Given a countable group G and a faithful G-flow X, we write
$\mathrm {Aut}(X, G)$
for the group of homeomorphisms of X which commute with the G-action. When G is abelian,
$\mathrm {Aut}(X, G)$
contains a natural copy of G resulting from the G-action, but in general this need not be the case. Much is unknown about how the properties of X restrict the complexity of
$\mathrm {Aut}(X, G)$
; for instance, Cyr and Kra [Reference Cyr and Kra1] conjecture that when
$G = \mathbb {Z}$
and
$X\subseteq 2^{\mathbb {Z}}$
is a minimal,
$0$
-entropy subshift, then
$\mathrm {Aut}(X, \mathbb {Z})$
must be amenable. In fact, no counterexample is known even when restricting to any two of the three properties ‘minimal’, ‘
$0$
-entropy’ or ‘subshift’. In an effort to shed light on this question, Frisch and Tamuz [Reference Frisch and Tamuz3] define a probability measure
$\mu $
on X to be characteristic if it is
$\mathrm {Aut}(X, G)$
-invariant. They show that
$0$
-entropy subshifts always admit characteristic measures. More recently, Cyr and Kra [Reference Cyr and Kra2] provide several examples of flows which admit characteristic measures for nontrivial reasons, even in cases where
$\mathrm {Aut}(X, G)$
is nonamenable. Frisch and Tamuz asked (Question 1.5, [Reference Frisch and Tamuz3]) whether there exists, for any countable group G, some minimal G-flow without a characteristic measure. We give a strong affirmative answer.
Theorem 0.1. For any countably infinite group G, there is a free minimal G-flow X so that X does not admit a characteristic measure. More precisely, there is a free
$(G\times F_2)$
-flow X which is minimal as a G-flow and with no
$F_2$
-invariant measure.
We remark that the X we construct will not in general be a subshift.
Over the course of proving Theorem 0.1, there are two main difficulties to overcome. The first difficulty is a collection of dynamical problems we refer to as minimal subdynamics. The general template of these questions is as follows.
Question 0.2. Given a countably infinite group
$\Gamma $
and a collection
$\{\Delta _i: i\in I\}$
of infinite subgroups of
$\Gamma $
, when is there a faithful (or essentially free, or free) minimal
$\Gamma $
-flow for which the action of each
$\Delta _i$
is also minimal? Is there a natural space of actions in which such flows are generic?
In [Reference Zucker8], the author showed that this was possible in the case
$\Gamma = G\times H$
and
$\Delta = G$
for any countably infinite groups G and H. We manage to strengthen this result considerably.
Theorem 0.3. For any countably infinite group
$\Gamma $
and any collection
$\{\Delta _n: n\in \mathbb {N}\}$
of infinite normal subgroups of
$\Gamma $
, there is a free
$\Gamma $
-flow which is minimal as a
$\Delta _n$
-flow for every
$n\in \mathbb {N}$
.
In fact, what we show when proving Theorem 0.3 is considerably stronger. Recall that given a countably infinite group
$\Gamma $
, a subshift
$X\subseteq 2^\Gamma $
is strongly irreducible if there is some finite symmetric
$D\subseteq \Gamma $
so that whenever
$S_0, S_1\subseteq \Gamma $
satisfy
$DS_0\cap S_1 = \emptyset $
(i.e.,
$S_0$
and
$S_1$
are D-apart), then for any
$x_0, x_1\in X$
, there is
$y\in X$
with
$y|_{S_i} = x_i|_{S_i}$
for each
$i< 2$
. Write
$\mathcal {S}$
for the set of strongly irreducible subshifts, and write
$\overline {\mathcal {S}}$
for its Vietoris closure. Frisch, Tamuz and Vahidi-Ferdowsi [Reference Frisch, Tamuz and Ferdowsi5] show that in
$\overline {\mathcal {S}}$
, the minimal subshifts form a dense
$G_\delta $
subset. In our proof of Theorem 0.3, we show that the shifts in
$\overline {\mathcal {S}}$
which are
$\Delta _n$
-minimal for each
$n\in \mathbb {N}$
also form a dense
$G_\delta $
subset.
This brings us to the second main difficulty in the proof of Theorem 0.1. Using this stronger form of Theorem 0.3, one could easily prove Theorem 0.1 by finding a strongly irreducible
$F_2$
-subshift which does not admit an invariant measure. This would imply the existence of a strongly irreducible
$(G\times F_2)$
-subshift without an
$F_2$
-invariant measure. As not admitting an
$F_2$
-invariant measure is a Vietoris-open condition, the genericity of G-minimal subshifts would then be enough to obtain the desired result. Unfortunately, whether such a strongly irreducible subshift can exist (for any nonamenable group) is an open question. To overcome this, we introduce a flexible weakening of the notion of a strongly irreducible shift.
The paper is organized as follows. Section
$1$
is a very brief background section on subsets of groups, subshifts and strong irreducibility. Section
$2$
introduces the notion of a UFO, a useful combinatorial gadget for constructing shifts where subgroups act minimally; Theorem 0.3 answers Question 3.6 from [Reference Zucker8]. Section
$3$
introduces the notion of
$\mathcal {B}$
-irreduciblity for any group H, where
$\mathcal {B}\subseteq \mathcal {P}_f(H)$
is a right-invariant collection of finite subsets of H. When
$H = F_2$
, we will be interested in the case when
$\mathcal {B}$
is the collection of finite subsets of
$F_2$
which are connected in the standard left Cayley graph. Section 4 gives the proof of Theorem 0.1.
1. Background
Let
$\Gamma $
be a countably infinite group. Given
$U, S\subseteq \Gamma $
with U finite, then we call S a (one-sided) U-spaced set if for every
$g\neq h\in S$
we have
$h\not \in Ug$
, and we call S a U-syndetic set if
$US = \Gamma $
. A maximal U-spaced set is simply a U-spaced set which is maximal under inclusion. We remark that if S is a maximal U-spaced set, then S is
$(U\cup U^{-1})$
-syndetic. We say that sets
$S, T\subseteq \Gamma $
are (one-sided) U-apart if
$US\cap T = \emptyset $
and
$S\cap UT = \emptyset $
. Notice that much of this discussion simplifies when U is symmetric, so we will often assume this. Also, notice that the properties of being U-spaced, maximal U-spaced, U-syndetic and U-apart are all right invariant.
If A is a finite set or alphabet, then
$\Gamma $
acts on
$A^\Gamma $
by right shift, where given
$x\in A^\Gamma $
and
$g, h\in \Gamma $
, we have
$(g{\cdot }x)(h) = x(hg)$
. A subshift of
$A^\Gamma $
is a nonempty, closed,
$\Gamma $
-invariant subset. Let
$\mathrm {Sub}(A^\Gamma )$
denote the space of subshifts of
$A^\Gamma $
endowed with the Vietoris topology. This topology can be described as follows. Given
$X\subseteq A^\Gamma $
and a finite
$U\subseteq \Gamma $
, the set of U-patterns of X is the set
$P_U(X) = \{x|_U: x\in X\}\subseteq A^U$
. Then the typical basic open neighborhood of
$X\in \mathrm {Sub}(A^\Gamma )$
is the set
$N_U(X):= \{Y\in \mathrm {Sub}(A^\Gamma ): P_U(Y) = P_U(X)\}$
, where U ranges over finite subsets of
$\Gamma $
.
