56
1 From Riemann manifolds to Riemann manifolds
Box 1.24 (UMP of E
2
A 1 ,A 1 ,A 2 versus UMP of S
2
r ).
Left conformal coordinates:
Right conformal coordinates:
P = A 1 Λ , Q = A 1 ln tan
„
π
4
+
Φ
2
« »
1 − E sin Φ
1 + E sin Φ
– E/2
!
.
q = r ln tan
„
π
4
+
φ
2
«
, p = rλ . (1.193)
Left matrix of the metric G l :
Right matrix of the metric G r :
G l =
2
6
6
6
4
A
2
1 cos
2 Φ
1 − E 2 sin
2 Φ
0
0
A
2
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 3
3
7
7
7
5
.
"
r
2 cos
2 φ
0
0
r
2
#
= G r .
(1.194)
Left Jacobi matrix:
Right Jacobi matrix:
P Λ = A 1 , Q Λ = P Φ = 0 , Q Φ =
A 1 (1 − E
2 )
1 − E 2 sin
2 Φ
1
cos Φ
.
p λ = r , q λ = p φ = 0 , q φ =
r
cos φ
. (1.195)
Solution (the first problem).
Start from the conformal map that is defined in Box 1.24. The three conditions to be fulfilled are
given in Box 1.25 for “left UMP” and in Box 1.26 for “right UMP”. First, we write down the left and
right specified Korn–Lichtenstein equations, namely
G 11 /G 22 ,
G 22 /G 11 and
g 11 /g 22 ,
g 22 /g 11 ,
respectively. Indeed, by transforming {Q Λ , Q Φ , P Λ , P Φ } as well as {q λ , q φ , p λ , p φ } left and right, KL
1st and KL 2nd are satified. Second, we specialize the left and right integrability conditions of the
vector-valued Korn–Lichtenstein equations, namely the left and right Laplace–Beltrami equations,
by {G 11 , G 22 , P Λ , P Φ , Q Λ , Q Φ } of E
2
A 1 ,A 1 ,A 2
as well as {g 11 , g 22 , p λ , p φ , q λ , q φ } of S
2
r . Indeed, we have
succeeded that {P, Q} are left harmonic and {p, q} are right harmonic. Third, we prove left and right
orientation by computing the left and right Jacobians which turn out positive.
End of Solution (the first problem).
Solution (the second problem).
In any textbook of Differential Geometry, you will find the representation of the Gaussian curvature of
a surface in terms of conformal coordinates (isometric, isothermal). Let the left and the right matrix of
the metric be equipped with a conformally flat structure {G l = λ
2
l I 2 , G r = λ
2
r I 2 }, which is generated by
a left and a right conformal coordinate representation. Then the left and the right Gaussian curvature
are given by k l = −(1/2λ
2
l )∆ l ln λ
2
l = −(1/λ
2
l )∆ l ln λ l and k r = −(1/2λ
2
r )∆ r ln λ
2
r = −(1/λ
2
r )∆ r ln λ r
as well as ∆ l := D P P + D QQ = D
2
P + D
2
Q and D
2
p + D
2
q = D pp + D qq := ∆ r , where ∆ l and ∆ r represent
the left and the right Laplace–Beltrami operator. Let us apply this result in solving the second problem.
By means of Boxes 1.27, 1.28, and 1.29, we have outlined how to generate a conformally flat metric of
an ellipsoid-of-revolution and of a sphere. It is the classical Gauss factorization.
End of Solution (the second problem).
Solution (the third problem).
By means of the left and the right mapping equations, i. e. by means of {P = A 1 Λ, Q(Φ = 0) = 0}
and {p = rλ, q(φ = 0) = 0}, respectively, it is obvious that an equatorial arc of E
2
A 1 ,A 1 ,A 2
and
of S
2
r is mapped equidistantly. Similarly, the factor of conformality derived in Box 1.27 amounts to
λ
2
l (Φ = 0) = 1 for the left manifold and to λ
2
r (φ = 0) = 1 for the right manifold. Such a configuration
on the ellipsoidal as well as the spherical equator we call an isometry. Indeed, the postulate of an
equidistant mapping of the left or right equator constitutes a boundary condition for the left and
right Korn–Lichtenstein equations, namely to make their solution unique.
End of Solution (the third problem).
