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212 PAULO TIRAO
K ρ1
(0,−1,1) = Ker(B2 + I) ∩ Ker(B3 − I)
Im(B2 − I) ∩ Im(B3 + I)
we write down the kernels and the images in the canonical basis {e1, e2} of
Z 2 . We have
Ker(B1 + I) = 〈e1, e2〉 Im(B1 − I) = 〈2e1, 2e2〉
Ker(B2 + I) = 〈e1 − e2〉 Im(B2 − I) = 〈e1 − e2〉
Ker(B3 − I) = 〈e1 − e2〉 Im(B3 + I) = 〈e1 − e2〉,
from which it follows that
K ρ1
(−1,0,1) Z2 and K ρ1
(0,−1,1) = 0.
The computation of these invariants (Proposition 2.1) for all the representations
in Proposition 2.2 and for the representations µ and ν in (2.2)
are as simple as those performed above. Thus, we put together the results
as a lemma, omitting the details.
Lemma 2.3. Let ρ be a representation equivalent to χi or ρi, (1 ≤ i ≤ 3)
as in Proposition 2.2 or equivalent to µ or ν as in (2.2). Then,
(a) K ρ
(1,−1,−1) =
Z2,
0,
if ρ χ1, ρ2, ρ3, µ;
if ρ χ2, χ3, ρ1, ν;
(b) K ρ
(−1,1,−1) =
Z2,
0,
if ρ χ2, ρ1, ρ3, µ;
if ρ χ1, χ3, ρ2, ν;
(c) K ρ
(−1,−1,1) =
Z2,
0,
if ρ χ3, ρ1, ρ2, µ;
if ρ χ1, χ2, ρ3, ν;
(d) K ρ
(0,1,−1) =
Z2,
0,
if ρ χ2, ρ3, µ;
if ρ χ1, χ3, ρ1, ρ2, ν;
(e) K ρ
(0,−1,1) =
Z2,
0,
if ρ χ3, ρ2, µ;
if ρ χ1, χ2, ρ1, ρ3, ν;
(f) K ρ
(0,−1,−1) =
Z2,
0,
if ρ χ1, ρ3, µ;
if ρ χ2, χ3, ρ1, ρ2, ν;
(g) K ρ
(1,0,−1) =
Z2,
0,
if ρ χ1, ρ3, µ;
if ρ χ2, χ3, ρ1, ρ2, ν;
(h) K ρ
(−1,0,1) =
Z2,
0,
if ρ χ3, ρ1;
if ρ χ1, χ2, ρ2, ρ3, µ, ν;