Page 158 - Textos de Matemática Vol. 44

148 H. NAKAZATO, A. KOVACˇEC, N. BEBIANO AND J. DA PROVIDEˆNCIA
assume s ∈ S. We prove s ∈ [0,1]∂S. For s = 0 this is trivial. So let s ̸= 0.
Since S is compact, there exists t0 = max{t ∈ R>0 : ts ∈ S}. Let s0 = t0s.
Then(t0+ε)s̸∈S,while(t0−ε)s=t0−εs0 ∈Sforallsmallε>0,sinceS t0
is starshaped w.r.t. 0. Consequently s0 ∈ ∂S. Also, since s ∈ S, t0 ≥ 1. Hence s = 1 s0 ∈ [0,1]∂S.
t0
By [1], Theorem 3, Uni(3) is compact star-shaped set with respect to C0. So
by rephrasing the lemma it follows that
(2.1) Uni(3)={tC0 +(1−t)A:0≤t≤1,A∈∂Uni(3)}.
Now from [11], Propositions 2.2, 2.3, 2.4 it follows that the boundary ∂Uni(3) of Uni(3) in A(3) coincides with the set of Hadamard products of rotation matrices by themselves, that is,
(2.2) ∂Uni(3) = {(o2ij ) : (oij ) ∈ SO(3)}.
We assume again that D = diag(λ1,λ2,λ3) for some complex numbers λj. Then
the generic diagonal entry of UDU∗, U ∈ U(3), is expressed as
3
(UDU∗)ll = |ulj|2λj, l = 1,2,3.
j=1
Thus we have
(2.3) W3Π(D) = {( a1jλj)( a2jλj)( a3jλj) : A = (aij) ∈ Uni(3)}.
333 j=1 j=1 j=1
In the case λ1 +λ2 +λ3 = 0, which covers ours since 1+ω+ω2 = 0, the following homogeneity holds
331−t 33
( 3 +t|ulj|2)λj =t3 |ulj|2λj. l=1 j=1 l=1 j=1
Thus, using (2.1) in (2.3), W3Π(D) is star-shaped with respect to 0 and so we proved the first part of Theorem 1.1. In reality we found somewhat more, namely:
Fact 2.2 Let λ1 +λ2 +λ3 = 0 and D = diag(λ1, λ2, λ3). Then W3Π(D) = [0, 1]Ω, where
33
Ω={ o2ljλj :(olj)∈SO(3)}.
l=1 j=1