A note on subgroups of automorphism groups of full shifts
Abstract
We discuss the set of subgroups of the automorphism group of a full shift, and submonoids of its endomorphism monoid. We prove closure under direct products in the monoid case, and free products in the group case. We also show that the automorphism group of a full shift embeds in that of an uncountable sofic shift. Some undecidability results are obtained as corollaries.
1 Outline
Automorphism groups of full shifts were first studied in [2], and the transitive SFT case was later studied in more detail in [1]. More results on them, and in particular on the set of subgroups of these groups, were shown in [5]. In particular, interesting results were shown about the set of subgroups of . It is known for example that free groups, finite groups and finitely generated abelian groups can be embedded in this group [2, 1], and so can all locally finite residually finite countable groups [5]. Other examples are ‘graph groups’
where is arbitrary, fundamental groups of manifolds, and the free product of all finite groups [5]. Despite the long list of examples, to our knowledge there is no known characterization of the set of subgroups of .
In the absense of a characterization, we turn to closure properties. One closure property for this set of groups is proved explicitly in [5], namely closure under extensions by finite groups. Another one can be found by a direct application of the ideas of [1, 5], namely closure under direct sums, and we show this in Theorem 1. Our main result is that this set of groups is also closed under free products, Theorem 2, which answers a question I myself asked in [9].
We also show is contained in the automorphism group of every uncountable sofic shift. As corollaries of the results we obtain undecidability results for cellular automata on uncountable sofic shifts.
The ‘marker method’ (the use of unbordered words) usually plays an important role in the construction of cellular automata, and this note makes no exception. However, of even more importance in our constructions is the concept of ‘conveyor belts’. This technique is implicit in many constructions found in the literature, and variants of it are found for example in [1] and [5]. We make this idea more explicit.
While our main interest is in automorphism groups, we prove results for the whole endomorphism monoid when possible, since the results about monoids are strictly stronger, and the proofs are typically the same. In some results, however, the fact we consider groups instead of monoids is important. For example, we do not know whether the set of submonoids of the endomorphism monoid of a full shift is closed under free product.
We also show some undecidability results that follow as corollaries from combining existing undecidability results about periodicity from [4] with our results: in particular we obtain that given two reversible cellular automata on a full shift, it is undecidable whether they generate a free group.
2 Definitions
The letters , , and stand for finite alphabets. We write for the set of finite words over the alphabet , and the set of nonempty words. The concatenation of two words is written as or simply , and points or configurations can be written as infinite concatenations
where the ‘decimal point’ in need not be at the origin, but simply denotes some fixed position of the point.
If is any word, write for the reversed word .
Let be a finite alphabet. Then with the product topology is called a full shift. We define the shift by . A subshift is a topologically closed set satisfying .
If we write if . We also write and for words , with obvious meanings, and
A cellular automaton on a subshift is a continuous function that commutes with . Equivalently, it has a radius and a local rule such that for all .
We assume the reader is familiar with groups and monoids, but give the basic definitions to clarify our choice of boundary between properties and structure. A monoid is a countable set together with an associative multiplication operation such that there is an identity element satisfying for all . A group is a monoid where every element has an inverse satisfying . We think of the existence of as simply a property of the element , not as an operation. If and are monoids, a homomorphism is a map satisfying and
A submonoid of a monoid is a subset of that contains the identity element of and is closed under multiplication, so that obtains a monoid structure from with , and we write . More generally, we write if there is an embedding, or an injective homomorphism . A bijective homomorphism is called an isomorphism, and we write if and are isomorphic. A subgroup is a submonoid that is a group. If is a monoid, write for the class of isomorphism classes of its submonoids. We write for the class of isomorphism classes of its subgroups.
The endomorphism monoid of a subshift is the monoid of cellular automata on it under function composition, with , the identity map on . The automorphism group of is the subgroup of containing the invertible elements:
By compactness of , this is precisely the set of elements of that are bijective.
Our definitions are set up so that the following holds.^{1}^{1}1For this, it is (at least a priori) important that the identity element of a subgroup of is the identity map – thus, the identity element must be part of the structure of a monoid, rather than a property.
Lemma 1.
Let be a group such that . Then . In particular, if then .
The following lemma is useful for defining cellular automata.
Lemma 2.
Let be a subshift and let be a subset such that and . If is uniformly continuous and commutes with the shift, then there is a unique continuous map such that , and it is a cellular automaton.