A subshift
$X\subseteq A^\Gamma $
is U-irreducible if for any
$x_0, x_1\in X$
and any
$S_0, S_1\subseteq \Gamma $
which are U-apart, there is
$y\in X$
with
$y|_{S_i} = x_i|_{S_i}$
for each
$i< 2$
. If X is U-irreducible and
$V\supseteq U$
is finite, then X is also V-irreducible. We call X strongly irreducible if there is some finite
$U\subseteq \Gamma $
with
$X\ U$
-irreducible. By enlarging U if needed, we can always assume U is symmetric. Let
$\mathcal {S}(A^\Gamma )\subseteq \mathrm {Sub}(A^\Gamma )$
denote the set of strongly irreducible subshifts of
$A^\Gamma $
, and let
$\overline {\mathcal {S}}(A^\Gamma )$
denote the closure of this set in the Vietoris topology.
More generally, if
$2^{\mathbb {N}}$
denotes Cantor space, then
$\Gamma $
acts on
$(2^{\mathbb {N}})^\Gamma $
by right shift exactly as above. If
$k< \omega $
, we let
$\pi _k\colon 2^{\mathbb {N}}\to 2^k$
denote the restriction to the first k entries. This induces a factor map
$\tilde {\pi }_k\colon (2^{\mathbb {N}})^\Gamma \to (2^k)^\Gamma $
given by
$\tilde {\pi }_k(x)(g) = \pi _k(x(g))$
; we also obtain a map
$\overline {\pi }_k\colon \mathrm {Sub}((2^{\mathbb {N}})^\Gamma )\to \mathrm {Sub}((2^k)^\Gamma )$
(where
$2^k$
is viewed as a finite alphabet) given by
$\overline {\pi }_k(X) = \tilde {\pi }_k[X]$
. The Vietoris topology on
$\mathrm {Sub}((2^{\mathbb {N}})^\Gamma )$
is the coarsest topology making every such
$\overline {\pi }_k$
continuous. We call a subflow
$X\subseteq (2^{\mathbb {N}})^\Gamma $
strongly irreducible if for every
$k< \omega $
, the subshift
$\overline {\pi }_k(X)\subseteq (2^k)^\Gamma $
is strongly irreducible in the ordinary sense. We let
$\mathcal {S}((2^{\mathbb {N}})^\Gamma )\subseteq \mathrm {Sub}((2^{\mathbb {N}})^\Gamma )$
denote the set of strongly irreducible subflows of
$(2^{\mathbb {N}})^\Gamma $
, and we let
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
denote its Vietoris closure.
The idea of considering the closure of the strongly irreducible shifts has it roots in [Reference Frisch and Tamuz4]. This is made more explicit in [Reference Frisch, Tamuz and Ferdowsi5], where it is shown that in
$\overline {\mathcal {S}}(A^\Gamma )$
, the minimal subflows form a dense
$G_\delta $
subset. More or less the same argument shows that the same holds in
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
(see [Reference Glasner, Tsankov, Weiss and Zucker6]). Recall that a
$\Gamma $
-flow X is free if for every
$g\in \Gamma \setminus \{1_\Gamma \}$
and every
$x\in X$
, we have
$gx\neq x$
. The main reason for considering a Cantor space alphabet is the following result, which need not be true for finite alphabets.
Proposition 1.1. In
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
, the free flows form a dense
$G_\delta $
subset.
Proof. Fixing
$g\in \Gamma $
, the set
$\{X\in \mathrm {Sub}((2^{\mathbb {N}})^\Gamma ): \forall x\in X\, (gx\neq x)\}$
is open; indeed, if
$X_n\to X$
is a convergent sequence in
$\mathrm {Sub}((2^{\mathbb {N}})^\Gamma )$
and
$x_n\in X_n$
is a point fixed by g, then passing to a subsequence, we may suppose
$x_n\to x\in X$
, and we have
$gx = x$
. Intersecting over all
$g\in \Gamma \setminus \{1_\Gamma \}$
, we see that freeness is a
$G_\delta $
condition.
Thus, it remains to show that freeness is dense in
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
. To that end, we fix
$g\in \Gamma \setminus \{1_\Gamma \}$
and show that the set of shifts in
$\mathcal {S}((2^{\mathbb {N}})^\Gamma )$
where g acts freely is dense. Fix
$X\in \mathcal {S}((2^{\mathbb {N}})^\Gamma )$
,
$k< \omega $
and a finite
$U\subseteq \Gamma $
; so a typical open set in
$\mathcal {S}((2^{\mathbb {N}})^\Gamma )$
has the form
$\{X'\in \mathcal {S}((2^{\mathbb {N}})^\Gamma ): P_U(\overline {\pi }_k(X')) = P_U(\overline {\pi }_k(X))\}$
. We want to produce
$Y\in \mathrm {Sub}((2^{\mathbb {N}})^\Gamma )$
which is strongly irreducible, g-free and with
$P_U(\overline {\pi }_k(Y)) = P_U(\overline {\pi }_k(X))$
. In fact, we will produce such a Y with
$\overline {\pi }_k(Y) = \overline {\pi }_k(X)$
.
Let
$D\subseteq \Gamma $
be a finite symmetric set containing g and
$1_\Gamma $
. Setting
$m = |D|$
, consider the subshift
$\mathrm {Color}(D, m)\subseteq m^\Gamma $
defined by
$$ \begin{align*}\mathrm{Color}(D, m) := \{x\in m^\Gamma: \forall\, i < m\, [x^{-1}(\{i\})\text{ is}\ D\text{-}\text{spaced}]\}.\end{align*} $$
A greedy coloring argument shows that
$\mathrm {Color}(D, m)$
is nonempty and D-irreducible. Moreover, g acts freely on
$\mathrm {Color}(D, m)$
. Inject m into
$2^{\{k,\ldots ,\ell -1\}}$
for some
$\ell> k$
and identify
$\mathrm {Color}(D, m)$
as a subflow of
$(2^{\{k,\ldots ,\ell -1\}})^\Gamma $
. Then
$Y:= \overline {\pi }_k(X)\times \mathrm {Color}(D, m)\subseteq (2^\ell )^\Gamma \subseteq (2^{\mathbb {N}})^\Gamma $
, where the last inclusion can be formed by adding strings of zeros to the end. Then Y is strongly irreducible, g-free and
$\overline {\pi }_k(Y) = \overline {\pi }_k(X)$
.
2. UFOs and minimal subdynamics
Much of the construction will require us to reason about the product group
$G\times F_2$
. So for the time being, fix countably infinite groups
$\Delta \subseteq \Gamma $
. For our purposes,
$\Gamma $
will be
$G\times F_2$
, and
$\Delta $
will be G, where we identify G with a subgroup of
$G\times F_2$
in the obvious way. However, for this subsection, we will reason more generally.
