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Minkowski's Geometry of Numbers Geometry of numbers MAT4250 — Høst 2013 Minkowski’s geometry of numbers Preliminary version. Version +2✏ — 31. oktober 2013 klokken 12:04 1 Lattices Let V be a real vector space of dimension n.Byalattice L in V we mean a discrete, additive subgroup. We say that L is a full lattice if it spans V .Ofcourse,anylattice is a full lattice in the subspace it spans. The lattice L is discrete in the induced topology, meaning that for any point x L 2 there is an open subset U of V whose intersection with L is just x.Forthesubgroup L to be discrete it is sufficient that this holds for the origin, i.e., that there is an open neigbourhood U about 0 with U L = 0 .Indeed,ifx L,thenthetranslate \ { } 2 U 0 = U + x is open and intersects L in the set x . { } Aoftenusefulpropertyofalatticeisthatithasonlyfinitelymanypointsinany bounded subset of V .Toseethis,letS V be a bounded set and assume that L S is ✓ \ infinite. One may then pick a sequence xn of different elements from L S.SinceS { } \ is bounded, its closure is compact and xn has a subsequence that converges, say to { } x. Then x is an accumulation point in L;everyneigbourhoodcontainselementsfrom L distinct from x. The lattice associated to a basis If = v1,...,vn is any basis for V ,onemayconsiderthesubgroupL = Zvi of B { } B i V consisting of all linear combinations of the vi’s with integral coefficients. Clearly L P B is a discrete subspace of V ;IfonetakesU to be an open ball centered at the origin and having radius less than mini vi ,thenU L = 0 .HenceL is a full lattice in k k \ { } B V . If might very well happen that another basis 0 generates the same lattice as .If B B this occurs, the vectors v0 of lies in L and they can be expressed as linear combinations i B vj0 = aivi with coefficients in Z,andsymmetrically,thevectorsfrom are integral i B linear combinations of those from 0. Thus the transition matrix (aij) must lie in P B Gl(n, Z),andweobservethatitsdeterminantequals 1. ± It might also happen that the lattice generated by 0 is contained, but not neces- B sarily equal to L . Then the transition matrix (aij) still has integral coefficients, but B the determinant is an integer different from 1. The quotient group L /L is a finite ± B B0 group whose order equals det(aij) . This number is also called the index of L in L | | B B0 and written [L : L ]. B B The fundamental domain There is a particular subset of V associated to the lattice L It is the subset T V of real linear combinations of the vi’s whose coefficients are B ✓ —1— Geometry of numbers MAT4250 — Høst 2013 confined to the interval [0, 1 >,thatis T = a v 0 a < 1 . { i i | i } i X This set is called the the fundamental domain of L .Itdependsonthebasis ,soa B B lattice has many fundamental domain, one for each basis. Assume now that V comes equipped with a metric,thatisascalarproduct.We can then speak about metric properties of subsets in V ,likelength,areaandvolume, etc. The etc stands for the the n-dimensional volume—or just volume or measure for short—and we denote by µ(S) the measure of a subset S (since it is in fact the Lebegues measure). The sets we shall meet are extremely nice from a measure-theoretical point of view, so there is never a question about they being measurable. Assume that is an orthonormal basis. Any other basis may of course be E B expressed in terms of by means of the transition matrix M.Afundamentalpro- E perty of the measure on V ,isthatthevolumeofthefundamentaldomainT of the lattice L is equal to the absolute value of the determinant of M,thatis B µ(T )= det M . | | The translates of T + v of the fundamental domain by elements v L are disjoint 2 B and their union equals the whole of V ,sotheyformacovering of V . This is not an open covering since T is not open (neither close for that matter ), a part of the border is contained in T and a part is not. To see that the translate are disjoint, assume that aivi = ai0 vi + nivi, i i i X X X with ai,a0i [0, 1 >. Then ni = ai a0 < 1, 1 >.Soiftheni’s in addition are 2 − i 2 − integers, they must all be zero. That the translates cover V , follows since aivi = <ai >vi + [ai]vi = t + v, i i X X X where <t>(resp. [t])denotesthefractional(respintegralvalue)ofanumbert. Every lattice has a basis In