Zeros of Generalized Eulerian Polynomials

Transcription

1 Zeros of Generalized Eulerian Polynomials Carla D. Savage 1 Mirkó Visontai 2 1 Department of Computer Science North Carolina State University 2 Google Inc (work done while at UPenn & KTH). Geometric and Enumerative Combinatorics, IMA, 11/10/2014

2 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

3 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

4 Polynomials with (only) real zeros Combinatorics, algebra, geometry, analysis,... Surveys by: Stanley ( 86), Brenti ( 94), Brändén (2014+).

5 Combinatorial significance Consider a generating polynomial n a k x k, k=0 if it has only real zeros then the coefficients are known to be strongly log-concave: a 2 k ( n k)( n k) a k 1 ( n ) ( a k+1 n k 1 k+1 log-concave: a 2 k a k 1a k+1 unimodal: a 0 a m a n (if a k > 0) Other, geometrically inspired notions: γ-nonnegativity. )

6 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

7 Eulerian polynomials as generating polynomials For a permutation π = π 1... π n in S n, let des(π) = {i π i > π i+1 } denote the number of descents in π. Definition The Eulerian polynomial is S n (x) := π S n x des(π) = where n k = {π Sn des(π) = k}. n 1 k=0 n k x k,

8 Eulerian numbers: n Euler s triangle k k: n:

9 Eulerian numbers: n Euler s triangle k k: n: S 1 (x) = 1, S 2 (x) = 1 + x, S 3 (x) = 1 + 4x + x 2, S 4 (x) = x + 11x 2 + x 3,...

10 The zeros of S n (x) Theorem (Frobenius) S n (x) has only (negative and simple) real zeros.

11 The zeros of S n (x) Theorem (Frobenius) S n (x) has only (negative and simple) real zeros. Corollary For all n 1, the Eulerian numbers n n n,,..., 0 1 n 1 form a (strongly) log-concave, and hence unimodal sequence.

12 The zeros of S n (x) Theorem (Frobenius) S n (x) has only (negative and simple) real zeros. Corollary For all n 1, the Eulerian numbers n n n,,..., 0 1 n 1 form a (strongly) log-concave, and hence unimodal sequence. Most proofs of the theorem rely on the recurrence: S n+1 (x) = (1 + nx)s n (x) + x(1 x)s n(x).

13 Frobenius proof (via interlacing) S n+1 (x) = (1 + nx)s n (x) + x(1 x)s n(x)

14 Frobenius proof (via interlacing) S n+1 (x) = (1 + nx)s n (x) + x(1 x)s n(x) = (n + 1)xS n (x) + (1 x) (xs n (x))

15 Frobenius proof (via interlacing) x(1 + 4x + x 2 ) S n+1 (x) = (1 + nx)s n (x) + x(1 x)s n(x) = (n + 1)xS n (x) + (1 x) (xs n (x))

16 Frobenius proof (via interlacing) x(1 + 4x + x 2 ) S n+1 (x) = (1 + nx)s n (x) + x(1 x)s n(x) = (n + 1)xS n (x) + (1 x) (xs n (x))

17 Interlacing polynomials Theorem (Obreschkoff) f(x) and g(x) have interlacing zeros λf + µg has only real zeros for any λ, µ R. Problem with this method:

18 Interlacing polynomials Theorem (Obreschkoff) f(x) and g(x) have interlacing zeros λf + µg has only real zeros for any λ, µ R. Problem with this method: Does

19 Interlacing polynomials Theorem (Obreschkoff) f(x) and g(x) have interlacing zeros λf + µg has only real zeros for any λ, µ R. Problem with this method: Does not

20 Interlacing polynomials Theorem (Obreschkoff) f(x) and g(x) have interlacing zeros λf + µg has only real zeros for any λ, µ R. Problem with this method: Does not scale.

