Eulerian-Catalan Numbers

0.12 MB·Indonesian

by  Hoda Bidkhori

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Eulerian-Catalan Numbers
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2 Hoda Bidkhori

n
Department of Mathematics
a
North Carolina State University, Raleigh, NC, 27695
J
[email protected]
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Seth Sullivant

]
O Department of Mathematics
North Carolina State University, Raleigh, NC, 27695
C
[email protected]
.
h
t
a
m
Abstract
[

We show that the Eulerian-Catalan numbers enumerate Dyck permutations. We provide two
1
proofs for this fact, the first using the geometry of alcoved polytopes and the second a direct
v
combinatorial proof via an Eulerian-Catalan analogue of theChung-Feller theorem.
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0
1
1 Key words: Eulerian-Catalannumber,Dyckpermutation,Dyckpath,Ballotsequence.
.
1
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1 1. Introduction
1
:
v Let Am,n denote the Eulerian numbers, which count the number of permutations on n letters
i with m descents. The Eulerian-Catalan numbers are defined by
X
1
r EC = A .
a n n+1 n,2n+1
We choose to attach the name Catalan to these numbers since A is the central Eulerian
n,2n+1
number,andfortheirconnectiontotheCatalannumbers,whichwillbecomeapparentshortly.The
Eulerian-Catalan numbers appear in the Online Encyclopedia of Integer Sequences [2], however,
no combinatorial interpretation appears there and we could not find one in the literature.
TheEulerian-CatalannumberisclearlyalwaysanintegersincetheEuleriannumberssatisfythe
following relations
A =(n−m)A +(m+1)A and A =A for all m,n
m,n m−1,n−1 m,n−1 m,n n−m−1,n
which imply that EC =A +A =2A . Given a permutation w of [n], we associate a
n n−1,2n n,2n n,2n
0/1 sequence of length n−1, ad(w), where ad(w) =0 if w <w and ad(w) =1 if w >w .
i i i+1 i i i+1
We call ad(w) the ascent/descent vector of w. A 0/1 sequence is called a ballot sequence if every
initial string has at least as many zeroes as ones.
PreprintsubmittedtoElsevier 7January2011
Alternately, the permutation w defines a lattice path L(w) starting from (0,0) and with step
(1,0)ifiis anascent,andwith step(0,1)fora descent.Writing theentriesofw alongthevertices
ofthe pathproducesastandardyoungtableauofaborderstrip.We callapermutationw ∈S
2n+1
aDyck permutation ifandonlyifL(w) isa latticepathfrom(0,0)to(n,n),with allpoints onthe
pathsatisfyingy ≤x.By the usualcorrespondencebetweenballotsequencesandDyckpaths(see,
[9, Ex. 6.19]), a permutation w is a Dyck permutation if and only if ad(w) is a ballot sequence.
Let L be a lattice path from (0,0) to (n,n) using steps of (0,1) and (1,0). The exceedance of
L, denoted exc(L) is defined to be the number of i∈{0,...,n} such that there is a point (i,i′) in
L with i<i′. Hence, the Dyck paths are the lattice paths with exceedance 0. The main results of
this paper is the following:
Theorem 1.1. Fix j = 0,...,n. The number of permutations w ∈ S with n descents such
2n+1
that exc(L(w)) = j does not depend on j. As a consequence the number of Dyck permutations
w ∈S is the Eulerian-Catalan number EC .
2n+1 n
WeprovideadirectcombinatorialproofofTheorem1.1inSection4.Wealsoprovideageometric
proofofthefactthatthenumberofDyckpermutationsw∈S istheEulerian-Catalannumber
2n+1
EC . We define a polytope P and we show that its normalized volume is the Eulerian-Catalan
n 2,n
number. This is proved in Section 2. The polytope P turns out to be an alcoved polytope [6],
2,n
andhence its volumecanalsobe interpretedascountingpermutationswithcertainrestrictionson
its descent positions, which is explained in Section 3. Combining these two arguments yields the
resultthatthe numberofDyckpermutationsw ∈S isthe Eulerian-CatalannumberEC .We
2n+1 n
prove these results in a Fuss-Catalan generality which gives a combinatorialinterpretation for the
numbers 1 A as the number of (k−1)-Dyck permutations.
n+1 n,kn+k−1
2. Subdividing the Hypersimplex
The hypersimplex ∆(k,n) is the polytope
n
∆(k,n)= (x ,...,x )∈[0,1]n : x =k .
