A new shellability proof of an old identity of Dixon
Ruth Davidson 1 , Augustine O’Keefe 2 , and Daniel Parry 3
1
University of Illinois Urbana-Champaign
2
3
Connecticut College
University of Cologne
October 18, 2015: AMS Southeastern Sectional
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“Dixon’s Identity"-Some History and Context
Combinatorial identity (Dixon 1891):
!3 (
n
X
0 if n is odd, and
s n
(−1)
=
3n/2
n/2
s
(−1)
,
n/2,n/2,n/2
s=0
if n is even.
Fun history facts:
• Hard to find original manuscript: Messenger of Mathematics- ended
1929
• See MacMahon’s Combinatory Analysis (1915)
• Zillions of ways to prove it. Ex: MacMahon’s Master Theorem!
• Special case of multiple families of identities (hypergeometric and
"beyond")
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Reformulate Dixon’s Identity as a Shellability Exercise
n
X
n
(−1)s
s
s=0
!3
(
=
0 if n is odd, and
3n/2
(−1)n/2 n/2,n/2,n/2
,
if n is even.
• Find a shellable simplicial complex with face numbers fs−1 =
reduced Euler Characteristic
Pn
s=0 (−1)
s+1
n 3
n 3
s
and
s
ei count i-dimensional facets attaching along entire
• Betti numbers β
boundary
in shelling
Euler-Poincaré Relation:
Pd
Pd order. Use
i
ie
i=−1 (−1) fi =
i=−1 (−1) βi .
Story: Elkies –> Hersh/Stanton –> Hersh –> Me – - - - –> Me + Dan + Tina
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The right simplicial complex
Definition (Hersh). Let ∆(n) be the simplicial complex with ground set
{(i` , j` , k` ) | i` , j` , k` ∈ [n]} and faces given by collections
{(i1 , j1 , k1 ), . . . , (is , js , ks )} satisfying
i1 < i2 < · · · < is ,
j1 < j2 < · · · < js ,
k1 < k2 < · · · < ks .
Example. F = {(1, 3, 3), (5, 4, 5)} is a face of ∆(k ) for k ≥ 5, but F is a
facet of ∆(5). G = {(1, 3, 3), (5, 5, 5)} is not a facet of ∆(5) because G ⊂ H
where H = {(1, 3, 3), (4, 4, 4), (5, 5, 5)}.
3
3
• ∆(n) has ns (s − 1)-dimensional faces ⇐⇒ fs−1 = ns .
• Facets of ∆(n) satisfy
1. 1 ∈ {i1 , j1 , k1 }
2. n ∈ {is , js , ks }
3. min{i` − i`−1 , j` − j`−1 , k` − k`−1 } = 1 for all ` ∈ {2, . . . s}
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A Shelling Order for ∆(n)
Theorem.The lex order on the label sequences
(i1 , j1 , k1 , i2 , j2 , k2 , . . . , ir , jr , kr ) ∈ [n]3r of the facets of ∆(n) taken in descending
dimension induces a shelling of ∆(n).
Facet {(1, 1, 2), (2, 3, 3)} has label sequence (1, 1, 2, 2, 3, 3).
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Homology Facet Characterization
Proposition.The homology facets of ∆(n) are the facets
{(i1 , j1 , k1 ), . . . , (ir , jr , kr )} satisfying max{i1 , j1 , k1 } > 1 and
(i` − i`−1 ) + (j` − j`−1 ) + (k` − k`−1 ) > 3 for all ` ∈ {2, 3, . . . , r }.
Example: {(1, 1, 2), (2, 3, 3)} is a homology facet of ∆(3)
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Homology Facet Characterization
Proposition.The homology facets of ∆(n) are the facets
{(i1 , j1 , k1 ), . . . , (ir , jr , kr )} satisfying max{i1 , j1 , k1 } > 1 and
(i` − i`−1 ) + (j` − j`−1 ) + (k` − k`−1 ) > 3 for all ` ∈ {2, 3, . . . , r }.
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Homology Facet Characterization
Proposition.The homology facets of ∆(n) are the facets
{(i1 , j1 , k1 ), . . . , (ir , jr , kr )} satisfying max{i1 , j1 , k1 } > 1 and
(i` − i`−1 ) + (j` − j`−1 ) + (k` − k`−1 ) > 3 for all ` ∈ {2, 3, . . . , r }.
