some recent work on special polytopesrstan/transparencies/polytopes.pdf · descent polytopes denis...
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![Page 1: Some Recent Work on Special Polytopesrstan/transparencies/polytopes.pdf · Descent polytopes Denis Chebikin, Ph.D. thesis, M.I.T., 2008, and Richard Ehrenborg S [n 1] = f1;2;:::;n](https://reader035.vdocuments.mx/reader035/viewer/2022070112/6056c3a4a470941fb427e2b1/html5/thumbnails/1.jpg)
Two Poset Polytopes
Slides available at:www-math.mit.edu/∼rstan/transparencies/polytopes.pdf
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The order polytope
P: p-element poset, say P = t1, . . . , tp
Definition. The order polytope O(P) ⊂ Rp is defined by
O(P) = (x1, . . . , xp) ∈ Rp ∶ 0 ≤ xi ≤ 1, ti ≤P tj ⇒ xi ≤ xj.
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The order polytope
P: p-element poset, say P = t1, . . . , tp
Definition. The order polytope O(P) ⊂ Rp is defined by
O(P) = (x1, . . . , xp) ∈ Rp ∶ 0 ≤ xi ≤ 1, ti ≤P tj ⇒ xi ≤ xj.
Example. (a) If P is an antichain, then O(P) is the p-dimensionalunit cube [0,1]p .(b) If P is a chain t1 < ⋯ < tp, then O(P) is the p-dimensionalsimplex 0 ≤ x1 ≤ ⋯ ≤ xp ≤ 1.
![Page 4: Some Recent Work on Special Polytopesrstan/transparencies/polytopes.pdf · Descent polytopes Denis Chebikin, Ph.D. thesis, M.I.T., 2008, and Richard Ehrenborg S [n 1] = f1;2;:::;n](https://reader035.vdocuments.mx/reader035/viewer/2022070112/6056c3a4a470941fb427e2b1/html5/thumbnails/4.jpg)
The order polytope
P: p-element poset, say P = t1, . . . , tpDefinition. The order polytope O(P) ⊂ Rp is defined by
O(P) = (x1, . . . , xp) ∈ Rp ∶ 0 ≤ xi ≤ 1, ti ≤P tj ⇒ xi ≤ xj.
Example. (a) If P is an antichain, then O(P) is the p-dimensionalunit cube [0,1]p .(b) If P is a chain t1 < ⋯ < tp, then O(P) is the p-dimensionalsimplex 0 ≤ x1 ≤ ⋯ ≤ xp ≤ 1.
dimO(P) = p, since if ti1, . . . , tip is a linear extension of P , thenO(P) contains the p-dimensional simplex 0 ≤ xi1 ≤ ⋯ ≤ xip ≤ 1 (ofvolume 1/p!).
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Convex polytope
convex hull conv(X ) of a subset X of Rp: intersection of allconvex sets containing X
half-space in Rp: a subset H of Rp of the formx ∈ Rp ∶ x ⋅ v ≤ α for some fixed 0 ≠ v ∈ Rp and α ∈ R.
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Convex polytope
convex hull conv(X ) of a subset X of Rp: intersection of allconvex sets containing X
half-space in Rp: a subset H of Rp of the formx ∈ Rp ∶ x ⋅ v ≤ α for some fixed 0 ≠ v ∈ Rp and α ∈ R.
Classical theorem. Let P ⊆ R. The following two conditions areequivalent.
![Page 7: Some Recent Work on Special Polytopesrstan/transparencies/polytopes.pdf · Descent polytopes Denis Chebikin, Ph.D. thesis, M.I.T., 2008, and Richard Ehrenborg S [n 1] = f1;2;:::;n](https://reader035.vdocuments.mx/reader035/viewer/2022070112/6056c3a4a470941fb427e2b1/html5/thumbnails/7.jpg)
Convex polytope
convex hull conv(X ) of a subset X of Rp: intersection of allconvex sets containing X
half-space in Rp: a subset H of Rp of the formx ∈ Rp ∶ x ⋅ v ≤ α for some fixed 0 ≠ v ∈ Rp and α ∈ R.
Classical theorem. Let P ⊆ R. The following two conditions areequivalent.
P is a bounded intersection of half-spaces.
![Page 8: Some Recent Work on Special Polytopesrstan/transparencies/polytopes.pdf · Descent polytopes Denis Chebikin, Ph.D. thesis, M.I.T., 2008, and Richard Ehrenborg S [n 1] = f1;2;:::;n](https://reader035.vdocuments.mx/reader035/viewer/2022070112/6056c3a4a470941fb427e2b1/html5/thumbnails/8.jpg)
Convex polytope
convex hull conv(X ) of a subset X of Rp: intersection of allconvex sets containing X
half-space in Rp: a subset H of Rp of the formx ∈ Rp ∶ x ⋅ v ≤ α for some fixed 0 ≠ v ∈ Rp and α ∈ R.
Classical theorem. Let P ⊆ R. The following two conditions areequivalent.
P is a bounded intersection of half-spaces.
P = conv(X ) for some finite X ⊂ Rp.
![Page 9: Some Recent Work on Special Polytopesrstan/transparencies/polytopes.pdf · Descent polytopes Denis Chebikin, Ph.D. thesis, M.I.T., 2008, and Richard Ehrenborg S [n 1] = f1;2;:::;n](https://reader035.vdocuments.mx/reader035/viewer/2022070112/6056c3a4a470941fb427e2b1/html5/thumbnails/9.jpg)
Convex polytope
convex hull conv(X ) of a subset X of Rp: intersection of allconvex sets containing X
half-space in Rp: a subset H of Rp of the formx ∈ Rp ∶ x ⋅ v ≤ α for some fixed 0 ≠ v ∈ Rp and α ∈ R.
Classical theorem. Let P ⊆ R. The following two conditions areequivalent.
P is a bounded intersection of half-spaces.
P = conv(X ) for some finite X ⊂ Rp.
Such a set P is a convex polytope.
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Vertices
A vertex v of a convex polytope (or even convex set) P is a pointin P such that
v = λx + (1 − λ)y , x , y ∈ P, 0 ≤ λ ≤ 1⇒ x = y .
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Vertices
A vertex v of a convex polytope (or even convex set) P is a pointin P such that
v = λx + (1 − λ)y , x , y ∈ P, 0 ≤ λ ≤ 1⇒ x = y .