1 n i
( )
i=1
X
Itiswell-knownthatthenormalizedvolumeofthehypersimplexistheEuleriannumberA .
k−1,n−1
Stanley [8] provides a combinatorial proof of this fact by triangulating the hypersimplex.
Fix k,n ∈ N, and consider the hypersimplex ∆(n+1,k(n+1)). We define the polytope P
k,n
with the following inequalities:
kt
P = (x ,...,x )∈∆(n+1,k(n+1)): x ≤t, t=1,...,n .
k,n 1 k(n+1) s
( )
s=1
X
Remark 2.1. A0/1sequence iscalleda k-ballot sequence ifeveryinitialstring hasatleastk-times
as many 0’s as 1’s. Note that a 1-ballot sequence is an ordinary ballot sequence. The polytope
P is equalto the convexhull ofthe (k−1)-ballotsequences oflength k(n+2).This is shownin
k,n
work of the first author [3], where these polytopes are studied in the largercontext of lattice path
matroid polytopes. Lattice path matroids were introduced in [4] and the Catalan matroid [1] is a
special case. The polytope P is the Catalan matroid polytope. Corollary2.3 below implies that
2,n
the normalized volume of the Catalan matroid polytope is the Eulerian-Catalan number. We do
not need these details here, and refer the reader to [3].
For each i∈{0,...,n} define the polytope P ⊆∆(n+1,k(n+1)) by the inequalities
k,n,i
2
kt
P = (x ,...,x )∈∆(n+1,k(n+1)): x ≤t, t=1,...,n
k,n,i 1 k(n+1) ki+s
( )
s=1
X
where the indices are considered modulo k(n+1). For example, with k = 2 and n = 2, we get
three polytopes:
P ={(x ,...,x )∈∆(3,6):x +x ≤1, x +x +x +x ≤2}
2,2,0 1 6 1 2 1 2 3 4
P ={(x ,...,x )∈∆(3,6):x +x ≤1, x +x +x +x ≤2}
2,2,1 1 6 3 4 3 4 5 6
P ={(x ,...,x )∈∆(3,6):x +x ≤1, x +x +x +x ≤2}.
2,2,2 1 6 5 6 5 6 1 2
Note that P =P .
k,n,0 k,n
Theorem 2.2. Fix k,n∈N. The interiors of the polytopes P and P are disjoint if i6=j,
k,n,i k,n,j
and ∆(n+1,k(n+1))=∪n P .
i=0 k,n,i
Proof. Since the P are allaffinely isomorphic via a transformationthat permutes coordinates,
k,n,i
it suffices to show that an x∈intP does not belong to any other P . Since x∈intP it
k,n,0 k,n,i k,n,0
satisfies the inequality
x +···+x <i.
1 ki
Since we are in the hypersimplex, we always have
x +···+x =n+1.
1 k(n+1)
Combining these, we deduce that
x +···+x >n+1−i
ki+1 k(n+1)
which implies x∈/ P .
k,n,i
Toprovethat∆(n+1,k(n+1))=∪n+1P ,wemustshowthatanypointx∈∆(n+1,k(n+1))
i=1 k,n,i
belongs to one of the P . To do this, we consider the linear transformation
k,n,i
φ:Rk(n+1) →Rn+1
such that y = −1+ k(i+1) x where the coordinates on Rn+1 are y ,...,y . The image of the
i s=ki+1 s 0 n
hypersimplex ∆(k(n+1),n+1) is the polytope
P
Q(k,n)={y ∈[−1,k−1]n+1 :y +···+y =0}
0 n
and the image of P is the polytope
k,n,i
R ={y ∈Q(k,n):y +···+y ≤0,t=0,n−1}.
