A topological Maslov index for 3-graded Lie groups - ResearchGate

A topological Maslov index for 3-graded Lie groups 17
1
Since E :=
0
0
is a grading element, the grading is inner. On the Lie
−1
algebra level we have
a
τ
c
∗ b −a
=
d
c∗ b∗ −d∗
,
showing that τ reverses the grading. The corresponding Jordan triple product
in A ∼ = g+ is given by
On the group level we have
GL2(A) + =
Then
1 A
0 1
{x, y, z} = 1
2 (xy∗ z + zy ∗ x).
, GL2(A) 0 =
GL2(A) + GL2(A) 0 GL2(A) − =
and any matrix in this set decomposes as
−1
a b 1 bd
=
c d 0 1
we obtain
× A 0
0 A ×
and GL2(A) − =
a b
∈ GL2(A): d ∈ A
c d
×
,
a − bd −1 c 0
0 d
1 0
d−1
.
c 1
From
1 0
−w∗
1 z 1 z
=
1 0 1 −w∗ 1 − w∗
z
BG(z, w) =
1 0
.
A 1
∗ −1 ∗ 1 − z(1 − w z) (−w ) 0
0 1 − w∗ −1
∗ 1 − zw 0
=
z 0 (1 − w∗z) −1
.
Next we calculate dG on quasi-invertible unitary triples (s1, s2, s3). For
unitary elements z, w ∈ S quasi-invertibility means that 1 − w ∗ z = 1 − w −1 z is
invertible, which means that w −z is invertible. Therefore all differences sj −sk ,
j = k, are invertible. Since
Lemma II.5 leads to
(s1 − s2) + (s2 − s3) + (s3 − s1) = 0,
(1 − s1s ∗ 2 )(1 − s3s ∗ 2 )−1 (1 − s3s ∗ 1 )(1 − s2s ∗ 1 )−1 (1 − s2s ∗ 3 )(1 − s1s ∗ 3 )−1
= (1 − s1s −1
−1
2 )(1 − s3s2 )−1 (1 − s3s −1
−1
1 )(1 − s2s 3
= (s2 − s1)(s2 − s3) −1 (s1 − s3)(s1 − s2) −1 (s3 − s2)(s3 − s1) −1
1 )−1 (1 − s2s −1
= −(s1 − s2)(s2 − s3) −1 (s3 − s1)(s1 − s2) −1 (s2 − s3)(s3 − s1) −1 = −1
)(1 − s1s −1
3 )−1