Equivalently, let V = ⋂Y ∶ P=conv(Y )Y . Then V is the set ofvertices of P (necessarily finite).
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Vertices
A vertex v of a convex polytope (or even convex set) P is a pointin P such that
v = λx + (1 − λ)y , x , y ∈ P, 0 ≤ λ ≤ 1⇒ x = y .
Equivalently, let V = ⋂Y ∶ P=conv(Y )Y . Then V is the set ofvertices of P (necessarily finite).
Equivalently, v is a vertex of P if and only if there exists ahalf-space H such that v = P ∩H.
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Order ideals
An order ideal of a poset P is a set I ⊆ P such thatt ∈ I , s < t ⇒ s ∈ I .
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Order ideals
An order ideal of a poset P is a set I ⊆ P such thatt ∈ I , s < t ⇒ s ∈ I .
Let S ⊆ P = t1, . . . , tp. The characteristic vector χS of S isdefined by χS = (x1, x2, . . . , xp) such that
xi = 0, ti /∈ S1, ti ∈ S .
Write χS = (1 − x1,1 − x2, . . . ,1 − xp).
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Order ideals
An order ideal of a poset P is a set I ⊆ P such thatt ∈ I , s < t ⇒ s ∈ I .
Let S ⊆ P = t1, . . . , tp. The characteristic vector χS of S isdefined by χS = (x1, x2, . . . , xp) such that
xi = 0, ti /∈ S1, ti ∈ S .
Write χS = (1 − x1,1 − x2, . . . ,1 − xp).Theorem. The vertices of O(P) are the sets χI , where I is anorder ideal of P.
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An example
t
t t
t1 2
3 4
(1,1,1,1)
(0,1,1,1)
(1,0,1,1)
(0,0,1,1)
(1,0,1,0)
(0,0,0,1)
(0,0,1,0)
(0,0,0,0)
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Proof
Theorem. The vertices of O(P) are the sets χI , where I is anorder ideal of P.
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Proof
Theorem. The vertices of O(P) are the sets χI , where I is anorder ideal of P.
Proof. Clearly χI ∈ O(P), and χI is a vertex (e.g., since it is avertex of the binary unit cube containing O(P)). Also, any binaryvector in O(P) has the form χI .
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Proof
Theorem. The vertices of O(P) are the sets χI , where I is anorder ideal of P.
Proof. Clearly χI ∈ O(P), and χI is a vertex (e.g., since it is avertex of the binary unit cube containing O(P)). Also, any binaryvector in O(P) has the form χI .
Assume v ∈ O(P) is a vertex and v ≠ χI for some I .
Idea: show v = λx + (1 − λ)y for some 0 < λ < 1 and x , y ∈ O(P).
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Proof
Theorem. The vertices of O(P) are the sets χI , where I is anorder ideal of P.
Proof. Clearly χI ∈ O(P), and χI is a vertex (e.g., since it is avertex of the binary unit cube containing O(P)). Also, any binaryvector in O(P) has the form χI .
Assume v ∈ O(P) is a vertex and v ≠ χI for some I .
Idea: show v = λx + (1 − λ)y for some 0 < λ < 1 and x , y ∈ O(P).Let v = (v1, . . . , vp). Choose vi so 0 < vi < 1 (exists since the onlybinary vectors in O(P) have the form χI ). Choose ǫ > 0 sufficientlysmall. Let v− (respectively, v+) be obtained from v by subtracting(respectively, adding) ǫ to each entry equal to vi . Thenv−, v+ ∈ O(P) and
v =1
2(v− + v+). ◻
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Two remarks
Note. Can also prove the previous theorem by showing directlythat every v ∈ O(P) is a convex combination of the χI ’s (notdifficult).
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Two remarks
Note. Can also prove the previous theorem by showing directlythat every v ∈ O(P) is a convex combination of the χI ’s (notdifficult).
Note. Entire facial structure of O(P) can be described, butomitted here.
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Linear extensions
L(P): set of linear extensions σ∶P → 1, . . . ,p, i.e., σ is bijectiveand order-preserving.
e(P) ∶=#L(P)
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An example (from before)
ba
c d
1
2
3
4
a b c d
b a c d
a b d c
b a d c
b d a c
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Decomposition of O(P)
Let σ ∈ L(P), regarded as the permutation ti1, ti2 , . . . , tip of P (soσ(tij ) = j). Define
O(σ) = (x1, . . . , xn) ∈ Rp ∶ 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1,a simplex of volume 1/p! .
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Decomposition of O(P)
Theorem. The simplices O(σ) have disjoint interiors, andO(P) = ⋃σ∈L(P)O(σ).
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Decomposition of O(P)
Theorem. The simplices O(σ) have disjoint interiors, andO(P) = ⋃σ∈L(P)O(σ).Corollary. The volume vol(O(P)) of O(P) is equal to e(P)/p!.
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Decomposition of O(P)
Theorem. The simplices O(σ) have disjoint interiors, andO(P) = ⋃σ∈L(P)O(σ).Corollary. The volume vol(O(P)) of O(P) is equal to e(P)/p!.Proof of Corollary. O(P) is the union of e(p) simplices O(σ)with disjoint interiors and volume 1/p! each. ◻
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Proof of decomposition theorem
The interior of the simplex 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1 is given by0 < xi1 < xi2 < ⋯ < xip < 1. Thus the interiors of the O(σ)’s aredisjoint, since a set of distinct real numbers has a unique orderingwith respect to <. And clearly O(σ) ⊆ O(P).
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Proof of decomposition theorem
The interior of the simplex 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1 is given by0 < xi1 < xi2 < ⋯ < xip < 1. Thus the interiors of the O(σ)’s aredisjoint, since a set of distinct real numbers has a unique orderingwith respect to <. And clearly O(σ) ⊆ O(P).Now suppose that (x1, . . . , xp) ∈ O(P). Define i1, . . . , ip (notnecessarily unique) by
xi1 ≤ xi2 ≤ ⋯ ≤ xip
andtij < tik and xij = xik ⇒ j < k .
Then ti1, ti2 , . . . , tip is a linear extension σ of P , and(x1, . . . , xp) ∈ O(P). ◻
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An aside
Note. This decomposition of O(P) is actually a triangulation,i.e., the intersection of any two of the simplices is a common face(possibly empty) of both. (Easy to see.)
