3 Proof of the Lattice Point Theorem; Other Results 339Put
yj=(xj−xm+ 1 )/ 2 (j= 1 ,...,m).Then, by construction,yj∈Λ(j= 1 ,...,m). Moreoveryj∈C,sincex 1 ,...,xm+ 1 ∈
CandCis symmetric, and similarlyyj−yk=(xj−xk)/ 2 ∈C. Finally, since
y 1 >y 2 >···>ym>O,we haveyj=Oandyj=±ykifj=k.
The conclusion of Proposition 12 need no longer hold ifCis not compact and
λ(C)= 2 nmd(Λ). For example, takeΛ=Znand letCbe the symmetric convex set
C={x=(ξ 1 ,...,ξn)∈Rn:|ξ 1 |<m,|ξj|<1for2≤j≤n}.Then d(Λ)=1andλ(C)= 2 nm. However, the only nonzero points ofΛinCare the
2 (m− 1 )points(±k, 0 ,..., 0 )( 1 ≤k≤m− 1 ).
To provide a broader view of the geometry of numbers we now mention
without proof some further results. A different generalization of Minkowski’s lattice
point theorem was already proved by Minkowski himself. LetΛbe a lattice inRnand
letKbe a compact symmetric convex subset ofRnwith nonempty interior. ThenρK
contains no nonzero point ofΛfor smallρ>0 and containsnlinearly independent
points ofΛfor largeρ>0. Letμidenote the infimum of allρ>0suchthatρK
contains at leastilinearly independent points ofΛ(i = 1 ,...,n). Clearly the
successive minimaμi=μi(K,Λ)satisfy the inequalities
0 <μ 1 ≤μ 2 ≤···≤μn<∞.Minkowski’s lattice point theorem says thatμn 1 λ(K)≤ 2 nd(Λ).Minkowski’stheorem on successive minimastrengthens this to
2 nd(Λ)/n!≤μ 1 μ 2 ···μnλ(K)≤ 2 nd(Λ).The lower bound is quite easy to prove, but the upper bound is more deep-lying —
notwithstanding simplifications of Minkowski’s original proof. IfΛ = Zn,then
equality holds in the upper bound for thecube K ={(ξ 1 ,...,ξn)∈Rn :|ξi|≤
1forall∑ i}and in the lower bound for thecross-polytope K={(ξ 1 ,...,ξn)∈Rn:
n
i= 1 |ξi|≤^1 }.
IfKis a compact symmetric convex subset ofRnwith nonempty interior, we
define itscritical determinant∆(K)to be the infimum, over all latticesΛwith no
nonzero point in the interior ofK, of their determinants d(Λ). A latticeΛfor which
d(Λ)=∆(K)is called acritical latticeforK. It will be shown in§6 that a critical
lattice always exists.