# 435-495-1-PB - Cyclic Derangements Sami H Assaf Department...

• Test Prep
• 14

This preview shows page 0 - 1 out of 14 pages.

Subscribe to view the full document.

Unformatted text preview: Cyclic Derangements Sami H. Assaf∗ Department of Mathematics MIT, Cambridge, MA 02139, USA [email protected] Submitted: Apr 16, 2010; Accepted: Oct 26, 2010; Published: Dec 3, 2010 Mathematics Subject Classification: 05A15; 05A05, 05A30 Abstract A classic problem in enumerative combinatorics is to count the number of derangements, that is, permutations with no fixed point. Inspired by a recent generalization to facet derangements of the hypercube by Gordon and McMahon, we generalize this problem to enumerating derangements in the wreath product of any finite cyclic group with the symmetric group. We also give q- and (q, t)-analogs for cyclic derangements, generalizing results of Gessel, Brenti and Chow. 1 Derangements A derangement is a permutation that leaves no letter fixed. Algebraically, this is an element σ of the symmetric group Sn such that σ(i) 6= i for any i, or, equivalently, no cycle of σ has length 1. Geometrically, a derangement is an isometry in Rn−1 of the regular (n − 1)-simplex that leaves no facet unmoved. Combinatorially, these are matrices with entries from {0, 1} such that each row and each column has exactly one nonzero entry and no diagonal entry is equal to 1. Let Dn denote the set of derangements in Sn , and let dn = |Dn |. The problem of counting derangements is the quintessential example of the principle of Inclusion-Exclusion [20]: dn = n! n X (−1)i i=0 i! . (1) For example, the first few derangement numbers are 0, 1, 2, 9, 44, 265. From (1) one can immediately compute that the probability that a random permutation has no fixed points is approximately 1/e. Another exercise that often accompanies counting derangements is to prove the following two term recurrence relation for n &gt; 2, dn = (n − 1) (dn−1 + dn−2 ) , ∗ (2) Partially supported by NSF Mathematical Sciences Postdoctoral Research Fellowship DMS-0703567 the electronic journal of combinatorics 17 (2010), #R163 1 with initial conditions d0 = 1 and d1 = 0; see [20]. From (2) one can derive the following single term recurrence for derangement numbers, dn = ndn−1 + (−1)n . (3) Recently, Gordon and McMahon [13] considered isometries of the n-dimensional hypercube that leave no facet unmoved. Algebraically, such an isometry is an element σ of the hyperoctahedral group Bn for which σ(i) 6= i for any i. Combinatorially, the problem then is to enumerate n × n matrices with entries from {0, ±1} such that each row and column has exactly one nonzero entry and no diagonal entry equals 1. Gordon and McMahon derive a formula for the number of facet derangements similar to (1), an expression of facet derangements in terms of permutation derangements, and recurrence relations for facet derangements similar to (2) and (3). In Section 2, we consider elements σ in the wreath product Cr ≀Sn , where Cr is the finite cyclic group of order r and Sn is the symmetric group on n objects. A cyclic derangement is an element of Cr ≀ Sn with no fixed point. Denote the set of cyclic derangements of (r) (r) (r) (r) Cr ≀ Sn by Dn , and denote their number by dn = |Dn |. Combinatorially, dn is also the number of matrices with entries from {0, 1, ζ, . . . , ζ r−1}, where ζ is a primitive rth root of unity, such that each row and each column has exactly one nonzero entry and no (r) diagonal entry equals 1. We derive a formula for dn that specializes to (1) when r = 1 and to the Gordon-McMahon formula for facet derangements when r = 2. We also give (r) an expression for dn in terms of dn as well as a two recurrence relations specializing to (2) and (3) when r = 1. Gessel [12] introduced a q-analog for derangements of Sn , q-counted by the major index, that has applications to character theory [17]. In Section 3, we give a q, t-analog for cyclic derangements of Cr ≀ Sn q-counted by a generalization of major index and tcounted by signs that specializes