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Homology Facet Characterization
Proposition.The homology facets of ∆(n) are the facets
{(i1 , j1 , k1 ), . . . , (ir , jr , kr )} satisfying max{i1 , j1 , k1 } > 1 and
(i` − i`−1 ) + (j` − j`−1 ) + (k` − k`−1 ) > 3 for all ` ∈ {2, 3, . . . , r }.
Note: 3 is odd and
P1
i=−1
βei (∆(3)) = 0.
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Homology Facet Data
Values are homology facets with r vertices.
∆(2) ∆(3) ∆(4) ∆(5) ∆(6)
r =1
6
12
18
24
30
r =2
0
12
114
396
948
r =3
0
6
372
3138
r =4
0
0
540
r =5
0
0
0
r =6
r =7
| A. S. |
6
0
90
0
1680
∆(7)
36
1860
13704
12240
360
0
0
0
Bummer. Not in the OEIS...
Bummer?
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Generating functions for Homology Facets: Main Ideas
• Suppose i1 = 1, j1 , k1 > 1:
f (x, y , z) = x
∞ X
∞
X
2
2
y ji z ki =
ji =2 ki =2
xy 2 z 2
.
(1 − y )(1 − z)
• Supppose ii = 1, and ji = 1:
h(x, y , z) = xy
∞
X
ki =2
z ki =
xyz 2
.
1−z
• So
P(x, y , z) = f (x, y , z)+f (y , x, z)+f (z, y , x)+h(x, y , z)+h(x, z, y )+h(z, y , x)
1 − xyz
⇐⇒ P(x, y , z) = xyz
−1 .
(1 − x)(1 − y )(1 − z)
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Right GF = Two GFs from Two Types of Homology Facets
Add Case (1) min{ir , jr , kr } < n, to Case (2) (ir , jr , kr ) = (n, n, n):
χ(x, y , z) =
X
P(x, y , z)r (−1)r −1 + xyz
r =1
= (P(x, y , z) + xyz)
X
P(x, y , z)r −1 (−1)r −1
r =1
∞
X
P(x, y , z)r −1 (−1)r −1
r =1
=
∞
X
xyz(1 − xyz)
P(x, y , z)r −1 (−1)r −1
(1 − x)(1 − y )(1 − z)
r =1
xyz(1 − xyz)
=
(1 − x)(1 − y )(1 − z)
=
1
1 − xyz +
!
xyz(1−xyz)
(1−x)(1−y )(1−z)
xyz
(1 − x)(1 − y )(1 − z) + xyz
Hit it with MacMahon’s Master Theorem..... get
(
0 if n is odd, and
n n n
[x y z ] =
3n/2
,
(−1)n/2 n/2,n/2,n/2
if n is even.
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Related identities
Generalized ∆(n): Γp (n) with vertices given by sequences in [n]p for p ≥ 1,
and faces given by collections of vertices
{(i1,1 , . . . , i1,p ), (i2,1 , . . . , i2,p ), . . . , (ir ,1 , . . . , ir ,p )}
satisfying i`,a ∈ [n] for all ` ∈ [r ] and all a ∈ [p], and i`,a < i(`+1),a for all
` ∈ [r − 1] and all a ∈ [p]. Then Γp (n) has face numbers given by
!p
n
fs−1 =
s
for 0 ≤ s ≤ n.
Γ1 (n): (EC Vol 1 Exercise 1.3-f)
n
X
k =0
(−1)
k
n
k
!
= 0,
n ≥ 1;
(1)
Γ1 (n) corresponds to ∆n−1 with labeled vertices: homotopy type of a point.
Γ2 (n) corresponds to the identity
!2 (
n
X
0 if n is odd,
k n
(−1)
=
n
k
(−1)n/2 n/2
, if n is even.
.
(2)
k =0
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Summary
• We have recast an old identity of Dixon as the reduced Euler
characteristic of a non-pure shellable simplicial complex.
• We identified the homology facets of the shellable complex and counted
them using a generating function.
• Machine for generating new identities?
• More interesting: a way to peek “under the hood" about the Master
Theorem, connections between types of enumeration...
Thanks to Noam Elkies for challenging Patricia Hersh to prove this using
topology, to Patricia Hersh for suggesting this problem to me, and to Dennis
Stanton for many history lessons...
Thanks to Dan and Tina!
Funding: DMS-0954865 and DMS-1401591
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