**Theorem.** (Gurevich). *There exists a finite set of relations over the alphabet and a word such that if is any class of semigroups containing all finite nilpotent semigroups and , then the following algorithmic problem is undecidable. Given a word , it is undecidable whether for every homomorphism with .*

We recall here that a semigroup is nilpotent if for some and . Necessarily is a zero, that is, for all .

We’ll prove this theorem first over a much bigger alphabet. There is then a standard trick to convert the proof to a two letter alphabet that I mostly leave as an exercise.

There is a standard construction of Markov and Post that encodes a Turing machine into a finite presentation of a semigroup (we mostly follow Rotman’s book here). A Turing machine is essentially a rewriting system for instantaneous descriptions of the Turing machine configuration. One can view it as a semigroup presentation. One then adds some rules that rewrite all halting configurations to some fixed symbol. The only subtlety is that you need to show that the symmetry in semigroup derivations doesn’t break the encoding. This is handled by exploiting the determinism to obtain confluence of a sort.

Gurevich’s fundamental insight was that to obtain undecidability for finite semigroups, one needs to in fact encode recursively inseparable sets into a semigroup presentation rather than a non-recursive set. Two disjoint sets of words over some alphabet are said to be *recursively inseparable* if there is no recursive set such that and . Notice that neither nor can be recursive.

For simplicity we work here with Turing machines that have a two-way infinite tape and that in any transition either the tape head moves left or right, or the machine writes on the current cell. We assume all Turing machines have exactly two halting states.

Formally, a Turing machine has a finite state set , an alphabet , a blank symbol , an initial state , halting states and a transition function . Set . Then . For example means that if the current state is and the symbol under the tape head is then the machine changes to state and moves right. On the other hand, means that if the current state is and the symbol under the tape head is then the machine changes to state and replaces the under the tape head by a .

A Turing machine starts a computation in the initial state with a word written on the tape (and the rest of the tape blank) and the head above the first letter of the word. We assume that input strings are non-empty. The following is a standard result in the theory of computation.

**Theorem A.** *There exists a Turing machine such that the set of input words on which halts in state is recursively inseparable from the set of input words in which halts in state .*

From now one we fix a Turing machine as in Theorem A. Let . A word of the form with , and will be called an *instantaneous description* of the Turing machine or a *special word* for simplicity. One should interpret this word as saying that the machine is in state and the tape holds the word with the head over the . The representation of a tape configuration by special words is not quite unique because of possible padding by blanks at either side, but this will cause no problems.

We now define rewriting rules which simulate the action of the Turing machine . They come in 5 types.

(1) if for

(2) if for

(3) if for

(4) if for

(5) if for

We write if and where is one of the above rules. The reflexive-transitive closure of is denoted . The equivalence relation generated by is denoted . It will also be convenient to write if or .

Note that we have

By a *derivation* from to , we mean a sequence such that for each .

The following lemma says that the rewriting system associated to is confluent when restricted to special words. It is, in fact, nothing more than the determinism of , but it is the key trick to simulating a Turing machine with a semigroup.

**Lemma 1.** (Confluence)

- If is a special word and , then is special.
- If are special words and , then there is a special word such that

**Proof.** The first statement is clear because applying any of the rewriting rules in the forward or backward direction preserves speciality.

To prove the second statement, suppose that is a shortest derivation. We claim that there are no `peaks’ . Indeed, because is deterministic at most one rewriting rule can be applied to any special word. Thus and so is a shorter derivation.

It is now immediate that either , or there is an index such that .

Let where is a new symbol. We define a rewriting system over by retaining the rules of the form 1-5 above and adding the following 3 types of rules, which serve to collapse all special words that lead to the halting state .

(6) for

(7) for

(8)

Let be the rewriting system consisting of rules (1)-(8). We use , , and analogously to the case of the rewriting system . We need two technical lemmas.

**Lemma 2**

(a) for any .

(b) If is a special word, is not a special word and , then

- with
- .