21 Interlacing polynomials Theorem (Obreschkoff) f(x) and g(x) have interlacing zeros λf + µg has only real zeros for any λ, µ R. Problem with this method: Does not scale. D n+2 (x) = (n(1 + 5x) + 4x)D n+1 (x) + 4x(1 x)d n+1 (x) +((1 x) 2 n(1 + 3x) 2 4n(n 1)x(1 + 2x))D n(x) (4nx(1 x)(1 + 3x) + 4x(1 x) 2 )D n(x) 4x 2 (1 x) 2 D n(x) +(2n(n 1)x(3 + 2x + 3x 2 ) + 4n(n 1)(n 2)x 2 (1 + x))d n 1 (x) +(2nx(1 x) 2 (3 + x) + 8n(n 1)x 2 (1 x)(1 + x))d n 1 (x) +4nx 2 (1 x) 2 (1 + x)d n 1 (x).

22 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

23 Compatible polynomials Definition The polynomials f 1 (x),..., f m (x) over R are compatible, if all their conic combinations, i.e., the polynomials m c i f i (x) for all c 1,..., c m 0 i=1 have only real zeros.

24 Compatible polynomials Definition The polynomials f 1 (x),..., f m (x) over R are compatible, if all their conic combinations, i.e., the polynomials m c i f i (x) for all c 1,..., c m 0 i=1 have only real zeros. Remark (Chudnovsky Seymour) f 1 (x),..., f m (x) are compatible if and only if f 1 (x),..., f m (x) have a common interleaver g(x).

25 Compatible polynomials Definition The polynomials f 1 (x),..., f m (x) are pairwise compatible if for all 1 i < j m. f i (x) and f j (x) are compatible

26 Compatible polynomials Definition The polynomials f 1 (x),..., f m (x) are pairwise compatible if for all 1 i < j m. f i (x) and f j (x) are compatible Lemma (Chudnovsky Seymour) The polynomials f 1 (x),..., f m (x) are compatible if and only if they are pairwise compatible.

27 Advantage of compatible polynomials Can handle nonnegative sum of many polynomials. Enough to prove pairwise compatibility.

28 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

29 Inversion sequences an alternative way to represent S n Definition The inversion sequence of a permutation π = π 1 π n is an n-tuple e = (e 1,..., e n ), where e j = { i {1, 2,..., j 1} π i > π j } counts the number of inversions ending in the jth position.

30 Inversion sequences an alternative way to represent S n Definition The inversion sequence of a permutation π = π 1 π n is an n-tuple e = (e 1,..., e n ), where e j = { i {1, 2,..., j 1} π i > π j } counts the number of inversions ending in the jth position. Example (n = 3) π 1 π 2 π e 1 e 2 e

31 Inversion sequences an alternative way to represent S n Definition The inversion sequence of a permutation π = π 1 π n is an n-tuple e = (e 1,..., e n ), where e j = { i {1, 2,..., j 1} π i > π j } counts the number of inversions ending in the jth position. Example (n = 3) π 1 π 2 π e 1 e 2 e Variants known under different names: Lehmer code, inversion code, inversion table, etc.

32 Inversion sequences ascent statistic Definition For an inversion sequence e = (e 1,..., e n ) I n, let asc I (e) = {i {1,..., n 1} : e i < e i+1 }, denote the number of ascents in e.

33 Inversion sequences ascent statistic Definition For an inversion sequence e = (e 1,..., e n ) I n, let asc I (e) = {i {1,..., n 1} : e i < e i+1 }, denote the number of ascents in e. Example (n = 3) e 1 e 2 e 3 asc I (e)

34 Observation The ascent statistics over inversion sequences is Eulerian. Theorem (Savage Schuster) e I n x asci(e) = π S n x des(π).

35 Observation The ascent statistics over inversion sequences is Eulerian. Theorem (Savage Schuster) Example (n = 3) e I n x asci(e) = π S n x des(π). e 1 e 2 e 3 asc I (e) π 1 π 2 π 3 des(π)

36 Advantage of inversion sequences Easy recurrence, the change in the ascent statistic asc I (e) = {i {1,..., n 1} : e i < e i+1 }, only depends on the last entry.