to Gessel’s formula at r = 1 and t = 1. Generalizing results in Section 2, we show that the cyclic q, t-derangements satisfy natural q, t-analogs of (1), (2) and (3). These results also generalize formulas of Garsia and Remmel [11] who first introduced q-analogs for (2) and (3) for r = 1 using a different (though equidistributed) permutation statistic. The analogs we present are similar to results of Chow [5] and Foata and Han [9] when r = 2, though the statistic we use differs slightly. Brenti [1] gave another q-analog for derangements of Sn q-counted by weak excedances and conjectured many nice properties for these numbers that were later proved by Canfield (unpublished) and Zhang [22]. More recently, Chow [6] and Chen, Tang and Zhao [4] independently extended these results to derangements of the hyperoctahedral group. In Section 4, we show that these results are special cases of cyclic derangements of Cr ≀ Sn q-counted by a generalization of weak excedances. The proofs for all of these generalizations are combinatorial, following the same reasoning as the classical proofs of (1) and (2). This suggests that studying derangements in this more general setting is very natural. In Section 5, we discuss possible directions for further study generalizing other results for permutations derangements. the electronic journal of combinatorics 17 (2010), #R163 2 2 Cyclic derangements Let Sn denote the symmetric group of permutations of a set of n objects, and let Cr denote the cyclic group of order r. The wreath product Cr ≀ Sn is the semi-direct product (Cr )×n ⋊ Sn , where Sn acts on n copies of Cr by permuting the coordinates. Let ζ be a generator for Cr , e.g. take ζ to be a primitive rth root of unity. We regard an element σ ∈ Cr ≀ Sn as a word σ = (ǫ1 s1 , . . . , ǫn sn ) where ǫi ∈ {1, ζ, . . . , ζ r−1} and {s1 , . . . , sn } = {1, . . . , n}. In this section we show that all of the usual formulas and proofs for classical derangement numbers generalize to these wreath products. We begin with (1), giving the following formula for the number of cyclic derangements. The two proofs below are essentially the same, though the first is slightly more direct while the latter will be useful for establishing q and q, t analogs. Theorem 2.1. The number of cyclic derangements in Cr ≀ Sn is given by n d(r) n = r n! n X (−1)i i=0 r i i! . (4) Inclusion-Exclusion Proof. Let Ai be the set of σ ∈ Cr ≀ Sn such that σi = +1 · i. Then |Aj1 ∩ · · · ∩ Aji | = r n−i(n − i)!, since the positions j1 , . . . , ji are determined and the remaining n − i positions may be chosen arbitrarily. Therefore by the Inclusion-Exclusion formula, we have (r) D = |Cr ≀ Sn | − |A1 ∪ · · · ∪ An | n n X X = r n n! − (−1)i−1 |Aj1 ∩ · · · ∩ Aji | i=1 j1 &lt;···&lt;ji n n   X X n (−1)i (−1)i r n−i (n − i)! = r n n! = . i i! i r i=0 i=0 M¨obius Inversion Proof. For S = {s1 &lt; s2 &lt; · · · &lt; sm } ⊆ [n] and σ ∈ Cr ≀ SS , define the reduction of σ to be the permutation in Cr ≀ Sm that replaces ǫi si with ǫi i. If σ ∈ Cr ≀ Sn (r) has exactly k fixed points, then define dp(σ) ∈ Dn−k to be the reduction of σ to the non-fixed points. For example, dp(5314762) =reduction of 53172 = 43152 and any signs are carried over.  The map dp is easily seen to be an nk to 1 mapping of cyclic permutations with (r) exactly k fixed points onto Dn−k . Therefore n   X n (r) n r n! = d . (5) k n−k k=0 The theorem now follows by M¨obius inversion [20]. the electronic journal of combinatorics 17 (2010), #R163 3 An immediate consequence of Theorem 2.1 is that the probability that a random element of Cr ≀ Sn is a derangement is approximately e−1/r . This verifies the intuition that as n and r grow, most elements of Cr ≀ Sn are in fact derangements. Table 1 gives values (r) for dn for r 6 5 and n 6 6. (r) Table 1: Cyclic derangement numbers dn for r 6 5, n 6 6. r\n 0 1 2 3 4 5 6 1 1 0 1 2 9 44 265 2 1 1 5 29 233 2329 27949 3 1 2 12 116 1393 20894 376093 4 1 3 25 299 4785 95699 2296777 5 1 4 41 614 12281 307024 9210721 We also have the following generalization of [13](Proposition 3.2), giving a formula relating the number of cyclic derangements with the number of permutation derangements. Proposition 2.2. For r &gt; 2 we have d(r) n = n   X n i=0 i r i (r − 1)n−i di (6) where di = |Di | is the number of derangements in Si . Proof. For S ⊂ {1, 2, . . . , n} of size i, the number of derangements σ ∈ Cr ≀ Sn with |σ(j)| = j if and only if j 6∈ S is equal to di r i (choose a permutation derangement of these indices and a sign for each) times (r − 1)n−i (choose a nonzero sign for indices k such that  |σk | = k). There are ni choices for each such S, thereby proving (6). Gordon and McMahon [13] observed that for r = 2, the expression in (6) is precisely the rising 2-binomial transform of the permutation derangement numbers as defined by Spivey and Steil [18]. In general, this formula gives an interpretation for the mixed rising r-binomial transform and falling (r − 1)-binomial transform of the permutation derangements numbers. The following two term recurrence relation for cyclic derangements generalizes (2). We give two proofs of this recurrence, one generalizing the classical combinatorial proof of (2) and the other using the exponential generating function for cyclic derangements. Theorem 2.3. For n &gt; 2, the number of cyclic derangements satisfies (r) (r) d(r) n = (rn − 1)dn−1 + r(n − 1)dn−2 , (r) (7) (r) with initial conditions d0 = 1 and d1 = r − 1. the electronic journal of combinatorics 17 (2010), #R163 4 (r) Combinatorial Proof. For σ ∈ Dn , consider the cycle decomposition of underlying permutation |σ| ∈ Sn . There are three cases to consider. Firstly, if n is in a cycle of length (r) one, then there are r − 1 choices for ǫn 6= 1 and dn−1 choices for a derangement of the remaining n − 1 letters. If n is in a cycle of length two in |σ|, then ǫn may be chosen freely in r ways, there are (n − 1) choices for the other occupant of this two-cycle in |σ| and (r) dn−2 choices for a cyclic derangement of the remaining n − 2 letters. Finally, if n is in a cycle of length three or more, then there are r choices for ǫn , n − 1 possible positions for (r) n in |σ|, and dn−1 choices for a derangement of the remaining n − 1 letters. Combining these cases, we have (r) (r) (r) d(r) n = (r − 1)dn−1 + r(n − 1)dn−2 + r(n − 1)dn−1 , from which (7) now follows. Algebraic Proof. First note that for fixed r, e−x = 1 − rx X (−1)i ! X ! xi r j xj i! i&gt;0 j&gt;0 X X n (−1)i r j X xn xn = d(r) = n i i! n! n&gt;0 n&gt;0 i+j=n is the exponential generating function for the number of cyclic derangements. Denoting this function by D (r) (x), we compute  xn X (r) (r) (rn − 1)dn−1 + (rn − r)dn−2 n! n n X X (r) X (r) xn X (r) xn x (r) x =r dn−1 − dn−1 + r dn−2 −r dn−2 (n − 1)! n! (n − 1)! n! Z Z ZZ = rxD (r) (x) − D (r) (x) + rx D (r) (x) − r D (r) (x) = D (r) (x), from which the recurrence now follows. Finally, we have the following simple recurrence relation generalizing (3). Corollary 2.4. For n &gt; 1, the number of cyclic derangements satisfies (r) n d(r) n = rndn−1 + (−1) , (8) (r) with initial condition d0 = 1. This recurrence follows by induction from the formula in Theorem 2.1 or the two term recurrence in Theorem 2.3, though it would be nice to have a direct combinatorial proof similar to that of Remmel [16] for the case r = 1. the electronic journal of combinatorics 17 (2010), #R163 5 3 Cyclic q, t-derangements by major index Gessel [12] derived a q-analog for the number of permutation derangements as a corollary to a generating function formula for counting permutations in Sn by descents, major index and cycle structure. In order to state Gessel’s formula, we begin by recalling the q-analog of a positive integer i given by [i]q = 1 + q + · · · + q i−1. In the same vein, we also have [i]q ! = [i]q [i − 1]q · · · [1]q , where [0]q ! is defined to be 1. For a permutation σ ∈ Sn , the descent set of σ, denoted by Des(σ), is given by Des(σ) = {i | σ(i) &gt; σ(i + 1)}. MacMahon [14] used the descent set to define a fundamental permutation statistic, called the major index