Gruber P. Convex and Discrete Geometry

508 Geometry of Numbers
We will show that
(15) S = V(G ′ ).
S = � v ∈ V(G ′ ) : lim
n→∞ ρ(n)
v �= 0 � .
For suppose not. Then, by a calculation similar to the one that led to (11), we see
that
(16) � � (n)
ϑv ρ � = π
2 #E�G ∧ (S) � − π(#S − l)
v∈S
v∈S
− 1
2
�
αi + �
vi ∈S
vw∈E(G ′ )
v∈S,w�∈S
arctan ρ(n) w
ρ (n)
v
By the definition of S, the last sum tends to 0 as n →∞. Thus
(17) � � (n)
ϑv ρ � →− π � � � � ∧ 1 �
2#S − #E G (S) − l − 2 + (π −αi) − π.
2
2
Since π − αi > 0fori = 1,...,k, and � (π − αi) = 2π, it follows from (17) that,
in case when (13) holds, � � �
ϑv ρ (n) � : v ∈ S � < 0 for all sufficiently large n. The
same is true if (14) holds with inequality. If (14) holds with equality, it is also true
since then l < k. (To see the latter suppose that l = k. Since k ≥ 3 and #S ≥ l it
then follows that l = k = #S = 3 and thus S ={v1,v2,v3}, #E � G∧ (S) � = 0.) The
remaining case, #S = 1, trivially yields the same conclusion. Taking into account
(10), we have
� � (n)
ϑv ρ � > 0 for all sufficiently large n.
v�∈S
By considering a suitable subsequence and renumbering, if necessary, it follows that
�
there is a vertex v �∈ S with ϑv ρ (n) � > 0 for all n. Then ρ (n)
v = 1 for all n by the
definition of R. Hence v ∈ S by the definition of S. This contradiction concludes the
proof of (15).
Let ρ = limn→∞ ρ (n) . Since the ϑv(·) are continuous, ρ ∈ R. We now show
that
(18) µ(ρ) = 0.
For, suppose that, on the contrary, µ(ρ) > 0. Then by (10),
∅ �= T = � v ∈ V(G ′ ) : ϑv(ρ) < 0 � � V(G ′ ).
Let σv = λρv for v ∈ T and σv = ρv for v �∈ T and given λ, where 0