Blaga P. Lectures on the differential geometry of - tiera.ru

4.16. Principal directions and curvatures 159
Since n ′ 2 ⊥ n2 (since n2 has constant length), it follows from here and from the previous
equation that n ′ 2 ⊥ r′ or, in other words, that there is a k2 ∈ R such that
n ′ 2 = −k2r ′ ,
i.e. γ is a line of curvature on the surface S 2, too. �
Consequence. The meridians and the parallels on a revolution surface are lines of curvature.
Proof. Let γ be the rotating curve and S the resulted revolution surface. A meridian
is obtained by intersecting a plane Πm, passing through the rotation axis and S . If p ∈
Πm ∩ S , then the unit normal of the surface S , n(p), lies on the plane Πm, therefore the
normal of the surface and the normal of the plane Πm make up a constant angle, equal
to π
2 . As for the plane the second fundamental form vanishes identically, the same is
true for the shape operator, which means that all the plane curves are lines of curvature.
It follows, that, in particular, the meridian is, also, a line of curvature in the plane Πm,
which implies, using the Joachimstahl theorem, that it is a line of curvature also for the
surface S .
A parallel is an intersection curve between the surface S and a plane Πp, passing
through a point of the curve γ, perpendicular to the rotation axis. It is obvious, from
symmetry reasons, that along a parallel we should have ∠( n, Πp) = const and we apply
again the reasoning we used before. �
4.16.1 The determination of the lines of curvature
As we saw earlier, the lines of curvature are curves on a surfaces whose tangent vectors
are eigenvectors of the shape operator. Therefore, before showing how one can find the
lines of curvature, we will indicate a way to find the eigenvectors of the shape operator.
Lemma. Let r : U → R 3 be a local parameterization of an oriented surface S . A
tangent vector v = v1r ′
u + v2r ′
v has principal direction iff
�
�
�
�v
�
�
�
�
�
2 2 −v1v2 v2 �
�
�
1 �
E F G
�
�
� = 0. (4.16.4)
�
�
D D ′ D ′′
Proof. Since v has principal direction iff it is an eigenvector of the shape operator A, i.e.
A(v) = λ · v, it follows that v has principal direction iff A(v) × v = 0. But, from the