### DIFFERENTIAL GEOMETRY COURSE

17. CURVATURE LINES

Definition 146 : In a point on the surface there are generally two directions where normal curvature is extreme, namely those of the principal axes of the indicatrix of Dupin.
We call them principal directions with corresponding principal curvatures. A curvature line is a curve on the surface whose direction is principal in each of its points.

Explanation 147: In planar points and navel points each direction is principal, so on spheres and planes is each curve a curvature line.
In general, however, the curvature lines form an orthogonal net.

Proposition 148: The differential equation for the net of the curvature lines is (with 'Einstein notation convention')

(ah - ah) duα duβ = 0.

Proof: In every direction, determined by the quotient du2/du1, take the point (du1,du2) on the ellipse a1 1 du12 + 2 a1 2 du1 du2 + a2 2 du22 = 1.
Then we can find the principal directions with the help of the theorem of Lagrange:
we find the extremes of k = h1 1 du12 + 2 h1 2 du1 du2 + h2 2 du22 under the condition a1 1 du12 + 2 a1 2 du1 du2 + a2 2 du22 = 1 in directions (du1,du2) where
(2 h1 1 du1 + 2 h1 2 du2, 2 h1 2 du1 + 2 h2 2 du2) is a scalar multiple of (2 a1 1 du1 + 2 a1 2 du2, 2 a1 2 du1 + 2 a2 2 du2).
So we find the principal directions with (h1 1 du1 + h1 2 du2)(a1 2 du1 + a2 2 du2) = (h1 2 du1 + h2 2 du2)(a1 1 du1 + a1 2 du2).
This is the very equation of the proposition.

Definition 149 : We call the product of the principal curvatures k1 and k2 total curvature, so ktot = k1.k2.
We call the sum of the principal curvatures mean curvature.

Proposition 150 : We find the principal curvatures with the equation det(hi j - k ai j) = 0.

Proof: For the principal directions (du1,du2) (that are not (0,0)) there must exist λ such that
(h1 1 - λa1 1) du1 + (h1 2 - λa1 2) du2 = 0,
(h1 2 - λa1 2) du1 + (h2 2 - λa2 2) du2 = 0.
But then λ is principal curvature: for if we multiply the first equation with du1 and the second with du2, and take the sum of both equations, we find that k = λ satisfies the equation of Meusnier (see proposition 140).

Proposition 151 : The total curvature satisfies ktot = det(hi j)/det(ai j). In connection with definition 134, the sign of ktot tells us what kind of point we have.

Proof: This is a simple consequence of the fact that the quadratic equation from proposition 150 is k2(det(ai j) + k(...) + det(hi j) = 0.

Explanation 151A : According to a proposition of Gausz (which we will not prove here and which he himself called 'theorema egregium'), the total curvature is invariant under distortion.
So for the surfaces that are isometric to a plane, total curvature is 0 everywhere (so each point is parabolic).

Proposition 152 : The parameter lines are curvature lines if and only if h1 2 and a1 2 are 0 everywhere.

Proof: If the parameter lines are curvature lines, they form an orthogonal net, so a1 2 = 0. Then the differential equation of proposition 148 becomes: a1 1 h2 1 du1 du1 + (...) du1 du2 - a2 2 h1 2 du2 du2 = 0.
This is the differential equation du1 du2 = 0 for the net of the parameter lines if and only if a1 1 h2 1 = a2 2 h1 2 = 0, so (considering regular points) if h1 2 = 0.

Problem 153 : Determine the helicoid surfaces (see 125) whose meridian curves are curvature lines.
(Solution: besides all surfaces of revolution, the only helicoid surfaces that meet the requirement are those with f(u) = + cu √(c2-y2)/y dy ; then the generating curve (u,0,f(u)) is called tractrix.)

Problem 154 : Determine the curvature lines of the torus.

Problem 155 : Show that the meridian curves and circles of latitude are the curvature lines of surfaces of revolution.

Note 156 : If we have three pencils of surfaces such that through each point of ℜ3 there goes one surface out of each pencil, whilst each pair out of these three surfaces is perpendicular, we call the system of three pencils tripelorthogonal.
There's a theorem of Dupin that says the intersection line of each pair of these three surfaces is always a curvature line.