# Does a Bijective Commutative transformation on a vector of angles exist?

I have a problem where I have two vectors a and b representing a list of angles.

I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)

For example: a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify

a.b doesn't qualify because I don't have a metric to compare square angles.

a-b doesn't qualify because it is not commutative

Do you think this is possible?

What I have tried that didn't work:

T(a,b) = {cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)}

And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.

This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.

Thank you!

Notation convention: the $$k$$th element of an array, such as $$a$$, will be denoted with a subscript, as in $$a_k$$.
If you're working with functions of the form $$c_k=f(a_k,b_k)$$ for some symmetric $$f$$, this is unavoidable. We can swap $$a_1$$ and $$b_1$$, leave all of the other $$a_k$$ and $$b_k$$ alone, and the $$c_k$$ will still be the same. Repeat with $$d_k=g(a_k,b_k)$$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $$k$$. Here's an idea:
\begin{align*}c_1 = a_1+b_1\quad d_1=\cos(a_1-b_1) &\\ c_2 = a_2+b_2\quad d_2=\cos(a_2-b_2) &\quad e_2=\cos(a_1+a_2-b_1-b_2)\\ c_3 = a_3+b_3\quad d_3=\cos(a_3-b_3) &\quad e_3=\cos(a_1+a_2+a_3-b_1-b_2-b_3)\end{align*} and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $$((x,y),(u,v))$$ and $$((x,v),(y,u))$$ since $$|x+y-u-v|\neq |x+v-u-y|$$ in general.