Definition 2.1. Let
$\Delta \subseteq \Gamma $
be countably infinite groups. A finite subset
$U\subseteq \Gamma $
is called a
$(\Gamma , \Delta )$
-UFO if for any maximal U-spaced set
$S\subseteq \Gamma $
, we have that S meets every right coset of
$\Delta $
in
$\Gamma $
.
We say that the inclusion of groups
$\Delta \subseteq \Gamma $
admits UFOs if for every finite
$U\subseteq \Gamma $
, there is a finite
$V\subseteq \Gamma $
with
$V\supseteq U$
which is a
$(\Gamma , \Delta )$
-UFO.
As a word of caution, we note that the property of being a
$(\Gamma , \Delta )$
-UFO is not upwards closed.
The terminology comes from considering the case of a product group, that is,
$\Gamma = \mathbb {Z}\times \mathbb {Z}$
and
$\Delta = \mathbb {Z}\times \{0\}$
. Figure 1 depicts a typical UFO subset of
$\mathbb {Z}\times \mathbb {Z}$
.
Figure 1 Sighting in Roswell; a
$(\mathbb {Z}\times \mathbb {Z}, \mathbb {Z}\times \{0\})$
-UFO subset of
$\mathbb {Z}\times \mathbb {Z}$
.
Proposition 2.2. Let
$\Delta $
be a subgroup of
$\Gamma $
. If
$|\bigcap _{u \in U} u \Delta u^{-1}|$
is infinite for every finite set
$U \subseteq \Gamma $
, then
$\Delta \subseteq \Gamma $
admits UFOs. In particular, if
$\Delta $
contains an infinite subgroup that is normal in
$\Gamma $
, then
$\Delta \subseteq \Gamma $
admits UFOs.
Proof. We prove the contrapositive. So assume that
$\Delta \subseteq \Gamma $
does not admit UFOs. Let
$U \subseteq \Gamma $
be a finite symmetric set such that no finite
$V \subseteq \Gamma $
containing U is a
$(\Gamma , \Delta )$
-UFO. Let
$D \subseteq \Delta $
be finite, symmetric and contain the identity. It will suffice to show that
$C = \bigcap _{u \in U} u D u^{-1}$
satisfies
$|C| \leq |U|$
.
Set
$V = U \cup D^2$
. Since V is not a
$(\Gamma , \Delta )$
-UFO, there is a maximal V-spaced set
$S \subseteq \Gamma $
and
$g \in \Gamma $
with
$S \cap \Delta g = \varnothing $
. Since S is V-spaced and
$u^{-1} C^2 u \subseteq D^2 \subseteq V$
, the set
$C_u = (u S) \cap (C g)$
is
$C^2$
-spaced for every
$u \in U$
. Of course, any
$C^2$
-spaced subset of
$C g$
is empty or a singleton, so
$|C_u| \leq 1$
for each
$u \in U$
. On the other hand, since S is maximal we have
$V S = \Gamma $
, and since
$S \cap \Delta g = \varnothing $
we must have
$C g \subseteq U S$
. Therefore,
$|C| = |C g| = \sum _{u \in U} |C_u| \leq |U|$
.
In the spaces
$\overline {\mathcal {S}}(k^\Gamma )$
and
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
, the minimal flows form a dense
$G_\delta $
. However, when
$\Delta \subseteq \Gamma $
is a subgroup, we can ask about the properties of members of
$\overline {\mathcal {S}}(k^\Gamma )$
and
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
viewed as
$\Delta $
-flows.
Definition 2.3. Given a subshift
$X\subseteq k^\Gamma $
and a finite
$E\subseteq \Gamma $
, we say that X is
$(\Delta , E)$
-minimal if for every
$x\in X$
and every
$p\in P_E(X)$
, there is
$g\in \Delta $
with
$(gx)|_E = p$
. Given a subflow
$X\subseteq (2^{\mathbb {N}})^\Gamma $
and
$n\in \mathbb {N}$
, we say that X is
$(\Delta , E, n)$
-minimal if
$\overline {\pi }_n(X)\subseteq (2^n)^\Gamma $
is
$(\Delta , E)$
-minimal. When
$\Delta = \Gamma $
, we simply say that X is E-minimal or
$(E, n)$
-minimal.
The set of
$(\Delta , E)$
-minimal flows is open in
$\mathrm {Sub}(k^\Gamma )$
, and
$X\subseteq k^\Gamma $
is minimal as a
$\Delta $
-flow iff it is
$(\Delta , E)$
-minimal for every finite
$E\subseteq \Gamma $
. Similarly, the set of
$(\Delta , E, n)$
-minimal flows is open in
$\mathrm {Sub}((2^{\mathbb {N}})^\Gamma )$
, and
$X\subseteq (2^{\mathbb {N}})^\Gamma $
is minimal as a
$\Delta $
-flow iff it is
$(\Delta , E, n)$
minimal for every finite
$E\subseteq \Gamma $
and every
$n\in \mathbb {N}$
.
In the proof of Proposition 2.4, it will be helpful to extend conventions about the shift action to subsets of
$\Gamma $
. If
$U\subseteq \Gamma $
,
$g\in G$
and
$p\in k^U$
, we write
$g{\cdot }p\in k^{Ug^{-1}}$
for the function where given
$h\in Ug^{-1}$
, we have
$(g{\cdot }p)(h) = p(hg)$
.
Proposition 2.4. Suppose
$\Delta \subseteq \Gamma $
are countably infinite groups and that
$\Delta \subseteq \Gamma $
admits UFOs. Then
$\{X\in \overline {\mathcal {S}}(k^\Gamma ): X \text { is minimal as a}\ \Delta -\text {flow}\}$
is a dense
$G_\delta $
subset. Similarly,
$\{X\in \overline {\mathcal {S}}(2^{\mathbb {N}})^\Gamma : X\text { is minimal as a}\ \Delta -\text {flow}\}$
is a dense
$G_\delta $
subset.
Proof. We give the arguments for
$k^\Gamma $
, as those for
$(2^{\mathbb {N}})^\Gamma $
are very similar.
It suffices to show for a given finite
$E\subseteq \Gamma $
that the collection of
$(\Delta , E)$
-minimal flows is dense in
$\overline {\mathcal {S}}(k^\Gamma )$
. By enlarging E if needed, we can assume that E is symmetric.