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An example
t
t t
t
t
t
t
1
2 34
6 5
7
.1
.3 .3
.7
.2
.3
.7
Three “compatible” linear extensions:
t1, t4, t2, t3, t6, t5, t7
t1, t4, t3, t2, t6, t5, t7
t1, t4, t2, t6, t3, t5, t7
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Why volume 1/p! ?
The simplices 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1 all have the same volume(for any permutation i1, . . . , ip of 1, . . . ,p), since they differ only bya permutation of coordinates.
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Why volume 1/p! ?
The simplices 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1 all have the same volume(for any permutation i1, . . . , ip of 1, . . . ,p), since they differ only bya permutation of coordinates.
Let P be a p-element antichain. Thus L(P) =Sp (allpermutations of 1, . . . ,p). Moreover, O(P) is a unit cube so hasvolume 1. By the decomposition theorem, it is a union of p!simplices 0 ≤ xi1 ≤ ⋯ ≤ xip ≤ 1, all with the same volume. Thus thevolume of each simplex is 1/p!.
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Ehrhart theory
P : a d -dimensional convex polytope in Rp with integer vertices.
nP = nx ∶ x ∈ P, n ≥ 1int(P): interior of P
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Ehrhart theory
P : a d -dimensional convex polytope in Rp with integer vertices.
nP = nx ∶ x ∈ P, n ≥ 1int(P): interior of PFor n ≥ 1, define
i(P,n) = # (nP⋂Zp)
i(P,n) = # (n int(P)⋂Zp) .
i(P,n) is the Ehrhart polynomial of P.
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An example
P 3Pi(P,n) = (n + 1)2i(P,n) = (n − 1)2 = i(P,−n)
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Main results of Ehrhart theory
i(P,n) is a polynomial in n of degree d = dimP for n ≥ 1, andhence is defined for all n ∈ Z
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Main results of Ehrhart theory
i(P,n) is a polynomial in n of degree d = dimP for n ≥ 1, andhence is defined for all n ∈ Z
i(P,0) = 1
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Main results of Ehrhart theory
i(P,n) is a polynomial in n of degree d = dimP for n ≥ 1, andhence is defined for all n ∈ Z
i(P,0) = 1(Ehrhart’s law of reciprocity) i(P,n) = (−1)d i(P,−n)
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Main results of Ehrhart theory
i(P,n) is a polynomial in n of degree d = dimP for n ≥ 1, andhence is defined for all n ∈ Z
i(P,0) = 1(Ehrhart’s law of reciprocity) i(P,n) = (−1)d i(P,−n)
Proofs: see e.g. EC1, §4.6.2, or Beck-Robins, Computing theContinuous Discretely.
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The order polynomial
P: p-element poset
For n ≥ 1, define the order polynomial ΩP(n) of P by
ΩP(n) =#f ∶P → 1, . . . ,n ∣ s ≤P t ⇒ f (s) ≤Z f (t) .
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The order polynomial
P: p-element poset
For n ≥ 1, define the order polynomial ΩP(n) of P by
ΩP(n) =#f ∶P → 1, . . . ,n ∣ s ≤P t ⇒ f (s) ≤Z f (t) .
ΩP(1) = 1
ΩP(2) = # order ideals of P
= # vertices of O(P)Ω(p-chain,n) = ((n
p)) = (n + p − 1
p)
Ω(p-antichain,n) = np
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Strict order polynomial
For n ≥ 1, define the strict order polynomial ΩP(n) of P by
ΩP(n) =#f ∶P → 1, . . . ,n ∣ s <P t ⇒ f (s) <Z f (t) .
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Strict order polynomial
For n ≥ 1, define the strict order polynomial ΩP(n) of P by
ΩP(n) =#f ∶P → 1, . . . ,n ∣ s <P t ⇒ f (s) <Z f (t) .
ΩP(1) = 0 unless P is an antichain
Ω(p-chain,n) = (np)
Ω(p-antichain,n) = np
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Polynomiality
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!. (Thus ΩP(n) determines e(P).)
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Polynomiality
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!. (Thus ΩP(n) determines e(P).)Proof. es : number of surjective order-preserving mapsP → 1, . . . , s (so es = 0 if s > p).
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Polynomiality
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!. (Thus ΩP(n) determines e(P).)Proof. es : number of surjective order-preserving mapsP → 1, . . . , s (so es = 0 if s > p).
To obtain f ∶P → 1, . . . ,n order-preserving, choose 1 ≤ s ≤ p,then choose an s-element subset S of 1, . . . ,n is (n
s) ways, and
finally choose a surjective order-preserving map P → S in es ways.
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Polynomiality
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!. (Thus ΩP(n) determines e(P).)Proof. es : number of surjective order-preserving mapsP → 1, . . . , s (so es = 0 if s > p).
To obtain f ∶P → 1, . . . ,n order-preserving, choose 1 ≤ s ≤ p,then choose an s-element subset S of 1, . . . ,n is (n
s) ways, and
finally choose a surjective order-preserving map P → S in es ways.
⇒ ΩP(n) = p
∑s=1
es(ns).
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Polynomiality
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!. (Thus ΩP(n) determines e(P).)Proof. es : number of surjective order-preserving mapsP → 1, . . . , s (so es = 0 if s > p).
To obtain f ∶P → 1, . . . ,n order-preserving, choose 1 ≤ s ≤ p,then choose an s-element subset S of 1, . . . ,n is (n
s) ways, and
finally choose a surjective order-preserving map P → S in es ways.
⇒ ΩP(n) = p
∑s=1
es(ns).
Now (ns) is a polynomial in n of degree s and leading coefficient
1/s!. Moreover, ep = e(P) (clear). The proof follows. ◻
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Polynomiality (cont.)
Similarly:
Theorem. ΩP(n) is a polynomial in n of degree p and leadingcoefficient e(P)/p!.
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Ehrhart polynomial of O(P)
integer points in nO(P): integer solutions (a1, . . . ,ap) to0 ≤ xi ≤ n, ti ≤P tj ⇒ xi ≤R xj
Define f (ti) = ai . Thusf ∶P → 0,1, . . . ,n, ti ≤P tj ⇒ ai ≤Z aj ,
so i(O(P),n) = ΩP(n + 1) (since #0,1, . . . ,n = n + 1).
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Order polynomial reciprocity
strict order-preserving map f ∶P → N: s <P t ⇒ f (s) <Z f (t)
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Order polynomial reciprocity
strict order-preserving map f ∶P → N: s <P t ⇒ f (s) <Z f (t)points (x1, . . . , xp) in int(O(P)):
0<xi<1, ti <P tj ⇒ xi<xj
points (x1, . . . , xp) in n int(O(P)):0 < xi < n, ti <P tj ⇒ xi < xj.
