On the joint distribution of descents and signs of permutations

We study the joint distribution of permutations by descents and sign. This has an application to riffle shuffling: for large decks the sign is close to random after a single shuffle. We derive generating functions for the Eulerian distribution refined according to sign, and use them to give two proofs of central limit theorems for positive and negative Eulerian numbers.


Introduction
The distribution of descents over all permutations is known as the Eulerian distribution, and the number of permutations of n with a given number of descents is known as an Eulerian number. The Eulerian numbers are ubiquitous in combinatorics; see [14] for an entire book devoted to Eulerian numbers, as well as various refinements and generalizations of them.
In this paper we study one such refinement, namely the joint distribution of descents and sign, which has been studied, e.g., by Tanimoto [17], and more recently by Dey and Sivasubramanian [5]. This work is also in some sense an extension of work of Loday [13] and Desarménien and Foata [4], who studied "signed" Eulerian numbers.
1.1. Basic definitions. For clarity, we now provide some definitions. First, we define a permutation w to be a bijection [n] → [n], which we write in one-line notation: w = w(1)w(2) · · · w(n). We let S n denote the set of all permutations of [n]. A descent of a permutation is a position i such that w(i) > w(i + 1). We let des(w) denote the number of descents of w, i.e., for w ∈ S n , des(w) = |{1 ≤ i ≤ n − 1 : w(i) > w(i + 1)}|.
If a permutation has k descents, then it has n − 1 − k ascents, while its reversal, ← − w = w(n)w(n − 1) · · · w(1) has k ascents. Thus, the permutations with k descents are in bijection with the permutations having k ascents. Note also that if w has k descents, the descent positions partition w into k + 1 maximally increasing runs. For example, the permutation w = 4|3|126|5 has des(w) = 3, and the descent positions (indicated with vertical bars) partition w into 4 increasing runs.
The Eulerian numbers are denoted n k , 1 ≤ k ≤ n, and they count the number of permutations with k increasing runs, i.e., with k − 1 descents. That is, n k = |{w ∈ S n : des(w) = k − 1}|.  Table 1. Triangle of the Eulerian numbers n k , the number of permutations in S n with k − 1 descents.
The first few rows of the Eulerian triangle are shown in Table 1.
The Eulerian polynomials are the generating functions for the rows of this triangle, i.e., the generating function for permutations according to the descent statistic: The sign of a permutation is 1 if it can be written as a product of an even number of transpositions; otherwise the sign is −1. The sign of a permutation is well-defined, and denoted by sgn(w). It is also related to other statistics for permutations, e.g., where c(w) is the number of cycles of w, and inv(w) is the number of inversions of w. Let S + n denote the set of permutations of positive sign (also known as "even" permutations), and let S − n = S n − S + n denote the set of permutations with negative sign, i.e., S + n = {w ∈ S n : sgn(w) = 1} and S − n = {w ∈ S n : sgn(w) = −1}. We now define the positive Eulerian number, denoted n k + , to be the number of permutations in S + n with k − 1 descents and define the negative Eulerian number, denoted n k − , to be the number of permutations in S − n with k − 1 descents, i.e., n k The first few rows of the positive and negative Eulerian numbers are shown in Tables 2 and  3.
We define the positive Eulerian polynomial, A + n (t), and the negative Eulerian polynomial, A − n (t), to be the generating functions for the positive and negative Eulerian numbers, respectively. That is, For example, A + 3 (t) = t+2t 2 , A − 3 (t) = 2t 2 +t 3 , A + 4 (t) = t+5t 2 +5t 3 +t 4 , and A − 4 (t) = 6t 2 +6t 3 . Throughout the paper, when referring generically to both the positive Eulerian numbers and the negative Eulerian numbers, or to their corresponding polynomial generating functions, we will write "±-Eulerian numbers" or "±-Eulerian polynomials" and use the notation n k ± and A ± n (t).
1.2. Main results. This paper contains three theorems and a conjecture, and each theorem is proved in more than one way. The conjecture, while interesting in its own right, would provide a third proof of one of the theorems. Among the many identities for Eulerian numbers is the following power series identity: Our first main result for the ±-Eulerian numbers is a similar identity, given in the following theorem.
Theorem 1.1 (Generating function identity). For all n ≥ 1, We will provide two different proofs of this result in Section 2. We will also draw two interesting conclusions from Theorem 1.1. The first of these is a central limit theorem for the distribution of ±-Eulerian numbers. Theorem 1.2 (Limiting distribution). The distribution of the coefficients of A ± n (t) is asymptotically normal as n → ∞. For n ≥ 4, these numbers have mean (n + 1)/2 and for n ≥ 6, these numbers have variance (n + 1)/12.
