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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, 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 $2$.

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?

and more generally

Question 2. Which strongly polycyclic groups admit a midpointed translation-invariant convex geometry?

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.

Question 3. What other midpointed translation-invariant convex geometries does $\mathbb{Z}^d$ admit?

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  • $\begingroup$ OK, I think you're defining somewhat discrete convex geometries (by the condition that the closure of a finite subset should be finite). $\endgroup$
    – YCor
    Dec 15, 2019 at 9:08
  • $\begingroup$ That is the intention, in particular since the closure is finite, there are arbitrarily large convex sets. Sorry if I'm being cryptic. $\endgroup$
    – Ville Salo
    Dec 15, 2019 at 9:09
  • $\begingroup$ It's just that the motivation is apparent only after reading the question, and the discreteness is not stated in the link. But I think I got it now. $\endgroup$
    – YCor
    Dec 15, 2019 at 9:15
  • $\begingroup$ I added a paragraph that states the motivation and lists the main properties. $\endgroup$
    – Ville Salo
    Dec 15, 2019 at 9:25

1 Answer 1

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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 $\mathfrak{g}$.

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  • $\begingroup$ Very nice. So the convex sets of the Lie algebra turn into convex sets of the group because nilpotency length two means translation only distorts them affinely. Plus the fact midpoints are actually given by vector halving... by an algebra accident? I checked this "accident", i.e. the "Computed in the Lie algebra, ..." claim for the Heisenberg group (in the stupid way, by just computing with real matrices). It would be interesting to see how you did it up to length four, and also whether it's an accident. I don't know much about the Lie algebra-Lie group correspondence. $\endgroup$
    – Ville Salo
    Dec 16, 2019 at 9:08
  • $\begingroup$ This answer is acceptable, but maybe I'll give it a day or two in case someone's doing all polycyclics as we speak :) $\endgroup$
    – Ville Salo
    Dec 16, 2019 at 9:13
  • $\begingroup$ The BCH formula endows every nilpotent real Lie algebra with a group structure, given by $x\ast y=x+y+\frac12[x,y]+\dots$. There are few terms of degree $\le 4$, and suddenly many more in degree 5. I've checked the equality $x\ast \Big(\frac12((-x)\ast y)\Big)=\frac12(x+y)$ until degree 4. $\endgroup$
    – YCor
    Dec 16, 2019 at 9:15
  • $\begingroup$ Ok, I see! Surely they must be the same then... maybe that'd be worth another MO question. (To clarify, I understand that it won't help with convex sets because rank 2 is used to get that the BCH formula is affine.) $\endgroup$
    – Ville Salo
    Dec 16, 2019 at 9:25
  • $\begingroup$ Actually I'd check degree 5 before asking :) It should be straightforward with a software computing in the free Lie algebra. $\endgroup$
    – YCor
    Dec 16, 2019 at 9:42

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