37 Advantage of inversion sequences Easy recurrence, the change in the ascent statistic asc I (e) = {i {1,..., n 1} : e i < e i+1 }, only depends on the last entry. Lend themselves to generalizations.

38 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

39 Generalized inversion sequences Recall some facts about the inversion sequences: I n = {(e 1,..., e n ) Z n 0 e i < i} = {0} {0, 1} {0, 1,..., n 1}.

40 Generalized inversion sequences Recall some facts about the inversion sequences: I n = {(e 1,..., e n ) Z n 0 e i < i} = {0} {0, 1} {0, 1,..., n 1}. Definition For a given sequence s = (s 1,..., s n ) N n, let I n (s) denote the set of s-inversion sequences by I (s) n = {(e 1,..., e n ) Z n 0 e i < s i }.

41 Generalized inversion sequences Recall some facts about the inversion sequences: I n = {(e 1,..., e n ) Z n 0 e i < i} = {0} {0, 1} {0, 1,..., n 1}. Definition For a given sequence s = (s 1,..., s n ) N n, let I n (s) denote the set of s-inversion sequences by I (s) n = {(e 1,..., e n ) Z n 0 e i < s i }. I (s) n = {0,..., s 1 1} {0,..., s 2 1} {0,..., s n 1}.

42 The ascent statistic on s-inversion sequences Savage and Schuster extended the definition of the ascent statistic to s-inversion sequences.

43 The ascent statistic on s-inversion sequences Savage and Schuster extended the definition of the ascent statistic to s-inversion sequences. Definition For e = (e 1,..., e n ) I (s) n, let { asc I (e) = i {0,..., n 1} : e i < e } i+1, s i s i+1 where we use the convention e 0 = 0 (and s 0 = 1).

44 The ascent statistic on s-inversion sequences Savage and Schuster extended the definition of the ascent statistic to s-inversion sequences. Definition For e = (e 1,..., e n ) I (s) n, let { asc I (e) = i {0,..., n 1} : e i < e } i+1, s i s i+1 where we use the convention e 0 = 0 (and s 0 = 1). Fact: The case s i = i agrees with usual inversion sequences. whenever 0 e k < k, for all k. e i < e i+1 e i i < e i+1 i + 1,

45 Examples The ascent statistic on s-inversion sequences e 1 e 2 e 3 e 1 e 2 e 3 Two examples for the sequence s = (2, 4, 6)

46 Examples The ascent statistic on s-inversion sequences e = (0, 3, 4) e = (1, 1, 2) e 1 e 2 e 3 e 1 e 2 e 3 Two examples for the sequence s = (2, 4, 6)

47 Examples The ascent statistic on s-inversion sequences e = (0, 3, 4) e = (1, 1, 2) e 0 e 1 e 2 e 3 e 0 e 1 e 2 e Two examples for the sequence s = (2, 4, 6)

48 Examples The ascent statistic on s-inversion sequences e = (0, 3, 4) with asc I (e) = 1. e = (1, 1, 2) with asc I (e ) = e 0 e 1 e 2 e 3 e 0 e 1 e 2 e Two examples for the sequence s = (2, 4, 6)

49 Examples The ascent statistic on s-inversion sequences e = (0, 3, 4) with asc I (e) = 1. e = (1, 1, 2) with asc I (e ) = = 0 2 < 3 4 > Two examples for the sequence s = (2, 4, 6) 0 1 < 1 2 > 1 4 <

50 s-eulerian polynomials Recall that S n (x) = π S n x des(π) = e I (s) n x asc I(e), when s = 1, 2,..., n.

51 s-eulerian polynomials Definition (s-eulerian polynomials) For an arbitrary sequence s = s 1, s 2,..., let E (s) n (x) := x asci(e). e I (s) n Recall that S n (x) = π S n x des(π) = e I (s) n x asc I(e), when s = 1, 2,..., n.