**Proof.** By (6) and (7) we have . This proves (a). For (b), observe that rewriting rules (1)-(5) preserve speciality when applied in either direction, whereas rules (7) and (8) have non-special words on both sides and so preserve non-speciality. Rewriting rule (6) preserves speciality in the backward direction but can remove it in the forward direction if and have the form above.

**Lemma 3.*** If is special, then one of the following mutually exclusive statements holds.*

- implies that is special.

**Proof.** Since is not a special word, it suffices to show that if (2) does not hold, then (1) does. So assume with not special. Let be a derivation. Then because is special and is not, there is a least with not special. By Lemma 2(b), we have and for some . By Lemma 1, there exists a special word such that . But is a halting state and so does not appear on the left hand side of rules (1)-(5). Thus and so by Lemma 2(a).

We can now show that the recursively inseparable sets are encoded in the semigroup presented by the rewriting system .

**Corollary 4.*** The following hold.*

(a)

(b) .

**Proof.**

(b) If , then for some and . Lemma 2(a) shows that and so .

Conversely, if , then by Lemma 3 we have . Since the left hand side of each of the rewriting rules (6)-(8) contains , we conclude that with and . Thus .

(a) If , then for some and . Suppose conversely that for some . First note that because is halting, no rewriting rule from can be applied to and so Lemma 3 implies that all words equivalent to are special. Moreover, none of these words can contain by Lemma 2(a), since otherwise we could derive . It follows that . Therefore, by Lemma 1 there is a special word such that . But again, we must have because is halting. Thus .

As the set is not recursive, we conclude that the word problem is undecidable in the semigroup presented by the rewriting system .

**Corollary 5*** There is a finitely presented semigroup with undecidable word problem. In fact, there is a fixed word such that it is undecidable whether another word is equal to in .*

To prove Gurevich’s theorem, we need to show that certain congruence classes are finite. Let be the semigroup with presentation . Let denote the congruence class of the word .

**Lemma 6.** *If , then is finite.*

**Proof.** Let be the number of steps for the Turing machine to halt on input (in state ). By Lemma 2, Lemma 3 and Corollary 4, each word in is special and does not contain the symbol . So if , then . By Lemma 1 this means there is a special word such that . Note that is an instantaneous description of the Turing machine at some point during its computation on input and hence we can derive in at most steps from . Since each of rules (1)-(5) either preserves length or increases it by one, we have that .

We can now use ideal quotients (or Rees quotients) to obtain suitable finite nilpotent quotients of . If , then is the set of all factors of . So if and only if for some .

We say a subset of is *saturated* if and implies . It is easy to check that if is saturated, then so is .

**Corollary 7.*** If , then is finite and saturated.*

An immediate consequence of Lemma 6 and Corollary 7 is the following.

**Proposition 8.** *Let and let . Then*

- is an ideal of .
- is a finite nilpotent semigroup.
- If is the canonical projection then and .

We can now prove a weakened form of the main result.

**Theorem B.** *There exists a finitely presented semigroup and an element such that if is any class of semigroups containing all finite nilpotent semigroups, then it is undecidable given a word whether, for every homomorphism with , one has that where is the element of represented by a word .
*

We sketch now the proof of how to obtain an example with two generators. Here we follow Albert, Baldinger and Rhodes, who proved a stronger result that Gurevich in which a relatively free semigroup in a finitely based variety is constructed. Let be an alphabet. Then Murskii defines an injective homomorphism by . If is a rewriting system over , define to be the rewriting system over with rules whenever is a rule of .

The following lemma is straightforward combinatorics on words.

**Lemma 9.** (Murski).* If and with , then with , .*

From this one easily deduces that if then if and only if where . It is then straightforward to prove Gurevich’s theorem. Namely, let and be constructed as above and consider the semigroup presentation with constructed from via as above. It is routine to verify that all results from Corollary 5 and onward now apply if we replace each word over with its image under .

