### DIFFERENTIAL GEOMETRY COURSE

16. NORMAL CURVATURE

Definition 139 : Let x(s) = x(u1(s),u2(s)) be a curve on the surface x(u1,u2).
Then, with respect to the surface and the point we consider, the normal curvature of this curve is k = (x ..) . N (where x .. is the curvature vector in the direction of the principal normal n, and N the normal with length 1 on the surface in the direction of xu1xu2).
So k = κ n . N, where κ is the curvature of the curve (in the usual sense of the word).

Proposition 140 (MEUSNIER) :
Curves through a given point on the surface that have the same tangent in that point, also have the same normal curvature in that point, namely
k = (h1 1(du1)2 + 2 h1 2 du1 du2 + h2 2(du2)2) / (a1 1(du1)2 + 2 a1 2 du1 du2 + a2 2(du2)2).

Proof : (x ..).N = (xu1 u1 .. + xu2 u2 .. + xu1u1 (u1 .)2 + 2 xu1u2 (u1 . u2 .) + xu2u2 (u2 .)2) . N = h1 1(du1)2 + 2 h1 2 du1 du2 + h2 2(du2)2.

Explanation 140A : The curvature of the intersection curve of the surface and a plane through N is the normal curvature in that direction, because k = (x ..) . N = κ N . N = κ, since the principal normal is then also surface normal.

Explanation 141 : From 140 we also see
+1 = h1 1((du1/ds)(1/√|k|))2 + 2 h1 2 du1/ds(1/√|k|) du2(1/√|k|) + h2 2((du2/ds)(1/√|k|))2.
So we also find the indicatrix by taking in each direction the point at distance 1/√|k| (indeed, xu1du1/ds + xu2du2/ds is a vector with length 1).
So k=0 gives the asymptotic directions.
We call a curve on the surface that has in each point an asymptotic direction an asymptotic line of the surface. A simple example of an asymptotic line is a straight line on the surface.
So the net of the asymptotic lines is given by h1 1(du1)2 + 2 h1 2 du1 du2 + h2 2(du2)2 = 0.

Proposition 142 : If a curve on the surface has an asymptotic direction in some point, then in that point the osculating plane and the tangent plane coincide (or the curve is a straight line).

Proof: The osculating plane has vector representation xo + λ x . + μ x .. with N perpendicular to x . and (since k=0) also to x .. .

Proposition 143 : A surface is a plane if and only if the second fundamental form is identically 0 everywhere (k=0 in each point in each direction).

Proof : Check by calculation that the second fundamental form of a plane (u,v,a+bu+cv) is identically zero in each point.
Inversely, suppose the normal curvature is everywhere 0 in each direction.
From xui . N = 0 we find xuiuj . N = -xui . Nuj, so using hi j = 0 we find that xui is perpendicular to Nuj for i,j=1,2.
Since N . N = 1 we have N . Nuj = 0, so Nuj is perpendicular to xu1, xu2 and N, so Nuj = 0 (j=1,2), so N is constant.

Definition 144 : A planar point is a point where the second fundamental form is zero (h1 1 = h1 2 = h2 2 = 0, so k=0 in each direction).

Problem 145 : For which surface of revolution holds that the indicatrix is an orthogonal hyperbola in each regular point? (See 138.)
Answer: the catenoid, surface of revolution of the catenary.

Extra Problem :
Consider the surface x(u,v) = (u2, uv, v2).
a) Describe the parameter lines: what kind of figures are they?
b) Give the first fundamental form and an equation for the set of points on the surface where the parameter lines are perpendicular to each other. What kind of figure forms this set?
c) Give the second fundamental form and the indicatrix of Dupin. Prove that each non-singular point is parabolic. So what kind of figure is the indicatrix? Give the asymptotic lines by means of a parameter representation.
d) Give the differential equation for the orthogonal trajectories of the u-lines. Notice this equation is homogeneous. So substite u = vw, and separate the variables. Find the orthogonal trajectories by means of a relation between u and v.
e) Give an equation for the surface and prove it is a half of a cone by giving the top and a direction curve. Check by calculation that the tangent plane is constant along a straight line on the cone.
f) Show that λ(t2,t,1) is another parametrisation of the same surface. Find the orthogonal trajectories of the rules (straight lines on the surface). Explain.