Successive minima need not be attained by a basis #
Let L = ℤ⁵ + ℤ · (½, …, ½) in EuclideanSpace ℝ (Fin 5), measured by the Euclidean unit ball.
Every nonzero vector of L has norm at least 1: a point of L has all coordinates in ℤ or all
coordinates in ½ + ℤ, and in the second case its squared norm is at least 5 / 4. The standard
basis vectors e 0, …, e 4 lie in L and have norm 1, so all five successive minima of L are
1, attained by the e i. But the e i span ℤ⁵, which has index two in L, so no basis of L
consists of them.
Cassels' basis lemma ZLattice.exists_basis_gauge_le_max_one_mul, applied to the e i, still
produces a basis b of L with ‖b i‖ ≤ max 1 ((i + 1) / 2). An explicit such basis is
e 0, e 1, e 2, e 3, (½, …, ½), whose last vector has norm √5 / 2 ≤ 5 / 2.
Main results #
TauCeti.GaugeBasisExample.one_le_norm_of_mem_lattice: nonzero vectors ofLhave norm≥ 1.TauCeti.GaugeBasisExample.relIndex_stdLattice_lattice:ℤ⁵has index two inL.TauCeti.GaugeBasisExample.not_exists_basis_eq_single: thee iare not a basis ofL.TauCeti.GaugeBasisExample.exists_basis_norm_le: the basis given by Cassels' lemma.TauCeti.GaugeBasisExample.norm_basis_le: the explicit basis meets the same bounds.
References #
- J. W. S. Cassels, An Introduction to the Geometry of Numbers, Springer, 1959, Chapter V, Lemma 8.
The point (½, …, ½) of EuclideanSpace ℝ (Fin 5).
Instances For
The generators e 0, e 1, e 2, e 3, (½, …, ½) of L.
Instances For
The standard lattice ℤ⁵, spanned by e 0, …, e 4.
Instances For
The lattice L = ℤ⁵ + ℤ · (½, …, ½).
Instances For
L is discrete, since it is spanned by the real basis gen.
Every point of L has its last coordinate in ½ℤ, and all its coordinates differ from the
last one by integers.
A point lies in L exactly when its last coordinate is in ½ℤ and all its coordinates differ
from the last one by integers.
A point of L lies in ℤ⁵ exactly when its last coordinate is an integer.
The minima witnesses span a sublattice of index two. ℤ⁵ has index two in L, with
(½, …, ½) representing the nontrivial coset.
The minima witnesses are not a basis. No basis of L consists of the vectors
e 0, …, e 4, although they realize all five successive minima of L.
The basis e 0, e 1, e 2, e 3, (½, …, ½) of L.