]]>Many books, papers and lecture notes on representation theory make the following assertion: the first order theory of finite dimensional modules over the free algebra on two generators over a field can interpret the word problem for finitely generated groups and hence is undecidable. For example, type “wild representation type undecidability” into Google and see what comes up.

The sources usually cited for this result in fact prove the undecidability of the word problem for groups is encoded in the first order theory of **all** -modules, not just the finite dimensional modules. In fact the group algebra of a finitely presented group with undecidable word problem is the key module used in the proof and this is **never** finite dimensional. Willard used Slobodoskoi’s undecidability of the uniform word problem for finite groups to prove that if is a finite field, then the first order theory of finite dimensional -modules is undecidable.

My goal here is to put on the web a proof that the first order theory of a finite dimensional vector space with two distinguished endomorphisms is undecidable (over any field). There are two tools in this proof: Malcev’s theorem on residual finiteness of finitely generated linear semigroups and Gurevich’s theorem that the uniform word problem for finite semigroups is undecidable. No claim is made here to great originality. I am sure there are experts who know everything I am writing.

A result of Prest implies you can interpret the first order theory of two endomorphisms of a finite dimensional vector space into the finite dimensional representation theory of any finite acyclic quiver of wild representation type (as well as most examples of finite dimensional algebras of wild type). So the first order theory of finite dimensional representations of any wild type quiver is undecidable.

We consider a first order language in which we have a binary relation symbol =, a binary operation symbol +, a unary operation – (think negation), two unary function symbols and a constant . We do not include unary function symbols for scalars because we can already express undecidable statements in this limited language!

If consists of a vector space over a field together with two endomorphisms of , then we can interpret variables as elements of and in the usual way. We interpret and by and . So for instance, if and only if is invertible.

If is a field, we denote by the set of all sentences in our first order language that are true in every model with a finite dimensional -vector space and . Our goal is to prove is undecidable, that is, there is no algorithm to determine whether a sentence in our first order language is true in all finite dimensional -vector spaces with a pair of distinguished endomorphisms.

Recall that a semigroup is *residually finite* if, for all with , there is a homomorphism with a finite semigroup and . The following theorem of Malcev is usually stated for groups but the proof actually gives the result for semigroups.

**Theorem 1.** (Malcev)

*Every finitely generated semigroup of matrices over a field is residually finite.*

As a consequence, we obtain the following corollary.

**Corollary 2.** *Suppose is a finitely generated semigroup and . Then cannot be separated by homomorphisms into finite semigroups if and only if cannot be separated by finite dimensional representations over .*

*Proof.* If is a homomorphism with finite and , then the composition of with the regular representation of (where denotes the result of adding an identity to ) gives a finite dimensional representation separating and .

Conversely, if is a representation with , then because is residually finite by Malcev’s theorem, we can find a homomorphism with finite such that .

Gurevich proved that the uniform word problem for finite semigroups is undecidable. More precisely, he proved that there is a finitely presented semigroup such that it is undecidable given words over whether and can be separated by homomorphisms into finite semigroups, i.e., one cannot decide if is a consequence of the relations in finite semigroups.

**Theorem 3.** *For any field , the theory of finite dimensional vector spaces over with two distinguished endomorphisms is undecidable.*

*Proof.* Let be the semigroup above from Gurevich’s theorem. Let be words over and consider the sentence One has that if and only if , induces a representation of . Thus for all with finite dimensional over if and only if cannot be separated by finite dimensional representations over . Corollary 2 then implies that this occurs if and only if cannot be separated by homomorphisms into finite semigroups. It follows from Gurevich’s theorem that we cannot decide whether . This proves the theorem.

One can similarly, prove the undecidability of the theory of a finite dimensional vector space together with two distinguished automorphisms using the exact same proof scheme as above, but using Slobodoskoi’s theorem in place of Gurevich’s theorem.

