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Let $P$ be a convex polytope (or more generally convex body, I suppose) in $V=\mathbb{R}^n$. For each $v\in \mathbb{P}V$, we define an involution $\tau_v\colon P\to P$ by setting $\tau_v(p)$ to be the reflection of $p$ within the line segment in $P$ passing through $p$ in direction $v$.

Let $p$ be a point in the interior of $P$. For each $k \geq 0$, let $\mu_k$ denote the probability distribution of $\tau_{v_k} \tau_{v_{k-1}} \cdots \tau_{v_1} p$ on $P$, where $v_1,\ldots,v_k \in \mathbb{P}V$ are chosen independently according to the standard distribution on real projective space. Does this sequence of distributions $\mu_0,\mu_1,\ldots$ converge to some measure in the limit (I assume so…)? What measure does it converge to (it is not uniform, is it...)? In general, is there somewhere these kind of random walks are studied?

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  • $\begingroup$ Alternatively we could also draw the $v_i$ from some finite set: e.g., the coordinate axes. $\endgroup$
    – Sam Hopkins
    Sep 19, 2021 at 18:31
  • $\begingroup$ Wait, is not the uniform measure invariant by Cavalieri principle? $\endgroup$
    – Fedor Petrov
    Sep 19, 2021 at 20:16
  • $\begingroup$ Not sure if I understand the question correctly, but it seems to me that the uniform measure on $P$ is invariant with respect to $\tau_v$ for every direction $v$ (because the uniform measure on an interval is invariant under reflections), and it is fairly straightforward to see that only the uniform measure has this property (if $n \geqslant 2$), and that each point of $P$ can be reached from any other point in a finite number of steps (again if $n \geqslant 2$). So I would conjecture that the sequence in question is in fact equidistributed in $P$. $\endgroup$
    – Mateusz Kwaśnicki
    Sep 19, 2021 at 20:18
  • $\begingroup$ @MateuszKwaśnicki : A disk and a straight-line segment are exceptions to the sequence in question being equidistributed in $P$. $\endgroup$
    – Iosif Pinelis
    Sep 19, 2021 at 20:22
  • $\begingroup$ @IosifPinelis: Ah, right, of course! There are in fact more exceptions, like the square (or anything else symmetric enough) with $p$ at its center of symmetry. Thanks for pointing this out. $\endgroup$
    – Mateusz Kwaśnicki
    Sep 19, 2021 at 20:41

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This is not a complete answer, but hopefully someone can complete it.

The answer is incomplete because it relies on the following assumption:

Suppose there is a compact topological group $G$ of maps from $P$ into $P$ such that (i) $G$ contains all the values of $\tau_{v_1}$ and (ii) the support of the distribution of $\tau_{v_1}$ (over $G$) is not contained in any coset of any closed proper normal subgroup of $G$.

Let $\pi$ denote the distribution of each of $\tau_{v_i}$'s. Then the distribution of $\tau_{v_k}\cdots\tau_{v_1}$ is the convolution $\pi^{*k}$.

So, by Theorem 3.2.4, the distribution of $\tau_{v_k}\cdots\tau_{v_1}$ will converge to the normalized Haar measure on $G$.

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  • $\begingroup$ Condition (ii) is not really a problem (it's just some kind of aperiodicity). It is compactness that is really the issue. $\endgroup$
    – R W
    Sep 19, 2021 at 22:21
  • $\begingroup$ @RW : I too think compactness is the main issue. $\endgroup$
    – Iosif Pinelis
    Sep 19, 2021 at 23:10

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