What is the group-like structure on $x^2+y^2+z^2-2xyz=1$?

Question

(Background: this is inspired by Chebyshev polynomials and expanding a function as a Chebyshev series.)

Solving for $ z $ gives $$ z=xy \pm \sqrt{(1-x^2)(1-y^2)}, $$ where $-1\leq x,y \leq 1$. Now choose the positive square root and call it $ x\oplus y$. (Alternatively, we can define $x\oplus y=\cos(\arccos(x)+\arccos(y))$.)

We can check commutativity, identity ($1$), inverses ($-x$). Associativity is might be a bit harder, but I reckon it works.

The defining equation is a cubic, hence, it reminds me of the group structure of an elliptic curve, but an elliptic curve has two variables and the group law is defined by intersection with a line, while this equation has three variable, one is the "sum" of the other two.

Question Has anyone seen a structure like this before?

Answer 1

This is better understood by considering the roots of unity. Suppose $\, u_n = r_n + s_n\sqrt{-1} \,$ for $\, n=1,2,3 \,$ be three roots of unity such that $\, u_3 = u_1 u_2. \,$ Then $\, r_1^2 + r_2^2 + r_3^2 - 2 r_1 r_2 r_3 = 1. \,$ The same equation is true if $\, u_3 = u_1/u_2 \,$ or $\, u_3 = u_2/u_1. \,$ Or more symmetrically if $\, u_1u_2u_3 = 1.\,$ The reason is that given only the real part of a root of unity, there are two possible values for the imaginary part. The roots of unity form a multiplicative group, but if we only have the real part, the group operation is two-valued.

The case of addition of points on elliptic curves is very appropriate. The group operation there is related to three points on a line, but not in the obvious way. That is, the third point is not the sum of the other two, but rather the negative of the sum. Another way to state it is that the sum of all three points is the identity. Also, given only the $\,x\,$ coordinate of a point, there are two values for the $\,y\,$ coordinate. There is some similarity between "adding" points on the unit circle and points on an elliptic curve.

Answer 2

You also can have
• $(2,3,6\pm\sqrt{24})$ so all three are greater than 1.
• $(-2,-3,6\pm\sqrt{24}$
• $(1,y,y)$
• $(-1,y,-y)$
Remember $\cos(a+b)=\cos a\cos b -\sin a\sin b$ so the choice of square-root might get tricky.

Answer 3

At first glance it seems to be the same as Cayley's nodal cubic surface. Although there it is described by the (projective) equation $$wxy+xyz+wxz+wyz=0.$$ How, then, do obtain my version of it (if, in fact, they are the same.)

Here is a page which seems to derive my equation from Cayley's.

Answer 4

Added: for real $x,y,z \geq 1$ we get all $$x = \cosh t, \; \; y = \cosh u, \; \; z = \cosh \; (t+u).$$ Actually, as $\cosh$ is even, we could demand $t+u+v=0$ with solution $$x = \cosh t, \; \; y = \cosh u, \; \; z = \cosh v.$$ Easy to check that this satisfies $x^2 + y^2 + z^2 = 1 + 2 xyz.$ Multiplying the variables by $i$ preserves the identity, while $\cosh it = \cos t,$ so we get $$x = \cos t, \; \; y = \cos u, \; \; z = \cos \; (t+u)$$ for $|x|,|y|,|z| \leq 1.$

Not sure yet whether having, say, $z \geq 1$ forces both $x,y \geq 1.$ Notice that there is no benefit to having two negative values, we can just negate both.. Of course, you might like negative numbers. Matter of taste. We might have the entire surface with these and then $(-x,-y,-z)$ for the $\cosh$ part. Note the gradient being $0$ at the point $(1,1,1)$ of the surface.

There are three involutions; by alternating these one may travel around the surface.

$$ (x,y,z) \mapsto ( 2yz-x,y,z) $$ $$ (x,y,z) \mapsto (x,2zx-y,z) $$ $$ (x,y,z) \mapsto (x,y,2xy-z) $$

Not sure why you are interested in $|x| \leq 1.$ For any $t$ we get a solution $$ (t,t,1) , $$ then the third involution takes us to $$ (t,t,2t^2 - 1) $$ A different involution ( and re-ordering) takes us to $$ ( t, 2t^2 - 1, 4 t^3 - 3 t ) $$ which reminds me of $(\cos \theta, \; \cos 2 \theta, \; \cos 3 \theta)$ but also the more useful $$(\cosh w, \; \cosh 2 w, \; \cosh 3 w).$$ This aspect is very successful: given integers $1 \leq m < n,$ and positive real $w,$ we get an ordered solution $$ ( \cosh mw, \; \cosh nw, \; \; \cosh \, (m+n)w ) $$ which explains the repetitions of numbers such as $5042,$ which is $\cosh 7w$ when $\cosh w = 2.$

Here are some solutions with positive integer entries that are distinct:

26  7  2
97  26  2
99  17  3
244  31  4
362  97  2
362  26  7
485  49  5
577  99  3
846  71  6
1351  362  2
1351  97  7
1921  244  4
2024  127  8
2889  161  9
3219  1933  1727
3363  577  3
3363  99  17
3510  2145  1998
3551  3287  2025
3614  3218  1663
3970  199  10
4015  3409  1727
4095  3569  3087
4097  3203  2947
4127  3417  3047
4237  4115  2095
4247  4183  2177
4299  4149  4095
4446  3130  927
4754  4665  582
4801  485  5
5042  1351  2
5042  362  7
5042  97  26