**Update.** I’m thinking of writing a future post with a proof of Gurrvich’s theorem. His proof shows something stronger than what I have stated, which in particular allows one to avoid the use of Malcev’s theorem in proving the undecidability of the first order theory of a finite dimensional vector space with two endomorphisms.

In fact, using the details of Gurevich’s proof I think one can replace fields by any base ring with identity (commutative is not necessary). If one so desires one can even assume the endomorphisms are nilpotent of index 2.

]]>

Let be a monoid and a commutative ring with unit. Then a mapping is called a *representative function* if it is a linear combination of matrix coefficients of some representation . Equivalently, is representative if the -submodule of generated by is contained in a finitely generated -submodule. The representative functions form a -algebra with pointwise operations. In fact is properly speaking a bialgebra and is the dual to the bialgebra .

Now let be a finite set and let be the free monoid on . We abbreviate to . We can identify a mapping with the formal power series . Schützenberger defined a power series to be recognizable if the corresponding mapping is representative and proved a power series is recognizable if and only if it is rational (i.e., is in the smallest -subalgebra of containing all polynomials and closed under inversion of power series with non-zero constant term). We remark that the product of representative functions corresponds to the Hadamard product of power series and not the (Cauchy) product.

The support of a rational series is the set of with .

**Pin’s question. *** Is there an algorithm which given a rational power series over (expressed as a linear combination of matrix coefficients of a representation), determines whether the support of is a regular languagae (i.e., recognized by a finite automaton)?*

The answer is no. This is not surprising because it is known to be undecidable whether the support is all words. The proof is based on the following idea, which goes back to Samuel Eilenberg in his Volume A of Automata, Machines and Languages. Recall that if are homomorphisms of free monoids (assumed finitely generated here), then the equalizer is the submonoid . The famous Post correspondence problem says that it is undecidable whether is the trivial submonoid. In Theorem 5.2 of this paper it is shown by Salomaa that it is both undecidable whether is regular and whether it is context-free. Basically, given , he constructs such that if is trivial, then is trivial and otherwise is not context-free. Note that always is context sensitive. Since is a submonoid of , we may always assume without loss of generality that .

Here is Eilenberg’s result, although he did not state it this way.

**Theorem (Eilenberg).*** Let be homomorphisms. Then there is a rational power series over with coefficients in whose support is the complement of .*

*Proof.* The proof relies on a faithful representation of over . My favorite is the following one: given by and . If you think of as representing the bit 0 and as representing the bit 1, then where is the binary number represented by . See Exercise 5.2.

Now let be homomorphisms. Then are -linear representations of . As the representative functions form a ring with pointwise operations, we obtain the representative function . Clearly, if and only if , if and only if . Thus the support of the rational series is the complement of , as required.

**Corollary. ***It is undecidable whether the support of a -rational power series is: *

*all words;**regular;**co-context free.*

**Update.** Stefan Göllar has pointed out to me that this result is proved (in essentially the same way) in D. Kirsten and K. Quaas, *Recognizablity of the support of recognizable series over the semiring of integers is undecidable*, Inf. Process. Lett. 111(10):500-502 (2011). The main difference is that the authors of this paper don’t seem to be aware of Salomaa’s result and so they reprove it.

**Update.** I’ve since discovered that this questions is an exercise in A. Salomaa and M. Soittola, “Automata-theoretic aspects of formal power series”. Texts and Monographs in Computer Science. *Springer-Verlag, New York-Heidelberg,* 1978. Presumably, they had in mind a reduction to Hilbert’s tenth problem since this is used for all the decidability results of that section.

First let me recall the definition of an oriented matroid in terms of covectors. Let with product given by

for all . We define a unary operation, written , that fixes and switches (note that negation is an order two automorphism).

If is any set (always finite in this post), then becomes a unary monoid via pointwise operations. The identity will be denoted and the involution by negation. Elements of are called *covectors*. If are covectors, then their *separation set* consists of all such that , that is it consists of the coordinates which have opposite signs. One should verify that if , then and that commute if and only if .

