For printing - MSP

(i) K ρ
(−1,0,−1) =
(j) K ρ
(−1,1,0) =
(k) K ρ
(1,−1,0) =
(l) K ρ
(−1,−1,0) =
PRIMITIVE Z2 ⊕ Z2-MANIFOLDS 213
Z2, if ρ χ2, ρ1, ρ3, µ;
0, if ρ χ1, χ3, ρ2, ν;
Z2, if ρ χ2, ρ1;
0, if ρ χ1, χ3, ρ2, ρ3, µ, ν;
Z2, if ρ χ1, ρ2, µ;
0, if ρ χ2, χ3, ρ1, ρ3, ν;
Z2, if ρ χ3, ρ1, ρ2, µ;
0, if ρ χ1, χ2, ρ3, ν.
Corollary 2.4. The representations µ and ν are not equivalent.
Proof. The result follows from Proposition 2.1 and Equation (a) in the previous
lemma.
Corollary 2.5. The representations µ and ν are indecomposable.
Proof. Being ν of rank 3, if decomposable, it must be ν χj1 ⊕ χj2 ⊕ χj3 ,
for 1 ≤ j1, j2, j3 ≤ 3 or ν χj1 ⊕ ρj2 , for 1 ≤ j1, j2 ≤ 3. But Equations
(a), (b) and (c) in Lemma 2.3 contradict both possibilities. The case of µ is
similar.
We had encountered 8 indecomposable representations of Z2 ⊕ Z2 having
no fixed points. The following theorem asserts that there are no more. Recall
that they are given by
(2.3)
B1 B2 B3
χ1 : (1) (−1) (−1)
χ2 : (−1) (1) (−1)
χ3 : (−1) (−1) (1)
ρ1 : −I J −J
ρ2 : J −I −J
ρ3 : J
−J
−1 1 −1
−I
1 −1 1
µ :
ν :
−1 0 0
0 1
1 0
−1 1 0
1 0
−1
0 −1
−1 0
−1 0 1
−1 0
−1
−1 0 1 1 −1 −1
−1 0 .
−1
Theorem 2.6. Let ρ be an indecomposable integral representation of Z2⊕Z2
having no fixed points. Then ρ is equivalent to one and only one of the
representations in (2.3).
The proof of Theorem 2.6 is elementary but not trivial. In order to make
the paper more readable, we wrote the proof in the forthcoming subsection.
By assuming Theorem 2.6 one can skip the following subsection without
losing the understanding of the whole paper.