Given that the Eulerian numbers are asymptotically normal, our central limit theorems (one for positive Eulerian numbers and one for negative Eulerian numbers) are not surprising. However our central limit theorems are also not trivial corollaries of existing results. While one proof of Theorem 1.2 will use Theorem 1.1, we provide another proof in Section 3. In fact we conjecture that the polynomials A ± n (t) have all roots real, which by Harper's method [15] would give a third proof of our central limit theorem. Conjecture 1.3 (Real roots). The ±-Eulerian polynomials have only real roots: A + n (t) has real roots for n ≥ 1 and A − n (t) has real roots for n ≥ 2. We have verified Conjecture 1.3 for n ≤ 100. We provide some remarks on the conjecture in Section 2.3.
The second conclusion that we can draw from Theorem 1.1 has to do with card shuffling.
Theorem 1.4 (Sign after a riffle shuffle). For a-shuffling starting at the identity, the probability of having sign 1 after k steps is equal to 1 2 + 1 2a k⌊n/2⌋ . By shuffling here we mean the Gilbert-Shannon-Reeds (GSR) model of riffle shuffling, the details of which will be provided in Section 4. In that section we also present two further proofs of Theorem 1.4, including one that follows immediately from work of Amy Pang. A similar result for shelf-shuffling machines is also discussed there, which says that (for ultimately trivial reasons) the probability of having sign 1 after one pass through a shelf-shuffling machine is exactly 1/2.

Identities for ±-Eulerian numbers and polynomials
In this section we will prove Theorem 1.1 and discuss some identities and recurrences for ±-Eulerian numbers and polynomials.
2.1. Proofs of Theorem 1.1. We will present two proofs of Theorem 1.1. The first proof uses a generating function for the joint distribution of descents and cycle structure studied by Fulman [10] and the second uses results of Desarménien and Foata [4], and Wachs [18], for the joint distribution of descents and inversions.
We begin the first proof by recalling an identity from [10, Theorem 1]: In this identity, the x i are indeterminates, n i (w) is the number of i-cycles in the permutation w, and where µ is the Möbius function of elementary number theory. While we do not make use of the fact here, the quantity f j,k counts the number of primitive necklaces of length j drawn from an alphabet of k letters. We will also make use of the following lemma, which is found Marshall Hall's group theory book [11], in connection with the commutator calculus on the free group.
Lemma 2.1. For any integer k ≥ 1, we have the following power series identity: The upshot for us comes from setting all x i = −1 and u = −u in Equation (3). Note , we have the following series identity: which is key in what follows.
First proof of Theorem 1.1. Since Equation (1) gives Thus by setting u = −u, we have the identity and we conclude that Taking the coefficient of u n on both sides proves Equation (5).
We now turn to the second proof of Theorem 1.1. For our second proof of Theorem 1.1, we will use the generating function for the refinement of the Eulerian polynomial that gives the joint distribution of inversions and descents: where inv(w) is the number of inversions of w, defined as the number of pairs (i, j) with i < j and w(i) > w(j).
Since the sign of w is sgn(w) = (−1) inv(w) , it follows that . Loday [13] initiated an investigation of the coefficients of A n (−1, t), which he and others, such as Desarménian and Foata [4], called "signed Eulerian numbers." Among other things, Desarménian and Foata prove that [4, Theorem 1]: The identities in (6) can be proved via manipulations of identities for A n (q, t) as q → −1.
Wachs [18] also gives a combinatorial proof of (6) with a sign-reversing involution. By adding or subtracting A m (t) = A + m (t) + A − m (t), with m = 2n or m = 2n + 1, to the equations in (6) we can now conclude . The second proof of Theorem 1.1 now follows from basic series manipulations.
Second proof of Theorem 1.1. Using the equations from (7) and the identity of Equation (1), we obtain Thus, for any n ≥ 1, as desired.

Symmetries and recurrences.
Having established our main series identity for the ±-Eulerian polynomials, we gather some interesting features of them now. First, we consider the homogeneous Eulerian polynomials where asc(w) denotes the number of ascents of w, as mentioned in the introduction. Similarly, we let Note that while A ± n (t) has degree n − 1 or n in t, this homogeneous version always has degree n − 1. For example, A + 3 (s, t) = s 2 + 2st, A − 3 (s, t) = 2st + t 2 , A + 4 (s, t) = s 3 + 5s 2 t + 5st 2 + t 3 , and A − 4 (s, t) = 6s 2 t + 6st 2 . Because ascents and descents are swapped under the involution that reverses a permutation, w → ← − w = w(n)w(n − 1) · · · w(1), we have that or equivalently, n k = n n + 1 − k .
That is to say, the rows of Table 1 are palindromic. It is also quite well-known that the Eulerian numbers satisfy the recurrence with initial conditions n 1 = n n = 1 for n ≥ 1. This identity can be explained bijectively, by considering whether the insertion of the letter n into a permutation in S n−1 leaves the number of descents unchanged or increases the number of descents by one. See, e.g., [14,Chapter 1].
The two-term recurrence of Equation (8) is neatly summarized with the polynomial recurrence where T is the linear operator See Section 7 of Brándën's article [3] for a discussion of sequences of polynomials defined by linear operators such as these. Now we present similar recurrences and symmetries for the ±-Eulerian numbers and polynomials.
Proposition 2.2 (Symmetries). For any n ≥ 1: In terms of coefficients, In other words, the rows of Tables 2 and 3 are palindromic for n ≡ 0, 1 (mod 4), while the rows of the two tables are mirror images when n ≡ 2, 3 (mod 4).
Proof. These symmetries are a consequence of the reversal involution previously discussed: w ↔ ← − w . It is straightforward to check that this map has the property that for any w ∈ S n , des( ← − w ) = n − 1 − des(w), while inv( ← − w ) = n 2 − inv(w). If n ≡ 0, 1 (mod 4), then n 2 is even, in which case inv( ← − w ) ≡ inv(w) (mod 2), and hence sgn( ← − w ) = sgn(w). Otherwise, if n ≡ 2, 3 (mod 4), we see that n 2 is odd, and therefore ← − w and w have opposite sign. This completes the proof.
To state the recurrence result, we first let Proposition 2.3 (Recurrences [5]). We have the following recurrences, for any m ≥ 1: In terms of coefficients, we have These recurrences can be explained combinatorially by considering the effect of inserting n + 1 into a permutation of length n, much as one proves the classical Eulerian recurrence in Equation (8). However, there are many cases to check carefully; we do not know of a simple argument. The full proof of Proposition 2.3 can be found in [5].

Real roots.
In the introduction, we presented Conjecture 1.3, which asserted that all the roots of the ±-Eulerian polynomials are real. It is known that the classical Eulerian polynomials are real rooted since Frobenius; see [3] for a thorough survey of similar sequences of polynomials.
From Equation (10), we find the univariate ±-Eulerian polynomials satisfy A ± n (t) = t n+1 A ± n (1/t) if n ≡ 0, 1 (mod 4) and A ± n (t) = t n+1 A ∓ n (1/t) if n ≡ 2, 3 (mod 4). This means the nonzero roots of A ± n (t) come in reciprocal pairs when n ≡ 0, 1 (mod 4), and when n ≡ 2, 3 (mod 4), the nonzero roots of A + n (t) are reciprocals of the nonzero roots of A − n (t). It would be lovely if the roots of A + n (t) and A − n (t) were interlacing. This is false in general. A fact one encounters in the study of real rootedness [3] is that the operator T preserves real roots. Thus, by Equation (13), we know that if A ± 2m (t) has real roots, then so does A ± 2m+1 (t). Therefore Conjecture 1.3 only needs to be proved in the even case. To this end, we can apply the recurrences of Proposition 2.3 twice in a row to see . It is not difficult to prove that the operators T s and T t preserve real-rootedness, and empirical evidence suggests that the pair of polynomials on the right are real and interlacing as well.

Central limit theorems
In this section we prove Theorem 1.2. In fact we will give two proofs of this central limit theorem, one using "analytic combinatorics" starting from the identity in Theorem 1.1, and the other using the "method of moments" in comparison with the usual Eulerian distribution.
For our first proof we start with the identity of Theorem 1.1 and use a modified Curtiss' theorem proved in Kim and Lee's paper [12, Proposition 2.2]. We state a version of that result in the following lemma. for all s ∈ I. Then X n converges in distribution to X.
We can express the moment generating function for the ±-Eulerian numbers n k ± explicitly. To be precise, we introduce a random variable W ± n which takes values in {1, · · · , n} with probabilities Note that W ± n is distributed as the descent plus one of a random permutation drawn uniformly at random from S ± n . Then, from Theorem 1.1, the moment generating function of W ± n takes the form Our task now shifts to estimating this function well enough to show that for all fixed s in an appropriately chosen interval I, the normalized moment generating function converges as n → ∞ to a moment generating function M X (s) for a normal distribution X with the desired mean and variance.
First proof of Theorem 1.2. We argue for positive Eulerian numbers; the argument for negative Eulerian numbers is almost identical. Fix a real number s > 0.
The asymptotic normality of W + n translates to the claim that the following normalized random variable converges in distribution to the normal distribution with zero mean and variance 1 12 . In view of Lemma 3.1 and the subsequent remark, it is sufficient to prove that the Laplace transform E[exp{−sZ + n }] converges to exp{ s 2 24 } as n → ∞ for each s in the interval I = (0, ∞). We henceforth write M + n (s) = E[exp{−sZ + n }]. Then we see that the prefactor of M + n (s) equals Given this representation, we estimate the summation part via approximation to integrals. Indeed, for t ∈ (0, 1) and a ≥ 1, we have a a−1 x n t x+1 dx ≤ a n t a ≤ a+1 a x n t x−1 dx.
Summing these over all a ∈ {1, 2, · · · }, we obtain Since s > 0, we have 0 < e −s/ √ n < 1, and setting t = e −s/ √ n+1 , we find that both t and 1 t have asymptotic form 1 + O(n −1/2 ). Thus we can replace the sum with an integral and our estimate now becomes Making the substitution u = sx/ √ n + 1 and using some easy comparisons, we find M + n (s) = e Our second proof relies on the following result about moments of the ±-Eulerian distributions.
Proposition 3.2. Let r be any positive integer. Then for ⌊n/2⌋ > r, the rth moment of the ±-Eulerian distribution equals the rth moment of the Eulerian distribution.
Proof. Instead of working with the rth moment, we can work with the rth falling moment, which can be computed from the relevant generating function by differentiating (with respect to t) r times and then setting t = 1.
Restating Equation (7), we have Now observe that if ⌊n/2⌋ > r, then differentiating r times and setting t = 1 gives 0. And by the product rule, differentiating the expression r times and setting t = 1 also gives 0. The theorem follows.
It is a well-known fact that the Eulerian numbers n k are asymptotically normal with mean (n + 1)/2 and variance (n + 1)/12. See, e.g., Bender [2]. Then by taking r = 1 in Proposition 3.2, we find that for n ≥ 4, averaging the number of descents over positive signed permutations gives (n−1)/2, and that averaging the number of descents over negative signed permutations also gives (n − 1)/2. Likewise, if r = 2, we find in both cases a variance of (n + 1)/12 when n ≥ 6. This gives our second proof of Theorem 1.2.
Second proof of Theorem 1.2. This is immediate from the method of moments, Proposition 3.2, and the fact that the Eulerian distribution is asymptotically normal with the claimed mean and variance.
We close this section by mentioning that if the polynomials A ± n (t) have only real roots, i.e., if Conjecture 1.3 holds, then Harper's method would give a third proof of Theorem 1.2.

Shuffling and sign
The mathematics of card shuffling is a lively topic; see [6] for a nice survey. In this section we study the sign of a permutation generated by the Gilbert-Shannon-Reeds (GSR) model of riffle shuffling. We derive a simple and striking exact formula for the chance of a given sign after a riffle shuffle, and this formula implies that (for large n), one shuffle suffices to randomize the sign of a permutation.
To begin we give some background on the GSR model of riffle shuffling, definitively studied in a lovely paper of Bayer and Diaconis [1], to which we refer the reader for further background.
Thus 0 ≤ j i ≤ n, a i=1 j i = n, and the j i have the same distribution as the number of balls in each box i if n balls are dropped at random into a boxes.
Given the j i , cut off the top j 1 cards, the next j 2 cards and so on, producing a packets (some possibly empty). Then drop cards one at a time, according to the rule that if there are A j cards in packet j, the next card is dropped from packet i with probability A i /(A 1 + · · ·+ A a ). This is done until all cards have been dropped.
When a = 2 this is a realistic model for how people shuffle cards. It turns out that an a 1 -shuffle followed by an a 2 -shuffle is equivalent to an (a 1 a 2 )-shuffle. Thus k iterations of a 2-shuffle is equivalent to a single 2 k -shuffle.
In what follows, we let P n,a (w) denote the probability of a permutation w after an a-shuffle started from the identity. In fact there is an explicit formula for this quantity, derived in [1]: (16) P n,a (w) = n + a − des(w −1 ) − 1 n /a n .
In particular, Equation (16) connects riffle shuffling with descents, showing that this probability depends only on the number of descents of the inverse permutation.