An *oriented matroid* with *ground set* is a unary submonoid satisfying the *separation axiom*:

- if and , then there exists such that and, for all , one has .

We call the *underlying unary monoid* of the oriented matroid.

The primary example comes from hyperplane arrangements, as discussed in this post. Let us recall here the construction of the set of covectors associated to an arrangement. Let us agree that the sign of a real number is , or according to whether it is zero, positive or negative. Then there is a natural mapping that replaces each entry of a vector by its sign.

Now if is a hyperplane arrangement in a finite dimensional real vector space given as the zero sets of forms , then we have a linear mapping given by . The set of covectors turns out to satisfy the oriented matroid axioms.

Indeed, and . If and , then trivially . Else choose on the line segment sufficiently close to so that and have the same sign whenever . This can be done because the are continuous. Then , as is easily checked. Thus is a unary submonoid. To verify the separation axiom, let with . Write and . Then takes on opposite signs at and so for some . Then one checks that taking satisfies the requirements of the separation axiom.

Oriented matroids which come from hyperplane arrangements are called *realizable*. Not all oriented matroids are realizable. Non-realizable matroids are still interesting to people in discrete geometry. One reason is because of the Bohne-Dress theorem.

The reader should refer back to this post for the definition of a zonotope and the connection with hyperplane arrangements. If is a hyperplane arrangement with associated zonotope , then a *zonatopal tiling* of is a polyhedral cell decomposition of such that each cell is a zonotope whose edges are parallel to edges of . The Bohne-Dress theorem states that zonotopal tilings of are in bijection with single-element liftings of the oriented matroid . Roughly speaking a single-element lifting of an oriented matroid is an oriented matroid obtained by adding one element to the ground set and whose contraction at that element returns the original matroid.

Now that we have the background out of the way, let me explain how isomorphism of oriented matroids is defined.

A *reorientation* of an oriented matroid is what you get by choosing and changing the sign of each covector of at . Obviously, this operation does not change the isomorphism class of the underlying unary monoid.

An element is called a *loop* of an oriented matroid if for all . Obviously removing or adding loops results in an oriented matroid with an isomorphic underlying unary monoid. Two elements are called *parallel* if either for all or for all . Being parallel is clearly an equivalence relation. Adding or removing elements parallel to a given element of obviously does not change the isomorphism class of the underlying unary monoid.

An oriented matroid is called a *relabeling* of if it can be obtained from by iteration of the following operations: adding/removing loops; adding/removing parallel elements; renaming the elements of .

Oriented matroids and are considered *isomorphic* if one can be obtained from the other by a series of relabellings and reorientations. Obviously isomorphic oriented matroids have isomorphic underlying unary monoids. The point of this post is to sketch a proof of the converse.

**Theorem 1.** *Oriented matroids and are isomorphic if and only if and are isomorphic unary monoids.*

It seems to me that this is a more concise and natural definition of isomorphism in this context. I have not been able to find it in the literature and it is not 100% trivial

The key idea is to show that “being” the underlying unary monoid of an oriented matroid is in some sense intrinsic in that one can recover up to isomorphism.

All unary monoids will have their identity denoted by and their unary operation by negation. Let us define a *functional* on a unary monoid to be a homomorphism of unary monoids . The zero functional is defined in the obvious way. If is a functional, then defined by is also a functional because negation is an automorphism. Clearly .

Define the *spectrum* of to be the set of non-zero functionals on and define the *Gelfand transformation* by

If is an oriented matroid, then for each we have an associated functional given by . Note that if and only if is a loop. One has that are parallel if and only if .

The proof of Theorem 1 relies on the following result, which I conjectured in this mathoverflow question.