52 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x):

53 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x),

54 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x), s = (2, 4,..., 2n): the type B Eulerian polynomial, B n (x),

55 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x), s = (2, 4,..., 2n): the type B Eulerian polynomial, B n (x), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x),

56 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x), s = (2, 4,..., 2n): the type B Eulerian polynomial, B n (x), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1},

57 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x), s = (2, 4,..., 2n): the type B Eulerian polynomial, B n (x), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1}, s = (k + 1, 2k + 1,..., (n 1)k + 1): the 1/k-Eulerian polynomial, x exc π (1/k) cyc π,

58 s-eulerian polynomials and why do we care Special cases of s-eulerian polynomials, E (s) n (x): s = (1, 2,..., n): the Eulerian polynomial, S n (x), s = (2, 4,..., 2n): the type B Eulerian polynomial, B n (x), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1}, s = (k + 1, 2k + 1,..., (n 1)k + 1): the 1/k-Eulerian polynomial, x exc π (1/k) cyc π, s = (1, 1, 3, 2, 5, 3, 7, 4,..., 2n 1, n): the descent polynomial for the multiset {1, 1, 2, 2,..., n, n}.

59 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

60 On the zeros of s-eulerian polynomials The theorem of Frobenius can be generalized to the following.

61 On the zeros of s-eulerian polynomials The theorem of Frobenius can be generalized to the following. Theorem (Savage, V.) For any sequence s of nonnegative integers, the s-eulerian polynomials E (s) n (x) = x asc I(e) have only real zeros. e I (s) n

62 Proving more is sometimes easier... Instead of working with E (s) n (x) = e I n (s) working with the partial sums P (s) n,k (x) := (e 1,...,e n 1,k) I (s) n x asc I(e) we will be x asc I(e 1,...,e n 1,k).

63 Proving more is sometimes easier... Instead of working with E (s) n (x) = e I n (s) working with the partial sums P (s) n,k (x) := (e 1,...,e n 1,k) I (s) n x asc I(e) we will be x asc I(e 1,...,e n 1,k). Clearly, s E (s) n 1 n (x) = P (s) n,k (x). k=0

64 Proving more is sometimes easier... Instead of working with E (s) n (x) = e I n (s) working with the partial sums Clearly, P (s) n,k (x) := (e 1,...,e n 1,k) I (s) n s E (s) n 1 n (x) = P (s) n,k (x). k=0 x asc I(e) we will be x asc I(e 1,...,e n 1,k). IDEA: P (s) n,i (x) are compatible = E(s) n (x) has only real zeros.

65 A simple recurrence P (s) n+1,i (x) = l 1 j=0 sn 1 n,j (x) + P (s) n,j (x) xp (s) j=l

66 A simple recurrence P (s) n+1,i (x) = j s < i n s n+1 xp (s) n,j (x) + j s i n s n+1 P (s) n,j (x)

67 Back to the proof of the main result Again, prove something stronger Theorem (Savage, V.) Given a sequence s = {s i } i 1, for all 0 i j < s n, (i) P (s) n,i (x) and P(s) n,j (x) are compatible

68 Back to the proof of the main result Again, prove something stronger Theorem (Savage, V.) Given a sequence s = {s i } i 1, for all 0 i j < s n, (i) P (s) n,i (x) and P(s) n,j (x) are compatible, and (ii) xp (s) n,i (x) and P(s) n,j (x) are compatible.

69 Back to the proof of the main result Again, prove something stronger Theorem (Savage, V.) Given a sequence s = {s i } i 1, for all 0 i j < s n, (i) P (s) n,i (x) and P(s) n,j (x) are compatible, and (ii) xp (s) n,i (x) and P(s) n,j (x) are compatible. Corollary P (s) n,0 (x), P(s) n,1 (x),..., P(s) n,s n 1 (x) are compatible.