k,n,i i i+t
Notethatapointx∈∆(n+1,k(n+1))belongstoP ifandonlyifitsimageinQ(k,n)belongs
k,n,i
to R . Furthermore, the argument in the preceding paragraph implies that the R(n,k,i) have
k,n,i
disjoint interiors.Hence, it suffices to show that Q(k,n)=∪n R . Since the inequalities −1≤
i=0 k,n,i
y ≤k−1 are common to all the polytopes, it suffices to show that the plane
i
H(n)={y ∈Rn+1 :y +···+y =0}
0 n
can be decomposed is the union of the cones
C(n,i)={y ∈H(n):y +···+y ≤0,t=0,n−1}.
i i+t
For i=0,...,n, let v =−e +e . The cones C(n,i) are simplicial, and it is straightforward to
i i−1 i
see that the generators of C(n,i) are {v ,...,v }\{v }. The vectors v ,...,v also span H(n),
0 n i 0 n
and v +···+v = 0. This implies that the union of the cones spanned by the sub-collections of
0 n
n vectors is all of H(n), which is what we needed to show.
3
Corollary 2.3. The normalized volume of P is 1 A .
k,n,i n+1 n,kn+k−1
Proof. Since ∆(n+1,k(n+1))=∪n P and their relative interiors are disjoint, we have that
i=0 k,n,i
n
vol(∆(n+1,k(n+1)))= volP .
k,n,i
i=0
X
As the cyclic shift ofcoordinateswhichsends P to P is volume preserving,andvol(∆(n+
k,n,0 k,n,i
1,k(n+1)))=A , we deduce the desired formula.
n,kn+k−1
3. Alcoved Polytopes
Fix integers k,n ∈ N. For 0 ≤ i < j ≤ n, let b ,c be integers, with b ≤ c . The alcoved
ij ij ij ij
polytope defined by this data is:
n
P(k,n,b,c):= (x ,...,x )∈Rn : x =k,b ≤x +···+x ≤c ,0≤i<j ≤n .
1 n i ij i+1 j ij
( )
i=1
X
AlcovedpolytopeswerestudiedindetailbyLamandPostnikov[6].Alcovedpolytopeshavenatural
triangulationsintounimodularsimpliceswhicharethemselvesalcovedpolytopesandwhicharein-
dexedbypermutations;thesesimplicesarecalledalcoves.Inthespecialcase,whereP(k,n,b,c)⊆
∆(k,n), Lam and Postnikov give an explicit description of the simplices involved in the alcove
triangulation, and hence a combinatorialformula for the volume of these alcoved polytopes.
Let W(k,n,b,c) ⊂ S be the set of permutations w = w w ···w ∈ S satisfying the
n−1 1 2 n−1 n−1
following conditions:
(i) w has k−1 descents.
(ii) The sequence w ···w has at least b descents. Furthermore, if w ···w has exactly b
i j ij i j ij
descents, then w <w .
i j
(iii) The sequence w ···w has at most c descents. Furthermore, if w ···w has exactly c
i j ij i j ij
descents, then we must have that w >w .
i j
In the above conditions we assume that w =0.
0
Theorem 3.1. [6] The normalized volume of P(k,n,b,c)⊆∆(k,n) is equal to |W(k,n,b,c)|.
We apply Theorem 3.1 to give a combinatorialformula for the volume of the polytope P . As
k,n
discussed previously, a lattice path from (0,0) to (nk,n) using (0,1) and (1,0) steps is called a
k-Dyck path if every point on the path satisfies y ≤ 1x. When L(w) is a k-Dyck path, w is called
k
a k-Dyck permutation. A 0/1 sequence is called a k-ballot sequence if each initial string has at
least k times as many 0’s as 1’s. Note that w is a k-Dyck permutation precisely where ad(w) is a
k-ballot sequence.