**Lemma 1.** *Let be an oriented matroid. If , then there exists such that .
*

A geometric argument was sketched by Hugh Thomas in his answer based on induction on rank and using contractions of oriented matroids. I extracted the key ingredient of his proof to construct an algebraic proof using tope graphs in my answer. The interested reader can find the details here. I will sketch the proof for realizable oriented matroids (i.e., hyperplane arrangements) here. The general case follows the same argument, but one must first replace the oriented matroid by an isomorphic one which is simple (i.e., has no loops or parallel elements) and then replace the word“chamber” by “tope”.

Let be a hyperplane arrangement in . Then the complement of the arrangement is a finite union of polyhedral cones called *chambers.* Two chambers are *adjacent* if they are separated by a single hyperplane. A *gallery* between two chambers is a sequence of adjacent chambers starting from and ending in . A key fact about hyperplane arrangements is that any two chambers are connected by a gallery. More generally, each non-empty intersection of halfspaces of the hyperplanes from is a polyhedral cone called a *face* of the arrangement. The faces are in bijection with the covectors of the corresponding oriented matroid (the signs tell you on which side of each hyperplane the face lies) and we do not distinguish between them notationally.

Now if is a functional on , then since it is non-zero and the chambers make up the minimal ideal of , it follows that does not vanish on any chamber. Fix a chamber . Then . Thus if we consider a gallery from to , we find adjacent chambers along this gallery with . The covectors associated to adjacent chambers differ in exactly one entry corresponding to the hyperplane separating them. Reorienting if necessary, we may assume that , and that belongs to the positive halfspace associated to and to the negative halfspace. We claim that agrees with . Let be the facet separating . Then the covector associated to has in position and agrees with in all other positions. From and one obtains that If is any other face, then and depending on whether , respectively. From this one easily deduces that .

We can now prove Theorem 2, which has Theorem 1 as an immediate consequence because it implies that an oriented matroid can be reconstructed up to isomorphism from its underlying unary monoid.

**Theorem 2.** *Let be a unary monoid. Then is isomorphic to the underlying unary monoid of an oriented matroid if and only if: *

*the Gelfand transformation is injective;**satisfies the separation axiom, that is, is an oriented matroid.*

*Moreover, if is an oriented matroid, then and are isomorphic oriented matroids.*

*Proof.* Trivially, if 1 and 2 are satisfied, then is isomorphic to the underlying unary monoid of an oriented matroid. If is an oriented matroid, then the functionals with separate points of and so the Gelfand transformation is injective. Lemma 1 immediately implies that if we remove from exactly one element of each parallel class , then, up to reorientation and renaming the elements, we get exactly the same oriented matroid as by removing from all loops and all but one element from each parallel class. This completes the proof.

Let me start with zonotopes. A *zonotope* is a linear projection of a hypercube. In other words, it is an image of under some linear map. Alternatively, one can define a zonotope as a Minkowski sum of line segments. In this case is the image of under Zonotopes are convex polytopes and each face of a zonotope is a zonotope (it is an image of a face of the hypercube).

Zonotopes are dual to central hyperplane arrangements. Let be a central hyperplane arrangement in a finite dimensional real inner product space . We assume that they are give by linear forms , for , with . The dual zonotope is the Minkowski sum in .

Associated to each point is a sign vector (also called a covector in oriented matroid theory). If , then is according to whether is positive, negative or zero. The sign vector of is then given by . For example, if is the coordinate hyperplane arrangement in , then the sign vector of just records the sign of each of its coordinates. The dual zonotope in this case is just the -cube .

Another important hyperplane arrangement is the braid arrangement in . It consists of the hyperplanes , with . Sign vectors are in bijection with ordered set partitions of . For instance, the ordered set partition of corresponds to all sign vectors of points with . The dual zonotope to the braid arrangement is the *permutahedron*. It is the convex hull of all vectors in of the form with .

The faces of are in bijection with sign vectors of the arrangement . Using we can think of vectors in as functionals on . Then one can check that the functionals associated to two vectors are maximized on the same face of if and only of they have the same sign vector. If is a sign vector of the arrangement, the corresponding face of is the Minkowski sum , which is again a zonotope!

