1 Introduction
We prove in this paper that every permutation of a Cartesian product of two finite sets can be performed by first permuting on the left, then on the right, then on the left. This is Theorem 2, and its proof is a reduction to Hall’s marriage theorem. For two countably infinite sets, every permutation of the Cartesian product can be performed either by permuting leftrightleftright or rightleftrightleft (sometimes only one of these orders works). This is Theorem 5. Its proof is elementary set theory, a Hilbert’s Hotel type argument. We also prove that for the direct product of two finitedimensional vector spaces, every linear automorphism can be written by a composition of three linear automorphisms, the first of which only modifies the left coordinate, the second the right coordinate and the third the left coordinate. This is Theorem 6. In Theorem 8, we show that a similar result does not hold for general selfhomeomorphisms of the plane.
It turns out that the results on finite sets and vector spaces have been independently proved several times. See Section 2 for this and other related results.
The study of such alternations arises from the unpublished draft [13] where this optimal number of alternations was studied for automorphism groups of subshifts. Here, we note that this notion applies to the automorphism group of a product object in any category, and study it in some categories of interest. We call the optimal length of a leftright alternation the alternation diameter (see below for more detailed definitions).
The alternation diameter is interesting as a general concept since the automorphism group of a product object always contains the automorphism groups of the left and right components (in a natural way) – and more generally what we call the left and right groups (defined below) –, but in addition it can contain many “entirely new” automorphisms that can only be understood globally, and do not easily reduce to the study of the left or right component of the product separately. In the case that the alternation diameter is bounded for a particular product object and the left and right groups are easy to describe, we get a handle on the elements of the automorphism group, at least as a set. Of course, in complicated categories we cannot expect this to happen very generally, but it can be a helpful technique in the study of automorphism groups of individual objects when it succeeds.
In this note we consider some simple categories where alternation diameter is actually globally bounded over the whole category (though in these simple cases it does not really help in understanding the automorphism group of a product), and show by examples that the left and right groups, not surprisingly, do not in general generate the automorphism group of a product object in more complex categories.
In terms of the alternation diameter, our results are the following:
Theorem 1.
The category

FinSet of finite sets and functions has alternation diameter ,

CountSet of countable sets and functions has alternation diameter ,

FinVect of finitedimensional vector spaces and linear maps has alternation diameter ,
Our examples of categories where alternation diameter is undefined (meaning that left and right alternations do not generate all automorphisms) are the following: The plane has, for slightly nontrivial reasons, undefined alternation diameter in the category Top of topological spaces and continuous functions. The square has undefined alternation diameter in Top for trivial reasons. In Pos, the product posets and where is the diamond, i.e. the poset with Hasse diagram , have undefined alternation diameter for (different) trivial reasons. (By a more Posspecific proof, Maximilien Gadouleau has shown that in fact has undefined alternation diameter for every finite poset .)
We now give some more detailed definitions (see [10] for basic examples and notions of category theory^{1}^{1}1Other than Section 4 and Section 5.2, we are not concerned with the “theory”, mainly terminology and concepts.). In the category of sets and functions a product object is just the Cartesian product of and , and the automorphism group of an object is just the full permutation group on that set. In concrete categories where products behave this way, for a product object and its automorphism group , we define the Left group , namely the subgroup of containing those automorphisms of that modify only the left coordinate, i.e. satisfy . Define the Right group analogously. Our precise statements about sets are that in the category FinSet of finite sets, , and in the category CountSet of countable sets, but generally ).
In less Setlike categories one can define and in terms of commutative diagrams (see Section 4 for the diagrams): let be the categorytheoretic product of and , and and the projections defining . Define as the group of automorphisms satisfying , and those satisfying . The groups obtained, as well as the subgroups and , do not depend on the choice of the product (up to isomorphism of the triples ).
In categories of sets, it is convenient to think of as indices into a (possibly infinite) matrix, with indexing the rows and the columns. This is the suggested convention for mental pictures and is the one used in the proofs. Then is the coLumn group that performs a permutation in each column separately (independently of each other) and is the Row group that performs independent permutations on rows. Then in FinSet states precisely that any by matrix containing each element of exactly once can be turned into the matrix by first permuting each row, then each column, then each row.
