Consider the following surface of revolution:
$$S(t,\theta)=(t,f(t)\cos\theta,f(t)\sin\theta)$$
To calculate its principal curvatures, first we calculate the coordinate tangent vectors, then normalize their cross product to get the unit normal vector $N$. Finally, we calculate the shape operator using $S\partial_i=-D_{\partial_i}N$. This is how I find principal curvatures for two dimensional, parameterized surfaces.
But if I revolve the graph $(t,f(t))$ along $x$-axis in $R^{n+1}$, giving the surface,
$$S(t,w)=(t,f(t)w)=(t,f(t)\cos\theta_1,f(t)\sin\theta_1\cos\theta_2,\cdots)$$
where $w$ is a direction in $S^n$, how do I calculate its shape operator? In particular, I am interested in its sectional and mean curvatures.
My intuition says that the principal curvatures should be the same in all angular directions.
Best Answer
You don't want to do this by brute force computation. Your intuition is correct: The principal directions are the tangent vector to the profile curve (meridian) and any basis vectors for the tangent space to the sphere.
You can verify this by looking at the covariant derivative of the unit normal in these directions. As you move along a profile curve, you are considering a plane curve and the covariant derivative of the normal must be orthogonal to the normal, hence tangent to the curve. Next, by rotational symmetry, as you move along a latitude sphere, you can see that the normal vector to the hypersurface has two components, one in the direction of the normal to the sphere and one pointing along the first axis; by symmetry, the magnitudes in those directions are constant. Thus, when you compute the covariant derivative in the direction of $V$, tangent to the latitude sphere, the first component gives you a multiple of the $V$ and the second component drops out.
Using the usual Meusnier's Theorem analysis that works for a surface in $\Bbb R^3$ [see, for example, pp. 51-2 of my text] you can figure out the principal curvatures. Up to sign, the principal curvature in the direction of the profile curve is the curvature of the curve. More interestingly, by Meusnier's Theorem, the normal curvature in any direction $V$ tangent to the sphere will be the $1/f(t)$ (the curvature of a great circle with tangent vector $V$) multiplied by $\cos\phi$, where $\phi$ is the angle between the normal to the hypersurface and the principal normal of that great circle. The principal normal points toward the center of the sphere, and so with a drawing you can check that (again up to supplements) $\phi$ is the angle between $(-f'(t),1)$ and $(0,1)$ in $\Bbb R^2$. That is, $\cos\phi = 1/\sqrt{1+f'(t)^2}$.
Finally, then, the principal curvatures (up to sign) of your hypersurface are $\dfrac{f''(t)}{(1+f'(t)^2)^{3/2}}$ and $\dfrac 1{f(t)\sqrt{1+f'(t)^2}}$ (with multiplicity $n-1$).