Proposition 3.2. The normalized volume of P is equal to the number of permutations w ∈
k,n
S such that L(w) is a (k−1)-Dyck path. Equivalently, the volume equals the number of
kn+k−1
permutations w ∈S such that ad(w) is a (k−1)-ballot sequence.
kn+k−1
Proof. ThepolytopeP ⊂∆(n+1,k(n+1))isanalcovedpolytopesinceitsdefininginequalities
k,n
are 0≤x ≤1, x +···+x =n+1 and
i 1 k(n+1)
kt
x ≤t, t=1,...,n.
s
s=1
X
4
Applying Theorem 3.1, we see that the volume of P is the number of permutations w =
k,n
w ···w ∈S satisfying the following conditions:
1 kn+k−1 kn+k−1
(i) w has n descents.
(ii) w ···w has at most i−1 descents, for 1≤i≤n.
1 ik
These conditions are satisfied if and only if ad(w) is a (k−1)-ballot sequence.
Combining Corollary 2.3 and Proposition 3.2 we deduce:
Theorem 3.3. The number of permutations w ∈S suchthat L(w) is a (k−1)-Dyck path is
kn+k−1
1 A . In particular, the number of Dyck permutations in S is the Eulerian-Catalan
n+1 n,kn+k−1 2n+1
number EC .
n
4. Exceedances of Lattice Paths and Eulerian-Catalan Numbers
In this section we prove Theorem 1.1 which gives us a combinatorialinterpretation of Eulerian-
Catalan number in terms of certain permutation statistics. This result is related to a classic proof
that the Catalan numbers enumerate Dyck paths [5]. First we provide a combinatorial proof of
Theorem 1.1. Later, we give a geometric proof of Theorem 1.1 for the case j = 1 (and hence also
for j =n−1). In Sections 2 and 3, we saw a geometric proof for the cases j = 0 (and hence also
for j =n). It is an interesting problem to provide geometric proofs for other cases.
Proof of Theorem 1.1. Consider a lattice path P with (0,1) and (1,0) steps from (0,0) to (n,n).
Define c(P) = (c ,...,c ), where c is the number of horizontal steps of P at height y = i. The
0 n i
cyclicpermutations C =(c ,...,c )ofc(P)arealldistinct, andforeachthereis aunique path
j j j−1
P from(0,0)to (n,n)sothat c(P )=C . The number ofexceedancesofthe paths P ,...,P are
j j j 0 n
the numbers 0,1,...,n in some order. These results are known as the Chung-Feller Theorem [5].
Consider a permutation W =w ···w with n descents. We have one of the following cases:
1 2n+1
(i) The cyclic permutation (w ···w ) has n cyclic descents.
1 2n+1
(ii) The cyclic permutation (w ···w ) has n+1 cyclic descents.
1 2n+1
The cyclic descents of a (cyclic) permutation also include the possibility of a descent at the last
position w , which occurs when w <w .
2n+1 1 2n+1
In the case (w ···w ) has n cyclic descents, we consider distinct indexes 1 = i < ··· <
1 2n+1 0
i ≤ 2n+1 so that w w is not a cyclic descent in the cyclic permutation (w ···w ).
n ik−1 ik 1 2n+1
Now, consider the n+1 permutations obtain by cyclic shifting the permutation W, starting at
w ,...,w .WedenotethesepermutationsbyW ,...,W .ThepermutationsW ,...,W allhave
i0 in 0 n 0 n
n descents, whereas all other cyclic shiftings of W have n−1 descents. As in the Chung-Feller
theorem, we define c(L(W ))= (c ,...,c ) where c is the number of horizontal steps at height i
0 0 n i
for a lattice path L(W ). The cyclic shiftings of c(L(W )), C =(c ,...,c ) are all distinct, and
0 0 j j j−1
for each there is a unique path P from (0,0) to (n,n) so that c(P ) = C . It is easy to see that
j j j
P =L(W ). By the Chung-Feller theorem,the lattice pathassociatedto these permutations have
j j
different number of exceedance 0,...,n. This shows that the n+1 cyclic shiftings of permutation
W which have n descents, have 0,...,n number of exceedances in some order.
Now, consider the case where the cyclic permutation (w ···w ) has n+1 cyclic descents.