In grouptheoretic terms, since and are subgroups, we can state the weaker fact equivalently as follows: the group is generated by , and its diameter with respect to the generating set is at most three (independently of and ). This diameter is in general what we call the alternation diameter of , or of the object having as automorphism group. The alternation diameter of a category is the least upper bound of alternation diameters of products.
This diameter can infinite for a single object , when is generated by and , but is not equal to a finite number of alternations (this is what happens in Top and Pos). In principle a category may also have infinite alternation diameter due to objects having arbitrarily large finite alternation diameters. If is not generated by and at all, we say the alternation diameter is undefined, and a category has undefined alternation diameter if some product in it does.
2 Existing related work
The case of finite sets, which started this paper, is inspired by [16], where it is shown that any permutation of , for three sets with , can be written as a composition of finitely many permutations where alternately only or is permuted. Our notion of alternation diameter in the category of finite sets is related to this definition, as it also means refers to “alternately permuting a product on the left and right”. The difference is that there is no communication coordinate (making it harder), but we allow the permutation to depend on the value on the right when permuting the value on the left, and vice versa (making it easier).
It turns out that the results about finite sets and vector spaces have been proved before in the context of memoryless computation: [3, Theorem 3] and [6, Theorem 2] are essentially the same result as Theorem 3. More related results on permutation groups can be found in [12]. The motivation and framework is ostensibly different, but the case of finite sets is proved in [2, Theorem 3.1] using a version of Hall’s theorem.
Theorem 6 is also known previously: the number of alternations needed for a product of length (which can be obtained from Theorem 6) can be found in [2, 3] (for a larger class of modules). It turns out that is not optimal at least for finite fields: in [11, Theorem 2.1] it is proved that for FinVect over a finite field the optimal number of alternations for a product of length is .
It should be possible to extract, from the results of [1], a natural category where alternation diameter is defined but infinite for some objects.
We do not know of previous work on alternation diameter in categories that are less obviously computationally relevant, in particular the results on countably infinite sets and homeomorphisms groups are new to the best of our knowledge.
In Section 3.4 we discuss a known related result in graph theory.
3 Bounded alternation diameter
3.1 Finite sets
In this section we look at symmetric groups for finite sets , i.e. automorphism groups of product objects in the category FinSet of finite sets and functions. Permutations act from the left and compose righttoleft, and we also write .
The following theorem is a wellknown corollary of Hall’s marriage theorem (see Lemma 4 for a proof):
Lemma 1.
Let be an by matrix over where all rows and columns sum to . Then cellwise, for some permutation matrix .
This naturally implies that every such matrix is a sum of permutation matrices, but we prefer to use the lemma directly.
Definition 1.
Let and be finite sets, and let . Define the coLumn group of as
and define the Row group symmetrically.
Theorem 2.
Let and be finite sets, and let . Then .
Proof.
We prove . The equality follows by symmetry.
A permutation of can be seen as an matrix containing each entry of exactly once. (Formula: .) We show that by showing that contains the identity map. In terms of the matrix , precomposing by , i.e. , corresponds to permuting the entries of the matrix by (in the obvious “forward” direction). Thus, our task is to turn into the matrix by first permuting the rows, then the columns, then the rows again.
Now, ignoring the component of every matrix entry, and supposing without loss of generality that and , we obtain an bymatrix over where every element of occurs exactly times.
To such a matrix , we associate an bymatrix defined as follows:
Then every row of sums to because is an by matrix, and every column of sums to because every appears times in . It follows from Lemma 1 that where is a permutation matrix.
We want to permute so that every column contains every symbol of exactly once. To do this, consider a row of , and let be the unique element such that . Then row of contains at least one copy of . On each row, move such an to the first column. Since is a permutation matrix, the elements moved to the first column are distinct, so the net effect of this is that the first column of contains each element of exactly once. Considering now the by matrix obtained from by deleting the first column, we observe that every element of appears exactly times. By induction, we obtain that can be permuted by an element of into a matrix where all columns contain each element of exactly once.
Now, apply an element of to sort each column so that the th row of contains only s.