fact, the previous collection L of examples of lattices is exhaustive—any full lattice B is of the type described there: Proposition 1 Let L be a full lattice in the real vector space V .Thenthereisabasis for V such that L = L . B B Proof: Since L spans V ,wemayfindabasis for V contained in L.LetL0 = L be C C the lattice spanned by . C —2— Geometry of numbers MAT4250 — Høst 2013 The main point of the proof, is that L0 is of finite index in L.Indeed,letT be the fundamental domain of L0. Now, L T is a finite set since L is discrete and T bounded. \ Any x L is of the form x = t + v with v L0 and t T ,andthereforet = x v is 2 2 2 − among the finitely many elements in T L,andx is congruent modulo L0 to one of \ the elements in T L. \ 1 The index m =[L : L0] kills the quotient L/L0 and therefore mL L0,orL m L0 1 1 ✓ ✓ if one wants. But L0 has the basis vi and is hence a finitely generated free abelian m { m } group. By a standard result about such groups, it follows that L is finitely generated and free, and thus has a basis. o Blichfeldt-Minkowski theory There is a whole industry producing results about the number of lattice points of agivenlatticeL in V lying in a given bounded set S.Onemay,forexample,ask for the asymptotic behavior of the cardinality of S L as a function of r (where \ s = rx x S ). This question goes back to Gauss and his circle problem, in { | 2 } which n =2and S is the unit circle. We take a slightly different point of view in this paragraph, merely asking for the intersection S L to be non empty—or more precisely \ having a non-zero element. We present a lemma, usually contributed to Hans Frederik Blichfeldt. He was an danish-american mathematician, born in the small village Iller close to Viborg on Jutland, whose parents emigrated to America in 1888. And we present what usually is called Minkowski’s theorem, a result due to the much more known Königsberg- and Göttingen-mathematician Hermann Minkowski. Blichfeldt’s lemma We work with at full lattice L in the real vector space V which is equipped with a metric, i.e., an inner product—you may have Rn with the standard inner product in mind. We choose a basis v1,...,vn for L and let T denote the corresponding fundamental domain. It is an absolutely obvious observation that if S V is a bounded subset and S T ,thenµ(S) µ(T ). Elaborating this, we obtain ✓ ✓ the following slightly more subtle statement, known as Blichfeldt’s lemma: Lemma 1 Assume that L V is a full lattice and that S V is a bounded set. If ✓ ✓ µ(S) >µ(T ),thentherearetwodifferentelementsx, y S such that x y L. 2 − 2 Proof: For any v L let Sv = S (T + v),thatis,thepartofthetranslateT + v 2 \ lying with in S. As the different translates T v form a disjoint covering of V as v − runs through the lattice L,onehasS = v L S (T + v). 2 \ Moving the set S (T + v) back into T we get the set \ S S = S (T + v) v = t T t + v S . v \ − { 2 | 2 } Two such sets share a common element t if and only if x = t + v and y = t + w both lie in S. Clearly their difference lies in L, so this is what we are aiming for. —3— Geometry of numbers MAT4250 — Høst 2013 Assume then that the Sv’s are disjoint, and let S0 = v L Sv. Then 2 S µ(S0)= µ(S )= µ(S (T + v)) = µ(S) >µ(T ), v \ X X contradicting the fact that S0 S. o ✓ For those acquainted with the world of manifolds and the general machinery of volume elements, there is a shorter way (which may be also reveals the mystery; once the machinery is in place, it is just the stupid statement we stated with) to present this proof: The quotient map V V/L is a local isometry, and the interior T is isometric ! to an open dense subset of V/L,henceµ(V/L)=µ(T ).Butifµ(S) >µ(T ),thesetS can not map invectively into V/L, hence two of its members differ by an element in L. For those who enjoy illustration, we have made a picture illustrating the first way of presenting the proof—the others can show it to their kids. The set S is the yellow square to the left. The four triangles to the right are the different intersection of S with the translates of T ,movedbacktoT . The quotient V/L is a torus. The quotient the map T V/L identifies the four corners of T ,theparallelpartsoftheborderare ! pairwise identified, and the four triangles are reunited to the yellow square.
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