and denoted by maj(σ), given by P maj(σ) = i∈Des(σ) i. Finally, recall MacMahon’s formula [14] for q-counting permutations by major index, X q maj(σ) = [n]q !. σ∈Sn Along these lines, define the q-derangement numbers, denoted by dn (q), by X dn (q) = q maj(σ) . (9) σ∈Dn Gessel showed that the q-derangement numbers for Sn are given by dn (q) = [n]q ! n X (−1)i i=0 [i]q ! i q (2) . (10) A nice bijective proof of (10) is given by Wachs in [21], where she constructs a descentpreserving bijection between permutations with specified fixed points and shuffles of two permutations and then makes use of a formula of Garsia and Gessel [10] for q-counting shuffles. Garsia and Remmel [11] also studied q-derangement numbers using the inversion statistic which is known to be equi-distributed with major index. Gessel’s formula was generalized to the hyperoctahedral group by Chow [5] with further results by Foata and Han [9] using the flag major index statistic. Faliharimalala and Zeng [8] recently found a generalization of (10) to Cr ≀ Sn also using flag major index Here, we give a different generalization to Cr ≀ Sn by q-counting with a different major index statistic and t-counting by signs. We begin with a generalized notion of descents derived from the following total order on elements of (Cr × [n]) ∪ {0}: ζ r−1n &lt; · · · &lt; ζn &lt; ζ r−1 (n−1) &lt; · · · &lt; ζ1 &lt; 0 &lt; 1 &lt; 2 &lt; · · · &lt; n (11) For σ ∈ Cr ≀ Sn , an index 0 6 i &lt; n is a descent of σ if σi &gt; σi+1 with respect to this total ordering, where we set σ0 = 0. Note that for σ ∈ Sn , this definition agrees P with the classical one. As with permutations, define the major index of σ by maj(σ) = i∈Des(σ) i. We also want to track the signs of the letters of σ, which we do with the statistic sgn(σ) defined by sgn(σ) = e1 + · · · + en , where σ = (ζ e1 s1 , . . . , ζ en sn ) . This is a generalization of the same statistic introduced by Reiner in [15]. the electronic journal of combinatorics 17 (2010), #R163 6 Remark 3.1. There is another total ordering on elements of (Cr ×[n])∪{0} that is equally as natural as the order given in (11), namely ζ r−1 n &lt; · · · &lt; ζ r−1 1 &lt; ζ r−2n &lt; · · · &lt; ζ1 &lt; 0 &lt; 1 &lt; 2 &lt; · · · &lt; n. (12) While using this alternate order will result in a different descent set and major index for (r) a given element of Cr ≀ Sn , the distribution of descent sets over Cr ≀ Sn and even Dn is the same with either ordering. In fact, there are many possible total orderings that refine (r) the ordering on positive integers and yield the same distribution over Cr ≀ Sn and Dn , since the proof of Theorem 3.2 carries through easily for these orderings as well. We have chosen to work with the ordering in (11) primarily to facilitate the combinatorial proof of Theorem 3.6. A first test that these statistics are indeed natural is to see that the q, t enumeration of elements of Cr ≀ Sn by the major index and sign gives X q maj(σ) tsgn(σ) = [r]nt [n]q !, (13) σ∈Cr ≀Sn which is a natural (q, t)-analog for r n n! = |Cr ≀ Sn |. Analogous to (9), define the cyclic (q, t)-derangement numbers by X d(r) (q, t) = q maj(σ) tsgn(σ) . n (14) (r) σ∈Dn (1) In particular, dn (q, t) = dn (q) as defined in (9). In general, we have the following (q, t)analog of (4) that specializes to (10) when r = 1. Theorem 3.2. The cyclic (q, t)-derangement numbers are given by d(r) n (q, t) = [r]nt [n]q ! n X (−1)i (2i ) q . [r]it [i]q ! i=0 (15) The proof of Theorem 3.2 is completely analogous to Wachs’s proof [21] for Sn which generalizes the second proof of Theorem 2.1. To begin, we define a map ϕ that is a sort of inverse to the map dp. Say that σi is a subcedant of σ if σi &lt; i with respect to the total order in (11), and let sub(σ) denote the number of subcedants of σ. For σ ∈ Cr ≀ Sm , let s1 &lt; · · · &lt; ssub(σ) be the absolute values of subcedants of σ. If σ has k fixed points, let f1 &lt; · · · &lt; fk be the fixed points of σ. Finally, let x1 &gt; · · · &gt; xm−sub(σ)−k be the remaining letters in [m]. For fixed