1 2n+1
For any permutation W =w ···w , we define: W =(w ··· ,w )=(2n+2−w )···(2n+
1 2n+1 1 2n+1 1
2−w ). If W has k exceedances, W has n−k exceedances, and if W has k cyclic descents,
2n+1
W has 2n+1−k cyclic descents. We consider i ,..c.,i sobthat wb w is a cyclic descent in W
0 n ik−1 ik
and therefore W does not have a cycliccdescent at w w . We consider the n+1 permutations
ik−1 ik
ocbtained by cyclic shifting the permutation W, starting with w ,...,w . We denote them by
i0 in
W0,...,Wn. Acs above,these are the only permutatibons obbtained by cyclic shifting W that have n
5
descents, all other shiftings having n+1 descents. As we see in the first case, we know that the
lattice path associated to L(W ) have different number of exceedance for j = 0,...,n. Therefore,
j
P =L(W )forj =0,...,nhavealldifferentnumbersofexceedancesfrom0,...,n insomeorder.
j j
Combining the above two rcesults,we know that among the 2n+1 cyclic shifts ofa permutation
w with n descents, exactly n+1 of them have n descents, and their associated lattice paths have
different numbers of exceedances 0,...,n. Therefore, the number of permutations w ∈S with
2n+1
n descent and exc(L(w)) = j are the same for j = 0,...,n. As the total number of permutations
w ∈S with n descents is A , we see that the number of permutations w ∈S with n
2n+1 n,2n+1 2n+1
descent and exc(L(w))=j is the Eulerian-Catalan number.
Note that the proof of Theorem 1.1 also gives a direct combinatorial proof for the fact that the
number of Dyck permutations in S is A +A . Indeed, the Dyck permutations are in
2n+1 n−1,2n n,2n
bijective correspondence with the set of cyclic permutation with either n or n+1 cyclic descents.
Cycling such a permutation until 2n+1 is at the end and deleting 2n+1 yields a bijection with
permutations in S with either n−1 or n descents.
2n
To conclude this section, we provide a geometric proof of Theorem 1.1 in the case of j = 1, in
the spirit of the proofs from Sections 2 and 3.
Proof of Theorem 1.1 with j =1. Consider the polytope P , which is given by inequalities:
2,n
2t
P = (x ,...,x )∈∆(n+1,2(n+1)): x ≤t, t=1,...,n .
2,n 1 2(n+1) s
( )
s=1
X
We call the inequality 2t x ≤ t the t-th inequality. Let T ⊆ [n] and consider the polytope
s=1 s
defined by flipping the t-th inequalities for t∈T:
P
2t 2t
P (T)= (x ,...,x )∈∆(n+1,2(n+1)): x ≥t,t∈T, x ≤t,t∈[n]\T .
2,n 1 2(n+1) s s
( )
s=1 s=1
X X
ApplyingTheorem3.1,weseethatthevolumeofP (T)isthenumberofpermutationsw ∈S
2,n 2n+1
such that L(w) has an exceedance in position t−1 if and only if t ∈ T (that is, there is a point
(t−1,s) in L(w) with s>t−1). Thus, to prove Theorem 1.1 it suffices to show that
EC = vol(P (T)). (4.1)
n 2,n
T⊆[nX]:|T|=j
We prove Eq.(4.1) in the case j =1.
To do this, we consider the linear transformation
φ:R2(n+1) →Rn+1
such that y = −1+ 2(i+1) x where the coordinates on Rn+1 are y ,...,y . The image of the
i s=2i+1 s 0 n
hypersimplex ∆(n+1,2(n+1)) is the following polytope
P
Q(2,n)={y ∈[−1,1]n+1 :y +···+y =0}
0 n
and the image of P (T) is the polytope
2,n
R (T)={y ∈Q(2,n):y +···+y ≥0,t∈T,y +···+y ≤0,t∈[n]\T}.
