The Möbius function of separable and decomposable permutations
Abstract.
We give a recursive formula for the Möbius function of an interval in the poset of permutations ordered by pattern containment in the case where is a decomposable permutation, that is, consists of two blocks where the first one contains all the letters for some . This leads to many special cases of more explicit formulas. It also gives rise to a computationally efficient formula for the Möbius function in the case where and are separable permutations. A permutation is separable if it can be generated from the permutation 1 by successive sums and skew sums or, equivalently, if it avoids the patterns 2413 and 3142. A consequence of the formula is that the Möbius function of such an interval is bounded by the number of occurrences of as a pattern in . We also show that for any separable permutation the Möbius function of is either 0, 1 or .
Key words and phrases:
Möbius function, pattern poset, decomposable permutations, separable permutations.1. Introduction
Let be the set of permutations of the integers . The union of all for forms a poset with respect to pattern containment. That is, we define in if there is a subsequence of whose letters are in the same order of size as the letters in . For example, , because 2,5,3 appear in the same order of size as the letters in . We denote the number of occurrences of in by , for example , since 243, 253 and 153 are all the occurrences of the pattern 132 in 24153.
A classical question to ask for any combinatorially defined poset is what its Möbius function is. For our poset this seems to have first been mentioned explicitly by Wilf [8]. The first result in this direction was given by Sagan and Vatter [5], who showed that an interval of layered permutations is isomorphic to a certain poset of compositions of an integer, and they gave a formula for the Möbius function in this case. A permutation is layered if it is the concatenation of decreasing sequences, such that the letters in each sequence are smaller than all letters in subsequent sequences. Further results were given by Steingrímsson and Tenner [7], who showed that the Möbius function is 0 whenever the complement of the occurrences of in contains an interval block, that is, when has a segment of two or more consecutive letters that form a segment of values, where none of these consecutive letters belongs to any occurrence of in . One example of such a pair is , where the letters do not belong to any occurrence of in . Steingrímsson and Tenner [7] also described certain intervals where the Möbius function is either 1 or .
In this paper, we focus on permutations that can be expressed as direct sums or skew sums of smaller permutations. A direct sum of two permutations and , denoted by , is the concatenation , where is obtained by incrementing each element of by . For example, 31426587 can be written as a direct sum 3142+2143. Similarly, a skew sum is the concatenation where is obtained by incrementing by .
A permutation that can be written as a direct sum of two nonempty permutations is decomposable. The decomposition of a permutation is an expression in which each summand is indecomposable. A permutation is separable if it can be obtained from the singleton permutation by iterating direct sums and skew sums (for an alternative definition see Section 2).
Our main result is a set of recurrences for computing the Möbius function when is decomposable. If is the decomposition of , then these recurrences express in terms of Möbius functions involving the summands .
In the special case when is separable, these recurrences provide a polynomialtime algorithm to compute . These recurrences also allow us to obtain an alternative combinatorial interpretation of the Möbius function of separable permutations, based on the concept of ‘normal embeddings’. This interpretation of generalizes previous results of Sagan and Vatter [5] for layered permutations.
Using these expressions of the Möbius function in terms of normal embeddings, we derive several bounds on the values of for and separable. In [7], Steingrímsson and Tenner conjectured that for permutations and avoiding the pattern 132 (or any one of the patterns 213, 231, 312) the absolute value of the Möbius function of the interval is bounded by the number of occurrences of in . We prove this conjecture for the more general class of separable permutations (for arbitrary and this bound does not hold in general). In particular, if has a single occurrence of then is either 1, 0 or . We also prove a generalization of another conjecture mentioned in [7], showing that for any separable permutation , is either 1, 0 or .
For a nonseparable decomposable permutation , our recurrences are not sufficient to compute the value of . Nevertheless, they allow us to give short simple formulas in many special cases.
For instance, suppose that is indecomposable and that is decomposable and of length at least 3. Then we show that can only be nonzero if all the blocks in the decomposition of are equal to the same permutation , except possibly the first and the last block, which may be equal to . In such cases, equals , where is the number of blocks of that are equal to .
As another simple example, our results imply that when and are permutations with decompositions and , with and both different from , then if , and if .
The paper is organized as follows: In the next section we provide necessary definitions. In Section 3 we present the main results, the recursive formulas for reducing the computation of the Möbius function of decomposable permutations to that of indecomposable permutations. Section 4 deals with the case of separable permutations and their normal embeddings. Finally, in Section 5 we mention some open problems, in particular questions about the topology of the order complexes of intervals in our poset, which we have not dealt with in the present paper.
2. Definitions and Preliminaries
An interval in a poset is the set . In this paper, we deal exclusively with intervals of the poset of permutations ordered by pattern containment.