n, ϕ(σ) is obtained from σ by the following replacements: ǫi si 7→ ǫi i fi 7→ i + sub(σ) xi 7→ n − i + 1. For example, for n = 8 we have ϕ(3, 2, −6, 5, 4, −1) = (7, 4, −3, 8, 2, −1). For disjoint sets A and B, a shuffle of α ∈ Cr ≀ SA and β ∈ Cr ≀ SB is an element of Cr ≀ SA∪B containing α and β as complementary subwords. Let Sh(α, β) denote the set of shuffles of α and β. Then we have the following generalization of [21](Theorem 2). the electronic journal of combinatorics 17 (2010), #R163 7 (r) Lemma 3.3. Let α ∈ Dn−k and γ = (sub(α) + 1, , . . . , sub(α) + k). Then the map ϕ gives ∼ a bijection {σ ∈ Cr ≀ Sn | dp(σ) = α} −→ Sh(ϕ(α), γ) such that Des(ϕ(σ)) = Des(σ) and sgn(ϕ(σ)) = sgn(σ). Proof. The preservation of sgn is obvious by construction. To see that the descent set is preserved, note that ǫi 6= 1 only if σi is a subcedant and the relative order of subcedants, fixed points and the remaining letters is preserved by the map. It remains only to show that ϕ is an invertible map with image Sh(ϕ(α), γ). For this, the proof of [21](Theorem 2) carries through verbatim thanks to the total ordering in (11). The only remaining ingredient to prove Theorem 3.2 is the formula of Garsia and Gessel [10] for q-counting shuffles. Though their theorem was stated only for Sn , the result holds in this more general setting. Lemma 3.4. Let α and β be cyclic permutations of lengths a and b, respectively, and let Sh(α, β) denotes the set of shuffles of α and β. Then   X a + b maj(α)+maj(β) sgn(α)+sgn(β) maj(σ) sgn(σ) q t = q t . (16) a q σ∈Sh(α,β) Proof of Theorem 3.2. For γ as in Lemma 3.3, observe maj(γ) = 0 = sgn(γ). Thus applying Lemma 3.3 followed by Lemma 3.4 allows us to compute X [r]nt [n]q ! = q maj(σ) tsgn(σ) = σ∈Cr ≀Sn n X X X q maj(σ) tsgn(σ) k=0 α∈D(r) dp(σ)=α n−k = n X X X q maj(σ) tsgn(σ) k=0 α∈D(r) σ∈Sh(ϕ(α),γ) n−k n X X n = q maj(α) tsgn(α) k q (r) k=0 α∈Dn−k = n   X n k=0 k (r) dk (q, t). q Applying M¨obius inversion to the resulting equation yields (15). Proposition 2.2 also generalizes, though in order to prove the generalization we rely on the (q, t)-analog of the recurrence relation given in Theorem 3.6 below. It would be nice to have a combinatorial proof as well. the electronic journal of combinatorics 17 (2010), #R163 8 Proposition 3.5. For r &gt; 2 we have d(r) n (q, t) = n   X n i i=0 n−i−1 Y [r]it ! ([r]t − q k ) di (q, t). q k=0 (17) The recurrence relation (7) in Theorem 2.3 also has a natural (q, t)-analog. Note that this specializes to the formula of Garsia and Remmel [11] in the case r = 1. The proof is combinatorial, though it would be nice to have a generating function proof as well. Theorem 3.6. The cyclic (q, t)-derangement numbers satisfy  (r) (r) n−1 d(r) dn−1 (q, t) + q n−1 [r]t [n − 1]q dn−2 (q, t), n (q, t) = [r]t [n]q − q (r) (18) (r) with initial conditions d0 (q, t) = 1 and d1 (q, t) = [r]t − 1. Proof. As in the combinatorial proof of Theorem 2.3, consider the cycle decomposition of underlying permutation |σ| ∈ Sn . We consider the same three cases, this time tracking the major index and sign. If n is in a cycle of length one, then the r − 1 choices for ǫn 6= 1 contribute t[r − 1]t , and there will necessarily be a descent in position n − 1, thus contributing q n−1 . This case then contributes (r) t[r − 1]t q n−1 dn−1 (q, t). If n is in a cycle of length two in |σ|, then ǫn is arbitrary contributing [r]t , and the n − 1 choices for the other occupant of the cycle will add at least n − 1 to the major index beyond the major index of the permutation with these two letters removed. This contributes a term of q n−1 [n − 1]q , making the total contribution (r) [r]t q n−1 [n − 1]q dn−2 (q, t). Finally, if n is in a cycle of length three or more, then each of the n − 1 possible positions for n in |σ| increases the major index by one, contributing a factor of [n − 1]q . The r choices for ǫn again contribute [r]t , giving a to...
View Full Document

• Fall '16
• Statistics, Sn, Electronic journal, cyclic derangements

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern