# Higher Dimensional Ellipsoids

An ellipse is a set of points in the plane which satisfy a certain equation, namely,

$\frac{(x - x_0)^2}{r_1} + \frac{(y - y_0)^2}{r_2} = k^2$

In this sense, an ellipse appears to just be a squashed circle. But there are dozens of ways to think about ellipses – as conic sections, for example – and in this post the idea I’d like to use is that of foci:

You can make an ellipse with two pins and a piece of string tied in a circle. Stick the two pins into a page and loop the string over them. Stretch the string out tight with a pen and draw all around. If you make sure to keep the string taut, you will have drawn an ellipse.

It’s a common enough geometric construction, and I’m sure many readers would know of it. I myself remember my dad showing it to me when I was a little boy. What it tells us is that an ellipse can be though of as the set of points whose total distance from two distinct points, the foci, is constant (plus the distance between those two points, but that’s always constant).

Naturally, a mathematician asks: “How can I generalise this idea?”

If we wanted to, we could perhaps pick more points in the plane and ask for the set of points at a constant total distance from all of them. Depending on how we decide to calculate this total distance, this potentially begins to differ from the string-and-pins construction, which can only produce a shape made up of arc segments of different two-foci ellipses (can you see why?). A natural distance calculation would be to sum over the distance to each focus, plus the distances between each pair of foci.

The thing is, with a potentially large number of foci placed arbitrarily in the plane, we could produce some pretty strange objects. A broad enough family so that discussing them here would be difficult. Ok, maybe not difficult, but not what I want to do right now.

So perhaps I’m looking for a different generalisation. Return to two foci. Could we ask perhaps for a constant total area of the triangle whose edges are the straight lines to the foci and the line between them?

No deal. Think: a triangle’s area is half its base times its height. The line between the foci (the base) is constant, so the height must be constant also. Thus we can only skew the triangle left or right and we get an infinite straight line. Not very interesting. But ‘area’ is on the right track.

Ok, I’ll stop being coy. What am I really doing? Dimension up! In three dimensions, pick three non-colinear points. Pretend now you have a sheet of cloth sewn into a sphere around them. Pinch a point on the cloth and pull it taut. Keeping the cloth taut, by moving the pinched point around we define a closed surface. This is the surface of constant cloth area, directly analogous to the ellipse, which had constant string length.

What is this surface?

It’s seen maybe easiest via the cross product of vectors. By rotations and scales of $\mathbb{R}^3$, we are always able to choose one focus to be at the origin, one focus to be at $x = 1$ and $y, z = 0$, and the third focus to have $z = 0$. Call these foci $p_0, p_1, p_2$ respectively.

Let $r$ be a generic point $(x, y, z)$ in threespace. Since the (magnitude of the) cross product gives the area of the parallelogram defined by two vectors, we are looking at the quantity

$\frac{1}{2}\|(r - p_0)\times (r - p_1)\|$

$+ \frac{1}{2}\|(r - p_1)\times (r - p_2)\|$

$+ \frac{1}{2}\|(r - p_2)\times (r - p_0)\|$

$+\frac{1}{2}\|(p_2 - p_0)\times (p_1 - p_0)\|$

being equal to some constant value $K$. Simplifying, this becomes the quantity

$\frac{1}{2}\sqrt{y^2 + z^2}$

$+ \frac{1}{2}\sqrt{((x_0 - 1)y - (x-1)y_0)^2 + ((x_0 - 1)^2 + y_0^2)z^2}$

$+ \frac{1}{2} \sqrt{(y_0 x + x_0 y)^2 + (x_0^2 + y_0^2)z^2}$

equalling $K - \frac{1}{2}y_0$, where we have set $p_2 = (x_0, y_0, 0)$. Plainly we would like the quantity $K - \frac{1}{2} y_0$ to be positive in order to avoid a degenerate surface (it must be at least non-negative, since it’s the sum of magnitudes of vectors in $\mathbb{R}^3$)

Of course this is a bounded surface, without boundary. It might not be smooth, though (it may have cusps). We want to know what this looks like. Using a certain popular mathematics package, I made a few level surface plots of the above function. I chose $K = 2$ since it was low enough to show small-scale structure but large enough to avoid most of the numerical problems.

This first one is of course with the unit-side equilateral triangle configuration for the foci.

Ellipsoid with 3 foci in an equilateral triangle

Looks weird, huh. Not sure it deserves the name ‘ellipsoid’. It kind of reminds me of a nut or a bivalve mollusc.

There may be some numerical issues here and there, but I tried to get as high a precision as possible. We should check the degenerate case of foci lying in a line. In this case, we should retrieve a cylinder, as this is the three-dimensional revolution of the surface area two-foci ‘ellipse’ we considered in two dimensions (which resulted in a straight line).

Degenerate ellipsoid; foci colinear.

When you make the triangle more interesting (without loss of generality, we can take $x_0, y_0 > 0$ through flip symmetry) you can get some pretty interesting stuff happening. This is the ellipsoid with foci in a right triangle of unit length legs.

Ellipsoid with foci in a right triangle

Another view of the right triangle ellipsoid.

And how about an obtuse triangle? This one has $(x_0, y_0) = (4, \frac{1}{2})$.

Ellipsoid with obtuse foci arrangement.

Another view of the obtuse case.

There are other ways that we can play with the base triangle to get more interesting shapes. For $y_0$ very small, regardless of the value of $x_0$, the foci are nearly colinear, and the resulting surface stretches out longer and smoother, approaching the cylinder as $y_0 \to 0$. Pushing $y_0$ out large starts squashing it in the other direction. Here is one with $(x_0, y_0) = (\frac{1}{2}, 5)$.

Ellipsoid with long isosceles foci configuration.

Interesting general features to note: three to four cusp-like peaks on each non-degenerate surface. Each surface also appears to have a ‘crease’, or ‘rim’ running in a complete loop around the surface, oscillating above and below the $xy$-plane.

Of course now we can look at more dimensions. You can see how this will generalise. We want $n$ non-degenerate ‘hyperfoci’ $p_0, p_1, \ldots, p_{n-1}$ in $\mathbb{R}^{n+1}$. Then we want the set of points such that the $(n+1)$-dimensional hypervolume defined by each point + the $n$ foci has constant $n$-dimensional hypersurface area.

It’s pretty easy to see that we can translate the first focus to the origin, and rotate so that each hyperfocus $p_k$ has nonzero components only in its first $k$ dimensions. In particualr, the $k$th compenent of $p_k$ must be non-zero to avoid degeneracy. We also have scaling available to us, meaning that without loss of generality we can take $p_1 = (1, 0, \ldots, 0)$. Then we take a point $X = (x_0, x_1, \ldots, x_{n-1})$ in our space and compute the hypervolumes defined by the vectors $(X - p_0), (X - p_1), \ldots, (X - p_{n-1})$.

I can’t draw these for you, sorry. Best I can do via blog is a 2D image of a 3D object.

Questions: parametric equation? Any applications? Any geometric connections – e.g., some kind of generalised conic section relationship? Et cetera. Go crazy.