Consider a nonempty open
$O\subseteq \overline {\mathcal {S}}(k^\Gamma )$
. By shrinking O and/or enlarging E if needed, we can assume that for some
$X\in \mathcal {S}(k^\Gamma )$
, we have
$O = N_E(X)\cap \overline {\mathcal {S}}(k^\Gamma )$
. We will build a
$(\Delta , E)$
-minimal shift Y with
$Y\in N_E(X)\cap \mathcal {S}(k^\Gamma )$
. Fix a finite symmetric
$D\subseteq \Gamma $
so that X is D-irreducible. Then fix a finite
$U\subseteq \Gamma $
which is large enough to contain an
$EDE$
-spaced set
$Q\subseteq U\cap \Delta $
of cardinality
$|P_E(X)|$
, and enlarging U if needed, choose such a Q with
$EQ\subseteq U$
. Fix a bijection
$Q\to P_E(X)$
by writing
$P_E(X) = \{p_g: g\in Q\}$
. Because X is D-irreducible, we can find
$\alpha \in P_U(X)$
so that
$(gq)|_E = p_g$
for every
$g\in Q$
. By Proposition 2.2, fix a finite
$V\subseteq \Gamma $
with
$V\supseteq UDU$
which is a
$(\Gamma , \Delta )$
-UFO. We now form the shift
$$ \begin{align*}Y = \{y\in X: \exists\text{ a max.}\ V\text{-}\text{spaced set}\ T\text{ so that }\forall g\in T\, (g{\cdot}y)|_U = \alpha\}.\end{align*} $$
Because
$V = UDU$
and X is D-irreducible, we have that
$Y\neq \emptyset $
. In particular, for any maximal V-spaced set
$T\subseteq \Gamma $
, we can find
$y\in Y$
so that
$(gy)|_U = \alpha $
for every
$g\in T$
. We also note that
$Y\in N_E(X)$
by our construction of
$\alpha $
.
To see that Y is
$(\Delta , E)$
-minimal, fix
$y\in Y$
and
$p\in P_E(Y)$
. Suppose this is witnessed by the maximal V-spaced set
$T\subseteq \Gamma $
. Because V is a
$(\Gamma , \Delta )$
-UFO, find
$h\in \Delta \cap T$
. So
$(hy)|_U = \alpha $
. Now, suppose
$g\in Q$
is such that
$p = p_g$
. We have
$(ghy)|_E = (g\cdot ((hy)|_U)|_E = p_g$
.
To see that
$Y\in \mathcal {S}(k^\Gamma )$
, we will show that Y is
$DUVUD$
-irreducible. Suppose
$y_0, y_1\in Y$
and
$S_0, S_1\subseteq \Gamma $
are
$DUVUD$
-apart. For each
$i< 2$
, fix
$T_i\subseteq \Gamma $
a maximal V-spaced set which witnesses that
$y_i$
is in Y. Set
$B_i = \{g\in T_i: DUg\cap S_i\neq \emptyset \}$
. Notice that
$B_i\subseteq UDS_i$
. It follows that
$B_0\cup B_1$
is V-spaced, so extend to a maximal V-spaced set B. It also follows that
$S_i\cup UB_i\subseteq U^2DS_i$
. Since
$V\supseteq UDU$
and by the definition of
$B_i$
, the collection of sets
$\{S_i\cup UB_i: i< 2\}\cup \{Ug: g\in B\setminus (B_0\cup B_1)\}$
is pairwise D-apart. By the D-irreducibility of X, we can find
$y\in X$
with
$y|_{S_i\cup UB_i} = y_i|_{S_i\cup UB_i}$
for each
$i< 2$
and with
$(gy)|_U = \alpha $
for each
$g\in B\setminus (B_0\cup B_1)$
. Since
$B_i\subseteq T_i$
, we actually have
$(gy)|_U = \alpha $
for each
$g\in B$
. So
$y\in Y$
and
$y|_{S_i} = y_i|_{S_i}$
as desired.
Proof of Theorem 0.3.
By Proposition 2.4, the generic member of
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma )$
is minimal as a
$\Delta _n$
-flow for each
$n\in \mathbb {N}$
, and by Proposition 1.1, the generic member of
$\overline {\mathcal {S}}((2^{\mathbb {N}})^\Gamma $
is free.
In contrast to Theorem 0.1, the next example shows that Question 0.2 is nontrivial to answer in full generality.
Theorem 2.5. Let
$G=\sum _{\mathbb {N}} (\mathbb {Z}/2\mathbb {Z})$
, and let X be a G flow with infinite underlying space. Then there exists an infinite subgroup H such that X is not minimal as an H flow.
Proof. We may assume that X is a minimal G-flow, as otherwise we may take
$H = G$
. We construct a sequence
$X\supsetneq K_0\supseteq K_1\supseteq \cdots $
of proper, nonempty, closed subsets of X and a sequence of group elements
$\{g_n: n\in \mathbb {N}\}$
so that by setting
$K = \bigcap _{\mathbb {N}} K_n$
and
$H = \langle g_n: n\in \mathbb {N}\rangle $
, then K will be a minimal H-flow. Start by fixing a closed, proper subset
$K_0\subsetneq X$
with nonempty interior. Suppose
$K_n$
has been created and is
$\langle g_0,\ldots ,g_{n-1}\rangle $
-invariant. As X is a minimal G-flow, the set
$S_n:= \{g\in G: \mathrm {Int}(gK_n\cap K_n)\neq \emptyset \}$
is infinite. Pick any
$g_n\in S_n\setminus \{1_G\}$
, and set
$K_{n+1} = g_nK_n\cap K_n$
. As
$g_n^2 = 1_G$
, we see that
$K_{n+1}$
is
$g_n$
-invariant, and as G is abelian, we see that
$K_{n+1}$
is also
$g_i$
-invariant for each
$i< n$
. It follows that K will be H-invariant as desired.
Before moving on, we give a conditional proof of Theorem 0.1, which works as long as some nonamenable group admits a strongly irreducible shift without an invariant measure. It is the inspiration for our overall construction.
Proposition 2.6. Let G and H be countably infinite groups, and suppose that for some
$k< \omega $
and some strongly irreducible flow
$Y\subseteq k^H$
that Y does not admit an H-invariant measure. Then there is a minimal G-flow which does not admit a characteristic measure.
Proof. Viewing
$Z = k^G\times Y$
as a subshift of
$k^{G\times H}$
, then Z is strongly irreducible and does not admit an H-invariant probability measure. The property of not possessing an H-invariant measure is an open condition in
$\mathrm {Sub}(k^{G\times H})$
; indeed, if
$X_n\to X$
is a convergent sequence in
$\mathrm {Sub}(k^{G\times H})$
and
$\mu _n$
is an H-invariant probability measure supported on
$X_n$
, then by passing to a subsequence, we may suppose that the
$\mu _n$
weak
$^*$
-converge to some H-invariant probability measure
$\mu $
supported on X. By Proposition 2.4, we can therefore find
$X\subseteq k^{G\times H}$
which is minimal as a G-flow and which does not admit an H-invariant measure. As H acts by G-flow automorphisms on X, we see that X does not admit a characteristic measure.
Unfortunately, the question of if there exists any countable group H and a strongly irreducible H-subshift Y with no H-invariant measure is an open problem. Therefore, our construction proceeds by considering the free group
$F_2$
and defining a suitable weakening of strongly irreducible subshift which is strong enough for G-minimality to be generic in
$(G\times F_2)$
-subshifts but weak enough for
$(G\times F_2)$
-subshifts without
$F_2$
-invariant measures to exist.
