Largest Cross Section of an n-Sphere

Question

For a regular sphere in $\mathbb{R}^3,$ i.e. $S^2,$ the largest cross-section of called the "great circle." A cross-section of an n-sphere is an (n-1)-sphere. Is there a similar notion for the largest cross-section of an n-sphere in general, perhaps the "great (n-1)-sphere?"

Answer 1

Yes. In fact, because the sphere is $SO(n)$ symmetric, we can always find a rotation so that the plane we are cross-sectioning with is parallel to the plane perpendicular to the 'vertical' direction. That is to say, we may think of $\mathbb{S}^n$ as the unit sphere in $\mathbb{R}^{n+1} = \mathbb{R}^n \times \mathbb{R}$, with a distinguished $z$-axis, and the hyperplane is of the form $z \equiv c$ for some $c$. Again because of symmetry over the $x_1,...,x_n$ plane, we may assume that $c \geq 0$. Now the only thing to do is to see that this is maximized at $c = 0$.

Hopefully you find it geometrically "obvious" that this is the case, but if not, here's a sketch of why. Clearly for $z \equiv 0$, we obtain the volume of the unit sphere of one dimension lower. This is clear because the equation of the sphere is $\sum_{i=1}^{n+1} x_i^2 = 1$, with $x_{n+1} = z$, and if that is $0$, well, that last term just disappears and this is the equation of a sphere of a dimension lower. If $z \equiv c > 0$, then we get the sphere $\sum_{i=1}^n x_i^2 = 1 - c^2$, which has smaller radius, and so the volume should be lower, since actually this is a sphere of radius $\sqrt{1-c^2}$.

In particular, we've proven that there are great hyperspheres, the ones that are rotations of the $z=0$ plane by arbitrary rotations in $\mathbb{R}^{n+1}$.