Against all odds I’m reading Pressley’s Elementary Differential Geometry while waiting for lessons to begin in October and I have to say I’ve been enjoying it quite a bit. When I had to study from this very book for an exam two years ago I found the subject so boring and dull… Strange how these things work!

Anyway, I’m trying to do some of the exercises and this is really cute:

Let be a unit-speed curve with curvature and torsion for all .

Show that is spherical, (i.e. it lies on the surface of a sphere) then:

.

Conversely if the previous equation holds then for some , where and .

Trying to solve this made me want to solve a slightly more general problem, what if the curve were lying on a quadric surface instead of a sphere?

I’m going to use the same notation the book uses, so for a curve parametrized at unit-speed we define the tangent vector , the unit normal and the binormal , these three vector-functions form the Frenet-Serret frame satisfying the well known formulae: and .

So, we begin by fixing a real symmetric matrix , vector and a real number defining the quadric . Suppose we have a curve parametrized at unit-speed lying on this quadric, i.e. satisfying at each value of the parameter the equation

We are going to take the derivative w.r.t. the arc-length parameter (denoted with a dot) on both sides for a number of times and see where it takes us.

. Using the fact that the matrix is symmetric and that the curve is parametrized at unit-speed we obtain:

(which means that ). Then or Differentiating again on both sides: .

On the left hand side, remembering that the tangent vector is orthogonal to , we are left with On the right hand side, after differentiating the last term we obtain . So the equation can be rearranged to: One last time: .

Recalling equation (2) we replace with to finally obtain:



As a sanity check it’s easy to see that for , the identity matrix, the quadric is a sphere and the equation just found reduces to the equation in Pressley’s exercise as and

Here’s an explicit example I managed to find after hours of derivatives and many many stupid mistakes.

First of all we must adjust the formula for non-unit speed curves as most parametrizations found in the wild are never unit speed. We will let denote the parameter of a curve on a quadric and be the natural parameter. The formula we found earlier:

Becomes, after applying the chain rule:

Where now the dots indicate a derivative with respect to !

We are going to use a hyperboloid of one sheet as our surface, with equation , parametrized via the patch . Our curve will be defined by .

First of all we are going to find all the derivatives:

with norm

with norm



And the cross product with norm .

From these, using the well known formulae it’s easy to find the curvature and torsion of

the curve: and .

Next we need to find the vectors of the Frenet-Serret frame:







Then, using the matrix that defines the quadric we can find the values and namely:







Now for the final test, with a lot of patience and some care one can verify the equation by hand or by inputting it piece by piece into Wolfram Alpha (or your computer algebra of choice).

The left hand side of the equation equals , the right hand side is more complicated and cannot be input whole into wolfram alpha!

Inside the parentheses we find and its derivative scaled by which equals .

Finally we subtract it by to find which, after multiplying, (thankfully!) equals what we found in the left hand side.

Even with computer assistance this took a lot of time to get right even though this is probably one of the simplest curves out there, explicit examples are hard and I’m so glad to be done with this!