Partial order on Grothendieck group of an abelian category In this article the authors define in 4.1 the Grothendieck group $\mathscr{G}(\mathcal{C})$ of an skeletally small abelian category $\mathcal{C}$ (skeletally small means that the class of isomorphism classes is actually a set) as the abelian group with one generator for each isomorphism class $[a]$ of objects $a \in \mathcal{C}$ and one relation $[b] = [a] + [c]$ if there is a short exact sequence
$ 0 \to a \to b \to c \to 0$ in $\mathcal{C}$.
They state that there is a translation invariant partial order on $\mathscr{G}(\mathcal{C})$ given by $[a] \leq [b]$ if there exists a monomorphism $a \to b$ in $\mathcal{C}$ (at least that's how I understand their notation).
So this partial order is only defined on generators but not on arbitrary elements in the group. So my first question is: How does it extend to a partial order on $\mathscr{G}(\mathcal{C})$? 
My idea is that it works something like $[a] - [b] \leq [c] - [d]$ if there are generators $[x],[y]$ with $[x] = [a] + [d]$ and $[y] = [c] + [b]$ and $[x] \leq [y]$.
My second question is: Are there good articles, books or other resources on partial orders on Grothendieck groups of abelian categories?
 A: It's not true that this always gives a translation invariant partial order on $\mathscr{G}(\mathcal{C})$. If it did, then unless $\mathscr{G}(\mathcal{C})=0$ the partial order could not be trivial, since $[a]\geq[0]$ for every object $a$ of $\mathcal{C}$.
But there are abelian categories whose Grothendieck groups are nontrivial torsion groups (see this question, for example), and so do not admit any nontrivial translation invariant partial order.
A: If you want to construct a translation invariant partial ordering on an abelian group $G$, it suffices to specify an a positive cone, i.e. a subset $H \subset G$ which satisfies


*

*$0 \in H$

*if $a, b \in H$ then $a+b \in H$

*if $a \in H$ and $-a \in H$ then $a = 0$
(see this for an explanation).
In your case the positive cone should be the free submonoid of $\mathcal{G}(\mathcal{C})$ generated by the elements $[b]-[a]$ for which there exists a monomorphism $a \hookrightarrow b$. Then 1. and 2. above are immediate.
As for 3., this is clear if $\mathcal{C}$ is a finite length category, meaning that every object has a finite composition series: in this case you can check that the Grothendieck group is actually the free abelian group $\mathbb{Z}S$ where $S$ is the set of isomorphism classes of simple objects in $\mathcal{C}$. Then the positive cone is simply $\mathbb{N}S$.
On the other hand, if $\mathcal{C}$ is not finite length, it seems that it still might be true that axiom 3 holds, but also could equally well be false in full generality. You need to show that for any finite indexing set $I$, any objects $(a_i)_{i \in I}$ and $(b_i)_{i \in I}$ in $\mathcal{C}$, and any bijection $\sigma: I \to I$ and any monomorphisms $a_i \hookrightarrow b_i$ and $b_i \hookrightarrow a_{\sigma(i)}$ for all $I$, we have that $\sum_I ([b_i] - [a_i]) = 0$. This seems like it could be doable by playing around with short exact sequences and the relation defining the Grothendieck group, but after a few tries I couldn't quite get it.
Maybe someone knows if this is true? See the prologue to this book for more discussion on this issue.
EDIT: As Jeremy's answer below shows, axiom 3 does not hold for a general abelian category.
