# What is the surface area an ellipse takes up on a sphere?

Take $$\Sigma$$ a $$k\times k$$ positive-definite real matrix and $$E$$ to be an associated ellipse: $$E:=\{(x_1,\dots, x_N): \frac{1}{N}\sum_n x_n^\dagger \Sigma x_n \leq 1\}.$$

Now take $$z$$ uniform on $$\mathbb{S}^{NK-1}$$ the unit sphere in $$\mathbb{R}^{NK}$$. For large $$N$$, then does the ratio of $$\mathbb{S}^{NK-1}$$'s surface area that $$z+E$$ occupies converge? What does it converge to?

If the ellipse is very small relative to the unit sphere then the surface area occupied by $$z+E$$ is roughly hyperplanar, and for very large dimension, most hyperplanes cut through something close to a sphere with radius-squared the geometric mean of $$E$$'s eigenvalues (I am guessing), or $$|E|^{1/{NK}}$$, so the area is proportional to this quantity. Is my guess correct?

Of course this does nothing for the general case, since $$E$$ could be big enough that one of its lobes passes through the entire unit sphere and encompasses area on the other side.

To the best of my understanding one must first find the distribution of balls in $$NK$$-real-projective-space generated by random placement of $$E$$ on the sphere's surface, and hope that this clusters onto some representative ball when $$N$$ is large.

• This is an interesting question that deserves a good answer. Since bumping a post to the front page via unnecessary editing is frowned upon, I'm gonna do this only once :P – Lee David Chung Lin Feb 16 at 17:02