3. Variants of strong irreducibility
In this section, we investigate a weakening of strong irreducibility that one can define given any right-invariant collection
$\mathcal {B}$
of finite subsets of a given countable group. For our overall construction, we will consider
$F_2$
and
$G\times F_2$
, but we give the definitions for any countably infinite group
$\Gamma $
. Write
$\mathcal {P}_f(\Gamma )$
for the collection of finite subsets of
$\Gamma $
.
Definition 3.1. Fix a right-invariant subset
$\mathcal {B}\subseteq \mathcal {P}_f(\Gamma )$
. Given
$k\in \mathbb {N}$
, we say that a subshift
$X\subseteq k^\Gamma $
is
$\mathcal {B}$
-irreducible if there is a finite
$D\subseteq \Gamma $
so that for any
$m< \omega $
, any
$B_0,\ldots , B_{m-1}\in \mathcal {B}$
, and any
$x_0,\ldots ,x_{m-1}\in X$
, if the sets
$\{B_0,\ldots , B_{m-1}\}$
are pairwise D-apart, then there is
$y\in X$
with
$y|_{B_i} = x_i|_{B_i}$
for each
$i< m$
. We call D the witness to
$\mathcal {B}$
-irreducibility. If we have D in mind, we can say that X is
$\mathcal {B}$
-D-irreducible.
We say that a subflow
$X\subseteq (2^{\mathbb {N}})^\Gamma $
is
$\mathcal {B}$
-irreducible if for each
$k\in \mathbb {N}$
, the subshift
$\overline {\pi }_k(X)\subseteq (2^k)^\Gamma $
is
$\mathcal {B}$
-irreducible.
We write
$\mathcal {S}_{\mathcal {B}}(k^\Gamma )$
or
$\mathcal {S}_{\mathcal {B}}((2^{\mathbb {N}})^\Gamma )$
for the set of
$\mathcal {B}$
-irreducible subflows of
$k^\Gamma $
or
$(2^{\mathbb {N}})^\Gamma $
, respectively, and we write
$\overline {\mathcal {S}}_{\mathcal {B}}(k^\Gamma )$
or
$\overline {\mathcal {S}}_{\mathcal {B}}((2^{\mathbb {N}})^\Gamma )$
for the Vietoris closures.
Remark.
-
1. If
$\mathcal {B}$
is closed under unions, it is enough to consider
$m = 2$
. However, this will often not be the case. -
2. By compactness, if
$X\subseteq k^\Gamma $
is
$\mathcal {B}$
-D-irreducible,
$\{B_n: n< \omega \}\subseteq \mathcal {B}$
is pairwise D-apart, and
$\{x_n: n< \omega \}\subseteq X$
, then there is
$y\in X$
with
$y|_{B_i} = x_i|_{B_i}$
. -
3. If
$\mathcal {B}\subseteq \mathcal {B}'$
, then
$\mathcal {S}_{\mathcal {B}'}(k^\Gamma )\subseteq \mathcal {S}_{\mathcal {B}}(k^\Gamma )$
and
$\mathcal {S}_{\mathcal {B}'}((2^{\mathbb {N}})^\Gamma )\subseteq \mathcal {S}_{\mathcal {B}}((2^{\mathbb {N}})^\Gamma )$
When
$\mathcal {B}$
is the collection of all finite subsets of H, then we recover the notion of a strongly irreducible shift. Again, we consider Cantor space alphabets to obtain freeness.
Proposition 3.2. For any right-invariant collection
$\mathcal {B}\subseteq \mathcal {P}_f(\Gamma )$
, the generic member of
$\overline {\mathcal {S}}_{\mathcal {B}}((2^{\mathbb {N}})^\Gamma )$
is free.
Proof. Analyzing the proof of Proposition 1.1, we see that the only properties that we need of the collections
$\mathcal {S}_{\mathcal {B}}(k^\Gamma )$
and
$\mathcal {S}_{\mathcal {B}}((2^{\mathbb {N}})^\Gamma )$
for the proof to generalize are that they are closed under products and contain the flows
$\mathrm {Color}(D, m)$
. If
$k, \ell \in \mathbb {N}$
an
$X\subseteq k^\Gamma $
and
$Y\subseteq \ell ^\Gamma $
are
$\mathcal {B}$
-D-irreducible and
$\mathcal {B}$
-E-irreducible for some finite
$D, E\subseteq \Gamma $
, then
$X\times Y\subseteq (k\times \ell )^\Gamma $
will be
$\mathcal {B}$
-
$(D\cup E)$
-irreducible. And as
$\mathrm {Color}(D, m)$
is strongly irreducible, it is
$\mathcal {B}$
-irreducible.
Now, we consider the group
$F_2$
. We consider the left Cayley graph of
$F_2$
with respect to the standard generating set
$A:= \{a, b, a^{-1}, b^{-1}\}$
. We let
$d\colon F_2\times F_2\to \omega $
denote the graph metric. Write
$D_n = \{s\in F_2: d(s, 1_{F_2}) \leq n\}$
.
Definition 3.3. Given n with
$1\leq n< \omega $
, we set
$$ \begin{align*}\mathcal{B}_n = \{D\in \mathcal{P}_f(F_2): \text{ connected components of}\ D\ \text{are pairwise}\ D_n\text{-}\text{apart}\}.\end{align*} $$
Write
$\mathcal {B}_\omega $
for the collection of finite, connected subsets of
$F_2$
.
Proposition 3.4. Suppose
$X\subseteq k^{F_2}$
is
$\mathcal {B}_\omega $
-irreducible. Then there is some
$n< \omega $
for which X is
$\mathcal {B}_n$
-irreducible.
Proof. Suppose X is
$\mathcal {B}_\omega $
-
$D_n$
-irreducible. We claim X is
$\mathcal {B}_n$
-
$D_n$
-irreducible. Suppose
$m< \omega $
,
$B_0,\ldots ,B_{m-1}\in \mathcal {B}_n$
are pairwise
$D_n$
-apart, and
$x_0,\ldots ,x_{m-1}\in X$
. For each
$i< m$
, we suppose
$B_i$
has
$n_i$
-many connected componenets, and we write
$\{C_{i,j}: j< n_i\}$
for these components. Then the collection of connected sets
$\bigcup _{i< m} \{C_{i,j}: j< n_i\}$
is pairwise
$D_n$
-apart. As X is
$\mathcal {B}_\omega $
-
$D_n$
-irreducible, we can find
$y\in X$
so that for each
$i< m$
and
$j< n_i$
, we have
$y|_{C_{i,j}} = x_i|_{C_{i,j}}$
. Hence,
$y|_{B_i} = x_i|_{B_i}$
, showing that X is
$\mathcal {B}_n$
-
$D_n$
-irreducible.
