Resin replica of the Cullinan Diamond (Credits: Wikipedia / Public domain)

### The Cullinan Diamond

Look at this picture of a man in his workshop. What can this be?

Joseph Asscher (Credits Wikipedia / Public domain)

This is 1908, and this is a picture of the first cut made in the biggest diamond ever found, the Cullinan. The giant stone was discovered three years earlier in 1905 in the Premier mine, South Africa. Its weight as a rough piece was of 3,106 carats (621.2 g). After a few intermediaries, the diamond was offered as a gift to King Edward VII in 1907, who entrusted it to a family of gemcutters in Amsterdam, The Royal Asscher Diamond Company, to be transformed into a jewel. The ambitious and perilous task of this gentleman is to cut the diamond, and to turn that rough, unpolished stone, into a beautiful, polished, shining jewel. In reality, it was finally split into several jewels. But you might wonder. Why are we telling you all this? What is the relationship with gravity field? Well, there is actually a common endeavor between those miners, gemcutters, and us space geodesists. We love to turn the rough into the refined. We share that same concern, and it also keeps us awake at night. See for yourself. We want to transform this unpolished rough gravity solution:

Unconstrained Cholesky solution

Into this beautiful polished gravity solution:

SVD solution

Who would have thought that space geodesy and the diamond industry may have things in common?

### The work of the gemcutter

The basic material behind both gravity field images is the same. How come are they so different then? So today we are going to shed a light on the craft that goes from one to the other. The work of the gemcutter. You already know a lot about it if you read our two previous articles of the series. In Part 2, we showed you exactly what components were left out during the truncation of eigenvalues. And now, let’s show you exactly what happens in terms of the least squares minimum.

### Parabolas, paraboloids and hyper-paraboloids

You all remember that gravity field coefficients are obtained by minimizing the sum of the least squares of the measurement residuals. If you look closely at what a least square minimization is, you will find that the concept amounts to finding the minimum of a parabola. In the calculations, you have to multiply a transposed vector by a symmetric definite positive matrix by the non-transposed vector. Something like tXNX. In one dimension, it is a parabola. In two dimensions, it is a paraboloid. In n dimensions, it is a hyper-paraboloid. These are all variants of the same animal. On the vertical axis is the sum of the squares of the residuals. In the horizontal plane is the parameter vector, the one you are trying to determine and adjust (in the case of spherical harmonics up to degree 90, this horizontal plane has 8291 dimensions). So let’s look in detail at the paraboloid. Diagonalizing the matrix consists of finding its natural axes. Along each axis, your formula is a simple parabola, such as ax2+bx+c. The leading coefficient (the “a”) gives you the steepness of the curve. In other words, is the parabola flat or steep? The eigenvectors are the natural axes, and the eigenvalues are the leading coefficients along each axis.

### Freestyle snowboard

Minimizing the least squares is making a move along your paraboloid. If you like freestyle snowboard, you know exactly what we are talking about. Just jump and slide. Start where you are, and slide to the lowest point of the curve. Just as in the pictures below. Mathematically, the move within the plane is the correction on the initial parameter (the gravity field coefficients). The move in the vertical direction is the improvement in residuals.

### How low can you go?

The lower the better. Or so they say. But is this really relevant? Let’s consider two problems.

- The linear approximation. Remember that the paraboloid is only an approximation. You previously linearized the problem with partial derivatives (the observation matrix), and the sum of the squares comes from the use of the normal matrix, which is the multiplication of the observation matrix by its transposed matrix. But the observation matrix is only valid close enough around the initial parameter. If you move too far away, your process doesn’t make any sense (unless you iterate several times, which is almost never the case for the spherical harmonic coefficients).
- Is the prize worth the effort? Have you noticed the difference between both sides? When the parabola is steep, you make a small correction on the parameters, and a big improvement on the residuals. When the parabola is flat, you make a big correction on the parameters, and a small improvement on the residuals. The first move is efficient, the second is not. It doesn’t bring much improvement to your cause. And besides, it may also get you out of the linearity zone. So, all this for nothing?

Actually, it’s even worse.

### Don’t sabotage yourself

Now is the time to closely examine what we are doing. Let’s pick up the images of the eigenvectors from the previous article. One with a high eigenvalue (steep parabola), and one with a low eigenvalue (flat parabola). The correction you add to your initial parameter along each direction is a real number multiplied by the vector that represents that direction. The real number is the amplitude of the horizontal move, and the vector is the eigenvector. Let’s have a close look on the images below. The appearance of the correction is the map, and the amplitude is the size of the horizontal arrow.

A big improvement in the residuals by adding a little bit of this vector

A barely noticeable improvement in the residuals by adding a LOT of that vector

In other words, the worse it is, the more you add. Are you nuts? Is it worth adding such a quantity of an ugly vector in order to barely improve your residuals by almost nothing? Don’t sabotage yourself. It is not necessary to add noise yourself in your solution. The least squares minimum is not the best solution. Along flat directions, the best thing to do is to do nothing.

### Your new problem, and the reward of your hard work

Now of course you just created a new problem for yourself. You know that you have to truncate the eigenvalues. But if you leave some directions out, how do you choose them? How do you craft the truncation? One threshold? A fixed number of eigenvalues or a percentage? Two thresholds? With a smooth gradual nullification of eigenvalues between them? Linear or exponential? You had a big pile of work, and now your pile is twice bigger. Will you shut that door, or engulf in it? Will you do like everybody else, or will you think out of the box? That’s a great question. As Don Meyer the basket-ball coach says: “Do you want to be safe and good or do you want to take a chance and be great?”. So, you are the jeweller. The cutter of diamonds. The one who turns the rough into the refined. That’s a daunting challenge. However, if you think, work, and persevere, you may get the reward of your hard work in the end.

The Cullinan, after the cut (one of the several jewels made)

### What next?

In the next part of the series, we will show you the mathematical equations. No diamonds, no pictures, no allegories. Only numbers and symbols, for your own delight. Stay tuned!