I have been trying to find the gaussian curvature of a pseudo sphere. I assumed the parametrization: X(u,v) = (cos(u)*sech(v), sin(u)*sech(v), u – tanh(u)). I know that it's a surface of revolution obtained by the tractrix. I could find it calculating the coefficients of the first and second form. But I observed that its curvature should be the product of the principal curvature, that is, the curvature of the circle of radius sech(v) and of the tractrix, that as I discovered, is csch(v). But I should find that this product is -1, that is the curvature of the pseudosphere. What is wrong with my reasoning?
[Math] Gaussian Curvature of a Pseudosphere.
curvaturedifferential-geometry
Best Answer
Your calculation of the curvature of the tractrix is fine. The problem comes from the other part: the plane in which the circle $(\cos{u}\sech{v},\sin{u}\sech{v},v-\tanh{v})$ (with constant $v$) lies is not one of the planes normal to the surface at the point where you want to measure the curvature (this is easy to see in the case of a surface with radius decreasing rapidly: the plane must depend on the gradient of the radial component, therefore.
Instead, we need to look at the curvature of a curve $\gamma(t)$ in the plane parallel to the surface normal, which is easily found to be $$ n=(\cos{u}\tanh{v},\sin{u}\tanh{v},\sech{v}). $$
If the curve is in both the surface and a plane parallel to this vector, it is easy to see that the normal vector to the curve must be parallel to the surface normal. (E.g., the curve's tangent vector is perpendicular to the surface normal by definition of the surface normal, the binormal is constant (and perpendicular to the plane and thus the surface normal) as the curve is planar, so the curve's normal vector, being perpendicular to both of these, lies in the plane and is perpendicular to the perpendicular of the surface normal, and hence parallel to it.)
Therefore, all we have to do is find $ (\gamma(t)-\gamma(0)) \cdot n$, where $\dot{\gamma}(0)$ is parallel to the circle $v= \text{const}$., and expand to get to the first nonzero term near t=0, which is easily seen to be: $$ \begin{align} (\gamma(t)-\gamma(0)) \cdot n &= \gamma'(0) \cdot n \, t + \gamma''(0) \cdot n \frac{t^2}{2} +O(t^3) \\ &= s'(0)T(0) \cdot n t + (s''(0)T(0)+ s'(0)^2\kappa(0) N(0) ) \cdot n \frac{t^2}{2} + O(t^3) \\ &= s'(0)^2 \kappa \frac{t^2}{2} + O(t^3), \end{align} $$ by the definition of curvature, where $s$ is the arclength parameter. The symmetry shows that we only need to do the calculation for $u=0$, so $$ n=(\tanh{v},0,\sech{v}), $$ and some boring calculation later, we find that $$ s'(0)^2 \kappa(0) = \sech{v}\tanh{v} \, (-U'(0)^2+V'(0)^2). $$ Of course, $s'(0)^2$ is the square of the length of $\gamma'(0)$, which is also easy to calculate, as $$ \lVert \gamma'(0) \rVert^2 = \sech^2{v} \, (U'(0)^2+\sinh^2{v} \, V'(0)^2) $$ (you can easily fill this in yourself, with enough differentiation and application of trigonometrical and hyperbolic identities). Therefore, $$ \kappa(0) = \sinh{v} \, \frac{-U'(0)^2+V'(0)^2}{U'(0)^2+\sinh^2{v} \, V'(0)^2}. $$ To find the principal curvatures, one has to maximise and minimise this homogeneous function, but in this case, it's easy, with the minimum obviously when $V'(0)=0$ (i.e., parallel to the circle), the maximum when $U'(0)=0$ (parallel to the tractrix), with values $-\sinh{u}$ and $1/\sinh{u}$ respectively. The latter you have already, and the product is $-1$ as it should be.