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Order polynomial reciprocity
strict order-preserving map f ∶P → N: s <P t ⇒ f (s) <Z f (t)points (x1, . . . , xp) in int(O(P)):
0<xi<1, ti <P tj ⇒ xi<xj
points (x1, . . . , xp) in n int(O(P)):0 < xi < n, ti <P tj ⇒ xi < xj.
Thus i(O(P),n) is the number of strict order-preserving mapsP → 1, . . . ,n − 1. Since ΩP(n) = i(O(P),n − 1) andi(O(P),n) = (−1)p i(O(P),−n), we get:
Corollary (reciprocity for order polynomials).ΩP(n) = (−1)pΩ(−n).
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Simple application
Corollary. Let ℓ be the length (one less than the number ofelements) of the longest chain of P. Then
ΩP(0) = ΩP(−1) = ⋯ = ΩP(−ℓ) = 0.
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Simple application
Corollary. Let ℓ be the length (one less than the number ofelements) of the longest chain of P. Then
ΩP(0) = ΩP(−1) = ⋯ = ΩP(−ℓ) = 0.Proof. If s0 < s1 < ⋯ < sm is a chain of length m andf ∶P → 1, . . . ,n is strictly order-preserving, then
f (s0) < f (s1) < ⋯ < f (sm).Thus n > m. ◻
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Simple application
Corollary. Let ℓ be the length (one less than the number ofelements) of the longest chain of P. Then
ΩP(0) = ΩP(−1) = ⋯ = ΩP(−ℓ) = 0.Proof. If s0 < s1 < ⋯ < sm is a chain of length m andf ∶P → 1, . . . ,n is strictly order-preserving, then
f (s0) < f (s1) < ⋯ < f (sm).Thus n > m. ◻
Many other interesting results on order polynomials, but no timehere!
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Mixed volumes
convex body: a nonempty, compact, convex subset X of Rp.(Given that X is convex, compact is the same as bounded.)
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Mixed volumes
convex body: a nonempty, compact, convex subset X of Rp.(Given that X is convex, compact is the same as bounded.)
Let α,β ≥ 0. The Minkowski sum αK + βL of two convex bodiesK ,L is given by
αK + βL = αx + βy ∶ x ∈ K , y ∈ L.
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An example
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Minkowski’s theorem
K ,L: convex bodies in Rp
α,β ≥ 0
Theorem (Minkowski). We have
vol(αK + βL) = p
∑i=0
(pi)Vi(K ,L)αp−iβi ,
for certain real numbers Vi(K ,L) ≥ 0, called the mixed volumesof K and L.
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Minkowski’s theorem
K ,L: convex bodies in Rp
α,β ≥ 0
Theorem (Minkowski). We have
vol(αK + βL) = p
∑i=0
(pi)Vi(K ,L)αp−iβi ,
for certain real numbers Vi(K ,L) ≥ 0, called the mixed volumesof K and L.
Note. V0(K ,L) = vol(K) (set α = 1, β = 0) and Vp(K ,L) = vol(L)(set α = 0, β = 1).
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An example
+ =
a = 3/√2, b = 4/√2
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An example
+ =
a = 3/√2, b = 4/√2
vol(αK + βL) = 12α2 + 27
2√2αβ
⇒ V0(K ,L) = 12, V1(K ,L) = 7
2√2, V2(K ,L) = 0
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Unimodality and log-concavity
a0,a1, . . . ,an: sequence of nonnegative real numbers
unimodal: a0 ≤ a1 ≤ ⋯ ≤ aj ≥ aj+1 ≥ ⋯ ≥ an for some j
log-concave: a2i ≥ ai−1ai+1, 1 ≤ i ≤ n − 1
no internal zeros: i < j < k , ai ≠ 0,ak ≠ 0⇒ aj ≠ 0
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Unimodality and log-concavity
a0,a1, . . . ,an: sequence of nonnegative real numbers
unimodal: a0 ≤ a1 ≤ ⋯ ≤ aj ≥ aj+1 ≥ ⋯ ≥ an for some j
log-concave: a2i ≥ ai−1ai+1, 1 ≤ i ≤ n − 1
no internal zeros: i < j < k , ai ≠ 0,ak ≠ 0⇒ aj ≠ 0
Note. Log-concave and no internal zeros ⇒ unimodal.
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Unimodality and log-concavity
a0,a1, . . . ,an: sequence of nonnegative real numbers
unimodal: a0 ≤ a1 ≤ ⋯ ≤ aj ≥ aj+1 ≥ ⋯ ≥ an for some j
log-concave: a2i ≥ ai−1ai+1, 1 ≤ i ≤ n − 1
no internal zeros: i < j < k , ai ≠ 0,ak ≠ 0⇒ aj ≠ 0
Note. Log-concave and no internal zeros ⇒ unimodal.
Proof. Suppose a0,a1, . . . ,an is not unimodal. Then for somei < j − 1, we have
ai > ai+1 = ai+2 = ⋯ = aj−1 < aj .
If no internal zeros, then 0 = ai+1 > 0. Then a2i+1 < aiai+2, so thesequence is not log-concave. ◻
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Height of an element in a linear extension
Let t ∈ P .
σ = s1, . . . , sp: linear extension of P
height of t in σ is k , if sk = t. Denoted htσ(t).
Ni = Ni(t): number of linear extensions σ of P for whichhtσ(t) = i . In other words
Ni =#σ ∈ L(P) ∶ σ(t) = i.
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An example
1 2
3 4
5
σ ht(4)1 2 3 4 5 42 1 3 4 5 41 2 4 3 5 32 1 4 3 5 32 4 1 3 5 21 2 3 5 4 52 1 3 5 4 5
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An example
1 2
3 4
5
σ ht(4)1 2 3 4 5 42 1 3 4 5 41 2 4 3 5 32 1 4 3 5 32 4 1 3 5 21 2 3 5 4 52 1 3 5 4 5
⇒ (N1,N2,N3,N4,N5) = (0,1,2,2,2) (log-concave)
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The Aleksandrov-Fenchel inequalities
Theorem (A.D. Aleksandrov, 1937–38, and W. Fenchel, 1936).For any convex bodies K ,L,⊂ Rp, we have
Vi(K ,L)2 ≥ Vi−1(K ,L)Vi+1(K ,L), 1 ≤ i ≤ p − 1.
