Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbb{T}$ denote the torus of revolution with the usual parametrization: $x(u,v) = ( (R + r\cos(u))\cos(v), (R + r\cos(u))\sin(v), r\sin(u) )$

Show that $\mathbb{T}$ has no umbilic points.

share|cite|improve this question

Simple calculation shows that $$x_u = ( -r\sin(u)\cos(v), -r\sin(u)\sin(v), r\cos(u)),$$ $$x_v = ( -(R + r\cos(u))\sin(v), (R + r\cos(u))\cos(v), 0),$$ $$x_{uu} = ( -r\cos(u)\cos(v), -r\cos(u)\sin(v), -r\sin(u)),$$ $$x_{uv} = ( r\sin(u))\sin(v), -r\sin(u))\cos(v), 0),$$ $$x_{vv} = ( -(R + r\cos(u))\cos(v), -(R + r\cos(u))\sin(v), 0).$$ These imply that the coefficients of the first fundamental form are $$E=x_u\cdot x_u=r^2, F=x_u\cdot x_v=0, G=x_v\cdot x_v=(R + r\cos(u))^2,$$ and the unit normal is given by $$n=\frac{u\times v}{\|u\times v\|}=( -\cos(u)\cos(v), -\cos(u)\sin(v), -\sin(u)).$$ Hence, the coefficients of the second fundamental form are $$e=x_{uu}\cdot n=r, f=x_{uv}\cdot n=0, g=x_{vv}\cdot n=(R + r\cos(u))\cos(u).$$ Therefore, the Gauss curvature $K$ and mean curvature $H$ are given by $$K=\frac{eg-f^2}{EG-F^2}=\frac{\cos(u)}{r(R + r\cos(u))}, H=\frac{eG+Eg-2fF}{EG-F^2}=\frac{R + 2r\cos(u)}{r(R + r\cos(u))}.$$

Now note that the principal curvatures $\kappa_1, \kappa_2$ are the roots of the quadratic equation $$\kappa^2-2H\kappa+K=0,$$ and by definition a point $x$ is umbilic if $\kappa_1=\kappa_2$. Therefore, $x$ is umbilic if and only if the above quadratic equation has double roots, which is equivalent to the discriminant $(-2H)^2-4K=0$, i.e. $H^2=K$. However, by using the above expressions for $K$ and $H$, we see that $$H^2-K=\frac{(R + 2r\cos(u))^2}{r^2(R + r\cos(u))^2}-\frac{\cos(u)}{r(R + r\cos(u))}=\frac{\frac{R^2}{4}+3(\frac{R}{2}+r\cos(u))^2}{r^2(R + r\cos(u))^2}>0.$$ This proves there does not exist $x$ such that $\kappa_1=\kappa_2$, i.e. $\mathbb{T}$ has no umbilic points.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.