In oriented matroid theory there is standard way to multiply sign vectors. First we define a monoid structure on as follows. is the identity element. The elements are left zeroes, that is and for all . Now we can view the set of -dimensional sign vectors as a monoid. A beautiful observation of Tits is that the set of sign vectors associated to

is a submonoid of . To see this let . Then where is the result of making a small movement on the line segment from to .

I want to give an action of the monoid on the zonotope by cellular maps.

Let’s consider first the coordinate hyperplane arrangement in . This case is very easy and explicit. Recall that and . Consider the following three functions : is the identity: is the constant map to ; and is the constant map to . Then the assignment gives an isomorphism of with a monoid of cellular mappings on .

I’ve not been able to generalize this explicit construction to arbitrary arrangements. Instead, I use the fact that a zonotope , like any polytope or regular CW complex, is homeomorphic to the geometric realization of the order complex of its face poset .

Moreover, if we give the CW structure whose closed balls are the geometric realizations of principal order ideals, then the above homeomorphism is a combinatorial isomorphism of CW complexes.

The beautiful fact is that if is a hyperplane arrangement, the face poset of the zonotope is Green’s -order on ! This is a standard fact in oriented matroid theory and can be easily checked from the above correspondence between sign vectors and faces. The reader should check that the -preorder on is an order and boils down to if and only if .

Now the -order is stable for left multiplication for any semigroup and so acts on the face poset of by order-preserving maps. Thus it acts on by cellular maps with respect to the CW structure defined above. As this is a combinatorially isomorphic CW complex with we have found the desired action.

I would very much like a direct description of this action which avoids going through face posets and order complexes. There should be something as simple as in the case of coordinate hyperplane arrangements!

One application of this action is that the augmented cellular chain complex for is a minimal length projective resolution of the trivial -module and so the cohomological dimension of is .

All this can be generalized to affine hyperplane arrangements, oriented matroids and complex hyperplane arrangements. This is part of forthcoming work with Margolis and Saliola.

]]>The question asks when a transformation monoid on a set can be the monoid of all continuous maps for some topology on . The OP observes that the monoid must contain all constant maps and asks what else, if anything is needed.

This question is equivalent to asking which monoids can be represented as endomorphism monoids of a topological space. Indeed, a transformation monoid containing all constant maps is the same thing as a monoid whose minimal ideal consists of left zeroes and acts faithfully on the left of it. (Note my functions act on the left.) So if for a topological space , then is the minimal ideal of with some topology.

This is not usually the kind of problem that appeals to me, but it is quite classical in semigroup theory. Nonetheless Google didn’t seem to give an answer. Instead it showed that people studied the related question of which monoids can be the set of all non-constant mappings on some topological space. The answer, not surprisingly, is all monoids .

A more interesting question to me is to restrict to the finite case and make it an algorithmic question. So here is my question.

**Question 1.** *What is the time complexity of deciding whether a finite monoid (whose minimal ideal is a left zero semigroup on which it acts faithfully) is the endomorphism monoid of a finite topological space as a function of the size of the minimal ideal?*

I have chosen to make the input the size of the minimal ideal rather than the size of the monoid because we want to think of the monoid as a transformation monoid.

I probably will not work seriously on this problem because I have too much else to do. I am hoping somebody has a student looking for a problem, who might find this one interesting.

Let’s make a small translation. A topology on a finite set is the same thing as a preorder on . Indeed, given a topology we can define the *specialization order* by Conversely, given a preorder, we can define a topology by taking the upper sets as the open sets. Continuous mappings correspond precisely to order-preserving mappings. So we have the following reformulation of Question 1.

**Question 2.** *What is the time complexity of deciding whether a finite monoid (whose minimal ideal is a left zero semigroup on which it acts faithfully) is the endomorphism monoid of a finite preordered set as a function of the size of the minimal ideal?*

The full transformation monoid is obviously the endomorphism of the discrete and of the indiscrete topology. So we shall always assume is a proper submonoid of the full transformation monoid on a finite set containing all the constant maps. Also we assume since the case is an easy exercise.