. Moreover, the sequence V0,V1, . . . ,Vp has no internal zeros.
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The Aleksandrov-Fenchel inequalities
Theorem (A.D. Aleksandrov, 1937–38, and W. Fenchel, 1936).For any convex bodies K ,L,⊂ Rp, we have
Vi(K ,L)2 ≥ Vi−1(K ,L)Vi+1(K ,L), 1 ≤ i ≤ p − 1.
. Moreover, the sequence V0,V1, . . . ,Vp has no internal zeros.
Proof is difficult.
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Chung-Fishburn-Graham conjecture
Theorem (conjecture of Fan Chung, Peter Fishburn, and RonGraham, 1980) For any p-element poset P and t ∈ P, we have
Ni(t)2 ≥ Ni−1(t)Ni+1(t), 2 ≤ i ≤ p − 1.
Moreover, the sequence N1, . . . ,Np has no internal zeros (so it isunimodal).
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Chung-Fishburn-Graham conjecture
Theorem (conjecture of Fan Chung, Peter Fishburn, and RonGraham, 1980) For any p-element poset P and t ∈ P, we have
Ni(t)2 ≥ Ni−1(t)Ni+1(t), 2 ≤ i ≤ p − 1.
Moreover, the sequence N1, . . . ,Np has no internal zeros (so it isunimodal).
Plan of proof. No internal zeros: easy combinatorial argument.
Log-concavity: find convex bodies (actually, polytopes) in Rp−1
such that Ni = Vi(K ,L) (up to multiplication by a positiveconstant). (Proof by wishful thinking)
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What are K and L?
Let P = t, t1, t2, . . . , tp−1.K = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 0 if ti < tL = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 1 if ti > t
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What are K and L?
Let P = t, t1, t2, . . . , tp−1.K = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 0 if ti < tL = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 1 if ti > tNote that K ,L ⊂ O(P − t).
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What are K and L?
Let P = t, t1, t2, . . . , tp−1.K = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 0 if ti < tL = (x1, . . . , xp−1) ∈ Rp−1 ∶ 0 ≤ xi ≤ 1, xi ≤ xj if ti ≤ tj , xi = 1 if ti > tNote that K ,L ⊂ O(P − t).Claim. Vi(K ,L) = Ni+1(t)(p − 1)!
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What is αK +βL?
αK + βL = set of all (x1, . . . , xp−1) ∈ Rp−1 such that ∶
ti ≤P−t tj ⇒ 0 ≤ xi ≤ xj ≤ α + β
ti <P t ⇒ xi ≤ β
ti >P t ⇒ xi ≥ β
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Proof that Vi(K ,L) = Ni+1(t)
Recall P = t1, . . . , tp−1, t. For α,β ≥ 0 let P = αK + βL. Foreach linear extension σ∶P → 1, . . . ,p, define
∆σ = (x1, . . . , xp−1) ∈ P ∶xi ≤ xj if σ(ti) ≤ σ(tj)xi ≤ β if σ(ti) < σ(t)
β ≤ xi ≤ α + β if σ(ti) > σ(t).
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Proof that Vi(K ,L) = Ni+1(t)
Recall P = t1, . . . , tp−1, t. For α,β ≥ 0 let P = αK + βL. Foreach linear extension σ∶P → 1, . . . ,p, define
∆σ = (x1, . . . , xp−1) ∈ P ∶xi ≤ xj if σ(ti) ≤ σ(tj)xi ≤ β if σ(ti) < σ(t)
β ≤ xi ≤ α + β if σ(ti) > σ(t).Easy to check (completely analogous the proof that O(P) is aunion of simplices O(σ)): ∆σ’s, for σ ∈ L(P), have disjointinteriors and union P.
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Proof that Vi(K ,L) = Ni+1(t)
Recall P = t1, . . . , tp−1, t. For α,β ≥ 0 let P = αK + βL. Foreach linear extension σ∶P → 1, . . . ,p, define
∆σ = (x1, . . . , xp−1) ∈ P ∶xi ≤ xj if σ(ti) ≤ σ(tj)xi ≤ β if σ(ti) < σ(t)
β ≤ xi ≤ α + β if σ(ti) > σ(t).Easy to check (completely analogous the proof that O(P) is aunion of simplices O(σ)): ∆σ’s, for σ ∈ L(P), have disjointinteriors and union P.
Note. ∆σ need not be a simplex because σ is a linear extension ofP , not P − t.
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Proof (cont.)
Let σ(t) = i , and define w ∈ Sp−1 by
σ(tw(1)) < σ(tw(2)) < ⋯ < σ(tw(p−1)).Then ∆σ consists of all (x1, . . . , xp−1) ∈ Rp−1 satisfying
0 ≤ xw(1) ≤ ⋯ ≤ xw(i−1) ≤ β ≤ xw(i) ≤ ⋯ ≤ xw(p−1) ≤ α + β.
This is a product of two simplices with volume
vol(∆σ) = αp−i
(p − i)!βi−1
(i − 1)! .
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An example
t 2
3 4
=1
1234 β ≤ x2 ≤ x3 ≤ x4 ≤ α + β
21 34 0 ≤ x2 ≤ β ≤ x3 ≤ x4 ≤ α + β
1243 β ≤ x2 ≤ x4 ≤ x3 ≤ α + β
21 43 0 ≤ x2 ≤ β ≤ x4 ≤ x3 ≤ α + β
241 3 0 ≤ x2 ≤ x4 ≤ β ≤ x3 ≤ α + β
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Conclusion of proof
∆σ is a product of two simplices with volume
vol(∆σ) = αp−i
(p − i)!βi−1
(i − 1)! .
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Conclusion of proof
∆σ is a product of two simplices with volume
vol(∆σ) = αp−i
(p − i)!βi−1
(i − 1)! .⇒ vol(P) = ∑
σ∈L(P)
vol(∆σ)
=p
∑i=1
Ni(t) αp−i
(p − i)!βi−1
(i − 1)!=
1
(p − 1)!p−1
∑i=0
Ni+1(t)(p − 1i)αp−1−iβi ,
so Vi(K ,L) = Ni+1(t)(p − 1)! . ◻
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The chain polytope
P = t1, . . . , tp as beforeDefinition. The chain polytope C(P) ⊂ Rp is defined by
C(P) = (x1, . . . , xp) ∈ Rp ∶ 0 ≤ xi , ∑ti ∈C
xi ≤ 1 for every chain
(or maximal chain) C of P.