Write for the set of permutations of , and for the set of all bijections . The free group with generators is written , and is the free group with a countably infinite set of generators.
The (external) direct product of two monoids is the monoid with operation . The free product of monoids and is defined up to isomorphism as follows: Let be a presentation, where each is a relation of the form where , and similarly let . Then
We also write for the direct product of copies of . There are obvious embeddings , and their direct limit is written as . The elements of the countable monoid are vectors of finite support with component values in . Write , and for the corresponding concepts for the free product.
3 Lemmas about submonoids
Lemma 3.
Let and be monoids. Then if and only if and .
In particular, to show that two groups have the same subgroups, we only need to show they contain each other as subgroups. We note that the set of subgroups of a group is not a complete invariant for group isomorphism: for example, free groups with different amounts of generators are nonisomorphic, but contain each other as subgroups.
In this note, we concentrate on automorphism groups of full shifts where is a finite alphabet of size at least . It is not known when two such groups and are isomorphic – in particular the case is open. Nevertheless, the set of subgroups is always the same:
Lemma 4.
Let be alphabets of size at least . Then
In particular, it follows that if . Lemma 4 follows directly from Lemma 7, which we prove in Section 7.
The interesting submonoids and groups are the infinite ones, as shown by the following result, which essentially already appears in [2]. The proof also illustrates the usefulness of Lemma 4.
Proposition 1.
Let . Then every finite monoid embeds in .
Proof.
Every finite monoid embeds in the transformation monoid of a finite set , and thus in . The result then follows from the previous lemma. ∎
We note that in general, if is closed under binary direct or free products, then it is closed under the corresponding countable products.
Lemma 5.
Let . Then the following are equivalent:

,

,

is closed under finite direct products.

,

is closed under countable direct products.
The analogous result is true for free products.
Proof.
We give the proof for direct products, the case of free products being similar. The equivalence of the first two conditions and the equivalence of the last two conditions are direct, as is the fact that the last two conditions imply the first two. We show that the third and fourth condition follow from the second.
For this, suppose that . Let and be the embeddings giving the embedding , that is, and for all . If this holds for two maps, we say the embedding conditions hold for them. For define inductively and . Then all maps are embeddings of into itself.
For , define . It is easy to show that the embedding conditions hold for and whenever . It follows that gives an embedding of into , and of into .
The claims for free products are proved analogously, but using a different embedding conditions, namely that the images of the embeddings satisfy no nontrivial relations.
∎
The importance of this lemma and Lemma 4 is that we do not need to worry about changing the alphabets of our full shifts when proving closure properties or about whether we use finite or countable products.
4 Conveyor belts
In our constructions, we will typically embed one automorphism group into another. In practise, this means that in the configurations of one full shift , we identify subsequences that code (parts of) configurations from another full shift . In these subsequences, we apply . To make this into a homomorphism from to , it is important to have natural behavior at the boundary between an area coding (part of) a configuration in , and an area containing something else. For this, we use conveyor belts.
Definition 1.
Let be any alphabet. A conveyor belt over is a word over the alphabet , that is, . Write for the set of conveyor belts over of length , that is . For a cellular automaton and , we define a function as follows: if , we decompose as for some , and we define
Applying a CA to a conveyor belt can alternatively be described as applying its local rule on the first track, applying its local rule in reverse on the second track, and gluing the tracks at the borders in the obvious way, as if the word were laid down on a conveyor belt: it is clear that this is essentially the same as applying the CA to a periodic point of even period. The following is then clear.
Lemma 6.
The map is a monoid homomorphism from to the monoid of functions on , and
is an embedding. Furthermore, the maps are uniformly continuous in the sense that there exists a radius such that for all and such that
we have .
The radius in the lemma can be taken to be just the usual radius of .
5 Direct products
Theorem 1.
If , is closed under direct products.
Proof.
By Lemma 4 and Lemma 5, it is enough to show that for two disjoint alphabets and , the monoid embeds into for some alphabet . We choose . To prove the claim, it is enough to give embeddings and such that for all and , and .
Let . Write for the set of points which are not left or right asymptotic to a point over or , in other words, points where all continuous runs over one of the subalphabets or are finite. If , then we can write
where and for all (and the decimal point need not be at the origin), and we define
(where the decimal point is in the same position as in ). It is clear that commutes with the shift on . From Lemma 6, it follows that is uniformly continuous on , and thus extends uniquely to a cellular automaton on . This gives a function from to . From Lemma 6, it easily follows that this is an embedding.
