Can the order on an ordered, cancellative monoid be extended to its Grothendieck group?

Suppose we have an ordered, cancellative monoid and we wish to apply the Grothendieck group construction to it. Can the total order be extended to the larger group?

Example: consider the ordered monoid $(\Bbb R \cup \{-\infty\},\triangledown)$, where $a \mathbin{\triangledown} b = \log(e^a + e^b)$. Since this monoid is cancellative, we can construct its Grothendieck group. Does the order on $\Bbb R \cup \{-\infty\}$ extend in a canonical way to this new group?

If the commutative monoid $M$ is cancellative, then its Grothendieck group can be defined as $K:=M\times M/\sim$ where $(x_1,x_2)\sim (y_1,y_2)$ iff $x_1+y_2=x_2+y_1$.
Now define $(m,n)>(0,0)$ iff $m>n$. (It is not hard to see that this does not depend on the representative of $(m,n)$ in $M \times M$.) Thus, in $K$ we must have $$(m,n)>(x,y)\ \iff\ (m,n)-(x,y)>0\ \iff\ (m,n)+(y,x)>0\\ \iff (m+y,\,n+x)>0\ \iff\ m+y>n+x$$ where we used $-(x,y)=(y,x)$ in $K$.
Your particular example is isomorphic to the additive monoid of $\Bbb R^{\ge0}$, and its natural extension will be (isomorphic to) $\Bbb R$.
• Note that proving transitivity, we seem to need that the order is total: thus $m+y>n+y$ implies $m>n$ (else we would be left with the case $m\le n$ when $m+y\le n+y$). – Berci Aug 31 '15 at 10:06