Computation and Properties of Maps
125
it is shown that ˆ
M
−1 is symplectic as well:
ˆ
M
−1
T ˆ
J ˆ
M
−1 = ˆ
J.
Now let us obtain the determinant of ˆ
M following Kauderer (page 10 in
[36]). Through a series of permutations, we can rewrite ˆ
J, which becomes
ˆ
J =
ˆ 0 ˆ
I
− ˆ
I ˆ 0
,
where ˆ
I is the n × n identity matrix. Furthermore, ˆ
M can be written as
ˆ
M =
ˆ
A ˆ
B
ˆ
C ˆ
D
,
where ˆ
A, ˆ
B, ˆ
C and ˆ
D are n × n matrices. The symplectic condition becomes
ˆ
A
T ˆ
C
T
ˆ
B
T ˆ
D
T
ˆ 0 ˆ
I
− ˆ
I ˆ 0
ˆ
A ˆ
B
ˆ
C ˆ
D
=
ˆ 0 ˆ
I
− ˆ
I ˆ 0
,
which leads to the relations
− ˆ
C
T ˆ
A + ˆ
A
T ˆ
C = ˆ 0,
− ˆ
C
T ˆ
B + ˆ
A
T ˆ
D = ˆ
I,
− ˆ
D
T ˆ
A + ˆ
B
T ˆ
C = − ˆ
I,
− ˆ
D
T ˆ
B + ˆ
B
T ˆ
D = ˆ 0.
(5.8)
Furthermore we need one more mathematical theorem, which is
det
ˆ
A ˆ 0
ˆ
C ˆ
D
= det ˆ
A · det ˆ
D = det( ˆ
A ˆ
D).
Hence we have
det
ˆ
I − ˆ
A
−1 ˆ
B
ˆ 0
ˆ
I
= 1.
Using this, we obtain
det
ˆ
A ˆ
B
ˆ
C ˆ
D
= det
ˆ
A ˆ
B
ˆ
C ˆ
D
· det
ˆ
I − ˆ
A
−1 ˆ
B
ˆ 0
ˆ
I
= det
ˆ
A
ˆ 0
ˆ
C ˆ
D − ˆ
C ˆ
A
−1 ˆ
B
= det ˆ
A det( ˆ
D − ˆ
C ˆ
A
−1 ˆ
B).
Furthermore, using the relations det ˆ
A = det ˆ
A
T , ˆ
C ˆ
A
−1 = ( ˆ
A
T )
−1 ˆ
C
T and
ˆ
A
T ˆ
D − ˆ
C
T ˆ
B = ˆ
I that have resulted from eq. (5.8), we obtain
det
ˆ
A ˆ
B
ˆ
C ˆ
D
= det ˆ
A
T det
ˆ
D − ( ˆ
A
T )
−1 ˆ
C
T ˆ
B
= det
ˆ
A
T ˆ
D − ˆ
C
T ˆ
B
= 1.