I'm interested in whether the finitely-generated discrete Heisenberg group admits a notion of "convex set". Below a formalization of what I need from the convex sets, in particular they should all be finite, should be arbitrarily large, the family should be translation-invariant, closed under intersections and satisfy anti-exchange. They should also have the (non-standard?) property that $g$ is always in the closure of $\{gh, gh^{-1}\}$.


Say a **convex geometry** $\mathcal{C} \subset 2^X$ is $\{\tau(S) \;|\; S \subset X \mbox{ finite}\}$ where $X$ is a set, and $\tau$ is closure operator which takes finite sets to finite sets, and additionally satisfies the **anti-exchange** axiom 
$$ a, b \notin \tau(A) \implies (a \notin \tau(A \cup \{b\}) \vee b \notin \tau(A \cup \{a\})). $$
(I took this from Wikipedia's page on [antimatroids](https://en.wikipedia.org/wiki/Antimatroid), it seems to be standard.)


If $\mathcal{C}$ is a convex geometry on a group $G$, we say it is **translation-invariant** (for the left action of the group on itself) if $C \in \mathcal{C} \implies \forall g \in G: gC \in \mathcal{C}$. Say a convex geometry on $G$ is **midpointed** if it satisfies $g \in \tau(\{gh, gh^{-1}\})$ for all $g, h \in G$.


(I wrote a proof sketch that a translation-invariant convex geometry is midpointed if and only if for all $C, D \in \mathcal{C}$ and for all $a \in C$ such that $C \setminus \{a\} \in \mathcal{C}$, there is a at most one $g \in G$ such that $gD \subset C$. This is the actual property that interests me, but it's perhaps less pretty on the first sight so I opted for the above.)


The sets $C \cap \mathbb{Z}^d$ where $C$ ranges over compact convex subsets of $\mathbb{R}^d$ form a midpointed translation-invariant convex geometry on the group $\mathbb{Z}^d$. If $G$ is not torsion-free, then it does not admit any midpointed convex geometry (an involution forces $1\_G$ and vice versa, so anti-exchange fails, and there's a similar argument for other $\mathbb{Z}\_p$). Neither does the abelian group $\mathbb{Z}[1/2]$ (because $0, 1$ necessarily has infinite closure), nor the finitely-generated metabelian group $\mathbb{Z}[1/2] \rtimes \mathbb{Z}$ where $\mathbb{Z}$ acts by multiplication by $2$.



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> Question 1. Does the discrete Heisenberg group of integer matrices of the form $\left( \begin{smallmatrix} 1 & a & c \\ 0 & 1 & c \\ 0 & 0 & 1 \end{smallmatrix}\right)$ admit a midpointed translation-invariant convex geometry?
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and more generally



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> Question 2. Which strongly polycyclic groups admit a midpointed translation-invariant convex geometry?
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I'm also interested in the following, though I have no direct use for it, and I imagine it's easier to find information about.



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> Question 3. What other midpointed translation-invariant convex geometries does $\mathbb{Z}^d$ admit?
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Convex sets on the discrete Heisenberg group
I claim that the answer is yes for the Heisenberg group, and more generally for every finitely generated torsion-free 2-step nilpotent group $\Gamma$.


Indeed, embed $\Gamma$ in its real Malcev closure $G$, and view $G=\mathfrak{g}$ as a Lie algebra, by identification through the exponential map. Recall that $G$ is uniquely divisible; in particular every $x\in G$ has a unique square root $x^{1/2}\in G$ (equal to $\frac12 x$ in $\mathfrak{g}$); define $\kappa(x,y)=x(x^{-1}y)^{1/2}$: this is a commutative binary law on $G$. Computed in the Lie algebra, it turns out to be equal to $\kappa(x,y)=\frac12(x+y)$ (I checked it for $G$ 4-step nilpotent and have not checked in higher nilpotency length.)


Define a subset $F$ of $\Gamma$ as "convex" if is the intersection of a convex subset of $\mathfrak{g}(=G)$ with $\Gamma$. Arbitrary intersection of convex subsets are convex, and hence every subset has a well-defined convex closure. 


To be convex is invariant under left translation. The reason is that the group law, as given by the BCH formula ($gh=g+h+\frac12[g,h]$), is affine as a function of $h$ for each fixed $g$.


I next claim that this satisfies the midpoint condition. Indeed, suppose $g,h\in \Gamma$, $gh,gh^{-1}\in F$ and $F$ is convex. Then $\kappa(gh,gh^{-1})=gh((gh)^{-1}(gh^{-1})^{1/2}=gh(h^{-2})^{1/2}=g\in\Gamma$, and it also equals $\frac12(gh+gh^{-1})$, so $g\in F$ by the definition.


Next I claim that the convex closure of every finite subset is finite. Indeed, given a finite subset $F$, the convex closure of $F$ is contained in the convex hull of $F$ (in the ordinary affine structure of $\mathfrak{g}$). The latter being bounded, it intersects $\Gamma$ in a finite subset.


Finally, the anti-exchange axiom follows from it being satisfied by the ordinary convexity in $\mathfrak{g}$.