We now construct a
$\mathcal {B}_\omega $
-irreducible subshift with no
$F_2$
-invariant measure. We consider the alphabet
$A^2$
and write
$\pi _0, \pi _1\colon A^2\to A$
for the projections. We set
$$ \begin{align*} X_{pdox} = \{x\in (A^2)^{F_2}:\, &\forall g, h\in F_2 \, \forall i, j< 2 \\ &(i, g)\neq (j, h)\Rightarrow \pi_i(x(g))\cdot g\neq \pi_j(x(h))\cdot h\}. \end{align*} $$
More informally, the flow
$X_{pdox}$
is the space of ‘
$2$
-to-
$1$
paradoxical decompositions’ of
$F_2$
using A. We remark that here, our decomposition need not be a partition of
$F_2$
; we just ask for disjoint
$S_0, S_1\subseteq F_2$
such that for every
$g\in G$
and
$i< 2$
, we have
$Ag\cap S_i\neq \emptyset $
. This is in some sense the prototypical example of an
$F_2$
-shift with no
$F_2$
-invariant measure.
Lemma 3.5.
$X_{pdox}$
has no
$F_2$
-invariant measure.
Proof. For
$u \in A^2$
set
$Y_u = \{x \in X_{pdox} : x(1_G) = u\}$
. Notice that if
$y \in Y_u$
,
$i < 2$
and
$x = \pi _i(u) y$
, then
$x(\pi _i(u)^{-1}) = y(1_G) = u$
. Consequently, if
$u, v\in A^2$
,
$x \in \pi _i(u) Y_u \cap \pi _j(v) Y_v$
then, since
$x \in X_{pdox}$
and
$$ \begin{align*}\pi_i(x(\pi_i(u)^{-1})) \pi_i(u)^{-1} = 1_G = \pi_j(x(\pi_j(v)^{-1})) \pi_j(v)^{-1},\end{align*} $$
we must have that
$(i, \pi _i(u)) = (j, \pi _j(v))$
, and hence also
$$ \begin{align*}\pi_{1-i}(u) = \pi_{1-i}(x(\pi_i(u)^{-1})) = \pi_{1-j}(x(\pi_j(v)^{-1})) = \pi_{1-j}(v).\end{align*} $$
Therefore,
$\pi _i(u) Y_u \cap \pi _j(v) Y_v = \varnothing $
whenever
$(i,u) \neq (j,v)$
.
If
$\mu $
were an invariant Borel probability measure on
$X_{pdox}$
, then we would have
$$ \begin{align*}2 \mu(X_{pdox}) = 2 \sum_{u \in A^2} \mu(Y_u) = \sum_{i<2} \sum_{u \in A^2} \mu(\pi_i(u) Y_u) \leq \mu(X)\end{align*} $$
which is a contradiction.
When proving that
$X_{pdox}$
is
$\mathcal {B}_\omega $
-irreducible, note that
$D_1 = A\cup \{1_{F_2}\}$
.
Proposition 3.6.
$X_{pdox}$
is
$\mathcal {B}_\omega $
-
$D_4$
-irreducible.
Proof. The proof will use a
$2$
-to-
$1$
instance of Hall’s matching criterion [Reference Hall7] which we briefly describe. Fix a bipartite graph
$\mathbb {G} = (V, E)$
with partition
$V = V_0\sqcup V_1$
. Given
$S\subseteq V_0$
, write
$N_{\mathbb {G}}(S) = \{v\in V_1: \exists u\in S (u, v)\in E\}$
. Then the matching condition we need states that if for every finite
$S\subseteq V_0$
, we have
$|N_{\mathbb {G}}(S)|\geq 2S$
, then there is
$E'\subseteq E$
so that in the graph
$\mathbb {G}':= (V, E')$
,
$d_{\mathbb {G}'}(u) = 2$
for every
$u\in V_0$
.
Let
$B_0,\ldots ,B_{k-1}\in \mathcal {B}_\omega $
be pairwise
$D_4$
-apart. Let
$x_0,\ldots ,x_{k-1}\in X_{pdox}$
. To construct
$y\in X_{pdox}$
with
$y|_{B_i} = x_i|_{B_i}$
for each
$i< k$
, we need to verify a
$2$
-to-
$1$
Hall’s matching criterion on every finite subset of
$F_2\setminus \bigcup _{i< k} B_i$
. Call
$s\in F_2$
matched if for some
$i< k$
, some
$g\in B_i$
and some
$j< 2$
, we have
$s = \pi _j(x_i(g))\cdot g$
. So we need for every finite
$E\in \mathcal {P}_f(F_2\setminus \bigcup _{i<k} B_i)$
that
$AE$
contains at least
$2|E|$
-many unmatched elements. Towards a contradiction, let
$E\in \mathcal {P}_f(F_2\setminus \bigcup _{i<k} B_i)$
be a minimal failure of the Hall condition.
In the left Cayley graph of
$F_2$
, given a reduced word w in alphabet
$A = \{a, b, a^{-1}, b^{-1}\}$
, write
$N_w$
for the set of reduced words which end with w. Now, find
$t\in E$
(let us assume the leftmost character of t is a) so that all of
$E\cap N_{at}$
,
$E\cap N_{bt}$
and
$E\cap N_{b^{-1}t}$
are empty. If any two of
$at$
,
$bt$
and
$b^{-1}t$
is an unmatched point in
$AE$
, then
$E\setminus \{t\}$
is a smaller failure of Hall’s criterion. So there must be some
$i< k$
, some
$g\in B_i$
and some
$j< 2$
, we have
$\pi _j(x_i(g))\cdot g \in \{at, bt, b^{-1}t\}$
. Let us suppose
$\pi _j(x_i(g))\cdot g = at$
. Note that since
$g\not \in E$
, we must have
$g\in \{bat, a^2t, b^{-1}at\}$
. But then since
$B_i$
is connected, we have
$D_1B_i\cap \{bt, b^{-1}t\} = \emptyset $
, and since the other
$B_q$
are at least distance
$5$
from
$B_i$
, we have
$D_1B_q\cap \{bt, b^{-1}t\} = \emptyset $
for every
$q\in k\setminus \{i\}$
. In particular,
$bt$
and
$b^{-1}t$
are unmatched points in
$AE$
, a contradiction.
We remark that
$X_{pdox}$
is not
$D_n$
-irreducible for any
$n\in \mathbb {N}$
. See Figure 2.
Figure 2 A pair of outgoing edges, drawn in solid red, is chosen at each of
$v_{00}$
,
$v_{01}$
,
$v_{10}$
and
$v_{11}$
. Edges which must consequently be oriented in a particular direction are indicated with dashed red arrows. Most importantly,
$v_{\varnothing }$
is forced to direct an edge to
$u_\varnothing $
. By considering the generalization of this picture for any length of binary string, we see that
$X_{pdox}$
cannot be
$D_n$
-irreducible for any
$n\in \mathbb {N}$
.
