Consider a regular unit speed curve $\alpha: (a,b) \to \Bbb R^3$. Then define the surface $S$ via the parametrization $x:(a,b)\times (-\pi,\pi)\to \Bbb R^3$ where $$x(u,v) = \alpha(t) + r(N(t)\cos v+B(t)\sin v)$$ where $N$ and $B$ are the unit normal and binormal to the curve alpha respectively. This defines a tube around the curve $\alpha$. We demand that $k(t) \leq \frac1r$ as is required for the surface to be regular. Now I want to find the principal curvatures. To this end I want to find the eigenvalues of the shape operator $-dN$ where $N : S \to \Bbb S^2$ is the Gauss map. Now I tried to compute the coefficients of the first and second fundamental form but this just turns into a nightmare. I found the following expressions that I think should be useful $$x_s = T(t) (1+rk\cos v) +r\tau (B(t)\cos v-N(t) \sin v) , $$ $$x_v = r(B(t)\cos v - N(t)\sin v)$$ where $T(t) = \alpha'(t)$ is the unit tangent vector to the curve $\alpha$.

I am pretty sure the principal curvatures are just the curvature of the circle at each points, which is $\frac{1}{r}$ and the curvature of $\alpha$ which is $k(t)$. My question is, how do I determine the linear map $dN$? I am pretty sure I am meant to be using the Frenet Serret frame to locally describe the tangent plane $T_pS$ to the surface. Then any vector in this space is a linear combination of $x_s$ and $x_v$. Then... yeah I am not sure. Any help would be much appreciated.

  • $\begingroup$ I'd suggest you start out with simple examples, like when your curve is a plane circle in three space. You will then see by inspection that your proposition ("the prinipal curvatures are just the curvature of the circle and the curvature of $\alpha$") is likely to not be correct. While the curvature of the circle defining the tube is a good candidate for one, the second principal curvature is more likely to be the curvature of the equidistant curves to the original plane curve in the plane in which it is sitting. $\endgroup$
    – Thomas
    Dec 9, 2014 at 21:13
  • $\begingroup$ Ah yes, I now see why that is incorrect, especially as points on the surface do not necessarily lie on the curve $\alpha$. Then could you help me with some way to find the shape operator I need to find the principal curvatures at each point? Thanks for your comment! $\endgroup$
    – Slugger
    Dec 9, 2014 at 21:18
  • 1
    $\begingroup$ ... what is $x_s$? $\endgroup$
    – janmarqz
    Dec 12, 2014 at 20:53

1 Answer 1


Note that the normal vector is nothing but $n(u,v)=N(u)\cos{v}+B(u)\sin{v}$. To see that write $x(u,v)=\alpha(u)+r\cdot n(u,v)$ then the tangent vectors are: $$x_u=T+r\cdot n_u, x_v=r\cdot n_v$$ Notice that $n\cdot n=1$ and $n\cdot T=0$ thus $n\cdot n_u=n\cdot n_v=0$ and $n\cdot x_u=n\cdot x_v=0$, which tells that $n$ is the unit normal vector, giving the Gauss map.

It is easy to compute the following using the Frenet formula: $$n_u=\cos{v}N_u+\sin{v}B_u=-k\cos{v}T+\tau\cdot n_v$$ $$x_u=T+r\cdot n_u=(1-rk\cos{v})T+r\tau\cdot n_v$$

Also it is easy to verify that {$T, n_v$} is an orthonormal basis of the tangent plane.

Hence $$dn(T)=n_udu(T)+n_vdv(T)$$ $$=\frac{1}{1-rk\cos{v}}n_u-\frac{\tau}{1-rk\cos{v}}n_v$$ $$dn(n_v)=n_udu(n_v)+n_vdv(n_v)=\frac{n_v}{r}$$

So $$dn(\begin{bmatrix}T \\n_v \end{bmatrix})=\begin{bmatrix}\frac{1}{1-rk\cos{v}} & -\frac{\tau}{1-rk\cos{v}} \\0 & \frac{1}{r} \end{bmatrix} \begin{bmatrix}n_u \\n_v \end{bmatrix}$$ $$=\begin{bmatrix}\frac{1}{1-rk\cos{v}} & -\frac{\tau}{1-rk\cos{v}} \\0 & \frac{1}{r} \end{bmatrix} \begin{bmatrix}-k\cos{v} & \tau \\0 & 1 \end{bmatrix} \begin{bmatrix}T \\n_v \end{bmatrix}$$ $$=\begin{bmatrix}-\frac{k\cos{v}}{1-rk\cos{v}} & 0 \\0 & \frac{1}{r} \end{bmatrix}\begin{bmatrix}T \\n_v \end{bmatrix}$$ Thus the Jacobian matrix of the Gauss map, with respect to the orthonormal basis, is given by $$\begin{bmatrix}-\frac{k\cos{v}}{1-rk\cos{v}} & 0 \\0 & \frac{1}{r} \end{bmatrix}$$ and the two principle curvatures are the diagonal elements.


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