Gruber P. Convex and Discrete Geometry

438 Geometry of Numbers
Let q(x) = � aik xi xk be a positive definite quadratic form q on E d , with arithmetric
minimum m and discriminant δ = det(aik). q is eutactic if the following
statement holds: let (bik) = (aik) −1 . Then
(b11, 2b12,...,2b1d, b22, 2b23,...,bdd),
the normal vector of the discriminant surface D(δ) at its point q, is a linear combination
with positive coefficients of the vectors
(u 2 1 , 2u1u2,...,2u1ud, u 2 2 , 2u2u3,...,u 2 d )
where ±(u1,...,ud) ∈ Zd ranges over the minimum vectors of q. These vectors
are the normal vectors of the facets of the Ryshkov polyhedron R(m) which contain
the boundary point q of R(m) and thus generate the normal cone of R(m) at q. This
seemingly strange definition is, in fact, perfectly natural, as will be clear from the
following basic theorem of Voronoĭ [1012] and its proof.
It seems that Coxeter [229] was the first to use the word eutactic in the present
context. Presumably, he wanted to express the fact that (bik) is well (=eu in Greek)determined
(=tactic) by the vectors (u2 1 ,...,2u1ud,...,u2 d ). See also Martinet [690].
Theorem 29.9. A positive definite quadratic form on E d is extreme if and only if it is
perfect and eutactic.
Proof (by means of the Ryshkov polyhedron). Let q be a positive definite quadratic
form on E d with coefficients aik, arithmetic minimum m and discriminant δ. For
the proof of the theorem, it is sufficient to show that the following statements are
equivalent:
(i) q is extreme.
(ii) A suitable neighbourhood of (a11, a12,...,add) in the polyhedron R(m)
is contained in the unbounded convex body determined by the smooth
and strictly convex surface D(δ) through (a11, a12,...,add).
(iii) (a11, a12,...,add) is the only point of R(m) in the tangent hyperplane
of D(δ) at (a11, a12,...,add), that is the hyperplane
�
v = (v11,...,vdd) ∈ E 1 2 d(d+1) : �
�
bik vik = d, vik = vki ,
where (aik) −1 = (bik).
(iv) q is perfect and eutactic.
Only (iii)⇔(iv) needs justification: Consider a point of a convex polyhedron and a
hyperplane through it. Then the following are equivalent (a) the hyperplane meets the
polyhedron only at this point and (b) this point is a vertex of the polyhedron and thus
the unique point contained in all facets through it, and the exterior normal vector
of the hyperplane is a linear combination with positive coefficients of the exterior
normal vectors of these facets, see Proposition 14.1. ⊓⊔
i,k