The Möbius function of an interval is uniquely defined by setting for all and requiring that
(1) 
for every . When , we define to be zero.
An equivalent definition is given by Philip Hall’s Theorem [6, Proposition 3.8.5], which says that
(2) 
where is the set of chains in that contain both and , denotes the length of the chain , and is the number of such chains of length in . A chain of length in a poset is a set of pairwise comparable elements . For details and further information, see [6].
The direct sum, , of two nonempty permutations and is the permutation obtained by concatenating and , where is with all letters incremented by the number of letters in . A permutation that can be written as a direct sum of two nonempty permutations is decomposable, otherwise it is indecomposable. Examples are , and 231, which is indecomposable. In the skew sum of and , denoted by , we increment the letters of by the length of to obtain and then concatenate and . For example, . We say that a permutation is skewindecomposable if it cannot be written as a skew sum of smaller permutations.
A decomposition of is an expression in which each summand is indecomposable. The summands will be called the blocks of . Every permutation has a unique decomposition (including an indecomposable permutation , whose decomposition has a single block ).
A permutation is separable if it can be generated from the permutation by iterated sums and skew sums. In other words, a permutation is separable if and only if it is equal to or it can be expressed as a sum or skew sum of separable permutations.
Being separable is equivalent to avoiding the patterns 2413 and 3142, that is, containing no occurrences of either. Separable permutations have nice algorithmic properties. For instance, Bose, Buss and Lubiw [2] have shown that it can be decided in polynomial time whether when and are separable, while for general permutations the problem is NPhard.
It is sometimes convenient to allow permutations to have zero length, while in other situations, the permutations are assumed to be nonempty. The unique permutation of length 1 is denoted by , and the unique permutation of length 0 is denoted by . We make it a convention that the permutation is neither decomposable nor indecomposable. In other words, whenever we say that a permutation is decomposable (or indecomposable), we automatically assume that is nonempty.
Suppose that is a nonempty permutation with decomposition . For an integer , we let denote the sum and let denote the sum . An empty sum of permutations is assumed to be equal to , so in particular . Any permutation of the form for some will be called a prefix of , and any permutation of the form is a suffix of . Note that , , and it is easily seen that for any .
3. The Main Results
Throughout this section, we assume that is a nonempty permutation with decomposition and that is a decomposable permutation (so and, in particular, is nonempty). The goal in this section is to prove a set of recurrences that allow us to express the Möbius function in terms of the values of the form , where is a block of and is a sum of consecutive blocks of . Note that may itself be indecomposable, in which case is equal to and .
There are two main recurrences to prove, dealing respectively with the cases and .
Proposition 1 (First Recurrence).
Let and be nonempty permutations with decompositions and , where . Suppose that . Let be the largest integer such that all the blocks are equal to , and let be the largest integer such that all the blocks are equal to . Then
Note that the suffixes , and in the statement of Proposition 1 may be empty. This first recurrence shows how to compute the Möbius function when starts with for some . The second recurrence takes care of the remaining cases, that is, when does not start with .
Proposition 2 (Second Recurrence).
Let and be nonempty permutations with decompositions and , where . Suppose that . Let be the largest integer such that all the blocks are equal to . Then
(3) 
Note that Propositions 1 and 2 remain true when all the direct sums are replaced with skew sums, and the decompositions are replaced with skew decompositions. To see this, it is enough to observe that if denotes the reversal of (i.e., is the permutation obtained by reversing the order of elements of ), then for any and , since and are isomorphic posets.
Before we prove the above two recurrences, we give three corollaries to provide some idea of how the second recurrence can be used. When we write we mean a sum with summands.
Corollary 3.
Let , and be as in Proposition 2, and suppose that is indecomposable, that is, . Then
Proof.
Since , Equation (3) takes the form
where the last equality follows from the fact that is equal to 0 whenever has more than one block.
We have when , when , and otherwise. In particular, can only be nonzero either when and , or when and . ∎
Corollary 3 implies that if is indecomposable and is decomposable, then almost always , since the two exceptions for given in the corollary are of a proportion that clearly goes to zero among decomposable permutations as their length goes to infinity.
Corollary 4.
With and as in Proposition 2, assume that and decompose into exactly two blocks, with and , and that . Then
Proof.
If, on the other hand, , then Equation (3) states that is equal to
Remark 5.
An obvious question to ask is whether the product formula , in the case when , is a result of the interval being isomorphic to the direct product of the intervals and . Although this seems to occur frequently, it does not hold in general.
The following corollary is an immediate consequence of Proposition 1 (the case when ).
Corollary 6.
Suppose and are permutations of length at least two, such that neither begins with 1. Then .