70 Back to the proof of the main result Again, prove something stronger Theorem (Savage, V.) Given a sequence s = {s i } i 1, for all 0 i j < s n, (i) P (s) n,i (x) and P(s) n,j (x) are compatible, and (ii) xp (s) n,i (x) and P(s) n,j (x) are compatible. Corollary P (s) n,0 (x), P(s) n,1 (x),..., P(s) n,s n 1 Corollary (x) + P(s) (x) + + P(s) P (s) n,0 n,1 (x) are compatible. n,s n 1 (x) has only real zeros.

71 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x).

72 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x).

73 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x). For i < j, we have l k. P (s) n+1,i = x (P(s) n,0 + + P(s) n,l 1 } {{ } ) + + P (s) n,k P(s) n,s n 1, l P (s) n+1,j = x (P(s) n,0 + + P(s) n,l P(s) n,k 1 } {{ } ) + + P (s) n,s n 1. k

74 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x). For i < j, we have l k. P (s) n+1,i = x (P(s) n,0 + + P(s) n,l 1 } {{ } ) + + P (s) n,k P(s) n,s n 1, l P (s) n+1,j = x (P(s) n,0 + + P(s) n,l P(s) n,k 1 } {{ } ) + + P (s) n,s n 1. k (i) P (s) n+1,i (x) and P(s) n+1,j (x) are compatible

75 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x). For i < j, we have l k. P (s) n+1,i = x (P(s) n,0 + + P(s) n,l 1 } {{ } ) + + P (s) n,k P(s) n,s n 1, l P (s) n+1,j = x (P(s) n,0 + + P(s) n,l P(s) n,k 1 } {{ } ) + + P (s) n,s n 1. k (i) cp (s) n+1,i (x) + dp(s) n+1,j (x) has only real zeros

76 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x). For i < j, we have l k. P (s) n+1,i = x (P(s) n,0 + + P(s) n,l 1 } {{ } ) + + P (s) n,k P(s) n,s n 1, l P (s) n+1,j = x (P(s) n,0 + + P(s) n,l P(s) n,k 1 } {{ } ) + + P (s) n,s n 1. k (i) cp (s) n+1,i (x) + dp(s) n+1,j (x) has only real zeros because { } { } }0 α<l { xp (s) n,α are compatible. (c + dx)p (s) n,β l β<k P (s) n,γ k γ<s n

77 Proof of compatibility Use induction. Base case: (x, 1) or (x, x) or (x 2, x). For i < j, we have l k. P (s) n+1,i = x (P(s) n,0 + + P(s) n,l 1 } {{ } ) + + P (s) n,k P(s) n,s n 1, l P (s) n+1,j = x (P(s) n,0 + + P(s) n,l P(s) n,k 1 } {{ } ) + + P (s) n,s n 1. k (i) cp (s) n+1,i (x) + dp(s) n+1,j (x) has only real zeros because { } { } }0 α<l { xp (s) n,α are pairwise compatible. (c + dx)p (s) n,β l β<k P (s) n,γ k γ<s n

78 Proof of compatibility (cont d) Now { { { xp n,α (s) }0 α<l (c + dx)p (s) n,β }l β<k are parwise compatible because: P (s) n,γ } k γ<s n

79 Proof of compatibility (cont d) Now { { { xp n,α (s) }0 α<l (c + dx)p (s) n,β }l β<k are parwise compatible because: P (s) n,γ } k γ<s n Two polynomials from the same set are compatible by IH(i).

80 Proof of compatibility (cont d) Now { { { xp n,α (s) }0 α<l (c + dx)p (s) n,β }l β<k are parwise compatible because: P (s) n,γ } k γ<s n Two polynomials from the same set are compatible by IH(i). xp (s) n,α and P (s) n,γ is compatible by IH(ii).

81 Proof of compatibility (cont d) Now { { { xp n,α (s) }0 α<l (c + dx)p (s) n,β }l β<k are parwise compatible because: P (s) n,γ } k γ<s n Two polynomials from the same set are compatible by IH(i). xp (s) n,α and P (s) n,γ is compatible by IH(ii). xp n,α (s) and (c + dx)p (s) n,β are compatible because (s) xp n,α, xp (s) n,β, P(s) n,β are pairwise compatible.

