a vector with $d$ zeros, and $\bm{1}$ denotes a vector with $d$ ones.

The first property that we require from $\mbox{MO-CR}[{\mathcal{E}},{\mathcal{F}}]$ is to be on a *multi-objective ratio scale*. Given ${\mathcal{E}},{\mathcal{F}}\subset{\mathbb{R}^d_{+}}$ and $\bm{r}\in{\mathbb{R}^d_{+}}$, the following shall hold. $$\begin{aligned}
\mbox{MO-CR}[{\mathcal{E}},{\mathcal{F}}] 
&\quad\subseteq& 
{\mathbb{R}^d}\label{eq:ratio:6}\\ 
\mbox{MO-CR}[\{\bm{0}\},{\mathcal{F}}] 
&\quad=& 
\{\bm{0}\}\label{eq:ratio:7}\\ 
\mbox{MO-CR}[\bm{r}\star{\mathcal{E}},{\mathcal{F}}] 
& \quad=& 
\bm{r}\star\mbox{MO-CR}[{\mathcal{E}},{\mathcal{F}}]\label{eq:ratio:8}\\ 
\mbox{MO-CR}[{\mathcal{E}},\bm{r}\star{\mathcal{F}}] 
& \quad=& 
\mbox{MO-CR}[{\mathcal{E}},{\mathcal{F}}]/\bm{r}\label{eq:ratio:9}\\ 
{\mathcal{E}}\subseteq{\mathcal{F}}& \quad\Leftrightarrow& 
\bm{1}\in \mbox{MO-CR}[{\mathcal{E}},{\mathcal{F}}]\label{eq:ratio:10} \end{aligned}$$ To fix these ideas one can think of $d=1$ and given two positive numbers $e,f$, to the properties of ratio $e/f$. Equation (\[eq:ratio:6\]) states that MO-CR is expressed in a multi-objective space. Equations (\[eq:ratio:7\]), (\[eq:ratio:8\]) and (\[eq:ratio:9\]) state that MO-CR is well-centered and sensitive on each objective to multiplications of outcomes, which is what we want. For instance, if ${\mathcal{E}}$ is three times better on objective $k$, then so is MO-CR. If there are two times more efficient opportunities in ${\mathcal{F}}$ on objective $k'$, then MO-CR is one half on objective $k'$. In other words, the efficiency of each objective independently reflects on MO-CR in a ratio-scale. Equation (\[eq:ratio:10\]) states that if all equilibria outcomes are efficient (i.e. ${\mathcal{E}}\subseteq{\mathcal{F}}$), then this amounts to $\bm{1}\in\text{MO-CR}[{\mathcal{E}},{\mathcal{F}}]$, i.e. the MO game is fully efficient.

These requirements rule out a set of first ideas. For instance, we 