2,n 0 t−1 0 t−1
Note that a point x ∈ ∆(n+1,2(n+1)) belongs to P (T) if and only if its image in Q(2,n)
2,n
belongstoR (T).Furthermore,alltheP (T)andQ (T)aredisjoint.Ifwecanfindacollection
2,n 2,n 2,n
of volume preserving linear transformations τ ,...,τ such that τ (R ({1})),...,τ (R ({n}))
1 n 1 2,n n 2,n
have disjoint interiors and such that
R (∅)=∪ τ (R ({i})),
2,n i∈[n] i 2,n
6
we will prove the theorem. This follows because it is straightforwardto lift the linear transforma-
tionsτ toR2(n+1)inawaythatwillyieldasimilarresultfortheP .Allourlineartransformations
i 2,n
will be coordinate permutations. Thus, we can ignore the common inequalities, −1≤y ≤1, and
i
consider the same questions in the plane
H(n)={y ∈Rn+1 :y +···+y =0}
0 n
for the cones
C (T)={y ∈H(n):y +···y ≥0,t∈T,y +···y ≤0,t∈[n]\T}.
n 0 t−1 0 t−1
Each of the cones C (T) is simplicial, generated by the set of rays
n
e −e :t∈T,−e +e :t∈[n]\T,
t−1 t t−1 t
where e denotes a standard unit vector.
i
For t ∈ [n], consider the linear transformation τ which sends y 7→y , where indices are
t i n−t+1+i
considered modulo n+1. This volume preserving linear transformation sends the generators of
C ({t}) to the vectors
n
−e +e and −e +e :t∈[n]\T.
0 n t−1 t
All of these rays belong to C (∅), hence τ (C ({t})) ⊆ C (∅). Furthermore, each of the cones
n t n n
τ (C ({t})) is generated by a facet of C (∅) together with the same interior ray −e +e . Hence,
t n n 0 n
the set of cones {τ (C ({t})),t = 1,...,n} form the facets of a polyhedral subdivision of C (∅)
t n n
which completes the proof.
5. Further Directions
Our results on Eulerian-Catalannumbers suggest a number of interesting problems.
(i) Both the Catalan numbers and the Eulerian numbers have numerous combinatorial inter-
pretations. Are there other interpretations of the Eulerian-Catalannumbers as enumerating
objects that are counted by the Eulerian numbers where a certain statistic is a Catalan
object?
(ii) BoththeCatalannumbersandEuleriannumbershaveq(andq,t)analogues.Dotheseextend
to the Eulerian-Catalan numbers?
(iii) Catalan numbers and Eulerian numbers have natural generalizations beyond the symmetric
group (i.e. to other types). Can these be extended to Eulerian-Catalannumbers?
(iv) Generalizing the geometric proof of Theorem 1.1 to arbitrary j suggests the existence of
interesting polyhedral decompositions of the cone of positive roots of a Weyl group.
Acknowledgement
Hoda Bidkhori was partially supported by the David and Lucille Packard Foundation. Seth
SullivantwaspartiallysupportedbytheDavidandLucillePackardFoundationandtheUSNational
Science Foundation (DMS 0954865).
References
[1] F.Ardila.TheCatalanmatroid,J. Combin. Theory Ser. A104(2003) 49–62.
[2] R.Bagula,SequenceA177042inTheOn-LineEncyclopediaofIntegerSequences(2010),publishedelectronically
athttp://eis.org.
[3] H.Bidkhori.Latticepathmatroidpolytopes,InPreparation.
7
[4] J. E. Bonin, A. Mier, and M. Noy. Lattice path matroids: enumerative aspects and Tutte polynomials. J.
Combin. Theory Ser. A104(2003) 63–94.
[5] K.L.ChungandW.Feller.Onfluctuations incoin-tossing.Proc. Nat. Acad. Sci. U. S. A.35,(1949). 605608.
[6] T.LamandA.Postnikov.Alcovedpolytopes,I,Discrete& Comput. Geom.,38(2007) 453–478.
[7] R.Stanley,Catalanaddendum (versionof21August2010),http://www-math.mit.edu/˜rstan/ec/catadd.pdf.
[8] R. Stanley. Eulerian partitions of a unit hypercube, in Higher Combinatorics (M. Aigner, ed.), Reidel,
Dordrecht/Boston, 1977,p.49.
[9] R.Stanley.Enumerative Combinatorics, vol.II,CambridgeUniversityPress,2002.
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