1 Preliminaries
By countable set, we mean injectable in . Let denote the empty word. For a countable alphabet, we note the set of finite words over . We also note for .
Let us note the complement of set . means that and is finite. means assuming that .
1.1 Computability
Computability problems are naturally defined over , but can easily be extended through subsets of it, cartesian products or disjoint union (by canonically injecting in sets of tuples). For example, if , then the set of tuples admits a simple injection into . Let us fix a (computable) countable set , that we can identify to integers.
Definition 1
Let us define the following reducibilities, for :

is Turingreducible to , , if: one can compute with oracle .

is enumerationreducible to , , if: from any and any integer , one can compute a finite set such that if and only if .

is positivereducible to , , if: from any , one can compute finitely many finite sets such that if and only if .

is manyonereducible to , , if: from any , one can compute some such that if and only if .

is oneonereducible to , , if, and is onetoone.
Oneone reducibility implies manyone reducibility, which in turns implies positivereducibility, which implies both Turingreducibility and enumerationreducibility.
Each reducibility induces a notion of equivalence : iff and . And each notion of equivalence induces a notion of degree : the degree of a set is its equivalence class for .
The join of and is the set such that iff and iff . It has the property that and for any reducibility previously defined.
See [2] for a reference on computabilitytheoretical reductions.
1.2 Monoids and groups
We will deal with countable monoids , where , is the free monoid generated by symbols from and is a monoid congruence^{1}^{1}1We could deal in the same way with semigroups, by prohibiting the empty word.. The monoid is always implicitly endowed with its generating set (later, some problems may depend on the presentation). Each element of the monoid is represented by a word , but the representation is not onetoone (except for the free monoid itself). We note if and is the natural quotient map.
It is also clear that the concatenation map, which from any two words outputs , which is one representative of the corresponding product, is computable. We say that is an effective group if, additionnally, there is a computable map such that .
The equality problem of , endowed with generating family , is the set of pairs , endowed with a natural enumeration so that we can consider it as a computability problem.
Remark 1

It is clear that the word problem is oneonereducible to the equality problem.

If is an effective group, then the word problem is actually manyoneequivalent to the equality problem.

The equality problems for endowed with two distinct finite generating sets are oneoneequivalent.

If is a submonoid of endowed with a generating set which is included in that of , then the equality problem in is oneonereducible to that of .

In particular, the equality problem in any finitely generated submonoid is oneonereducible to that of .
Nevertheless, there are countable groups whose word problem is computable when endowed with one generating family, and uncomputable when endowed with another one.
The word problem is known to be decidable if and only if the group is computable (see [3] for a proof in the finitely generated case), that is, it can be seen as a computable subset of over which the composition rule is a computable function (this implies that inversion is also a computable map).
1.3 Subshifts
Let be a finite alphabet with at least two letters, and a group (most of the following should be true if is a cancellative monoid though). A finite pattern over with support is a map . Depending on the context, note that, for , may either be an element of or a subpattern with support . If and is a pattern, we will note the pattern with support such that for all .
We are interested in , which is a Cantor set, when endowed with the prodiscrete topology, on which acts continuously by (left) shift: we note for and .
A subshift is a closed invariant subset . Equivalently, can be defined as the set avoiding a language , which is then called a (defining) forbidden language. If can be chosen finite, the subshift is called of finite type (SFT); if it can be chosen computably enumerable, it is called effective.
The language with support of subshift is the set ; the language of is , and its colanguage is the complement of it. The latter is a possible defining forbidden language. If , we define the corresponding cylinder .
Remark 2
induces a natural covering by . Its image set is a subshift over the free monoid. One can note the following.

.

The colanguage of the full shift is the same as that of the subshift : the set
of patterns that do not respect the monoid congruence.

Nevertheless, is a forbidden language defining .

The colanguage of every subshift is the set of patterns , , whose all extensions to configurations involve as a subpattern a pattern of either , or . In that case, by compactness, at least one such subpattern appears within a finite support , with , which depends only on .
Remark 3
Let be a monoid.

The equality problem in is positiveequivalent (and oneonereducible) to the colanguage of the full shift.

The colanguage of any subshift is enumerationreducible to the join of any defining forbidden language for and the equality problem of .
Proof

