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A left-invariant preordering on a group $G$ is a reflexive, transitive and a complete relation $\preceq$ on $G$ such that $x\preceq y$ implies $gx\prec gy$ for any $g$ (anticommutativity is not requred). A preordering is discrete if for some element $a\succ 1$ there is no elements $b$ such that $1\prec b\prec a$. The element $a$ is called a least positive element.

Let $G$ be the surface group of an oriented surface (for example, the free group $F_n$). The question is: which elements $a\in G$ are least minimal for some discrete preordering on $G$?

The answer is negative for $a\in G$ which are not "homotopically prime", i.e. $a=b^k$, $k>1$.

The answer is positive for $a$ which are "homologically prime", i.e. the projection $\bar a$ of $a$ is prime in the abelianization $G_{\mathrm{ab}}=G/[G,G]$. Indeed, there is a homomorphism $\bar\phi: G_{\mathrm{ab}}\to\mathbb Z$ such that $\bar\phi(\bar a)=1$. Then the corresponding homomorphism $\phi:G\to \mathbb Z$ defines the required preordering: $x\preceq y$ iff $\phi(x^{-1}y)\ge 0$.

What about the elements which do not belong to those two subsets?

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  • $\begingroup$ I assume "complete" means that for all $g,h$ we have $g\preceq h$ or $h\preceq g$. $\endgroup$
    – YCor
    Commented Sep 20, 2022 at 13:02
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    $\begingroup$ Call these elements "potentially minimal". If there is a torsion-free nilpotent quotient of $G$ in which $g$ maps to a primitive element of the center of $G$, then $g$ is potentially minimal (same argument). This applies, for instance, in a free group, to the commutator of two distinct generators. $\endgroup$
    – YCor
    Commented Sep 20, 2022 at 13:05
  • $\begingroup$ That's right: here "complete" means total (or connected) ordering. $\endgroup$
    – nim
    Commented Sep 21, 2022 at 14:11
  • $\begingroup$ We can give the following description of potentially minimal elements: let $a$ be a pm-element. Consider the subsgroup $H=\{g\in G\mid 1\sim g\}$ where $g\sim h$ means $g\preceq h$ and $h\preceq g$. Since $1\preceq g$ implies $1\preceq aga^{-1}$ (see arxiv:0808.2686), then $aHa^{-1}=H$. Define a new preordering: $g\preceq' h$ iff $g\preceq ha^p$ for some $p\ge 0$. Then the subgroup $H_1$ generated by $H$ and $a$ is $H_1=H<a>=\{g\in G\mid g\sim'1\}$. We can replace the ordering $\preceq$ with the lexicographic preordering $(\preceq', \preceq)$ which is discrete with a least minimal element $a$. $\endgroup$
    – nim
    Commented Sep 21, 2022 at 14:33
  • $\begingroup$ On the other hand, let $H_1\subset G$ be a subgroup such that: 1) there is a preordering $\preceq'$ on $G$ with $H_1=\{g\in G\mid g\sim'1\}$, i.e. an invariant total ordering of the coset $G/H_1$; 2) there is a normal subgroup $H\lhd H_1$ with $H_1/H\simeq\mathbb Z$. Then any element $a\in H_1$ which projects to a generator of $H_1/H$ is potentially minimal. $\endgroup$
    – nim
    Commented Sep 21, 2022 at 14:41

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Theorem. Let $G$ be the surface group of an oriented surface. Then an element $a$ is a least positive element for some discrete preordering on $G$ if and only if $a$ is not a proper power.

Proof. Let $a$ be not a proper power in $G$. Consider the normal closure $H=\langle\langle a\rangle\rangle$ of $a$. Hempel proved that the quotient group $G/H$ is locally indicable, hence, there is a left-ordering $\preceq_1$ on it.

The surface group $G$ is a quotient of a free group $F$ by a relation $w=[a_1,b_1]\cdots[a_n,b_n]$, $n\ge 0$. Let $\tilde H$ be the preimage of $H$ in $F$. The group $\tilde H$ is free and is not generated by $w$, hence, there is an epimorphism $\tilde\phi\colon \tilde H\to\mathbb Z$, such that $\tilde\phi(w)=0$. Then $\tilde\phi$ defines an epimorphism $\phi\colon H\to\mathbb Z$.

The subgroup $H$ is generated by elements $gag^{-1}$. Denote $d=gcd\{\phi(gag^{-1})\mid g\in G\}$. Since $\phi$ is an epimorphism, $d=1$. Then there exist $g_1,\dots, g_p\in G$ and $l_1,\dots, l_p\in\mathbb Z$ such that $\sum_{i=1}^p l_i\phi(g_iag_i^{-1})=1$. Consider the homomorphism $\phi'=\sum_{i=1}^p l_i\cdot\phi^{g_i}$ from $H$ to $\mathbb Z$ where $\phi^g(h)=\phi(ghg^{-1})$. Then $\phi'(a)=1$.

Now, define a preordering on $G$: $g_1\preceq g_2$ iff $g_1H\prec_1 g_2H$ or $g_1H=g_2H$ and $\phi'(g_1^{-1}g_2)\ge 0$. Then $a$ is a least positive element in the preordering $\preceq$.

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