# When is the minimal Martin boundary closed?

Let $\Gamma$ be a finitely generated group and $\mu$ a symmetric measure of finite support on $\Gamma$. Let $\partial_{M}\Gamma$ be the Martin boundary of $(\Gamma,\mu)$ and let $\partial^{min}_{M}\Gamma$ be the subset consisting of minimal harmonic functions.

My question is: what are some conditions on the group or measure that guarantee $\partial^{min}_{M}\Gamma$ is a closed subset of $\partial_{M}\Gamma$?

There are many examples (e.g. hyperbolic groups) where all points in the Martin boundary are minimal, but otherwise examples are hard to come by.

Is there an explicit example where $\partial^{min}_{M}\Gamma$ is not closed in $\partial_{M}\Gamma$?

• I don't think the definition of minimal here is correct – tmh Jun 16 '17 at 0:45
• Yes--thank you there was a typo in the first sentence--it should have said "minimal"instead of positive. – Yellow Pig Jun 16 '17 at 5:35
• You wrote two questions. The first doesn't seem to have a complete answer, and a partial answer you gave already in the next sentence (which I find somewhat self-contradictory). I answered the second question below. – Uri Bader Jun 16 '17 at 7:03

Here is how to build examples where $\partial^{\text{min}}_M\Gamma$ is a proper subset of its closure.

First note the following two facts:

1. Every positive harmonic function on $\Gamma$ is represented uniquely as a measure on $\partial^{\text{min}}_M\Gamma$.

2. Calling the unique measure on $\partial^{\text{min}}_M\Gamma$ representing the constant function $1$ on $\Gamma$ $\nu$, the measured space $(\partial^{\text{min}}_M\Gamma,\nu)$ is a model for the Poisson boundary of $(\Gamma,\mu)$.

Now take an amenable $\Gamma$ with a measure $\mu$ such that there exist non-constant bounded $\mu$-harmonic functions on $\Gamma$ (an explicit example would be the lamp-lighter group endowed with the uniform measure on the standard generating set). You will get two extra properties:

1. There exists a $\Gamma$-invariant measure on $\overline{\partial^{\text{min}}_M\Gamma}$, $\nu'$.

2. $\nu$ is not invariant.

Observe that $\nu'$ represents an invariant function on $\Gamma$ taking the value $1$ at the identity, thus the constant function $1$. Assuming $\partial^{\text{min}}_M\Gamma=\overline{\partial^{\text{min}}_M\Gamma}$ we will get $\nu'=\nu$ by the uniqueness of the presentation, 1. But by 4, $\nu\neq\nu'$. Therefor $\partial^{\text{min}}_M\Gamma\neq\overline{\partial^{\text{min}}_M\Gamma}$.

• How do you know that minimal harmonic functions are dense in the Martin boundary? – R W Jul 6 '17 at 23:42
• @RW, It was just a sloppy presentation - I replaced in my mind the Martin boundary with the closure of the minimal harmonics, but forgot to tell this to the reader. I fixed it now. Thanks for the remark. – Uri Bader Jul 7 '17 at 7:46
• Btw, a good online survey: math.wustl.edu/~sawyer/hmhandouts/martbrwf.pdf – Uri Bader Jul 7 '17 at 7:47

As it has been proved in http://www.ams.org/mathscinet-getitem?mr=2191210, if the set of minimal harmonic functions is closed, then the group action on it is topologically amenable. Groups which admit topologically amenable actions on compact spaces are called amenable at infinity (see Amenability at infinity). Therefore, minimal harmonic functions are not closed in the Martin boundary for any group which is not amenable at infinity.

There are also examples where the minimal Martin boundary is a proper set and is closed when the random walk is not symmetric. I don't know if you can find any such example with a symmetric one.

In a slightly different context, Hueber and Müller showed that the Martin boundary for a continuous random walk on the Heisenberg group $H^3(\mathbb{R})$, the Martin boundary is a disc, while the minimal Martin boundary is a circle (see "asymptotics for some Green kernels onthe Heisenberg group and the Martin boundary”. In:Mathematische Annalen283 (1989), pp. 97–119.)

Actually, if $\Gamma$ is a nilpotent group with some nice condition of generation (for instance finitely generated), positive harmonic functions are constant on the cosets of the commutator subgroup. That was proved by Margulis (see "positive harmonic functions on nilpotent groups”. In:Dok-lady Akademii nauk SSSR166 (1966). In Russian. English translation SovietMathematics Doklady, 7, 1966, 241-244, pp. 1054–1057.)

Thus, a positive harmonic function $f$ defines a positive harmonic function on the abelian group $\Gamma/[\Gamma,\Gamma]$. You can easily show that minimal harmonic functions on $\Gamma$ define minimal harmonic functions on the abelianized group, so that the minimal Martin boundary reduce to the minimal Martin boundary of an abelian group.

Now, if the random walk is non-centered on an abelian group, its minimal boundary is a sphere, so it is compact. This gives other examples taking nilpotent groups with a non-minimal Martin boundary (as the Heisenberg group, according to Hueber and Müller). Though, to my knowledge, few is known in general for nilpotent groups.

• The OP has explicitly asked about symmetric measures – R W Jul 7 '17 at 0:38