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ON THE SIGNATURE 41
Proposition 5.5. Consider ν ∈ a∗ int such that J(ν) has integral infinitesimal
character. If conditions (1)-(4) above are satisfied then ν =
α∈Σ + cαα
with cα = 1/2 or 0.
Proof. Assume that r = 1 and ɛ = 0, in other words that G = SO(p, q) with
p − q even. So s = p−q
2 ∈ Z. In this setting J(ν) has integral infinitesimal
character if and only if νi ∈ Z and
R(i) = #{n : νn = i} i ≥ 1
and
R(0) = 2#{n : νn = 0}.
Then by conditions (1)-(4), Formula (5.1) and assuming R(s) = 0 and there
exists j < s such that R(j) = 0 we can consider ν as follows
(5.6)
ν = (s + k, . . . , s + k, s + (k − 1), . . . , s+(k − 1), . . . . . . , s, . . . , s,
with 0 ≤ νi < s, i = 1, . . . , j.
νj, . . . , νj, . . . . . . , ν1, . . . ν1)
In order to complete the proof of this proposition, we will need the following:
Lemma 5.7. Take ν as above. Then
k − i + 1 ≤ q −
i
R(s + k − r).
r=0
Proof. Since R(s) = 0 we have, by condition (2) in Theorem 5.4 R(s+m) =
0, for m = 1, . . . , k. Hence, using that q = k r=0 R(s+k−r)+ j−1
p=0 R(νj−p)
we have
k − i + 1 ≤
k
j−1
R(s + k − r) + R(νj−p)
r=i+1
= q −
i
R(s + k − r).
r=0
So, the proof of the lemma is complete.
Proof Proposition 5.5. (Cont.) Recalling that each root ei ± ej has multiplicity
1 and el has multiplicity p − q = 2s, and defining
m−1 i=0 R(s + k − i), if m ≥ 1
T (m) =
0, if m = 0
and
S(n) =
p=0
n−1
i=0 R(νj−i), if n ≥ 1
0, if n = 0