# Labeling of axes in quotient space

It seems to be trivial, but I couldn't find anything.

I have a metric space, let us say $\mathbb{R}_{>0}^3$, and a equivalence relation $\sim$, let's say $(x_1,y_1,z_1)\sim(x_2,y_2,z_2)$ if $z_1=z_2$ and $\exists p>0$ such that $px_1=x_2$ and $py_1=y_2$. Then, the quotient space $\mathbb{R}_{>0}^3/\sim$ can be identified with $\mathbb{R}_{>0}^2$. So far, so good.

Now, I want to plot a given trajectory both in $\mathbb{R}_{>0}^3$ and $\mathbb{R}_{>0}^3/\sim$. As a representation of $\mathbb{R}_{>0}^3/\sim$ as $\mathbb{R}_{>0}^2$ I have chosen a 2D slice through, let's say, $y_2=const$. Then I projected all points of the trajectory on the 2D slice such that each point stays in its equivalence class, and then plotted that trajectory in $\mathbb{R}_{>0}^2$ as a representation of the trajectory in the quotient space.

My problem: What is a suitable labeling of the axes? One axis I would like to label something along the line of "$z$", but labeling the other "$x$" would be misleading. $[z]$ and $[x,y]$ also feels wrong.

Note: (i) Due to technical reasons I have to project on a slice with $y=const$. (ii) The axes labels should somehow tell something about their relationship with x,y, and z. (iii) The equivalence relation is in general more complex.

• The relation you describe is not even an equivalence relation at the moment. It can become one if you modify it a bit, but then the quotient is not $\mathbb{R}^2$. Sep 4 '15 at 6:42
• Sorry, the error with the equivalence relation sneaked in when I simplified my actual question. I guess allowing for x=y=z=0 was the problem. I removed that case by restricting x,y, and z to $\mathbb{R}_{>0}$ instead of $\mathbb{R}$. @Amitai : Is the question now correct? Sep 4 '15 at 8:59

As you say, one coordinate can be called $z$, and this is fine.
The other coordinate can be represented best by the ratio $x/y$ (or $y/x$).
• Thanks for the answer. However, as said in the question, this is a simplified example: In general, the equivalence relations looks like $(x_1,y_1,z_1)\sim(x_2,y_2,z_2)$, if $z_1=z_2$ and $\exists (p_1, p_2)\in P$ such that $p_1(x_1)=x_1$ and $p_2(x_2)=x_2$, with $P$ some set. I am also operating in more general spaces, i.e. $x/y$ is not even defined... I like your idea of using the projection rule as the label, but I somehow fail to generalize it. Sep 4 '15 at 11:01