Now, consider the action of this transformation on the original by matrix with entries in . After the transformation (by an element of ), the row contains only values of the form where . Since every entry appears exactly once in , the set of values on row is then precisely . We can now apply a final permutation in to permute all elements into their correct position, obtaining the matrix representing the identity permutation on . ∎
Let us make some additional observations. Consider an arbitrary product . Write for the group of permutations of that only modify the th component of their input. By induction on , Theorem 2 shows the following:
Theorem 3.
Let where are finite sets and let be as above. Then
If each has at least two elements, then no sequence of less than groups suffices.
Proof.
We first prove the formula for . The case is trivial, and is Theorem 2. Now, let and . Then any permutation of is in where and are defined with respect to the decomposition . Let be the corresponding decomposition. Then and are in because they do not modify the component of their input.
We can write as where each is a permutation of that modifies the component only if the component is equal to . Each permutation is in by induction, where are the groups corresponding to components of the product . Thus
and we can reorder the product to get since the permutations corresponding to distinct commute (as they have disjoint supports).
For the second claim, we show a stronger result: We cannot have
for any sequence where there are at least two indices that occur at most once. Suppose the contrary, and let , with , and and each occur only once. Suppose that .
For write for the set of elements of where the coordinate is equal to and the coordinate is equal to . We claim that if is in and fixes , then it maps no elements of into , which clearly proves the claim. To see this, observe that all of the groups except and leave the sets invariant. Thus, when gets its turn, elements of and have not yet been moved. Since fixes , the permutation cannot move any elements away from (since after this step, their coordinate will no longer change). But then elements of cannot be moved into by since , so after applying , elements of still have nonzero coordinate, which will no longer change. Thus cannot move them into . ∎
Corollary 1.
The category FinSet has alternation diameter .
One can extract a full characterization of from the proof of Theorem 2.
Lemma 2.
Let and be sets, and let . The set contains precisely those permutations such that for the natural projection map, for all , the map is bijective.
Proof.
Note that the set contains precisely the permutations that can be mapped to the identity by precomposing first with an element of , and then an element of . In matrix form, they are the by matrices over that can be turned into the matrix by first applying a permutation of columns (from ) and then a permutation of rows (from ).^{2}^{2}2To readers experiencing chiral confusion, we write some formulas: if we “first” apply and “then” , to a matrix , then formulaically we obtain
, which contains the identity matrix if and only if
.For sufficiency, observe that the above property is precisely the one that holds after the first application of in the proof of Theorem 2 – in terms of the matrix , it states that every column of contains every symbol of exactly once. The two following steps in that proof perform any permutation of using only this property, and do not use the finiteness of or .
For necessity, write in matrix form, , and suppose that for some , we can turn into the matrix by first applying and then . First, must be injective: otherwise, some column of contains both and for some . This still holds after applying , so at the time of the final application of , and are on the same column, thus cannot both be on the correct row . Second, must be surjective: if there exists and such that the th column of does not contain any , , then the same is true after applying . In particular the th column will necessarily contain some value , , on the th row, after the application of . ∎
One can give precise formulas for the sizes of each of the sets obtained by applying in various orders.
Definition 2.
When is an automorphism group, for , write .
Theorem 4.
Let be finite sets, . For any , is equal to one of the sets in and
Proof.
Since by Theorem 2 and and because and are groups, the claim about is true. The first formula comes from the fact that . By definition, , so since and are subgroups,
where since we choose an independent permutation of each of the columns of size . The formula follows by symmetry. ∎
The above theorem shows that at least three alternations are needed also in the stronger sense that for large enough. This is because is dwarfed by by a straightforward application of Stirling’s formula.
We note that Lemma 2 gives a characterization of permutations in . We have not investigated whether there is a simple formula for the cardinality in terms of and .
From Theorem 3 we immediately obtain an alternation diameter result for finitesupport permutations on infinite sets. Write for the group of finitesupport permutations of a set . Define as before, requiring that only the th coordinates of inputs are modified by elements of .
Corollary 2.
Let where are arbitrary sets and let be as above. Then
If each has at least two elements, then no sequence of less than groups suffices.
Proof.
Every permutation has finite support, thus finite projection of the support on the sets . Pick suitable finite initial segments of the sets, and for sufficiency apply Theorem 3, and for necessity its proof. ∎
3.2 Countably infinite sets
We now look at permutations with countably infinite support.
We recall a version of Hall’s theorem for infinite sets and include a proof sketch (see e.g. [5, 4] for details).
Here, graphs are undirected without selfloops, but may have multiple edges between two vertices. A graph is bipartite if its vertices can be partitioned into two nonempty sets in such a way that no edge goes between two vertices in or between two vertices in . Write for the open neighborhood of a subset of a graph, i.e. the set of vertices connected to a vertex in by an edge. A graph is locally finite if every vertex has finite degree, i.e. for all vertices . Write for a finite subset of . A matching in a bipartite graph, with a fixed bipartition of the vertices, is a to correspondence that matches a subset of the elements of injectively to a subset of . We write matchings as partial functions
Lemma 3.
Let be a locally finite bipartite graph where for each finite set of vertices , . Then admits a perfect matching.
Proof.
Let us call the vertices “left” or “right” depending on which side they are on, and as sets. The set of subsets of the edge set has a natural compact topology, namely the product topology on where is the set of edges and means the edge is included in the set. Matchings form a closed subset of this space. Since the graph is locally finite, for each vertex the set of matchings where is matched is clopen, thus compact. Thus, there exists a matching where a maximal set of left vertices is (injectively) matched with some vertex on the right, and for this maximal set, a maximal set of right vertices is matched.
If some left vertex is not matched in , let be the set of those vertices (left or right) and edges which are reachable by a path starting from where every second edge is part of the maximal matching . If is infinite, by König’s lemma we can find an infinite path, and swapping the edges that are in with those not in on this path, we add to but remove no vertex from , so was not maximal. If is finite, and some is not matched, then we can take a path from to and again swap the matching edges with nonmatching edges to add to without removing any matched left vertices. If is finite and all elements of are matched, then (because gives an injection from into ), a contradiction since all edges from are included in and thus .
We conclude that , i.e. matches every left vertex with a right vertex. Suppose then that is not surjective. Then perform the above argument with the roles of left and right reversed, and observe that we also never unmatched a matched right vertex when modifying our matching. Alternatively, one can construct a leftsurjective and rightsurjective matching separately and apply the Cantor–Schröder–Bernstein argument. ∎
In the following, we use the matrix terminology, though indexing by infinite sets. The meaning should be clear.
Lemma 4.
Let be any set and any by matrix over . If every row and column sums to (in particular, cofinitely many entries have value on every row and column), then for some permutation matrix .
Proof.
Construct the bipartite graph with a copy of “on the left” and another copy of “on the right”. Include an edge from left to right if . Then this graph satisfies the assumptions of the previous lemma: Clearly it is locally finite and bipartite. Consider any finite set of left vertices . We have
Similarly, for any we have . The previous lemma gives a perfect matching, i.e. a permutation matrix . ∎
Theorem 5.
Let be sets. If , then
If both and are infinite, then and . If is infinite and , then . If is finite, then .
Proof.
In the first claim, if and are finite, then this is Theorem 2, and if only one of them is finite, this follows from the last claim (which we prove last). We thus consider the case that both are infinite.
As usual, we consider the by matrix over and the cellwise projection with values in , representing a permutation of . We will compose with elements of in that order. Again we use the standard left action, which formulaically is precomposition with inverse, and in terms of matrices corresponds to directly permuting the entries.
We prove the following:

if any cofinite set of columns of contains infinitely many distinct values of , i.e.
then ,

if the dual claim holds, then ,

either the claim or its dual holds.
We first prove the third item. Suppose that does not satisfy the first item. Then a finite set of columns of contains all values from a cofinite subset of . Then those columns of must contain all pairs with , so takes every possible value in these finitely many columns. This clearly cannot happen in finitely many rows and finitely many columns. Thus the dual claim of the first item holds.
Thus we only need to prove the first item, as the second is symmetric, and by the third claim, these together give .
Assume then that any cofinite set of columns of contains infinitely many distinct values of . We describe what happens to the matrices and after each step (we do not rename them after each step). Our plan is the following:

After applying , every row of has infinitely many distinct values, and every that appears on infinitely many columns also appears on infinitely many rows.

After an application of , all columns of have exactly one copy of each .

By Lemma 2, another application of finishes the proof.
Assume . In the first step, we modify each column at most once, and then freeze it and never modify it again. We go through , and on the th turn, we modify a finite set of columns and then freeze them. What we ensure on the th turn is that the first rows of all contain at least distinct values that appear frozen columns, and that each with which appears on infinitely many columns also appears on at least distinct rows on frozen columns.
No trick is needed, we just do it: After a finite number of steps, we have seen only frozen finitely many columns, and contains infinitely many values in any cofinite set of columns, so we never run out of fresh values to move to rows needing them, thus we can indeed make sure each row contains more and more distinct values, and if appears on infinitely many columns, then we can make this choice infinitely many times.
In the limit, in the compact topology of cellwise convergence of the matrix entries, clearly the resulting matrix still describes a permutation (we performed a permutation at most once on each column, so clearly the transformation is columnwise bijective, thus bijective). Every row contains infinitely many distinct values since for any row and any , on the th turn we made sure the th row contains at least values.
In the second step, we again construct the permutation column by column, but now we are permuting rows instead of columns. We modify each row infinitely many times, but with smaller and smaller supports, and take the limit of the process. Note that the set of matrices representing injective maps is closed with respect to this topology, so the limit is automatically injective (if welldefined). Matrices representing surjective maps are not closed, so our matrix may fail to be surjective in the limit if we are careless. To ensure surjectivity, we fix an enumeration of , and say the index of is the such that . For surjectivity, it is enough that each appears in the limit, and for this it is enough that from some point on, we no longer move pairs with indices up to .
To ensure that we get a limit in the second step, we modify each column at most once. We go through , and on the th turn, we permute the rows in such a way that the th column is not modified for , the th column contains exactly one copy of each after the permutations, and the th pair is moved to column unless it is already in some column .
Call/color an element of the th column red if the row containing it has not yet been permuted on the th turn, and green otherwise. We begin the turn by moving the th pair to column if it is among columns (and color it green, so that at this point we have at most one green entry). We now perform a backandforth argument where we modify each row at most once by alternating the following steps:

Pick the next that has not yet been moved into the column (i.e. all occurrences of in it are red), move that into the column (possibly it was already there), color it green and freeze the row.

Let be the first row with a red symbol and move any fresh (that does not yet appear as a green symbol in the th column) into it.
The first type of move in the backandforth is always possible: If appeared on only finitely many columns initially (thus also after the first step), in which case it appeared on infinitely many distinct rows, and since exactly copies of are on frozen columns , there are still infinitely many unfrozen copies of on infinitely many distinct rows. If appeared on infinitely many columns before the first step, then after the first step appears on infinitely many rows, thus on the th turn there are again still unfrozen copies of on infinitely many distinct rows.
The second type of move in the backandforth is always possible: All rows contain infinitely many distinct symbols , and at any point of the process we have only finitely many green symbols on the column.
This concludes the construction, as Lemma 2 applies to the resulting matrix.
For the second claim, we observe that no matrix with projection
where the symbols are elements of , is in . Namely, the first application of is useless, as the set of matrices of this form is invariant under . The following application of will move at most one nonzero into each row. Then already the top left by block necessarily contains at least two zeroes on some column, so Lemma 2 does not apply. Note that matrices of this form do exist since .
For the third claim, let , . Pick a bijection . Consider a permutation mapping , , . Writing again as an by matrix , the row contains exactly one element which should be in the row in the end, and the row contains only elements that belong to it. We have : After the application of to , we still have exactly one element of the form (namely ) in the row. If it is in the column, then the column contains two elements of this form, and thus after the application of , the column contains such an element in some row . Therefore after applying we still have at least one element of the form in the row, so we have not turned to the identity.
Corollary 3.
The category CountSet has alternation diameter .
Since in FinSet we had alternation diameter and , while in CountSet we have alternation diameter and but not or , one notes that “alternation diameter” indeed loses some information. One could instead mimic quantifier hierarchies and define , , and inductively , , . Then in FinSet, the alternation hierarchy collapses on the level , while the one for CountSet collapses at the join .
3.3 Finitedimensional vector spaces
Besides Set, an obvious place to look for categorywide diameter bounds is the category of finitedimensional vector spaces. The first reason is that it is a category where all objects and morphisms behave nicely. The second is the intuition that dimension can often replace cardinality. We find that alternating diameter is indeed , as in FinSet.
Fix a field and let be the category of finitedimensional vector spaces over .
Theorem 6.
The category has alternation diameter . More precisely, let and be in . Then satisfies .
Proof.
Let . The claims and , are symmetric, so we only prove the first.
Consider an arbitrary matrix in block representation with four blocks, of widths and heights and , respectively (the “by block” of size by on the top left). Now applying automorphisms in (from the left) amounts to row operations that do not modify the bottom blocks, and modifies the bottom blocks only. We need to turn into the identity matrix with an element of .
If are the rows of , then their restriction to their first coordinates has full rank. It is easy to see that then there is a finite sequence of row operations that do not affect the bottom rows – that is, an element of – left multiplication by which turns the top left block into the identity matrix.
Next, apply an element of to turn the bottom left block into zeroes, and then the bottom right block into the identity matrix. Finally, apply an element of
to turn the top left block into the allzero matrix. We have shown that
can be turned into the identity matrix by multiplying it by an element of , thus .To see that this is optimal, it is enough to consider and show that does not satisfy .
Algebraically, it is easy to see that . Namely, after applying a row operation that modifies only the second row (element of ) to the identity matrix, the first row is still so the second row cannot be , and thus an application of cannot turn the resulting matrix into . ∎
In the case , one can also verify geometrically by staring intently at a square.
Special linear groups also have alternation diameter in an obvious sense, by the same proof as for . (This fact does not directly fit the framework of this paper, in that the author does not know whether can be seen (in a natural way) as the automorphism group of a product object in a category.)
3.4 Graphs
We mention a related result from graph theory. The box product , sometimes called the Cartesian product of and defined by
(though it is not the categorytheoretic product in the usual category of simple graphs) admits unique prime decompositions for finite connected graphs, and automorphisms of are essentially entirely determined by and (and a bit of counting) in the sense that if we decompose and into their prime factors, every automorphism consists of a permutation of the factors (with respect to a fixed identification of isomorphic factors), followed by separate permutations of the factors. In this sense, connected graphs with respect to box product have “bounded alternation diameter up to reordering of prime factors”. See [8] for details. For some related observations see Remark 1 and Section 6.
4 Left and right groups
In this section we perform the (rather trivial) diagram chasing and algebra required to show that and “make sense”, i.e. are actually subgroups, and are independent of the choice of the product object .
We give the diagrammatic definition of these subgroups. For a product object with defining projections and , write and write for the set of elements such that the leftmost diagram below commutes in Figure 1 (resp. for the set of elements such that the rightmost diagram commutes).
It is easy to see (by gluing diagrams, or by algebra) that and are submonoids of under composition. To see that they are groups, note that if and then . From this, it follows (symmetrically ) is indeed a subgroup of .
For an object in a category , the over category above , denoted , is the category whose objects are morphisms in (or simply the morphisms themselves), and morphisms from to are morphisms in such that the following diagram commutes:
C rrh drf & & D dlg
& B &
Applying some geometric transformations to this diagram reveals a similarity with and , and we can make the observation that the above proof that is a group actually shows that is the automorphism group of the morphism as an object of the over category above . Similarly is the automorphism group of in the over category above .
To see that the choice of the product object does not matter, suppose is another product of and , and the defining projections. By the universal property, there is a unique isomorphism such that the following diagram commutes:
& C [swap]ldπ_A’ rdπ_B’ ddϕ &
A & & B
& A ×B luπ_A [swap]ruπ_B &
Then is an isomorphism between and and thus gives an isomorphism of their automorphism groups in the over category , which as discussed are the groups (corresponding to the two different choices of the product object). The same applies to .
We summarize the discussion into a theorem. Define a group triple to be a triple of groups such that . We say two group triples and are isomorphic is an isomorphism such that .