82 Proof of compatibility (cont d) Now { { { xp n,α (s) }0 α<l (c + dx)p (s) n,β }l β<k are parwise compatible because: P (s) n,γ } k γ<s n Two polynomials from the same set are compatible by IH(i). xp (s) n,α and P (s) n,γ is compatible by IH(ii). xp n,α (s) and (c + dx)p (s) n,β are compatible because (s) xp n,α, xp (s) n,β, P(s) n,β are pairwise compatible. (c + dx)p (s) n,β and P(s) n,γ are compatible because P (s) n,β, xp(s) n,β, P(s) n,γ are pairwise compatible.

83 Proof of compatibility (cont d) (i) Thus, P (s) n+1,i (x) and P(s) n+1,j (x) are compatible because { { } }0 α<l }l β<k { xp (s) n,α are pairwise compatible. (c + dx)p (s) n,β P (s) n,γ k γ<s n

84 Proof of compatibility (cont d) (i) Thus, P (s) n+1,i (x) and P(s) n+1,j (x) are compatible because { { } }0 α<l }l β<k { xp (s) n,α are pairwise compatible. (c + dx)p (s) n,β P (s) n,γ k γ<s n

85 Proof of compatibility (cont d) (i) Thus, P (s) n+1,i (x) and P(s) n+1,j (x) are compatible because { { } }0 α<l }l β<k { xp (s) n,α are pairwise compatible. (ii) xp (s) n+1,i (x) and P(s) shown in a similar way. (c + dx)p (s) n,β n+1,j P (s) n,γ k γ<s n (x) are also compatible and can be

86 Proof of compatibility (cont d) (i) Thus, P (s) n+1,i (x) and P(s) n+1,j (x) are compatible because { { } }0 α<l }l β<k { xp (s) n,α are pairwise compatible. (c + dx)p (s) n,β P (s) n,γ k γ<s n (ii) xp (s) n+1,i (x) and P(s) n+1,j (x) are also compatible and can be shown in a similar way.

87 Proof of compatibility (cont d) (i) Thus, P (s) n+1,i (x) and P(s) n+1,j (x) are compatible because { { } }0 α<l }l β<k { xp (s) n,α are pairwise compatible. (c + dx)p (s) n,β P (s) n,γ k γ<s n (ii) xp (s) n+1,i (x) and P(s) n+1,j (x) are also compatible and can be shown in a similar way. Q.E.D.

88 Outline Intro Polynomials with (only) real zeros Eulerian polynomials Toolbox Compatible polynomials Inversion sequences s-eulerian polynomials Generalized inversion sequences Proving real zeros via compatible polynomials Consequences Summary

89 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results:

90 The Eulerian polynomials have only real zeros One proof for all The fact that E (s) n (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius),

91 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti),

92 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x) (Steingrímsson),

93 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x) (Steingrímsson), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1} (Diaconis Fulman),

94 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x) (Steingrímsson), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1} (Diaconis Fulman), s = (k + 1, 2k + 1,..., (n 1)k + 1): the 1/k-Eulerian polynomial, x exc π (1/k) cyc π (Brenti, Brändén, Ma Wang),

95 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x) (Steingrímsson), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1} (Diaconis Fulman), s = (k + 1, 2k + 1,..., (n 1)k + 1): the 1/k-Eulerian polynomial, x exc π (1/k) cyc π (Brenti, Brändén, Ma Wang), s = (1, 1, 3, 2, 5, 3, 7, 4,..., 2n 1, n): the descent polynomial for the multiset {1, 1, 2, 2,..., n, n} (Simion)

96 The Eulerian polynomials have only real zeros One proof for all The fact that E n (s) (x) has only real zeros implies several results: s = (1, 2,..., n): type A, S n (x), (Frobenius), s = (2, 4,..., 2n): type B, B n (x), (Brenti), s = (k, 2k,..., nk): the descent polynomial for the wreath products, G n,r (x) (Steingrímsson), s = (k, k,..., k): the ascent polynomial for words over a k-letter alphabet {0, 1, 2,..., k 1} (Diaconis Fulman), s = (k + 1, 2k + 1,..., (n 1)k + 1): the 1/k-Eulerian polynomial, x exc π (1/k) cyc π (Brenti, Brändén, Ma Wang), s = (1, 1, 3, 2, 5, 3, 7, 4,..., 2n 1, n): the descent polynomial for the multiset {1, 1, 2, 2,..., n, n} (Simion) have only real zeros.

