From Riemann Manifolds to Riemann Manifolds - Map Projections: Cartographic Information Systems - page 52

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 Gauss ( 1822 , 1844 )
and applied to the conformal mapping of

2 onto 2

E S

r . We shall come back to this point-of-view

after the proof.

Proof (first part).

(i) ⇒ (ii) .

U 1 T G l U 2 = u 1 T G r u 2 ⇔

Ψ l = Ψ r → cos Ψ l =cos Ψ r ⇔

d u 1 J r G l J r d u 2 = d S 1

d s 1 d u 1 G r d u 2 d S 2

d u 1 C r d u 2 = λ 1 d u 1 G r d u 2 λ 2

⇔

d s 2 ⇔

⇔ (1.158)

λ 1 = λ 2 = λ ( u 0 ) , C r = λ 2 ( u 0 )G r q. e. d. ⇔

d U 1 J l G r J l d U 2 = d s 1

d S 1 d U 1 G l d U 2 d s 2

u 1 T G r u 2 = U 1 T G l U 2 ⇔ cos Ψ r =cos Ψ l ⇔

d S 2 ⇔

(1.159)

⇔ Λ 1 = Λ 2 = Λ ( U 0 ) , C l = Λ 2 ( U 0 )G l q. e. d.
(i)

⇐ (ii) .

cos Ψ l = U 1 T G l U 2 = ds 1

J r G l J r =C r = λ 2 ( u 0 )G r , λ − 1 = λ − 2 = λ − 1

cos Ψ l = u 1 G r u 2 =cos Ψ r
orientation is preserved

u 1 J r G l J r d u 2 d s 2

⇒ ⇔ (1.160)

d S 1 d S 2

⇔ Ψ l = Ψ r q. e. d.

End of Proof.

Proof (second part).

(iii) .
Left eigenvalue problem:

(ii) ⇒

Λ 2 ( U 0 )= Λ 1 = Λ 2
K ( U 0 )= K 1 = K 2

.

C l = Λ 2 ( U 0 )G l , E l = K ( U 0 )G l ⇔

(1.161)

Right eigenvalue problem:

λ 2 ( u 0 )= λ 1 = λ 2
κ ( u 0 )= κ 1 = h 2

.

C r = λ 2 ( u 0 )G r , E r = κ ( u 0 )G r ⇔ (1.162)

(iii) .
Λ 1 = Λ 2 = Λ 2 ( U 0 ) , F T − l diag[ Λ 1 ,Λ 2 ]F − l =C l , F T − l F − l =G l ⇒

(ii) ⇐

C l = Λ 2 ( U 0 )G l ,

(1.163)

λ 1 = λ 2 = λ 2 ( u 0 ) , F T − r diag[ λ 1 ,λ 2 ]F − r =C r , F T − r F − r =G r C r = λ 2 ( u 0 )G r .

The statements for the quantities E l , E r , E l G − l , E r G − r ,K,κ,Λ, and λ
follow in the same way.

⇒

End of Proof (second part).