The main purpose of this section is to prove Theorem 1.4, which establishes a formula for the probability of having a permutation with sign 1 after k iterations of a-shuffling starting from the identity permutation 123 · · · n. Letting P + n,a k denote this probability (and P − n,a k its complementary probability), we will show (17) P + n,a k = 1 2 + 1 2a k⌊n/2⌋ . We will give two proofs of this fact, and sketch a third using unpublished work of Amy Pang. One of these again uses the power series identity in Lemma 2.1 as we did in Section 2.1. The other builds off of the identity in (2) from Theorem 1.1.
To get the first proof started, we recall the following series identity from Diaconis, Mc-Grath, and Pitman [8, Proposition 5.6]: which is very similar to Fulman's identity (3) we used in Section 2.1. As before the quantity n i (w) counts the number of i-cycles of w and f j,a counts the number of primitive necklaces of length j on an alphabet of a letters. Also as in Section 2.1, we recall that sgn(w) = (−1) n+c(w) , so that replacing u with −u and setting the x i equal to (−1) yields: which is our jumping off point for what follows.
First proof of Theorem 1.4. The coefficient of u n on the left hand side of Equation (19) is i.e., the difference in probabilities between positively and negatively signed permutations. By application of Lemma 2.1 as in the first proof of Theorem 1.1, we can show the right hand side of Equation (19) is equal to and taking the coefficient of u n in this series gives 1/a ⌊n/2⌋ . We have shown that P + n,a − P − n,a = 1 a ⌊n/2⌋ . Since P + n,a + P − n,a = 1, it follows that P + n,a = 1 2 + 1 2a ⌊n/2⌋ . Since k a-shuffles is the same as one a k shuffle, the result follows.
Our second proof builds from the identity in Theorem 1.1.
Taking the coefficient of t a on both sides gives w∈S + n n + a − des(w) − 1 n = a n + a ⌈n/2⌉ 2 .
We now sketch a third proof of Theorem 1.4. As we learned from Diaconis, this is a simple consequence of a result of Amy Pang (unpublished and prior to our work) stating that f (w) = sgn(w) is a right eigenfunction of the a-shuffle Markov chain with eigenvalue a −⌊n/2⌋ . The proof of Pang's result uses a heavy dose of Hopf algebras, along the lines of [9].
Third proof of Theorem 1.4. Letting P denote the transition operator of the riffle shuffling Markov chain, one has that P acts on functions f by P (f )[x] = y∈Sn P (x, y)f (y), and more generally, Letting id denote the identity permutation, and taking f to be the sign function, since f = sgn is a right eigenfunction with eigenvalue a −⌊n/2⌋ , it follows that P k sgn[id] = sgn(id) a k⌊n/2⌋ = 1 a k⌊n/2⌋ . On the other hand, P k sgn[id] = y∈Sn P k (id, y) sgn(y), which is simply the chance of sign 1 after k steps minus the chance of sign −1 after k steps. So this difference is equal to 1/(a k⌊n/2⌋ ), and the result follows as in the first proof.
We finish this section with two further remarks on shuffling and sign.
Remark 4.1 (Time to randomness). It is well known [1] that order 3 2 log 2 (n) many 2-shuffles are necessary and sufficient to mix a deck of n cards. However if one is only interested in certain features of the deck (for instance the number of fixed points of a permutation), then randomness can occur much earlier. Theorem 1.4 shows that (for large n), the sign of a permutation becomes random after one step.
Remark 4.2 (Shelf-shuffling machines and sign). Another model for card shuffling is the shelf-shuffling machine as studied in [7]. These machines are used in casinos with m = 10. One of the main findings of [7] is that a single use of a shelf-shuffler does not adequately mix 52 cards, but that two iterations do adequately mix 52 cards. One may wonder what effect shelf-shuffling machines have on the sign of a permutation. It turns out that the probability that sgn(w) = 1 after one pass through a shelf-shuffling machine is exactly 1/2 for all n ≥ 2.
This follows because, as found in [7], the probability of obtaining the permutation w ∈ S n when using a shelf shuffler with m shelves and an n card deck is where val(w) is the number of valleys w(i − 1) > w(i) < w(i + 1). That is, the probability of reaching a particular permutation depends only on the number of valleys in that permutation. However, for any number of valleys, say k, we have an equal number of positively and negatively signed permutations with that many valleys, i.e., |{w ∈ S + n : val(w) = k}| = |{w ′ ∈ S − n : val(w ′ ) = k}|. This is easily seen under the involution that swaps n and n − 1; neither of these elements can be in a valley and swapping them only changes the number of inversions by one. For example, w = 34812765 ↔ 34712865 = w ′ .