Both recurrences (Propositions 1 and 2) are proved using arguments that involve cancellation between certain types of chains in the poset of permutations. Let us first introduce some useful notation. For a chain of permutations let denote the length of , which is one less than the number of elements of . The weight of , denoted by , is the quantity . If is any set of chains, then the weight of is defined by
Recall that is the set of all the chains from to that contain both and . We know that , by Philip Hall’s Theorem.
For a chain and a permutation , we let denote the chain . The chain is defined analogously.
3.1. Proof of the first recurrence
Let us now turn to the proof of Proposition 1. Suppose that , , , , , and are as in the statement of the proposition. For a permutation , define the degree of , denoted by , to be the largest integer such that can be expressed as for some (possibly empty) permutation . In particular, we have and .
Let be a chain of permutations. We say that a permutation is the pivot of the chain , if for each , and for each . In other words, the pivot is the element of the chain with minimum degree, and if there are more elements of minimum degree, the pivot is the largest of them.
Let denote the permutation . Obviously and . We partition the set of chains into three disjoint subsets, denoted by , and , and we compute the weight of each subset separately. A chain belongs to if its pivot is the permutation , the chain belongs to if its pivot is the permutation , and belongs to otherwise. We now separate the main steps of the proof into independent claims.
Claim 7.
If (so ), then is empty. Otherwise, .
Proof.
Obviously, if , then no chain from to can have for pivot, because the pivot must have minimal degree among the elements of the chain. Thus, is empty.
Assume now that . We show that there is a lengthpreserving bijection between the set of chains and the set of chains . Indeed, take any chain , and create a new chain . Then is a chain from to , and since every element of has degree at least , while has degree exactly , we see that is the pivot of . Hence .
On the other hand, if is any chain from , we see that each element of has degree at least , because has degree and is the pivot of . Thus, every element is of the form for some , and hence there exists a chain such that . Since is clearly injective and lengthpreserving, we conclude that , as claimed. ∎
Claim 8.
If (so ), then is empty. Otherwise, .
Proof.
If then cannot be the pivot of any chain containing and is empty.
Assume now that . We will describe a parityreversing bijection between the set of chains and the set of chains . Take a chain . Define a new chain by . Notice that is a chain from to whose pivot is and whose length is equal to the length of . Define the chain by . Then the chain belongs to and has length . It is again easy to see that is a bijection between and , which shows that
as claimed. ∎
Claim 9.
.
Proof.
We construct a parityreversing involution . Let be a chain from , let be its pivot, and let be the successor of in . By definition of , is not equal to , so is well defined. From the definition of a pivot, we know that . Let us distinguish two cases:

If , we define a new chain by . Note that in this case, we know that is different from , because otherwise would be equal to , contradicting the definition of . Thus, . Note that is a pivot of .

If , then we easily deduce that (recall that ). We then define a new chain , in which the new element is inserted between and .
The mapping is easily seen to be an involution on the set which preserves the pivot and maps oddlength chains to evenlength chains and vice versa. This shows that , as claimed. ∎
3.2. Proof of the second recurrence
It remains to prove Proposition 2. The proof is again based on cancellation among the chains from to . Before stating the proof, we need more terminology and several lemmas.
Let , and be any permutations. We say that is a tight subpermutation of , denoted by , if but is not contained in . We say that a chain is tight if for every . Let be the set of all the tight chains from to .
The following simple properties of tightness follow directly from the definitions, and they are presented without proof.
Lemma 10.
For arbitrary permutations , , and , we have if and only if .
Lemma 11.
If is a nonempty indecomposable permutation, and if and are arbitrary permutations, then if and only if .
The next lemma shows the relevance of tightness for the computation of .
Lemma 12.
Let be a permutation with decomposition . Let be a nonempty indecomposable permutation, and let be any permutation.

If , then .

If , then .
Proof.
Let us first deal with the first claim of the lemma. Let us define to be the set of all the chains from to that are not tight. The first part of the lemma is equivalent to saying that . To prove this, we find a parityreversing involution on the set .
Consider a chain . Since is not tight, there is an index such that . Fix the smallest such value of . We distinguish two cases: either , or .
If , define a new chain
On the other hand, if , define a new chain
Note that, since we assume that and that is indecomposable, we know that is not equal to . Moreover, in the chain the element is not a tight subpermutation of its successor in the chain. Thus, we see that is a chain from . It is easy to see that is an involution, and that it reverses the parity of the length of the chain, showing that . This proves the first part of the lemma.
Let us prove the second part. Assume that , that is, . Consider a chain from to , and let be the elements of . We say that the chain is almost tight if its second largest element is equal to and if for each . Note that an almost tight chain is never tight, because is not a tight subpermutation of .
We partition the set into three disjoint sets , , and , where is the set of tight chains, is the set of almost tight chains, and contains the chains that neither tight nor almost tight.
Consider again the mapping defined in the proof of the first part of the lemma. This mapping, restricted to the set , is easily seen to be a parityreversing involution on , which shows that . This means that .