4. The construction
Our goal for the rest of the paper is to use
$X_{pdox}$
to build a subflow of
$(2^{\mathbb {N}})^{G\times F_2}$
which is free, G-minimal and with no
$F_2$
-invariant measure. In what follows, given an
$F_2$
-coset
$\{g\}\times F_2$
, we endow this coset with the left Cayley graph for
$F_2$
using the generating set A exactly as above. We extend the definition of
$\mathcal {B}_n$
to refer to finite subsets of any given
$F_2$
-coset.
Definition 4.1. Given n with
$1\leq n\leq \omega $
, we set
$$ \begin{align*}\mathcal{B}_n^*= \{D\in \mathcal{P}_f(G\times F_2): \text{ for each}\ F_2-\text{coset}\ C, D\cap C\in \mathcal{B}_n\}.\end{align*} $$
Given
$y\in k^{G\times F_2}$
and
$g\in G$
, we define
$y_g\in k^{F_2}$
where given
$s\in F_2$
, we set
$y_g(s) = y(g, s)$
. If
$X\subseteq k^{F_2}$
is
$\mathcal {B}_n$
-irreducible, then the subshift
$X^G\subseteq k^{G\times F_2}$
is in
$\mathcal {S}_{\mathcal {B}_n^*}$
, where we view
$X^G$
as the set
$\{y\in k^{G\times F_2}: \forall g\in G\, (y_g\in X)\}$
. In particular,
$(X_{pdox})^G$
is
$\mathcal {B}^*_4$
-irreducible. By encoding
$(X_{pdox})^G$
as a subshift of
$(2^m)^{G\times F_2}$
for some
$m\in \mathbb {N}$
and considering
$\tilde {\pi }_m^{-1}((X_{pdox})^G)\subseteq (2^{\mathbb {N}})^{G\times F_2}$
, we see that there is a
$\mathcal {B}_4^*$
-irreducible subflow of
$(2^{\mathbb {N}})^{G\times F_2}$
for which the
$F_2$
-action doesn’t fix a measure. It follows that such subflows constitute a nonempty open subset of
$\Phi := \overline {\bigcup _n \mathcal {S}_{\mathcal {B}_n^*}((2^{\mathbb {N}})^{G\times F_2})}$
. Combining the next result with Proposition 3.2, we will complete the proof of Theorem 0.1.
Proposition 4.2. With
$\Phi $
as above, the G-minimal flows are dense
$G_\delta $
in
$\Phi $
.
Proof. We show the result for
$\Phi _k:= \overline {\bigcup _n \mathcal {S}_{\mathcal {B}_n^*}(k^{G\times F_2})}$
to simplify notation; the proof in full generality is almost identical.
We only need to show density. To that end, fix a finite symmetric
$E\subseteq G\times F_2$
which is connected in each
$F_2$
-coset. It is enough to show that the
$(G, E)$
-minimal subshifts are dense in
$\Phi _k$
. Fix some nonempty open
$O\subseteq \Phi _k$
. By enlarging E and/or shrinking O, we may assume that for some
$n< \omega $
and
$X\in \mathcal {S}_{\mathcal {B}_n^*}(k^{G\times F_2})$
that
$O = \{X'\in \Phi _k: P_E(X') = P_E(X)\}$
. We will build a
$(G, E)$
-minimal subshift
$Y\subseteq k^{G\times F_2}$
so that
$P_E(Y) = P_E(X)$
and so that for some
$N< \omega $
, we have
$Y\in \mathcal {S}_{\mathcal {B}_N^*}(k^{G\times F_2})$
.
Recall that
$D_n\subseteq F_2$
denotes the ball of radius n. Fix a finite, symmetric
$D\subseteq G\times F_2$
so that
$\{1_G\}\times D_{2n}\subseteq D$
and X is
$\mathcal {B}_n^*$
-D-irreducible. Find a finite symmetric
$U_0\subseteq G$
with
$1_G\subseteq U_0$
and
$r< \omega $
so that upon setting
$U = U_0\times D_r\subseteq G\times F_2$
, then U is large enough to contain an
$EDE$
-spaced set
$Q\subseteq G$
with
$EQ\subseteq U$
. As X is
$\mathcal {B}_n^*$
-D-irreducible, there is a pattern
$\alpha \in P_U(X)$
so that
$\{(g\alpha )|_E: g\in Q\} = P_E(X)$
.
Let
$V\supseteq UD^2U$
be a
$(G\times F_2, G)$
-UFO. We remark that for most of the remainder of the proof, it would be enough to have
$V\supseteq UDU$
; we only use the stronger assumption
$V\supseteq UD^2U$
in the proof of the final claim. Consider the following subshift:
$$ \begin{align*}Y = \{y\in X: \exists\text{ a max.}\ V\text{-spaced set}\ T\ \text{so that }\forall g\in T\, (gy)|_U = \alpha\}.\end{align*} $$
The proof that Y is nonempty and
$(G, E)$
-minimal is exactly the same as the analogous proof from Proposition 2.4. Note that by construction, we have
$P_E(Y) = P_E(X)$
.
We now show that
$Y\in \mathcal {S}_{\mathcal {B}_N^*}(k^{G\times F_2})$
for
$N = 4r+3n$
. Set
$W = DUVUD$
. We show that Y is
$\mathcal {B}_N^*$
-W-irreducible. Suppose
$m< \omega $
,
$y_0,\ldots ,y_{m-1}\in Y$
and
$S_0,\ldots ,S_{m-1}\in \mathcal {B}_N^*$
are pairwise W-apart. Suppose for each
$i< m$
that
$T_i\subseteq G\times F_2$
is a maximal V-spaced set which witness that
$y_i\in Y$
. Set
$B_i = \{g\in T_i: DUg\cap S_i\neq \emptyset \}$
. Then
$\bigcup _{i< m} B_i$
is V-spaced, so enlarge to a maximal V-spaced set
$B\subseteq G\times F_2$
.
For each
$i< m$
, we enlarge
$S_i\cup UB_i$
to
$J_i\in \mathcal {B}_n^*$
as follows. Suppose
$C\subseteq G\times F_2$
is an
$F_2$
-coset. Each set of the form
$C\cap Ug$
is connected. Since
$S_i\in \mathcal {B}_N^*$
, it follows that given
$g\in B_i$
, there is at most one connected component
$\Theta _{C, g}$
of
$S_i\cap C$
with
$Ug\cap \Theta _{C, g} = \emptyset $
, but
$Ug\cap D_n\Theta _{C, g}\neq \emptyset $
. We add the line segment in C connecting
$\Theta _{C, g}$
and
$Ug$
. Upon doing this for each
$g\in B_i$
and each
$F_2$
-coset C, this completes the construction of
$J_i$
. Observe that
$J_i\subseteq D_{n-1}S_i\cap UB_i$
.
Claim. Let C be an
$F_2$
-coset, and suppose
$Y_0$
is a connected component of
$S_i\cap C$
. Let Y be the connected component of
$J_i\cap C$
with
$Y_0\subseteq Y$
. Then
$Y\subseteq D_{2r+n}Y_0$
. In particular, if
$Y_0\neq Z_0$
are two connected components of
$S_i\cap C$
, then
$Y_0$
and
$Z_0$
do not belong to the same component of
$J_i\cap C$
.