Theorem 7.
Let be a category and let be a product of objects and with defining projections and . Then the sets and defined by the diagrams in Figure 1 are subgroups of under composition. The resulting triple does not depend on the choice of the product , up to isomorphism of group triples.
5 Undefined alternation diameter
5.1 Topological spaces
It seems that, not surprisingly, in typical categories with a lot of structure, the alternation diameter is undefined for the whole category, that is, left and right automorphisms do not generate all others. We give in Top a nontrivial example (the plane) of undefined alternation diameter, and also a trivial example (the square ), and in Pos we exhibit an object with has trivial left and right groups, but nontrivial (though not far from trivial) automorphism group.
In the category Top of topological spaces and continuous functions, the automorphism group has undefined alternation diameter, since the left border and the top border can be exchanged by an automorphism, but this obviously cannot be done by or . We state this as a metalemma.^{3}^{3}3We add “meta” to distinguish this from the lemma which would be obtained by replacing “nice” by the best possible list of necessary properties, which can be deduced from the proof.
Proof.
We have by the definition of , and since is definable. Since is a group action and , we must have . Similarly, and . The flip is in by the universal property of (see Lemma 5 for a diagrammatic deduction). Since it does not (setwise) stabilize , the subgroups do not generate it, thus they do not generate . ∎
Corollary 4.
The category Top has undefined alternation diameter.
For homogenous spaces the question is more interesting. Let us show that also has undefined alternation diameter, by showing that rowwise and columnwise homeomorphisms cannot untangle sufficiently wild homeomorphisms in finite time.
Theorem 8.
The automorphism group has an element that is not in for any .
Proof.
To agree with our matrix convention ( permutes the Rows), draw the axes of the plane so that the second is the horizontal axis (lefttoright) and the first axis is vertical (topdown).
Consider a homeomorphism . Let be the unit speed path from to , and consider the image of this path, which is a path from to . Then is a path from to for any homeomorphism .
Now, cut out a small compact neighborhood of and consider the sequence in which traverses the rays in cardinal directions eminating from , before it first enters (ignoring repeated crossings of the same ray). Let , be the (finite) word thus obtained. By the intermediate value theorem, between occurrences of and there is an occurrence of or .^{4}^{4}4The word may depend on the choice of , and there need not be a best possible choice for which this word is the longest. What is important is that some choice gives a long word. Formally, one can consider the set of all words that correspond to some choice of to obtain a more canonical invariant.
Now consider for some , and consider the corresponding word , computed up to the neighborhood . Observe that changes either the orientation of all rows, or none of the rows. Then in , we have at least as many alternations between and as in . Similarly, an application of cannot decrease the number of alternations between and .
It follows that if the path from to is mapped by to a path having as a subsequence of the word corresponding to some choice of a small neighborhood of , the corresponding spiral in each homeomorphism in has a corresponding word with subsequence , thus is not the identity map, as the identity map preserves , and does not spiral with respect to any choice of .
Of course, the points are not in any way special. By including such spirals of all finite diameters in our homeomorphism by twisting horizontal paths from to , we obtain a homeomorphism with undefined alternation diameter. One can also have infinitely many twists around the same point: through the usual identification the homeomorphism defined by
is not in for any , as any appears as a subword of corresponding to a small enough choice of around the origin. See Figure 2 for a visualization of this homeomorphism. ∎
By a more careful analysis, one can construct homeomorphisms with word norm from the identity map in the homeomorphism group of w.r.t. the generators . It follows that for , is not equal to for any . We do not have a global understanding of this group .
5.2 Posets
In many categories, there are even easier ways to find undefined alternation diameter than Metalemma 1. We now prove that under rather general assumptions, the flip automorphism used from Metalemma 1 is not generated by and . We show how to apply them to some (finite) examples in Pos.
Lemma 5.
Let be a category and . If is trivial and the first and second canonical projections are distinct, then has undefined alternation diameter.
Proof.
Let be the defining left projection and be the defining right projection. Define the flip automorphism as follows: let and define a left and right projection, respectively , by . The universal property yields a morphism satisfying , . By the assumption, , so .
From the existence of we see that if
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