oneonereducibility: one can computably map each word to a unique pattern over involving two different symbols. By Point 2 of Remark 2, this pattern is in the colanguage of the full shift if and only if .
positivereducibility (with all s being singletons): from each pattern , one can compute the set of pairs such that . By Point 2 of Remark 2, is in the colanguage if and only if one of these pairs is an equality pair in . 
Consider the set of locally inadmissible patterns, that involve a subpattern either from the forbidden language or from . From any pattern , one can enumerate all of its subpatterns and all of their shifts, i.e. all patterns such that there exists with and for every . This shows that is enumerationreducible to the join of the forbidden language and , the latter being equivalent to the equality problem, by the previous point. It remains to show that the colanguage of is enumerationreducible to .
From any pattern and any , one can compute some including , in a way that and (for example take the union of with balls in the Cayley graph). Then, one can compute the set of extensions of to , i.e. patterns with support whose restriction over is . By Point 4 of Remark 2, if and only if there exists with such that all extensions of to are in ; and in particular this should happen for some , which precisely means that . ∎
It results that, in some sense, one expects most subshifts to have a colanguage at least as complex as the equality problem in the underlying monoid.
1.4 Homomorphisms
Let and be subshifts. Denote the set of homomorphisms (continuous shiftcommuting maps) from to , and the set of bijective ones (conjugacies). We also note the monoid of endomorphisms of , and the group of its automorphisms.
If is finitely generated, then homomorphisms correspond to block maps (or cellular automata), thanks to a variant of the CurtisHedlundLyndon theorem [4].
Theorem 1.1
Let be finitely generated. A map from subshift into subshift is a homomorphism if and only if there exist a radius and a block map such that for every and , (where the latter has to be understood with the obvious reindexing of the argument).
Let us order the block maps by increasing radius , and then by lexicographic order, so that we have a natural bijective enumeration (because , and are finite). This gives in particular a surjective enumeration and in general, a partial surjective enumeration . In general, . It is a nontrivial problem to ask whether is computable (this is the case for the full shift when ), but not the topic of the present paper. Obtaining a bijective enumeration for would be easily achieved by enumerating each block map only for its smallest possible radius. Nevertheless, trying to achieve a bijective enumeration in general for , or even for , is a process that would depend on the colanguage of the subshift (we want to avoid two block maps that differ only over the colanguage), which may be uncomputable.
Even when is an effective group, need not be an effective group!
For the rest of the paper, let us assume that is an effective group. More precisely, all results could be interpreted as reductions to a join with a problem representing the composition map of the group, and sometimes to an additional join with a problem representing the inversion.
2 Equality problem is not too hard
Remark 4
Two distinct block maps representing an endomorphism of actually represent the same one if and only if for every pattern , .
The equality problem is at most as complex as the language.
Theorem 2.1
The equality problem in is positivereducible to .
Proof
One can directly apply Remark 4, by noting that it is easy to transform each block map into an equivalent one, so that the resulting two block maps have the same radius (the original maximal one, by ignoring extra symbols). ∎
Of course, this remains true for the equality problem in . Since positivereducibility implies both Turingreducibility and enumerationreducibility, we get the following for the lowest classes of the arithmetic hierarchy (which was already known; see [5]).
Corollary 1

The equality problem is decidable, in the endomorphism monoid of any subshift with computable language (for instance 1D sofic subshift, 1D substitutive subshift, minimal effective subshift, twoway spacetime diagrams of a surjective cellular automaton…).

The equality problem is computably enumerable, in the endomorphism monoid of any effective subshift (for instance multidimensional sofic subshift, substitutive subshift, limit set of cellular automaton…).
3 Automorphism groups with hard equality problem
The purpose of this section is to prove a partial converse to Theorem 2.1: a subshift for which the two problems involved are equivalent, however complex they are.
Let and be subshifts. For and , let us define the controlled map as the homomorphism over such that if ; otherwise. Denote also the projection to the first component, and the shift of the first component with respect to element : for every .
Remark 5

.

If is a group and , then .

.

is injective if and only if is a permutation.

if is (locally) permutable, i.e. for all , if we define by , for , then .

From Remark 4, is the identity over if and only if or is the trivial permutation over letters appearing in .
Example 1
Examples of permutable subshifts are the full shift on or, if and , the sunnysideup defined by forbidding every pattern which involves two occurences of . We have seen that the colanguage of the former is positiveequivalent to the word problem in . The language of the latter can be easily proven to be manyoneequivalent to the word problem in (as essentially noted in [6, Prop 2.11]), hence yielding a kind of jump for the colanguage.
If , let us denote the cycle mapping to , to , to , and any other element to itself. The following lemma corresponds essentially to [7, Lemma 18].
Lemma 1
Suppose has at least 5 distinct elements . Let be a pattern, , and . Then , where and .
Proof
If , then , where is the involution that swaps and on the one hand, and on the other hand; otherwise . If , then , where is the involution that swaps and on the one hand, and on the other hand; otherwise . Since , one can see that if , then . Now if , then we see that , and , so that we get the stated result. ∎
Theorem 3.1
Let be a subshift and an permutable subshift for every , where . Then is oneonereducible to the word problem in the subgroup of automorphisms of generated by and for , and .
Proof
Consequently, subshifts can have finitely generated groups with equality problem as complex as their colanguage, as formalized by the following corollary. In that case, the equality problem of the whole automorphism group is as complex also.
Corollary 2

If and are as in Theorem 3.1, then is oneoneequivalent to the word problem in (a finitely generated subgroup of) .

For every subshift over a finitely generated group , there exists a countabletoone extension such that is oneoneequivalent to the word problem in (a finitely generated subgroup of) .

For every subshift over a finitely generated group , there exists a full extension such that is oneoneequivalent to the word problem in (a finitely generated subgroup of) .

Every Turing degree contains the word problem in (a finitely generated subgroup of) , for some 2D SFT .

There exists a 2D SFT for which the word problem in (a finitely generated subgroup of) is undecidable.
Proof
Note that the number of generators can be decreased if we want to reduce only the language whose support is spanned by a subgroup. For instance 2D SFTs are already known to have (arbitrarily ) uncomputable 1D language. Indeed, our automorphisms do not alter the layer, so that their parallel applications to all traces with respect to a subgroup is still an automorphism.
Among the open questions, we could wonder whether there is a natural class of SFT (irreducible, with uncomputable language, at least over ) whose colanguage could be proven reducible to the word problem in the automorphism group. This could require to encode the whole cartesian product of Theorem 3.1 inside such subshifts. Another question would be to adapt our construction while controling the automorphism group completely so that it is finitely generated.
Acknowledgements
This research supported by the Academy of Finland grant 296018.
We thank Ville Salo for some discussions on commutators, on the open questions, and for a very careful reading of this preprint.
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