22
1 From Riemann manifolds to Riemann manifolds
Box 1.11 (Eckert II, the third problem).
G l = R
2
»
cos
2 Φ 0
0
1
–
,
C l according to Box 1.10.
(1.81)
Left general eigenvalue problem:
˛
˛ C l − Λ
2 G l
˛
˛ = 0 ⇔ Λ
2
1,2 = Λ
2
+,− =
1
2
„
tr
ˆ
C l G
−1
l
˜ ±
q `
tr
ˆ
C l G
−1
l
˜´ 2 − 4det
ˆ
C l G
−1
l
˜
«
,
(1.82)
`
C l G
−1
l
´
11
= c 11 G
−1
11 = +
2
3π
4 − 3 sin |Φ|
cos 2 Φ
,
`
C l G
−1
l
´
12
= c 12 G
−1
22 = −
1
π
Λ cos Φ signΦ ,
`
C l G
−1
l
´
21
= c 21 G
−1
11 = −
1
π
Λ cos Φ signΦ ,
`
C l G
−1
l
´
22
= c 22 G
−1
22 = +
3
2π
cos
2 Φ
4 − 3 sin |Φ|
`
Λ
2 + π
2 ´
,
(1.83)
det
ˆ
C l G
−1
l
˜
= 1 , tr
ˆ
C l G
−1
l
˜
=
2
3π
4 − 3 sin |Φ|
cos 2 Φ
+
3
2π
cos
2 Φ
4 − 3 sin |Φ|
`
Λ
2 + π
2 ´
.
(1.84)
Λ 1 Λ 2 = 1:
Λ
2
1 Λ
2
2 = det
ˆ
C l G
−1
l
˜
= 1 .
(1.85)
Left eigencolumns:
(i)
√ :=
q
G 11 (c 22 − Λ
2
1 G 22 ) 2 + G 22 c
2
12
(G 12 = 0) ,
2
4
F 11
F 22
3
5 =
1
√
2
6
6
4
3
2π
R
2
cos
2 Φ
4 − 3 sin |Φ|
`
Λ
2 + π
2 ´ − Λ
2
1 R
2
1
π
R
2 Λ cos Φ signΦ
3
7
7
5 ;
(1.86)
(ii)
√ :=
q
G 22 (c 11 − Λ
2
2 G 11 ) 2 + G 11 c
2
12
(G 12 = 0) ,
2
4
F 12
F 21
3
5 =
1
√
2
6
6
4
1
π
R
2 Λcos Φ signΦ
2
3π
R
2 (4 − 3 sin |Φ|) − Λ
2
2 R
2 cos
2 Φ
3
7
7
5 .
(1.87)
Box 1.12 (Eckert II, the fourth problem).
Λ 1 = Λ 2 ⇔
`
tr
ˆ
C l G
−1
l
˜´ 2 = 4det
ˆ
C l G
−1
l
˜
,
(1.88)
det
ˆ
C l G
−1
l
˜
= 1 ,
(1.89)
`
tr
ˆ
C l G
−1
l
`
Λ = 0, Φ = 0
´˜´ 2 =
„
64 + 9π
2
24π
« 2
= 4det
ˆ
C l G
−1
l
˜
= 4 ⇒ Λ 1 = Λ 2 .
(1.90)