97 Combinatorics of Coxeter groups The descents can be defined in a more general setting. Definition (Björner Brenti) Let S be a set of Coxeter generators, m be a Coxeter matrix, and W = S : (ss ) m(s,s ) = id, for s, s S, m(s, s ) < be the corresponding Coxeter group. Given a pair (W, S) and σ W, let l W (σ) be the length of σ in W with respect to S.

98 Eulerian polynomials for Coxeter groups Definition For W a finite Coxeter group, with generator set S = {s 1,..., s n } the descent set of σ W is D W (σ) = {s i S : l W (σs i ) < l W (σs i )}. W(x) = x DW(σ). σ W

99 Eulerian polynomials for Coxeter groups Definition For W a finite Coxeter group, with generator set S = {s 1,..., s n } the descent set of σ W is D W (σ) = {s i S : l W (σs i ) < l W (σs i )}. W(x) = σ W x D W(σ). Conjecture (Brenti) Eulerian polynomials W(x) for all Coxeter groups W have only real zeros.

100 Eulerian polynomials for Coxeter groups Theorem (Brenti) The Eulerian polynomial for type B n and for all the exceptional Coxeter groups has only real zeros.

101 Eulerian polynomials for Coxeter groups Theorem (Brenti) The Eulerian polynomial for type B n and for all the exceptional Coxeter groups has only real zeros. Observation: Eulerian polynomials are multiplicative. Enough to consider irreducible groups. Type D n is the last remaining piece of the puzzle. (Verified up to n = 100.)

102 Eulerian polynomials for type D n Conjecture (Brenti) Eulerian polynomials for type D n have only real zeros.

103 Eulerian polynomials for type D n Conjecture (Brenti) Eulerian polynomials for type D n have only real zeros. 1. Why did type D n resist so far?

104 Eulerian polynomials for type D n Conjecture (Brenti) Eulerian polynomials for type D n have only real zeros. 1. Why did type D n resist so far? 2. Motivating question raised by Krattenthaler at SLC (Strobl): Why don t you apply your method to type D n?

105 Answer to the first question Why did type D n resist so far? Combinatorial definition is not as pretty. Think of elements of B n (resp. D n ) as signed (resp. even-signed) permutations. The definition of descents: des B (σ) = {i σ i > σ i+1 } {0 σ 1 > 0} des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0} No nice recurrence. The only recurrence (due to Chow) is rather complicated.

106 Answer to the first question Why did type D n resist so far? Combinatorial definition is not as pretty. Think of elements of B n (resp. D n ) as signed (resp. even-signed) permutations. The definition of descents: des B (σ) = {i σ i > σ i+1 } {0 σ 1 > 0} des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0} No nice recurrence. The only recurrence (due to Chow) is rather complicated. D n+2 (x) = (n(1 + 5x) + 4x)D n+1 (x) + 4x(1 x)d n+1 (x) +((1 x) 2 n(1 + 3x) 2 4n(n 1)x(1 + 2x))D n(x) (4nx(1 x)(1 + 3x) + 4x(1 x) 2 )D n(x) 4x 2 (1 x) 2 D n(x) +(2n(n 1)x(3 + 2x + 3x 2 ) + 4n(n 1)(n 2)x 2 (1 + x))d n 1 (x) +(2nx(1 x) 2 (3 + x) + 8n(n 1)x 2 (1 x)(1 + x))d n 1 (x) +4nx 2 (1 x) 2 (1 + x)d n 1 (x).