We symmetrically construct a homomorphism from to by rewriting the contiguous segments over . It is clear that these mappings commute, and since fixes all symbols from and all symbols from , only the identity map is in the image of both embeddings. ∎
Corollary 1.
If , is closed under direct products.
6 Free products
In this section, we prove our main theorem:
Theorem 2.
If , is closed under free products.
In the direct product case, we had two kinds of conveyor belts, and we applied the and maps completely independently on both of them. To embed a free product, the idea is to have these conveyor belts talk to each other, so that any alternating product of elements from the embedding and the embedding can transmit information arbitrarily far over an alternating sequence of conveyor belts and conveyor belts, as long as the lengths and contents of these belts are chosen suitably.
For this, we increase the size of our alphabet, and add a component that allows us to transmit any kind of modification of the left end of an conveyor belt to any kind of modification of the right end of a conveyor belt on its left, and vice versa. Let us proceed to the details.
Proof.
By Lemma 4, it is enough to show how to embed the free product of and into , where and are two disjoint alphabets with the same cardinality and is an alphabet of our choosing. We choose an arbitrary abelian group structure on both and . Let , and choose the alphabet
Similarly as in the proof of Theorem 1, for each , we will define a CA so that embeds into as a subgroup . An embedding of into is defined symmetrically by swapping the roles of and , and will be clear from the construction. To show they generate the free product of and in , we must show there are no nontrivial relations between elements of and . Because the roles of and are symmetric, this will follow from showing that for any
where , and for even and for odd , we have .
Let . Let be the set of points where contiguous runs over each of the subalphabets , and are finite. The set is shiftinvariant, so to define cellular automata on , it is enough to define shiftinvariant uniformly continuous maps on . Let thus , so that
where for all , , or , and and are over different alphabets. We write
where for all . If , we let . If , we apply to the conveyor belt as in the proof of Theorem 1, and let .
Finally, let us define for . Suppose First, for . If (that is, either or but ) or , we also let . If and , we let
where addition is performed with respect to the abelian group structure of . It is easy to see that this defines on , and since the map defined is uniformly continuous and shiftcommuting, we can extend it in a unique way to a CA . If , it has an inverse , and it is easy to see that . It follows that , because the only extension of the identity map on to a CA on is the identity map.
For , the map is defined symmetrically, modifying the rightmost symbol of a word over as a function of the leftmost symbol of a word over to the right of it when separated by the single symbol , by the same formula, but using the abelian structure of instead of that of , and the function instead of .
We now show that and indeed give the free product of and . Suppose thus that
where , and for even and for odd . We need to show . Suppose the minimal radius of is . We show that depends on at least the cell where . Clearly this will imply that .
For this, we will define two points
where , , if is even, if is odd, , for , and for all . The choices of and are arbitrary, and for all . We will choose the words and symbols carefully so that , while for all . Write for the interval where occurs in .
Let be even, and let and be words of length with for such that . We then have or , and if we also suppose or . If , is called a left dependence and otherwise a right dependence.
If is a left dependence, define the words by
where is arbitrary, and by the same formula using the word . The word is a conveyor belt version of where the center of is at the left end of and the leftmost coordinate is at the right end of .
The important property of and is that they agree apart from their rightmost coordinates, but after applying to the conveyor belt, the images differ in the leftmost coordinate, that is,
If is a right dependence, we produce words and with this property in a similar fashion.
Choose so that and
For odd coordinates , we choose words and similarly, switching the roles of and .
We let , . The words will be chosen inductively in such a way that
and
for .
To see this is possible, observe that information travels only from right to left over symbols when maps from are applied, and a map will move information over at most symbols in . Thus, when applying our maps to the points , we will automatically have for all . It follows that it is enough to show that the can be chosen so that for all : by the choice of the symbols , we will then automatically have .
But naturally we can choose such words by induction on , since each of the maps is reversible and information travels only to the left over symbols in . ∎
In [1], it is shown that embeds in for a full shift , and thus also the twogenerator free group does. More generally, it is known that every free product of finitely many finite groups embeds in for a full shift . In [5], this is attributed to R. C. Alperin. Combining the previous theorem and Proposition 1 gives a new proof of this result.