Proof. Let
$L = \{x_j: j< \omega \}\subseteq C$
be a ray with
$x_0\in Y_0$
and
$x_j\not \in Y_0$
for any
$j\geq 1$
. Then
$\{j< \omega : x_j\in J_i\cap C\}$
is some finite initial segment of
$\omega $
. We want to argue that for some
$j\leq 2r+n+1$
, we have
$x_j\not \in J_i\cap C$
. First, we argue that if
$x_n\in J_i\cap C$
, then
$x_n\in UB_i$
. Otherwise, we must have
$x_n\in D_{n-1}S_i$
. But since
$x_n\not \in D_{n-1}Y_0$
, there must be another component
$Y_1$
of
$S_i\cap C$
with
$x_n\in D_nY_1$
. But this implies that
$Y_0$
and
$Y_1$
are not
$D_{2n-1}$
-apart, a contradiction since
$2n-1\leq 4r-3n = N$
.
Fix
$g\in B_i$
with
$x_n\in Ug$
. Let
$q< \omega $
be least with
$q> n$
and
$x_q \not \in U_g$
. We must have
$q\leq 2r+n+1$
. We claim that
$x_q\not \in J_i\cap C$
. Towards a contradiction, suppose
$x_q\in J_i\cap C$
. We cannot have
$x_q\in UB_i$
, so we must have
$x_q\in D_{n-1}S_i$
. But now there must be some component
$Y_1$
of
$S_i\cap C$
with
$x_q\in D_{n-1}Y_1$
. But then
$D_{2r+2n}Y_0\cap Y_1\neq \emptyset $
, a contradiction as
$Y_0$
and
$Y_1$
are
$D_N$
-apart. This concludes the proof that
$Y\subseteq D_{2r+n}Y_0$
.
Now, suppose
$Y_0\neq Z_0$
are two connected components of
$S_i\cap C$
. Then
$Y_0$
and
$Z_0$
are N-apart. In particular,
$Z_0\not \subseteq D_{2r+n}Y_0$
, so cannot belong to the same connected component of
$J_i\cap C$
as
$Y_0$
.
Claim.
$J_i\in \mathcal {B}_n^*$
.
Proof. Fix an
$F_2$
-coset C and two connected components
$Y\neq Z$
of
$J_i\cap C$
. By the previous claim, each of Y and Z can only contain at most one nonempty component of
$S_i\cap C$
. The claim will be proven after considering three cases.
-
1. First, suppose each of Y and Z contain a nonempty component of
$S_i\cap C$
, say
$Y_0\subseteq Y$
and
$Z_0\subseteq Z$
. Then since
$Y_0$
and
$Z_0$
are
$D_{4r+3n}$
-apart, the previous claim implies that Y and Z are
$D_n$
-apart. -
2. Now, suppose Y contains a nonempty component
$Y_0$
of
$S_i\cap C$
and that Z does not. Then for some
$g\in B_i$
, we have
$Z = Ug\cap C$
. Towards a contradiction, suppose
$D_nY\cap Ug \neq \emptyset $
. Let
$L = \{x_j: j\leq M\}$
be the line segment connecting Y and
$Ug$
with
$L\cap Y = \{x_0\}$
and
$L\cap Ug = \{x_M\}$
. We must have
$M\leq n$
. We cannot have
$x_0\in UB_i$
, so we must have
$x_0\in D_{n-1}S_i$
. This implies that
$x_0\in D_{n-1}Y_0$
. We cannot have
$x_0\in Y_0$
, as otherwise, we would have connected
$Y_0$
and
$Ug\cap C$
when constructing
$J_i$
. It follows that for some
$h\in B_i$
, we have that
$x_0$
is on the line segment
$L' = \{x_j': j\leq M'\}$
connecting
$Y_0$
and
$Uh\cap C$
, and we have
$M'\leq n$
. But this implies that
$Ug\cap D_{2n}Uh\neq \emptyset $
, a contradiction since
$V\supseteq UDU$
and
$D\supseteq D_{2n}$
. -
3. If neither Y nor Z contain a component of
$S_i\cap C$
, then there are
$g\neq h\in B_i$
with
$Y = Uh\cap C$
and
$Z = Ug\cap C$
. It follows that Y and Z are
$D_n$
-apart.
Claim. Suppose
$i\neq j< m$
. Then
$J_i$
and
$J_j$
are D-apart.
Proof. We have that
$J_i\subseteq D_{n-1}S_i\cup UB_i$
, and likewise for j. As
$UB_i\subseteq U^2DS_i$
and as
$D\supseteq D_{2n}$
, we have
$J_i\subseteq U^2DS_i$
, and likewise for j. As
$S_i$
and
$S_j$
are W-apart and as
$V\supseteq UDU$
, we see that
$J_i$
and
$J_j$
are D-apart.
Claim. Suppose
$g\in B\setminus \bigcup _{i< m} B_i$
. Then
$Ug$
and
$J_i$
are D-apart for any
$i< m$
.
Proof. As
$g\not \in B_i$
, we have
$U_g$
and
$S_i$
are D-apart. Also, for any
$h\in B$
with
$g\neq h$
, we have that
$Ug$
and
$Uh$
are D-apart. Now, suppose
$DUg\cap J_i\neq \emptyset $
. If
$x\in DUg\cap J_i$
, then on the coset
$C = F_2x$
, x must belong on the line between a component of
$S_i\cap C$
and
$Uh$
for some
$h\in B_i$
. Furthermore, we have
$x\in D_{n-1}Uh$
. But since
$D_{2n}\subseteq D$
, this contradicts that
$Ug$
and
$Uh$
are
$D^2$
-apart (using the full assumption
$V\supseteq UD^2U$
).
We can now finish the proof of Proposition 4.2. The collection
$\{J_i: i< m\}\cup \{Ug: g\in B\setminus (\bigcup _{i< m} B_i)\}$
is a pairwise D-apart collection of members of
$\mathcal {B}_n^*$
. As X is
$\mathcal {B}_n^*$
-D-irreducible, we can find
$y\in X$
with
$y|_{J_i} = y_i|_{J_i}$
for each
$i< m$
and with
$(gy)|_U = \alpha $
for each
$g\in B\setminus (\bigcup _{i< m} B_i)$
. As
$J_i\supseteq UB_i$
and since
$B_i\subseteq T_i$
, we actually have
$(gy)|_U = \alpha $
for each
$g\in B$
. As B is a maximal V-spaced set, it follows that
$y\in Y$
and
$y|_{S_i} = y_i|_{S_i}$
as desired.
Competing interest
The authors have no competing interest to declare.
Funding statement
J.F. was supported by NSF Grant DMS-2102838. B.S. was supported by NSF Grant DMS-1955090 and Sloan Grant FG-2021-16246. A.Z. was supported by NSF Grant DMS-2054302 and NSERC Grants RGPIN-2023-03269 and DGECR-2023-00412.
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