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The chain polytope
P = t1, . . . , tp as beforeDefinition. The chain polytope C(P) ⊂ Rp is defined by
C(P) = (x1, . . . , xp) ∈ Rp ∶ 0 ≤ xi , ∑ti ∈C
xi ≤ 1 for every chain
(or maximal chain) C of P.Example. (a) If P is an antichain, then C(P) = O(P), thep-dimensional unit cube [0,1]p .(b) If P is a chain t1 < ⋯ < tp, then C(P) is the p-dimensionalsimplex xi ≥ 0, x1 +⋯+ xp ≤ 1.
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Two notes
Note. If P is a chain t1 < ⋯ < tp, then define φ∶O(P) → C(P) byφ(x1, . . . , xp) = (x1, x2 − x1, . . . , xp − xp−1). Then φ is a linearisomorphism from O(P) to C(P). It preserves volume since thelinear transformation has determinant 1 (lower triangular with 1’son the diagonal).
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Two notes
Note. If P is a chain t1 < ⋯ < tp, then define φ∶O(P) → C(P) byφ(x1, . . . , xp) = (x1, x2 − x1, . . . , xp − xp−1). Then φ is a linearisomorphism from O(P) to C(P). It preserves volume since thelinear transformation has determinant 1 (lower triangular with 1’son the diagonal).
Note. dimC(P) = p, since the cube [0,1/p]p is contained in C(P).
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Vertices of C(P)
Recall that if S ⊆ P , then
χS = (x1, . . . , xp) ∶ xi = 0, ti /∈ S1, ti ∈ S .
Theorem. The vertices of C(P) are the sets χA, where A is anantichain of P.
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An example
t
t t
t1 2
3 4
(0,0,0,0)
(1,0,0,0)
(0,1,0,0)
(0,0,1,0)
(0,0,0,1)
(1,1,0,0)
(1,0,0,1)
(0,0,1,1)
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Proof
Theorem. The vertices of C(P) are the sets χA, where A is anantichain of P.
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Proof
Theorem. The vertices of C(P) are the sets χA, where A is anantichain of P.
Proof. Clearly χA ∈ O(P), and χA is a vertex (e.g., since it is avertex of the binary unit cube containing C(P)). Also, any binaryvector in C(P) has the form χA.
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Proof
Theorem. The vertices of C(P) are the sets χA, where A is anantichain of P.
Proof. Clearly χA ∈ O(P), and χA is a vertex (e.g., since it is avertex of the binary unit cube containing C(P)). Also, any binaryvector in C(P) has the form χA.
Assume v ∈ C(P) is a vertex and v ≠ χA for some A.
Idea: show v = λx + (1 − λ)y for some 0 < λ < 1 and x , y ∈ C(P).
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Proof
Theorem. The vertices of C(P) are the sets χA, where A is anantichain of P.
Proof. Clearly χA ∈ O(P), and χA is a vertex (e.g., since it is avertex of the binary unit cube containing C(P)). Also, any binaryvector in C(P) has the form χA.
Assume v ∈ C(P) is a vertex and v ≠ χA for some A.
Idea: show v = λx + (1 − λ)y for some 0 < λ < 1 and x , y ∈ C(P).Let v = (v1, . . . , vp), v ≠ χA. Let Q = ti ∈ P ∶ 0 < vi < 1. Let Q1
be the set of minimal elements of Q and Q2 the set of minimalelements of Q −Q1.
Easy: if v is a vertex then Q1,Q2 ≠ ∅.
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Proof (cont.)
Defineε = minvi ,1 − vi ∶ ti ∈ Q1 ∪Q2.
Define y = (y1, . . . , yp), z = (z1, . . . , zp) ∈ Rp by
yi =
⎧⎪⎪⎪⎨⎪⎪⎪⎩vi , ti /∈ Q1 ∪Q2
vi + ε, ti ∈ Q1
vi − ε, ti ∈ Q2,
zi =
⎧⎪⎪⎪⎨⎪⎪⎪⎩vi , ti /∈ Q1 ∪Q2
vi − ε, ti ∈ Q1
vi + ε, ti ∈ Q2,
Easy to see y , z ∈ C(P). Since y ≠ z and v = 1
2(y + z), it follows
that v is not a vertex of C(P). ◻
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Order ideals and antichains
Recall: vertices of O(P): χI , where I is an order ideal
vertices of C(P): χA, where A is an antichain
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Order ideals and antichains
Recall: vertices of O(P): χI , where I is an order ideal
vertices of C(P): χA, where A is an antichain
Posets 101: if P is a finite poset, then there is a bijection f fromorder ideals of P to antichains of P
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Order ideals and antichains
Recall: vertices of O(P): χI , where I is an order ideal
vertices of C(P): χA, where A is an antichain
Posets 101: if P is a finite poset, then there is a bijection f fromorder ideals of P to antichains of P
f (I) = maximal elements of If −1(A) = s ∈ P ∶ s ≤ t for some t ∈ A
= ⋂order ideals J
A⊆J
J.
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Order ideals and antichains
Recall: vertices of O(P): χI , where I is an order ideal
vertices of C(P): χA, where A is an antichain
Posets 101: if P is a finite poset, then there is a bijection f fromorder ideals of P to antichains of P
f (I) = maximal elements of If −1(A) = s ∈ P ∶ s ≤ t for some t ∈ A
= ⋂order ideals J
A⊆J
J.
Thus there is a (canonical) bijection V (O(P))→ V (C(P)). Whatabout other faces?
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Example of order ideal ↔ antichain bijection
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Example of order ideal ↔ antichain bijection
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The X -poset
a b
c
d e
O(P): 0 ≤ a,b (2 equations), d , e ≤ 1 (2 equations), a ≤ c , b ≤ c ,c ≤ d , c ≤ e, so 8 facets (maximal faces)
C(P): 0 ≤ a,b, c ,d , e (5 equations), a + c + d ≤ 1, a + c + e ≤ 1,b + c + d ≤ 1, b + c + e ≤ 1, so 9 facets
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The X -poset
a b
c
d e
O(P): 0 ≤ a,b (2 equations), d , e ≤ 1 (2 equations), a ≤ c , b ≤ c ,c ≤ d , c ≤ e, so 8 facets (maximal faces)
C(P): 0 ≤ a,b, c ,d , e (5 equations), a + c + d ≤ 1, a + c + e ≤ 1,b + c + d ≤ 1, b + c + e ≤ 1, so 9 facets
Hence O(P) and C(P) are not combinatorially equivalent (i.e.,they don’t have isomorphic face posets)
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Length two posets
Exercise. If P has no 3-element chain, then there is a bijectivelinear transformation τ ∶O(P)→ C(P) of determinant 1. ThusO(P) and C(P) are combinatorial equivalent and have the sameEhrhart polynomial (and hence the same volume).