107 Answer to the second question Why don t you apply your method to type D n? Short answer: Does not work.

108 Answer to the second question Why don t you apply your method to type D n? Long answer: Does not work out of the box.

109 Answer to the second question Why don t you apply your method to type D n? Long answer: Does not work out of the box. Remedy: Try harder and use some tricks!

110 Answer to the second question Why don t you apply your method to type D n? Long answer: Does not work out of the box. Remedy: Try harder and use some tricks! Trick 1 Look at 2D n (x) instead of D n (x).

111 Answer to the second question Why don t you apply your method to type D n? Long answer: Does not work out of the box. Remedy: Try harder and use some tricks! Trick 1 Look at 2D n (x) instead of D n (x). Trick 2 Find an ascent statistic for 2D n (x).

112 Answer to the second question Why don t you apply your method to type D n? Long answer: Does not work out of the box. Remedy: Try harder and use some tricks! Trick 1 Look at 2D n (x) instead of D n (x). Trick 2 Find an ascent statistic for 2D n (x). Trick 3 Believe in your method!

113 Trick 1 Getting rid of parity Recall, des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0}. Proposition For n 2, x desd σ = 2 x desd σ. σ B n σ D n

114 Trick 1 Getting rid of parity Recall, des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0}. Proposition For n 2, x desd σ = 2 x desd σ. σ B n σ D n

115 Trick 2 A type D n ascent statistic Asc A (e) = { i e i i < e } i+1 i+1 Asc B (e) = { i e i i < e i+1 Asc D (e) = { i e i i < e i+1 i+1 } i+1 {0 e1 > 0} } { 0 e1 + e } Des A (σ) = {i σ i > σ i+1 } Des B (σ) = {i σ i > σ i+1 } {0 σ 1 > 0} Des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0}

116 Trick 2 A type D n ascent statistic 2D n (x) = e I 2,4,6,... n x Asc D(e). Asc A (e) = { i e i i < e } i+1 i+1 Asc B (e) = { i e i i < e i+1 Asc D (e) = { i e i i < e i+1 i+1 } i+1 {0 e1 > 0} } { 0 e1 + e } Des A (σ) = {i σ i > σ i+1 } Des B (σ) = {i σ i > σ i+1 } {0 σ 1 > 0} Des D (σ) = {i σ i > σ i+1 } {0 σ 1 + σ 2 > 0}

117 Putting all together A recursive proof for type D n D n (x) = 2n 1 i=0 D n,i (x).

118 Putting all together A recursive proof for type D n D n (x) = 2n 1 i=0 D n,i (x). Only one problem: base case does not hold. D 2,0 (x) = 1, D 2,1 (x) = D 2,2 (x) = x, D 2,3 (x) = x 2.

119 Putting all together A recursive proof for type D n D n (x) = 2n 1 i=0 D n,i (x). Only one problem: base case does not hold. D 2,0 (x) = 1, D 2,1 (x) = D 2,2 (x) = x, D 2,3 (x) = x 2. Also, D 3,0 (x),..., D 3,5 (x) are not compatible.

120 Trick 3 Leap of faith D 4,0 (x),..., D 4,7 (x) are compatible.

121 Trick 3 Leap of faith D 4,0 (x),..., D 4,7 (x) are compatible. By induction, are compatible for all n 4. D n,0 (x),..., D n,2n 1 (x) Corollary D n (x) = 2n 1 i=0 D n,i (x) has only real zeros.

122 Summary

123 Summary Unified proof of existing results, but also can be used to solve new problems (Brenti s type D conjecture).

124 Summary Unified proof of existing results, but also can be used to solve new problems (Brenti s type D conjecture). The method of compatible polynomials is a simple yet powerful method to prove real zeros.

125 Summary Unified proof of existing results, but also can be used to solve new problems (Brenti s type D conjecture). The method of compatible polynomials is a simple yet powerful method to prove real zeros. A reformulation, or even generalization (s-inversion sequences) often makes the problem easier.