46
1 From Riemann manifolds to Riemann manifolds
Before we present the sketches of proofs for the various conditions, it has to be noted that the
generalized Korn–Lichtenstein equations, which govern conformal mapping M
2
l → M
2
r , suffer from
the defect that they contain the unknown functions g 11 [u
λ (U
Λ )] and g 22 [u
λ (U
Λ )], and the reason is
that the mapping functions u
λ (U
Λ ) have to be determined. In case of {M
2
r , g µν } = {R
2 , δ µν }, the
corresponding Korn–Lichtenstein equations do not suffer since these functions do not appear. The
stated problem is overcome by representing the right Riemann manifold M
2
r by isometric coordinates
(also called conformal coordinates or isothermal coordinates) directly such that the quotient g 22 /g 11
is identical to one. This is exactly the procedure advocated by C. F. Gauss (1822, 1844) and applied
to the conformal mapping of E
2 onto S
2
r . We shall come back to this point-of-view after the proof.
Proof (first part).
(i) ⇒ (ii).
Ψ l = Ψ r → cos Ψ l = cos Ψ r ⇔ U
1
T G l U
2 = u
1
T G r u
2 ⇔
⇔ du
T
1 J
T
r G l J r du 2 =
dS 1
ds 1
du
T
1 G r du 2
dS 2
ds 2
⇔ du
T
1 C r du 2 = λ 1 du
T
1 G r du 2 λ 2 ⇔
⇔ λ 1 = λ 2 = λ(u 0 ) , C r = λ
2 (u 0 )G r q. e. d.
(1.158)
cos Ψ r = cos Ψ l ⇔ u
1
T G r u
2 = U
1
T G l U
2 ⇔ dU
T
1 J
T
l G r J l dU 2 =
ds 1
dS 1
dU
T
1 G l dU 2
ds 2
dS 2
⇔
⇔ Λ 1 = Λ 2 = Λ(U 0 ) , C l = Λ
2 (U 0 )G l q. e. d.
(1.159)
(i) ⇐ (ii).
⎡
⎢
⎣
cos Ψ l = U
1
T G l U
2 =
ds 1
dS 1
u
T
1 J
T
r G l J r du 2
ds 2
dS 2
J
T
r G l J r = C r = λ
2 (u 0 )G r , λ
−1
1 = λ
−1
2 = λ
−1
⎤
⎥
⎦ ⇒
cos Ψ l = u
1
T G r u
2 = cos Ψ r
orientation is preserved
⇔
⇔ Ψ l = Ψ r q. e. d.
(1.160)
End of Proof (first part).
Proof (second part).
(ii) ⇒ (iii).
Left eigenvalue problem:
C l = Λ
2 (U 0 )G l , E l = K(U 0 )G l ⇔
⎡
⎣
Λ
2 (U 0 ) = Λ
2
1 = Λ
2
2
K(U 0 ) = K
2
1 = K
2
2
⎤
⎦ .
(1.161)
Right eigenvalue problem:
C r = λ
2 (u 0 )G r , E r = κ(u 0 )G r ⇔
λ
2 (u 0 ) = λ
2
1 = λ
2
2
κ(u 0 ) = κ
2
1 = κ
2
2
.
(1.162)
(ii) ⇐ (iii).
Λ
2
1 = Λ
2
2 = Λ
2 (U 0 ) , F
T
l
−1 diag
Λ
2
1 , Λ
2
2
F
−1
l
= C l , F
T
l
−1 F
−1
l
= G l ⇒ C l = Λ
2 (U 0 )G l ,
λ
2
1 = λ
2
2 = λ
2 (u 0 ) , F
T
r
−1 diag
λ
2
1 , λ
2
2
F
−1
r = C r , F
T
r
−1 F
−1
r = G r ⇒ C r = λ
2 (u 0 )G r .
(1.163)
The statements for the quantities E l , E r , E l G
−1
l , E r G
−1
r , K, κ, Λ, and λ follow in the same way.
End of Proof (second part).
