DIFFERENTIAL GEOMETRY COURSE

13. *ISOGONAL NETS*

*Explanation 114 :* We can represent a curve on the surface * x*(u

When a family of curves is such that through every point of the surface there goes exactly one of these curves, and it is given by the equations φ(u

Now we can ask, for instance, for another system of curves on the surface such that through every point of the surface there goes exactly one curve of both systems, whilst these two curves intersect under a fixed angle.

Then we call the pair of systems an isogonal net. In the special case that this fixed angle is 90 degrees, we call it an orthogonal net and the two systems each other's orthogonal trajectories.

*Explanation 115 :* Let the curve * x*(u

Then we have for all t : φ(u

So the tangent vector

*Explanation 116 :* We find the isogonal trajectories of a system φ(u_{1},u_{2}) = λ on the surface * x*(u

let

cos(α) = (-φ_{u2}du_{1} a_{1 1} -φ_{u2}du_{2} a_{1 2} +φ_{u1}du_{1} a_{1 2}
+φ_{u1}du_{2} a_{2 2})/(√(a_{1 1}φ_{u2}^{2} - 2a_{1 2}φ_{u1}φ_{u2}
+ a_{2 2}φ_{u1}^{2}).√(a_{1 1}du_{1}^{2} + 2 a_{1 2}du_{1} du_{2} + a_{2 2}du_{2}^{2})).

When we solve this differential equation, we get the isogonal trajectories with angle α in the form ψ(u_{1},u_{2}) = μ.

In the special case that α is ninety degrees, the differential equation becomes

du_{1}/du_{2} = (a_{2 2}φ_{u1} - a_{1 2}φ_{u2})/(a_{1 1}φ_{u2} - a_{1 2}φ_{u1}).

*Problem 117* : Determine the orthogonal trajectories of the v-lines u=λ of (u+v,u-v,uv).

*Problem 118* : Determine the curves on the sphere that make a fixed angle α with the parallels of latitude (we call them the loxodromes of the sphere).

*Explanation 119 :* A quadratic equation in du_{1} and du_{2}, say
c_{1 1}du_{1}^{2} + 2 c_{1 2}du_{1} du_{2} + c_{2 2}du_{2}^{2} = 0, determines in each point (u_{1},u_{2}) two
directions (if the roots of the equation are real).

Each of both solutions du_{1}/du_{2} = ψ_{ui}(u_{1},u_{2}) (i=1,2) determines a system of curves on the surface, so the quadratic form
determines a net . For example, du_{1} du_{2} = 0 determines the net of the parameter lines.

*Proposition 120 :* The net c_{1 1}du_{1}^{2} + 2 c_{1 2}du_{1} du_{2} + c_{2 2}du_{2}^{2} = 0 on a surface
with first fundamental form a_{1 1}du_{1}^{2} + 2 a_{1 2}du_{1} du_{2} + a_{2 2}du_{2}^{2} is orthogonal if and only if
a_{1 1}c_{2 2} - 2 a_{1 2}c_{1 2} + a_{2 2}c_{1 1} = 0.

*Proof* : In 'each' point the quadratic equation c_{1 1} + 2 c_{1 2}du_{2}/du_{1} + c_{2 2}(du_{2}/du_{1})^{2} = 0 has two
solutions du_{2}/du_{1} = ψ_{1} and du_{2}/du_{1} = ψ_{2}, where ψ_{1}+ψ_{2} = -2c_{1 2}/c_{2 2} and
ψ_{1}ψ_{2} = c_{1 1}/c_{2 2}.

The requirement of orthogonality becomes
(__x___{u1} + __x___{u2}ψ_{1}).(__x___{u1} + __x___{u2}ψ_{2}) = 0,
so a_{1 1} - a_{1 2}(ψ_{1}+ψ_{2}) + a_{2 2}ψ_{1}ψ_{2} = 0, so
a_{1 1}c_{2 2} - 2 a_{1 2}c_{1 2} + a_{2 2}c_{1 1} = 0.