1 From Riemann manifolds to Riemann manifolds
Box 1.24 (UMP of E
2
A 1 ,A 1 ,A 2 versus UMP of S
2
r ).
Left conformal coordinates:
Right conformal coordinates:
P = A 1 Λ , Q = A 1 ln tan
„
π
4
+
Φ
2
« »
1 − E sin Φ
1 + E sin Φ
– E/2
!
.
q = r ln tan
„
π
4
+
φ
2
«
, p = rλ . (1.193)
Left matrix of the metric G l :
Right matrix of the metric G r :
G l =
2
6
6
6
4
A
2
1 cos
2 Φ
1 − E 2 sin
2 Φ
0
0
A
2
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 3
3
7
7
7
5
.
"
r
2 cos
2 φ
0
0
r
2
#
= G r .
(1.194)
Left Jacobi matrix:
Right Jacobi matrix:
P Λ = A 1 , Q Λ = P Φ = 0 , Q Φ =
A 1 (1 − E
2 )
1 − E 2 sin
2 Φ
1
cos Φ
.
p λ = r , q λ = p φ = 0 , q φ =
r
cos φ
. (1.195)
Solution (the first problem).
Start from the conformal map that is defined in Box 1.24. The three conditions to be fulfilled are
given in Box 1.25 for “left UMP” and in Box 1.26 for “right UMP”. First, we write down the left and
right specified Korn–Lichtenstein equations, namely
G 11 /G 22 ,
G 22 /G 11 and
g 11 /g 22 ,
g 22 /g 11 ,
respectively. Indeed, by transforming {Q Λ , Q Φ , P Λ , P Φ } as well as {q λ , q φ , p λ , p φ } left and right, KL
1st and KL 2nd are satified. Second, we specialize the left and right integrability conditions of the
vector-valued Korn–Lichtenstein equations, namely the left and right Laplace–Beltrami equations,
by {G 11 , G 22 , P Λ , P Φ , Q Λ , Q Φ } of E
2
A 1 ,A 1 ,A 2
as well as {g 11 , g 22 , p λ , p φ , q λ , q φ } of S
2
r . Indeed, we have
succeeded that {P, Q} are left harmonic and {p, q} are right harmonic. Third, we prove left and right
orientation by computing the left and right Jacobians which turn out positive.
End of Solution (the first problem).
Solution (the second problem).
In any textbook of Differential Geometry, you will find the representation of the Gaussian curvature of
a surface in terms of conformal coordinates (isometric, isothermal). Let the left and the right matrix of
the metric be equipped with a conformally flat structure {G l = λ
2
l I 2 , G r = λ
2
r I 2 }, which is generated by
a left and a right conformal coordinate representation. Then the left and the right Gaussian curvature
are given by k l = −(1/2λ
2
l )∆ l ln λ
2
l = −(1/λ
2
l )∆ l ln λ l and k r = −(1/2λ
2
r )∆ r ln λ
2
r = −(1/λ
2
r )∆ r ln λ r
as well as ∆ l := D P P + D QQ = D
2
P + D
2
Q and D
2
p + D
2
q = D pp + D qq := ∆ r , where ∆ l and ∆ r represent
the left and the right Laplace–Beltrami operator. Let us apply this result in solving the second problem.
By means of Boxes 1.27, 1.28, and 1.29, we have outlined how to generate a conformally flat metric of
an ellipsoid-of-revolution and of a sphere. It is the classical Gauss factorization.
End of Solution (the second problem).
Solution (the third problem).
By means of the left and the right mapping equations, i. e. by means of {P = A 1 Λ, Q(Φ = 0) = 0}
and {p = rλ, q(φ = 0) = 0}, respectively, it is obvious that an equatorial arc of E
2
A 1 ,A 1 ,A 2
and
of S
2
r is mapped equidistantly. Similarly, the factor of conformality derived in Box 1.27 amounts to
λ
2
l (Φ = 0) = 1 for the left manifold and to λ
2
r (φ = 0) = 1 for the right manifold. Such a configuration
on the ellipsoidal as well as the spherical equator we call an isometry. Indeed, the postulate of an
equidistant mapping of the left or right equator constitutes a boundary condition for the left and
right Korn–Lichtenstein equations, namely to make their solution unique.
End of Solution (the third problem).