1 From Riemann manifolds to Riemann manifolds
Before we present the sketches of proofs for the various conditions, it has to be noted that the
generalized Korn–Lichtenstein equations, which govern conformal mapping M
2
l → M
2
r , suffer from
the defect that they contain the unknown functions g 11 [u
λ (U
Λ )] and g 22 [u
λ (U
Λ )], and the reason is
that the mapping functions u
λ (U
Λ ) have to be determined. In case of {M
2
r , g µν } = {R
2 , δ µν }, the
corresponding Korn–Lichtenstein equations do not suffer since these functions do not appear. The
stated problem is overcome by representing the right Riemann manifold M
2
r by isometric coordinates
(also called conformal coordinates or isothermal coordinates) directly such that the quotient g 22 /g 11
is identical to one. This is exactly the procedure advocated by C. F. Gauss (1822, 1844) and applied
to the conformal mapping of E
2 onto S
2
r . We shall come back to this point-of-view after the proof.
Proof (first part).
(i) ⇒ (ii).
Ψ l = Ψ r → cos Ψ l = cos Ψ r ⇔ U
1
T G l U
2 = u
1
T G r u
2 ⇔
⇔ du
T
1 J
T
r G l J r du 2 =
dS 1
ds 1
du
T
1 G r du 2
dS 2
ds 2
⇔ du
T
1 C r du 2 = λ 1 du
T
1 G r du 2 λ 2 ⇔
⇔ λ 1 = λ 2 = λ(u 0 ) , C r = λ
2 (u 0 )G r q. e. d.
(1.158)
cos Ψ r = cos Ψ l ⇔ u
1
T G r u
2 = U
1
T G l U
2 ⇔ dU
T
1 J
T
l G r J l dU 2 =
ds 1
dS 1
dU
T
1 G l dU 2
ds 2
dS 2
⇔
⇔ Λ 1 = Λ 2 = Λ(U 0 ) , C l = Λ
2 (U 0 )G l q. e. d.
(1.159)
(i) ⇐ (ii).
⎡
⎢
⎣
cos Ψ l = U
1
T G l U
2 =
ds 1
dS 1
u
T
1 J
T
r G l J r du 2
ds 2
dS 2
J
T
r G l J r = C r = λ
2 (u 0 )G r , λ
−1
1 = λ
−1
2 = λ
−1
⎤
⎥
⎦ ⇒
cos Ψ l = u
1
T G r u
2 = cos Ψ r
orientation is preserved
⇔
⇔ Ψ l = Ψ r q. e. d.
(1.160)
End of Proof (first part).
Proof (second part).
(ii) ⇒ (iii).
Left eigenvalue problem:
C l = Λ
2 (U 0 )G l , E l = K(U 0 )G l ⇔
⎡
⎣
Λ
2 (U 0 ) = Λ
2
1 = Λ
2
2
K(U 0 ) = K
2
1 = K
2
2
⎤
⎦ .
(1.161)
Right eigenvalue problem:
C r = λ
2 (u 0 )G r , E r = κ(u 0 )G r ⇔
λ
2 (u 0 ) = λ
2
1 = λ
2
2
κ(u 0 ) = κ
2
1 = κ
2
2
.
(1.162)
(ii) ⇐ (iii).
Λ
2
1 = Λ
2
2 = Λ
2 (U 0 ) , F
T
l
−1 diag
Λ
2
1 , Λ
2
2
F
−1
l
= C l , F
T
l
−1 F
−1
l
= G l ⇒ C l = Λ
2 (U 0 )G l ,
λ
2
1 = λ
2
2 = λ
2 (u 0 ) , F
T
r
−1 diag
λ
2
1 , λ
2
2
F
−1
r = C r , F
T
r
−1 F
−1
r = G r ⇒ C r = λ
2 (u 0 )G r .
(1.163)
The statements for the quantities E l , E r , E l G
−1
l , E r G
−1
r , K, κ, Λ, and λ follow in the same way.
End of Proof (second part).