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Length two posets
Exercise. If P has no 3-element chain, then there is a bijectivelinear transformation τ ∶O(P)→ C(P) of determinant 1. ThusO(P) and C(P) are combinatorial equivalent and have the sameEhrhart polynomial (and hence the same volume).
Would like something similar for any finite poset, but a lineartransformation won’t work since O(P) and C(P) need not becombinatorially equivalent.
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Length two posets
Exercise. If P has no 3-element chain, then there is a bijectivelinear transformation τ ∶O(P)→ C(P) of determinant 1. ThusO(P) and C(P) are combinatorial equivalent and have the sameEhrhart polynomial (and hence the same volume).
Would like something similar for any finite poset, but a lineartransformation won’t work since O(P) and C(P) need not becombinatorially equivalent.
Note. For the characterization of P for which there is a bijectivelinear transformation τ ∶O(P)→ C(P) of determinant 1, see Hibiand Li, arXiv:1208.4029.
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The transfer map
Definition. Define the transfer map φ∶O(P) → C(P) as follows:if x = (x1, . . . , xp) ∈ O(P) and ti ∈ P , then φ(x) = (y1, . . . , yp),where
yi = minxi − xj ∶ ti covers tj in P.(If ti is a minimal element of P , then yi = xi .)
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An example
.1 .2 .3
.3 .5
.8
.1 .3
0 .2
.3
.2
f
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Transfer theorem
Theorem. (a) The transfer map φ is a continuous,piecewise-linear bijection from O(P) onto C(P).Note. Piecewise-linear means that we can express O(P) as afinite union O(P) = X1 ∪⋯∪ Xk such that φ restricted to each Xi
is linear.
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Transfer theorem
Theorem. (a) The transfer map φ is a continuous,piecewise-linear bijection from O(P) onto C(P).Note. Piecewise-linear means that we can express O(P) as afinite union O(P) = X1 ∪⋯∪ Xk such that φ restricted to each Xi
is linear.
(b) Let n ∈ P = 1,2, . . . and x ∈ O(P). Then nx ∈ Zp if and onlyif nφ(x) ∈ Zp.
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Transfer theorem
Theorem. (a) The transfer map φ is a continuous,piecewise-linear bijection from O(P) onto C(P).Note. Piecewise-linear means that we can express O(P) as afinite union O(P) = X1 ∪⋯∪ Xk such that φ restricted to each Xi
is linear.
(b) Let n ∈ P = 1,2, . . . and x ∈ O(P). Then nx ∈ Zp if and onlyif nφ(x) ∈ Zp.
Corollary. vol(C(P)) = vol(O(P)) = e(P) andi(C(P),n) = i(O(P),n) = ΩP(n + 1).
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Transfer theorem proof.
(a) The transfer map φ is a continuous, piecewise-linear bijectionfrom O(P) onto C(P).Proof. Continuity is immediate from the definition. Recall thedecomposition O(P) = ⋃σ∈L(P)O(σ), where σ = (ti1 , . . . , tip) and
O(σ) = (x1, . . . , xn) ∈ Rp ∶ 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1.Clearly φ is linear on each O(σ), so φ is piecewise-linear.
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Transfer theorem proof.
(a) The transfer map φ is a continuous, piecewise-linear bijectionfrom O(P) onto C(P).Proof. Continuity is immediate from the definition. Recall thedecomposition O(P) = ⋃σ∈L(P)O(σ), where σ = (ti1 , . . . , tip) and
O(σ) = (x1, . . . , xn) ∈ Rp ∶ 0 ≤ xi1 ≤ xi2 ≤ ⋯ ≤ xip ≤ 1.Clearly φ is linear on each O(σ), so φ is piecewise-linear.
Define ψ∶C(P) →O(P) by ψ(y1, . . . , yp) = (x1, . . . , xp), wherexj = maxyi1 + yi2 +⋯+ yik ∶ ti1 < ti2 < ⋯ < tik = tj.
Easy to check: ψφ(x) = x and φψ(y) = y for all x ∈O(P) andy ∈ C(P). Hence φ is a bijection with inverse ψ.
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An example (redux)
.1 .2 .3
.3 .5
.8
.1 .3
0 .2
.3
.2
f
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Conclusion of proof
Remains to show that nx ∈ Zp if and only if nφ(x) ∈ Zp. This isclear from the formulas for φ(x) and ψ(y).
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Conclusion of proof
Remains to show that nx ∈ Zp if and only if nφ(x) ∈ Zp. This isclear from the formulas for φ(x) and ψ(y).Alternatively, the restriction of φ to each O(σ) belongs toSL(p,Z), i.e., the matrix of of φ with respect to the standard basisof Rp is integral of determinant 1. (In fact, it’s lower triangularwith 1’s on the diagonal.)
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Interesting corollary
∆(P): set of chains of P (order complex)
Corollary. ΩP(n) (and hence e(P)) depends only on ∆(P).
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Interesting corollary
∆(P): set of chains of P (order complex)
Corollary. ΩP(n) (and hence e(P)) depends only on ∆(P).Proof. The set A of antichains of P depends only on ∆(P).Recall
C(P) = convχA ∶ A ∈ A.Thus C(P) depends only on ∆(P). Since ΩP(n + 1) = i(C(P),n),the proof follows. ◻
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An example where ∆(P) =∆(Q)
1 2
3 4 5
6 7
8 9
10
11
1 2
3
6
4
7
8 5
910
11
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An example where ∆(P) =∆(Q)
1 2
3 4 5
6 7
8 9
10
11
1 2
3
6
4
7
8 5
910
11
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Aside: when does ∆(P) =∆(Q)?
P: finite poset
autonomous subset Q ⊆ P : if t ∈ P −Q, then either:
t > s for all s ∈ Q
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Aside: when does ∆(P) =∆(Q)?