Furthermore, note that an almost tight chain from to consists of a tight chain from to followed by , and conversely, any tight chain from to can be extended to an almost tight chain from to by adding the element . Thus, we see that . This implies that , as claimed. ∎
The next lemma is an easy consequence of Lemma 12.
Lemma 13.
Let be a permutation with decomposition . Let be an indecomposable permutation, and let be any permutation. Let be the largest integer such that the blocks are all equal to . Then
Proof.
Before we proceed towards the proof of Proposition 2, we need to introduce more definitions. Let be a permutation with decomposition into indecomposable blocks, let be any permutation. Let be a chain of permutations, with elements . We express each element of the chain as a sum of two permutations, called head and tail, denoted respectively as and , with . The head and tail are defined inductively as follows: for , we have and we define and .
Suppose now that the head and tail of have been already defined, and let us define head and tail of . Let us put , and assume that has decomposition into indecomposable blocks. Let be the smallest integer such that . It then follows that . We define and . In other words, the tail of is its longest suffix that is contained in the tail of .
If the chain is clear from the context, we write and instead of and . Note that and .
We say that the chain of length is split if there is an index such that and . Such an index is then necessarily unique. The next lemma demonstrates the relevance of these notions.
Lemma 14.
Let be a permutation with decomposition such that . Let be an arbitrary permutation. Let be the set of all the chains from which are split and tight. Then .
Proof.
By the first part of Lemma 12, we know that is equal to , that is, to the total weight of all the tight chains from to . Define the set of all the tight, nonsplit chains from to .
To prove the lemma, we need to show that . To achieve this, we again use a parityreversing involution on the set . Consider a chain with elements . Since is not split, there must exist an index such that either

and , or

and .
Fix such an index as large as possible and distinguish two cases depending on which of the two abovementioned possibilities occur for this index .
Case (1). Assume that and . Let us write , , , and , so we have and . Define a permutation , and a new chain . Note that since is a tight chain, and in particular , we also have , and hence is a tight chain as well.
We need to prove that , which follows easily from the following claim.
Claim 15.
Each permutation of has the same head and tail in as in . The permutation has head and tail in .
Proof of Claim 15.
It is clear that the claim holds for all the permutations that are greater than .
It is also easy to see that the claim holds for . Indeed, the successor of in is the permutation , whose tail is . Since the tail of cannot be greater than and since , it follows that the tail of is and its head is .
Let us now consider the permutation . The successor of in is the permutation , and the successor of in is the permutation . Since the two successors have the same tail , and since the tail of a permutation only depends on the tail of its successor, we see that has the same tail (and hence also the same head) in as in .
From these facts, the claim immediately follows. ∎
We may now conclude that , and turn to the second case of the proof of the lemma.
Case (2). Assume now that and . Let us define , , , and . In particular, , , and . Define the chain .
We claim that is tight. To see this, it is enough to prove . Assume, for a contradiction, that . In any occurrence of inside , the prefix must occur inside , otherwise we get a contradiction with the assumption that is the tail of . This shows that , and hence , contradicting the assumption that is tight. To finish the proof of the lemma, we need one more claim.
Claim 16.
Each permutation of has the same head and tail in as in .
Proof of Claim 16.
It is enough to prove the claim for the permutation , because any other permutation of has the same successor in as in . For , the claim follows from the fact that the successor of in has the same tail as the successor of in . This completes the proof of the claim. ∎
We now see that even in this second case, belongs to .
Combining the two cases described above, we see that is a parityreversing involution of the set . This means that , and consequently, , as claimed. This completes the proof of the lemma. ∎
Finally, we can prove Proposition 2. Assume that is a permutation with decomposition and that is a permutation with decomposition , where and . Let be the largest integer such that all the blocks are equal to . Recall that our goal is to prove identity (3), which reads as follows:
Let be the set of tight split chains from to . From Lemma 14, we know that . For a chain , let be the tail of the element , which is the smallest element in the chain. By definition, is a suffix of , that is, it is equal to for some value of . Define, for each , the set of chains
The sets form a disjoint partition of . We will now compute the weight of the individual sets .
Claim 17.
Let be a chain from . Every element of has nonempty head. Consequently, is never equal to , and hence is empty.
Proof.
Suppose that has an element with empty head. Let be the largest such element. By definition, the element has head equal to , so . In particular, has a successor in , and has nonempty head. Let and be the head and tail of . By assumption, is nonempty, which means that . Moreover, , because is its own tail. This means that , which is impossible because the chains in are assumed to be tight.
This shows that every element of has nonempty head, and the rest of the claim follows directly. ∎
Claim 17 implies that , and hence . It remains to determine the value of for .
Fix an integer . Define , ,