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DIFFERENTIAL GEOMETRY COURSE

15. *THE SECOND FUNDAMENTAL FORM*

*Definition 132 :* Let __x__(u,v) be a surface, and __N__(u,v) the normal with length 1 (in the direction of __x___{u}⊗__x___{v}).

Let h_{i j} = __x___{uiuj} . __N__ (with u_{1} = u, u_{2} = v).

Then we call h_{1 1} (du)^{2} + 2 h_{1 2} du dv + h_{2 2} (dv)^{2} the second fundamental form.

*Explanation 133 :* Let α be the tangent plane in the point __x___{o} with parameter values u_{o} and v_{o}, and α ' the plane parallel to
α at a (little) distance γ from α, and let k be the intersection curve of α ' and the surface.

For the points on k we have __x__ = __x__(u,v) and (__x__ - __x___{o} + γ__N__) . __N__ = 0.

Using Taylor we find __x__ - __x___{o} = __x___{u} du + __x___{v} dv + (1/2) __x___{u u} (du)^{2} +
__x___{u v} du dv + (1/2) __x___{u v} (dv)^{2}.

Since __x___{u} and __x___{v} are perpendicular to __N__, the equation becomes in second order approximation:

((1/2) __x___{u u} (du)^{2} + __x___{u v} du dv + (1/2) __x___{u v} (dv)^{2}) . __N__ = - γ, so
h_{1 1} (du)^{2} + 2 h_{1 2} du dv + h_{2 2} (dv)^{2} = -2γ.

This is (in coordinates with respect to a not necessarily orthonormal base __x___{u} , __x___{v}) the equation of a conic, in which the right hand side
is the second fundamental form.

*Definition 134 :* We call the conic h_{1 1} (du)^{2} + 2 h_{1 2} du dv + h_{2 2} (dv)^{2} = 1 indicatrix of Dupin.

After the kind of the indicatrix we distinguish:

1. hyperbolic points: det(h_{i j}) negative. (Then we can resolve the form in linear factors, and Σ h_{i j} du dv = 0 gives the asymptotic directions,
see our course of projective geometry).

2. parabolic points: det(h_{i j}) = 0. (Indicatrix exists of two parallel lines; for example (2du + 3dv)^{2} = 1 gives du = __+__(1/2) - (3/2)dv, so
indicatrix __x___{u} (__+__(1/2) - (3/2)t) + __x___{v} t.

3. elliptic points: det(h_{i j}) positive. (Then we can't resolve the form and there are no asymptotes).

*Problem 135 :* Do you know a surface on which each point is navel point (umbilical point), that is: the indicatrix is in each point a circle?
Check this by calculation.

Do you know a surface on which each point is planar, that is: the second fundamental form is everywhere the 0-form? Check this by calculation.

*Problem 136 :* Determine the second fundamental form of the torus and check that det(h_{i j}) = R cos(φ) (r + R cos(φ)).

Deduce that the points at the inner side of this inner tube of a bicycle are hyperbolic, and the points at the outer side elliptic. The points on the circles φ = __+__π/2 are parabolic.

*Problem 137 :* Consider the right helicoid surface ((a+u)cos(v), (a+u)sin(v), bv), see 128.

Show that the parameter lines are asymptotice lines (that is: they have in each point an asymptotic direction), and form an orthogonal net, so each point is hyperbolic with an orthogonal hyperbola as
its indicatrix.

*Problem 138 :* Prove that a_{1 1}h_{2 2} - 2 a_{1 2}h_{1 2} + a_{2 2}h_{1 1} = 0 is the necessary and sufficient condition
so that in each hyperbolic point the indicatrix is an orthogonal hyperbola. (See proposition 120.)

answers

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