More About Symmetry

In Some Definitions we used a short sequence of operations to connect the expansion T(u) to the expansion B(u).

2ABCD11		11DCBA2
02ABCD11		011DCBA2
22ABCD11		211DCBA2

There, we started with the crosspiece and moved down the sides, and we used T(u) merely as an example to show that reading an expansion backward doesn't change the entry in the lower left corner of the corresponding matrix. Here, we'll start in the lower left and move clockwise, and we'll use T(u) in particular, as itself, to compute the whole matrix B(u). The question marks represent various values that we don't care about.

0.		start with T(u)	22ABCD11	2u + t	?
				u	u − t
1.		subtract 2	02ABCD11	t	?
				u	u − t
2.		take reciprocal	2ABCD11	u	u − t
				t	?
3.		reverse	11DCBA2	u	t
				u − t	?
4.		take reciprocal	011DCBA2	u − t	?
				u	t
5.		add 2	211DCBA2	3u − t	?
				u	t

That's the first of the two promised results. Note how it depends on the expansion T(u) starting with 2.

Let's also take care of the second of the two promised results now. I thought it would require some effort, but actually it's trivial. B(u) is a reflection of T(u), so according to The Discriminant, it has the same Markov constant. QED.

Before we go further, we should go back and understand why the matrix changes in the way that it does. We can get intuition from the fact that the two columns of a matrix are also the last two convergents of the corresponding expansion. So, for example, here's how the first column changes in step 1.

(2u + t)/u − 2 = t/u

Intuition is good, but it's important to back it up with facts and details.

1. If we add or subtract an integer, then we can read this equation forward to get the new expansion, …

   c + [a₀, a₁, a₂, a₃, … ] = [c+a₀, a₁, a₂, a₃, … ]

   … and we can truncate it and read it backward to see that all the convergents behave as expected. If we add or subtract a non-integer, though, then we can't even get started because we don't know the new expansion.

2. If we take the reciprocal of a positive number, then we get the new expansion by adding or removing a zero at the start, and all the convergents behave as expected as long as we match them up working backward from the last convergents. If we take the reciprocal of a negative number, though … well, fortunately we don't need to open that can of worms.

   To be more precise, for numbers greater than 1 we add a zero, and for numbers in the range (0,1) we remove a zero. What about the number 1 in between? Well, for the proper form [1] we add a zero, and for the improper form [0,1] we remove a zero. In other words, the two forms are reciprocals! That's not a matter of opinion, it's a fact that you can check by looking at the matrices. At first it seemed full of significance to me, as if I'd discovered some hidden property of rational numbers, but in the end I doubt it means anything. If you want to think further about it, I'd start by returning to the matrix groups in Equivalence. Note that that's the only time that taking the reciprocal connects a proper form to an improper form, at least for positive numbers.

3. Here I can't improve on what I said in Some Definitions.

   Reversing an expansion corresponds to taking the transpose of a matrix. (Why? See the middle of The Discriminant.)

4. Note that the net effect of steps 2–4 is to transpose the matrix across the other diagonal.

Now let's try a completely different sequence of operations inspired by Triangles and Sketch of a Proof.

0.		start with T(u)	22ABCD11	2u + t	?
				u	u − t
1.		proper-improper	22ABCD2	2u + t	?
		conversion		u	t
2.		same-denominator
		conversion
	a.	negate	(−3)11ABCD2	−2u − t	?
				u	t
	b.	add 5	211ABCD2	3u − t	?
				u	t

Compare the last row here to the last row in the previous table. The expansions seem to be different, but the matrices seem to be the same. What does it mean? Well, the values are certainly the same, (3u − t)/u. That value is a rational number, and a rational number has only two possible expansions, proper and improper. But, the expansions here both end with 2, so they're both proper, so they're the same!

That argument is simple and correct, but to me it sounds like a lot of hand-waving, so here's another argument. The expansions have the same length, so the matrices have the same determinant. (In fact, the length is even and the determinant is 1, but the argument doesn't depend on that.) We can compute the values of the question marks from the determinants and the other three entries, but the determinants are the same and the other three entries are the same, therefore all four entries are the same. In other words, the matrices are the same, and that means the expansions are the same.

So, either way, the expansions are the same. But, from there it follows that DCBA = ABCD, i.e., that the middle part has reflection symmetry!

As a reminder, that reflection symmetry is different from the third symmetry. I talked about that a little in Triangles—look for the part about a strange thing.

Again, we should go back and understand why the matrix changes in the way that it does.

1. Suppose we have an expansion that ends with a coefficient a_L−1 ≥ 2, and suppose we run the second algorithm on both the proper and improper forms.

   		…	aL−2	aL−1
   0	1	…	pL−2	pL−1
   1	0	…	qL−2	qL−1
   		…	aL−2	aL−1 − 1	1
   0	1	…	pL−2	p	pL−1
   1	0	…	qL−2	q	qL−1

   The convergents up to and including p_L−2/q_L−2 are obviously the same, and the final convergent p_L−1/q_L−1 has to be the same; the only difference is that the improper form contains an extra convergent p/q. We can find the value of p via the equation that involves the coefficient 1.

   pL−1	= p + pL−2

   In fact, p and p_L−2 are related by the involution that maps x to p_L−1 − x.

   p	= pL−1 − pL−2
   pL−2	= pL−1 − p

   The equations for q are identical. So, to convert a matrix from proper to improper form, we leave the first column alone and apply the corresponding involutions to the second column, and to convert a matrix from improper to proper form, we do exactly the same thing.

2. I introduced the same-denominator form in Triangles. At the end of that essay I mentioned the equation for negative numbers, but I didn't understand the full significance of it until now.

   −[a₀, a₁, a₂, a₃, … ] = [−a₀−1, 1, a₁−1, a₂, a₃, … ]

   There, to investigate same-denominator conversion, we ran the second algorithm on [a₀, a₁] and [a₀, 1, a₁−1]. The last two denominators were the same, but the last two numerators were different. Why did that matter?

   … the equation that defines the denominators only looks back two spaces, so if we take an arbitrary suffix and append it to the two expansions, we'll continue to get the same denominators.

   Here, to investigate negation, let's run the second algorithm on [a₀, a₁] and [−a₀−1, 1, a₁−1].

   		a0	a1
   0	1	a0	a0a1 + 1
   1	0	1	a1
   		−a0 − 1	1	a1 − 1
   0	1	−a0 − 1	−a0	−a0a1 − 1
   1	0	1	1	a1

   The last two denominators are the same, which can continue, and the last two numerators are negated, which can also continue! That shows that negation is more fundamental than same-denominator conversion, and also that intuition gives the right answer: to negate a matrix, we just negate the numerators in the first row. There are complications for matrices that represent integers, but I'll save the details for another essay, Details About Operations.

   Then, to get the same-denominator form, we can just add the integer 2a₀ + 1.

Note that the order of the two conversions doesn't matter.