TITLE: lattice with Voronoi cell inside a circle
QUESTION [5 upvotes]: This considers real-valued lattices in two dimensions.
I need to find the densest lattice $\Lambda$, i.e., the one with the smallest determinant of its generator matrix, such that the Voronoi cell of its dual lattice $\Lambda^\ast$ is fully contained in the unit disc.
I strongly believe that it should still be the hexagonal lattice (and numerical tests seem to validate this belief).
Is there a simple proof.

REPLY [4 votes]: Your conjecture is correct.  The condition on $\Lambda^*$ means that
the covering radius of $\Lambda^*$ is at most $1$, and it is known that
the best covering lattice is hexagonal, which makes $\Lambda$ hexagonal too.
One way to see that the hexagonal lattice is the best covering lattice
is as follows.  In two dimensions, the Voronoi region is always a 
centrally-symmetric hexagon (possibly degenerate to a rectangle)
inscribed in a circle whose radius $-$ besides being the covering radius $-$
is the circumcenter of a triangle with vertices $0,v,w$ for some
vectors $v,w$ that generate the lattice.  So in our setting 
the sparsity of $\Lambda^*$ is twice the area of a triangle inscribed
in a unit circle, and this area is maximized when the triangle is
equilateral.  This makes $\Lambda^*$, and thus also $\Lambda$,
hexagonal, QED.