P: finite poset
autonomous subset Q ⊆ P : if t ∈ P −Q, then either:
t > s for all s ∈ Q
t < s for all s ∈ Q
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Aside: when does ∆(P) =∆(Q)?
P: finite poset
autonomous subset Q ⊆ P : if t ∈ P −Q, then either:
t > s for all s ∈ Q
t < s for all s ∈ Q
t ∥ s (i.e., s and t are incomparable) for all s ∈ Q.
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Aside: when does ∆(P) =∆(Q)?
P: finite poset
autonomous subset Q ⊆ P : if t ∈ P −Q, then either:
t > s for all s ∈ Q
t < s for all s ∈ Q
t ∥ s (i.e., s and t are incomparable) for all s ∈ Q.
flip of an autonomous Q ⊆ P : keep all relations s < t unchangedunless both s, t ∈ Q. If s < t in Q, then change to s > t, and viceversa. (We are “dualizing” Q inside P .)
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Aside: when does ∆(P) =∆(Q)?
P: finite poset
autonomous subset Q ⊆ P : if t ∈ P −Q, then either:
t > s for all s ∈ Q
t < s for all s ∈ Q
t ∥ s (i.e., s and t are incomparable) for all s ∈ Q.
flip of an autonomous Q ⊆ P : keep all relations s < t unchangedunless both s, t ∈ Q. If s < t in Q, then change to s > t, and viceversa. (We are “dualizing” Q inside P .)
Special case: if Q = P , then the flip of Q gives P∗, the dual of P
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Flip example
1 2
3 4 5
6 7
8 9
10
11
1 2
3
6
4
7
8 5
910
11
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Flip theorem
Theorem (Dreesen, Poguntke, Winkler, 1985; implicit in earlierwork of Gallai and others). Let P and Q be finite posets. Then∆(P) =∆(Q) if and only if Q can be obtained from P by asequence of flips.
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Flip theorem
Theorem (Dreesen, Poguntke, Winkler, 1985; implicit in earlierwork of Gallai and others). Let P and Q be finite posets. Then∆(P) =∆(Q) if and only if Q can be obtained from P by asequence of flips.
Easy to see that flips preserve the order polynomial ΩP(n). Thisgives another proof that if ∆(P) =∆(Q) then ΩP(n) = ΩQ(n), soalso e(P) = e(Q) (Golumbic, 1980).
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Chain polytope analogue of N1(t), . . . ,Np(t).
Recall: for t ∈ P ,
Ni (t) =#σ ∈ L(P) ∶ σ(t) = i.Then Ni(t)2 ≥ Ni−1(t)Ni+1(t), and no internal zeros.
Proof based on Aleksandrov-Fenchel inequalities for polytopesrelated to O(P).
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Chain polytope analogue of N1(t), . . . ,Np(t).
Recall: for t ∈ P ,
Ni (t) =#σ ∈ L(P) ∶ σ(t) = i.Then Ni(t)2 ≥ Ni−1(t)Ni+1(t), and no internal zeros.
Proof based on Aleksandrov-Fenchel inequalities for polytopesrelated to O(P).Would like to “transfer” this result to C(P). What is the analogueof Ni(t)?
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Mi(t)
Given t ∈ P and σ ∈ L(P) with σ(t) = j , definespread
σ(t) = maxi ∶ σ−1(j − 1), σ−1(j − 2), . . . , σ−1(j − i) are
all incomparable with t.
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Mi(t)
Given t ∈ P and σ ∈ L(P) with σ(t) = j , definespread
σ(t) = maxi ∶ σ−1(j − 1), σ−1(j − 2), . . . , σ−1(j − i) are
all incomparable with t.Mi (t) =#σ ∈ L(P) ∶ spreadσ(t) = i.
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Mi(t)
Given t ∈ P and σ ∈ L(P) with σ(t) = j , definespread
σ(t) = maxi ∶ σ−1(j − 1), σ−1(j − 2), . . . , σ−1(j − i) are
all incomparable with t.Mi (t) =#σ ∈ L(P) ∶ spreadσ(t) = i.Theorem. Mi(t)2 ≥Mi−1(t)Mi+1(t), 1 ≤ i ≤ p − 1
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Mi(t)
Given t ∈ P and σ ∈ L(P) with σ(t) = j , definespread
σ(t) = maxi ∶ σ−1(j − 1), σ−1(j − 2), . . . , σ−1(j − i) are
all incomparable with t.Mi (t) =#σ ∈ L(P) ∶ spreadσ(t) = i.Theorem. Mi(t)2 ≥Mi−1(t)Mi+1(t), 1 ≤ i ≤ p − 1
Proof. “Transfer” the proof that Ni(t)2 ≥ Ni−1(t)Ni+1(t) (messydetails omitted).
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An example
1
2
3
4
σ spread(2)1 2 3 4 01 3 2 4 11 3 4 2 23 1 2 4 03 1 4 2 13 4 1 2 0
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An example
1
2
3
4
σ spread(2)1 2 3 4 01 3 2 4 11 3 4 2 23 1 2 4 03 1 4 2 13 4 1 2 0
⇒ (M0,M1,M2,M3) = (3,2,1,0)
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Decreasing property
Theorem. We have M0(t) ≥M1(t) ≥ ⋯ ≥Mp−1(t).
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Decreasing property
Theorem. We have M0(t) ≥M1(t) ≥ ⋯ ≥Mp−1(t).Proof. Let
Mi(t) = σ ∈ L(P) ∶ spreadt(σ) = i.Suffices to give a injection (1-1 correspondence)Mi(t)→Mi−1(t), 1 ≤ i ≤ p − 1.
Given a linear extension
σ = (tk1 , ..., tkj−1 , tkj = t, tkj+1 , . . . , tkp)of spread at least 1, map it to
(tk1 , ..., tkj−2 , t, tkj−1 , tj+1, . . . , tkp). ◻
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Decreasing property
Theorem. We have M0(t) ≥M1(t) ≥ ⋯ ≥Mp−1(t).Proof. Let
Mi(t) = σ ∈ L(P) ∶ spreadt(σ) = i.Suffices to give a injection (1-1 correspondence)Mi(t)→Mi−1(t), 1 ≤ i ≤ p − 1.
Given a linear extension
σ = (tk1 , ..., tkj−1 , tkj = t, tkj+1 , . . . , tkp)of spread at least 1, map it to
(tk1 , ..., tkj−2 , t, tkj−1 , tj+1, . . . , tkp). ◻END OF TOPIC 2