One possible algorithm is to compute all preorders on that are stable under and then see if some other functions preserve them, but that is hopelessly slow.

Let me remark that there are serious constraints on the group of units of . For example if is transitive on , then it is easy to see that the preorder is an equivalence relation. This is because if and , then for all . Then since is finite, we obtain eventually that .

Continuing the above example, since we are assuming is a proper submonoid of , we can assume this equivalence relation is neither the equality relation nor the universal relation. But then it follows that contains a non-constant, non-invertible mapping (crush a non-trivial equivalence class to a point). Thus cannot be just a transitive group and the constant mappings. This is our first “non-trivial” constraint.

The above discussion also shows that if is transitive, then it is is not primitive (a permutation group is primitive if it preserves no non-trivial equivalence relation). This is another restriction.

**Question 3.*** Which finite permutation groups 0n a finite set are the homeomorphism group of some topology on . What is the time complexity in deciding this (as a function of )?*

denotes the free monoid on and denotes the space of right infinite words. If , then denotes the set of all infinite concatenations of elements of .

**Lemma 1.** *If is periodic with periods , then is a period of *

*Proof. *Let be the shift operator. Then acts as a cyclic permutation of the orbit of and The result follows from basic group theory.

**Theorem 1.** *Suppose that are commuting words, i.e., Then are powers of a common word. *

*Proof.* This is a variation of a proof in Mark Sapir’s book, which he attributes to Victor Guba. Consider the infinite word . Then is periodic and both are periods of . Therefore is a period of . Thus if is the prefix of of length , then are powers of .

Theorem 1 is a consequence of the more general fact that two words which are not powers of a common word freely generated a free submonoid of We prove this via Jeffery Shallit’s version of the Fine and Wilf theorem.

**Theorem 2 (Shallit’s Fine and Wilf).** *Let and suppose there exist infinite words and such that initial segments of of length coincide. Then:*

- are powers of some word

*Proof.* The proof is by induction on . The base case is when one of the words is empty, which is trivial. Assume without loss of generality that . Then since have the same prefix of length , we deduce for some Note that because . Also note that . Let be the words obtained from respectively by removing the common initial segment . Then and are such that the initial segments of of length coincide. By induction, we obtain and hence , and also are powers of some word But then is also a power of This completes the proof.

I would love to see a proof of this theorem along the lines of the proof of Theorem 1 that avoids induction.

**Corollary 1 (Fine and Wilf ). ***If are infinite periodic words with periods and , respectively, and if the initial segments of of length coincide, then and is a period of .*

*Proof.* Apply Theorem 2 with as the initial segments of of lengths respectively.

The original proof is an elegant application of generating functions.

**Corollary 2. ***Let . Then the following are equivalent.*

- commute.
- are powers of some word .
- do not freely generate a free submonoid of .

*Proof.* Clearly 2 implies 1 implies 3. Suppose 3 occurs. Then we can find distinct words in such that . Since is cancellative, we may assume that begins with and begins with . Now apply Theorem 2 to the infinite words and and use that .

I work in semigroup theory. Semigroups have not enjoyed the greatest popularity in the mathematical community. For instance the nlab entry for semigroup suggests that many people think it is a form of “centipede mathematics.” (Think how many legs can you cut off a centipede and still have it walk.) My hope is that this blog will enlighten people about the beauty of semigroup theory and convince them that there is more to the subject than they first thought.

The branch of semigroup theory that has been the most fruitful, in my opinion, is finite semigroup theory. Finite semigroups are very naturally connected to finite state automata, leading to the beautiful field of algebraic automata theory. Such diverse illustrious mathematicians as Samuel Eilenberg, John Rhodes and Marcel-Paul Schützenberger have worked on the subject.

I also hope to touch on the representation theory of finite semigroups, which has found useful applications to finite Markov chains in recent years.

Enjoy!

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