . . (cid:12) where α= U−1 2n2(U−l). Especially if k =Un we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ck0...e∗1 ...... ck0...c∗k v10...v1∗ ...... 00... (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) PD(l=N11,k.−..n,(NU−Ul)) ≤KUY−1Nl−2n(Ul−l) (cid:12) (cid:12) l=1 (cid:12)(cid:12) 0 ... 0 0 ... vn−kvn∗−k (cid:12)(cid:12) and (cid:12) (cid:12) K because if we write e =c −x where x ∈L(c ,...,c ) D(N)≤ then v e∗ =v (c −xi )∗ =i v ci∗−v x∗i=0−0i+=1 0 for akll Nβ j i j i i j i j i i=1,...,n−1 and j =1,...,n−k. This gives that where β = U−1 2n(U−l). Proof:PLelt=u1s uselthe notation C =(c⊤ ,...,c⊤ )⊤ for |v |2...|v |2det(AA†)=|v |2...|v |2det(BB†) l l,1 l,n 1 n−k 1 n−k l=1,...,U. and hence det(AA†)=det(BB†). Let us first fix some small CU ∈ LU(NU). Now Lemma 2.2: Let V = Rn be an n-dimensional vector |CU| = O(1). Then write WU = {(x⊤1,...,x⊤n)⊤|xi ∈ space, N a given positive integer, and let c1,c2,...,cn ∈V cLo(mcUp,l1e,m..e.n,tcUan,nd)}π. Th:enMlet VU(C=) W→U⊥Vbe aitns oorrtthhooggoonnaall be a basis for V. Let also U be a k-dimensional subspace of U n×k U projection. V and π : V → U an orthogonal projection into U. We can choosea basis{π(c ),π(c ),...,π(c )}forU suchthatif A subspace VU has dimR(VU) = 2nk − dimR(WU) = v = a c +a c +j1···+aj2c with |ajk| ≤ N for all i then 2nk − 2n2 = 2n(k − n) so the image πU(LU−1(NU−1)) 1 1 2 2 n n i π(v)=b π(c )+b π(c )+···+b π(c )with|b |≤n2N fallsinto a 2n(n−k)-dimensionalhypercubewith side length 1 j1 2 j2 k jk i smaller or equal than (2nk)2N = O(N ) by lemma for all i. U−1 U−1 Proof: Since {c ,c ,...,c } is a basis for V we can 2.2 with coordinates having restricted length since projection choose a basis for U 1from2 the sent {π(c1),π(c2),...,π(cn)}. can only shrink. We also have |LU−1(NU−1)|=θ(NU2n−k1) so using the linearity of π and pigeon hole principle we have Without loss of generality we may assume that U π(c ),π(c ),...,π(c ) are linearly independent. We some CU−1 ∈LU−1(NU−1) such that 1 2 k say that they form a basis K . Letπ(ck+1)=d1π(c1)+d12π(c2)+···+dkπ(ck)forsome π (C )=O(2n(k−nvu) NU2n−(1k−n))=O(N−k−nn). d1,d2,...,dk ∈R. U U−1 ut NU2n−k1 U−1 If |d |≤1 for all i=1,2,...,k then let K =K . i 2 1 Now similarly build V = W⊥ for l = 0,...,U −2 πdk(+cO1jt)πhe(cr=wis−ed)dj+1leπ−t(dc|jd1+j)1|π+(>c−dd1j2)πb+(ec·2·a)·+m+−adx·k·im+·1+aπl(cc−odde)jj−fafi1ncπdie(tcnhjte.−a1Nb)soo+w- Lby(cU,1s,e.tt.in.,gcU,nW,cUU−−l1,1,U.−.=.l,cU−1{U,n(−x,l.⊤1.,..,.c.U,−xl⊤n,1),⊤.|.x.i,cU−l,∈n)}. dj k+1 dj j+1 dj k This gives dimR(VU−l) = 2nk − dimR(WU−l) = lutevaluesofcoefficientsontherighthandsidearesmalleror 2nk − 2n2(l + 1) = 2n(k − nl − n). And again we equalthatone.InthiscaseletK =(K \π(c ))∪{π(c )}. 2 1 j k+1 find C such that U−l−1 It is clear that K is a basis for U. 2 forNoalwl isim=ila1r,ly2,f.o.r.m,na−nkewanbdaswisriKtei+K1 u=sinKgnb−aks+is1 K=i πU−l(CU−l−1)=O(2n(k−nl−nvuu) NU2Nn−(2lk−n−k1nl−n))=O(NU−−kl−n−ln+1l−nn). {π(c ),π(c ),...,π(c )}. t U−l−1 j1 j2 jk whNeroew|dπl,(ic|l)≤=ndflo,1rπa(cllj1l)a+nddli,2πbe(ccaj2u)se+e·v·e·ry+tdiml,keπw(chjekn) BL=em(mπa(C2.1)⊤g,.iv.e.s,πth(aCt if )A⊤,C=⊤)⊤(Ct1⊤h,e.n..d,eCt(U⊤A)A⊤†)an=d we took some π(ch) off from the basis it then had a such det(BB†2) th1at is of sizUe U−1 U representation in the new basis that all the coordinates were absolutelysmallerorequalthanzero.Repeatingthisprocedure O((U−2N−k−nln+l−nn)2n)=O(U−1N−k2−n2n((UU−−ll))). at most n−k ≤ n times and using triangle inequality gives Y U−l−1 Y l thentheproperty.Henceifv=a c +a c +···+a c with l=0 l=1 1 1 2 2 n n |a |≤ N then π(v) = b π(c )+b π(c )+···+b π(c ) i 1 j1 2 j2 k jk with |b |≤n2N for all i. i Theorem 2.3: For a MIMO-MAC lattice code REFERENCES (L1(N1),L2(N2),...,LU(NU))ofU users,eachtransmitting [1] H.-F.Lu,J.Lahtonen, R. Vehkalahti, and C.Hollanti: Remarks onthe withntransmissionantennas,havingacodeoflengthk ≥Un, criteriaofconstructingMAC-DMToptimalcodespresentedattheIEEE Inf.TheoryWorkshop,Cairo,Egypt,2010. and each users lattice is of full rank r = 2kn we have a [2] J.Lahtonen,R.Vehkalahti,H.-F.Lu,C.Hollanti,andE.Viterbo:Onthe constant K such that DecayoftheDeterminantsofMultiuserMIMOLatticeCodespresented attheIEEEInf.TheoryWorkshop,Cairo, Egypt,Jan.2010. D(N ,...,N )≤KU−1N−k2−n2n((UU−−ll)) 1 U Y l l=1