Corollary 2.
If , every free product of finite groups embeds in .
7 Embeddings between endomorphism groups
In [5], it was shown that automorphism groups of full shifts embed in those of transitive SFTs. With essentially the same proof, we show that endomorphism monoids of full shifts embed in those of uncountable sofic shifts, equivalently in ones with positive entropy. The proof relies on a number of basic properties of sofic shifts, which can be found in [7, 6].
Lemma 7.
Let be any uncountable sofic shift, and any full shift. Then .
Proof.
It is easy to see that for a sofic shift, uncountability is equivalent to having positive entropy.
The syntactic monoid of a subshift is the monoid whose elements are equivalence classes of words in under the equivalence . Sofic shifts are characterized as the subshifts with a finite syntactic monoid. If has positive entropy, then its minimal SFT cover also does. Then has a transitive component with positive entropy, and its image in the covering map is a positiveentropy transitive sofic subshift of .
Let be such that if then for some with – such exists because is transitive and because its syntactic monoid is finite. Let be such that contains at least words of length representing the same element of the syntactic monoid of – such exists because the syntactic monoid of is finite, and because has positive entropy. Let be a synchronizing word in of some length , that is, such that
Such a word can be found in any sofic shift [6].
In every aperiodic infinite word, one can find unbordered words of arbitrary length [8]. Take any configuration in which is aperiodic and where appears syndetically, to find an unbordered word of length at least containing the word . Then is synchronizing, since it contains a synchronizing subword.
By the assumptions on and , there is a set of words where for some and such that two words in can only overlap nontrivially by sharing the subword , and all words in represent the same element of the syntactic monoid of . Since all words in occur in and is synchronizing, the language is contained in the language of .
Now, let , and fix a bijection . Given a CA , the embedding is now constructed as in the previous sections: if we have a maximal finite subword of the form
(note that subwords of two such words cannot overlap by the assumptions), we apply to the corresponding conveyor belt and let . We rewrite the word by
As in the previous proofs, it is easy to check that this gives an embedding of into . ∎
The converse is not true in general for positiveentropy sofic shifts :
Example 1.
The group is residually finite [1]. We show that need not be residually finite for a sofic shift . For this, let
Then contains a copy of the group of all permutations of with finite support, by permuting the (orbits of) isolated points. This group is not residually finite, so is not residually finite for any subshift . In particular, by letting be a positiveentropy sofic shift we obtain the result.
Even assuming transitivity, we are not aware of a general technique of embedding into for two sofic shifts, and this seems tricky to do even for particular examples.
Question 1.
Let and be two mixing SFTs. When do we have ?
8 Decidability
In this section, we briefly discuss some decidability corollaries for cellular automata that follow from the constructions. We fix the local rule of a CA as its computable presentation. This allows us to ask decidability questions about cellular automata. We start with a lemma that shows that the translation between cellular automata and their local rules is completely algorithmic. We omit the standard proof.
Lemma 8.
Let be a sofic shift. Then given a function , it is decidable whether defines a cellular automaton on , and if it does, we can compute a minimal radius and a local rule for .
Of course, an algorithm that minimizes the local rule implies that given two local rules and , it is decidable whether they define the same CA.
Let be a countable group, with a fixed computable presentation for the elements. The torsion problem is the problem of, given , deciding whether there exists such that .
Theorem 3 ([3]).
For some , the torsion problem of is undecidable.
Theorem 4 ([10]).
The torsion problem of is decidable for every zeroentropy sofic shift .
Combining the theorems with Lemma 7 gives the following.
Theorem 5.
Let be a sofic shift. Then the torsion problem of is decidable if and only if has zero entropy.
Another definition of the torsion problem is to decide, given a CA, whether it generates a copy of . Next, we discuss other problems of this type, omitting the easy proofs.
Proposition 2.
Given a finite set and a finite group , it is decidable whether .
Abelianness is also easy to check.
Proposition 3.
Given a finite set , it is decidable whether is abelian.
Nevertheless, combining the result of [3] with Theorem 1 and the fundamental theorem of abelian groups we see that it is impossible to check which abelian group is generated by a finite set of CA.
Proposition 4.
Let be any infinite finitelygenerated abelian group. Then given a finite set , it is undecidable whether .
Theorem 6.
Given , it is undecidable whether .
9 Acknowledgements
The author was supported by FONDECYT Grant 3150552.
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