Proof of the q-TSPP Conjecture
Christoph Koutschan (collaboration with Manuel Kauers and Doron Zeilberger)
Research Institute for Symbolic Computation (RISC) Johannes Kepler Universit¨atLinz, Austria
13.09.2010 S´eminaireLotharingien de Combinatoire 65 TCSP
CyclicallyTotally Symmetric Plane (3D Ferrers diagram) • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1
• πi,j = πj,i • cyclically symmetric 3D Ferrers diagram is invariant under the action of S3.
P
Partition
17 = 5+4+3+2+1+1+1 17 = 5+4+3+2+1+1+1
TCS
CyclicallyTotally Symmetric (3D Ferrers diagram)
• πi,j = πj,i • cyclically symmetric 3D Ferrers diagram is invariant under the action of S3.
PP
Plane Partition • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1 17 = 5+4+3+2+1+1+1
TCS
CyclicallyTotally Symmetric
• πi,j = πj,i • cyclically symmetric 3D Ferrers diagram is invariant under the action of S3.
PP
Plane Partition (3D Ferrers diagram) • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1 17 = 5+4+3+2+1+1+1
TC
CyclicallyTotally (3D Ferrers diagram)
• cyclically symmetric 3D Ferrers diagram is invariant under the action of S3.
SPP
Symmetric Plane Partition • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1
• πi,j = πj,i 17 = 5+4+3+2+1+1+1
T
Totally (3D Ferrers diagram)
• πi,j = πj,i
3D Ferrers diagram is invariant under the action of S3.
CSPP
Cyclically Symmetric Plane Partition • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1
• cyclically symmetric 17 = 5+4+3+2+1+1+1
C
Cyclically (3D Ferrers diagram)
TSPP
Totally Symmetric Plane Partition • two-dimensional array π = (πi,j)1≤i,j • πi,j ∈ N with finite sum P |π| = πi,j
• πi,j ≥ πi+1,j and πi,j ≥ πi,j+1
• πi,j = πj,i • cyclically symmetric 3D Ferrers diagram is invariant under the action of S3. TSPP Count
Y i + j + k − 1 Theorem: There are TSPPs in [0, n]3. i + j + k − 2 1≤i≤j≤k≤n
Example: (n = 2)
Y i + j + k − 1 2 3 4 5 = · · · = 5. i + j + k − 2 1 2 3 4 1≤i≤j≤k≤2 TSPP Count
Y i + j + k − 1 Theorem: There are TSPPs in [0, n]3. i + j + k − 2 1≤i≤j≤k≤n
Proof: See • John Stembridge, The enumeration of totally symmetric plane partitions, Advances in Mathematics 111 (1995), 227–243. • George E. Andrews, Peter Paule, and Carsten Schneider, Plane Partitions VI. Stembridge’s TSPP theorem, Advances in Applied Mathematics 34 (2005), 709–739. • Christoph Koutschan, Eliminating Human Insight: An Algorithmic Proof of Stembridge’s TSPP Theorem, Contemporary Mathematics 517 (2010), 219–230. Example: TSPP and orbits Example: TSPP and orbits Example: TSPP and orbits Example: TSPP and orbits Example: TSPP and orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits Orbits 36 Orbits q0 +q1 +q2 +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n. +q1 +q2 +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 + +q2 +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 + +q2 +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 + + +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 + + +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 + + +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 + + + +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 q3 + + + +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 q3 + + + +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 q3 + + + +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 q3 + + + + =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n
Let T (n) denote set of TSPPs with largest part at most n.
q0 q1 q2 q3 q4 Let T (n) denote set of TSPPs with largest part at most n.
q0 +q1 +q2 +q3 +q4 =
1 − q5 = 1 + q + q2 + q3 + q4 = 1 − q
X Y 1 − qi+j+k−1 q-TSPP conjecture: q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n
Y i + j + k − 1 Stembridge’s theorem: |T (n)| = i + j + k − 2 1≤i≤j≤k≤n All these problems had been solved, except one: q-TSPP.
The q-TSPP conjecture Conjectured independently by George Andrews and David Robbins (ca. 1983) Last surviving conjecture of the collection by Richard Stanley: A baker’s dozen of conjectures concerning plane partitions (1986) (alternating sign matrix conjecture, TSPP conjecture, etc.) The q-TSPP conjecture Conjectured independently by George Andrews and David Robbins (ca. 1983) Last surviving conjecture of the collection by Richard Stanley: A baker’s dozen of conjectures concerning plane partitions (1986) (alternating sign matrix conjecture, TSPP conjecture, etc.)
All these problems had been solved, except one: q-TSPP. Determinantal formulation
Also in Stanley’s paper, we find: Okada’s determinant
Soichi Okada: On the generating functions for certain classes of plane partitions, Journal of Combinatorial Theory, Series A (1989).
Rewrite “the sum of all minors” as a single determinant!
The q-TSPP conjecture is true if
2 Y 1 − qi+j+k−1 det (a ) = =: b . i,j 1≤i,j≤n 1 − qi+j+k−2 n 1≤i≤j≤k≤n where ! i+j−1 i + j − 2 i + j − 1 i ai,j := q + q +(1+q )δi,j −δi,j+1. i − 1 q i q Zeilberger’s holonomic ansatz Doron Zeilberger: The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations, Annals of Combinatorics (2007).
Problem: Given ai,j and bn 6= 0. Show det (ai,j)1≤i,j≤n = bn.
Method: “Pull out of the hat” a function cn,j and prove
cn,n = 1 (n ≥ 1), n X cn,jai,j = 0 (1 ≤ i < n), j=1 n X bn c a = (n ≥ 1). n,j n,j b j=1 n−1
Then det (ai,j)1≤i,j≤n = bn holds. 2n 2 nn n+j bn det(q (;aqi,j)n)1≤i,j≤n XX (−1) Mn,j n 2 2 aan,jn,jcn,j bn−1 (q ; qb)nn−1 bn−1 jj=1=1 | {z } =: cn,j n+n (−1) Mn,n cn,n = 1 bn−1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
nn n+j XX (−1)n+j Mn,j 0 = aai,ji,j(c−n,j1) Mn,j bn−1 jj=1=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n X n+j bn = det (ai,j)1≤i,j≤n = an,j (−1) Mn,j j=1 nn (q2n; q)2 XX b det (a n) aa (c−1)n+jM n (qn; q2i,j)2 1≤i,j≤n n,jn,j n,j n,j n jj=1=1
n+n (−1) Mn,n cn,n = 1 bn−1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
nn n+j XX (−1)n+j Mn,j 0 = aai,ji,j(c−n,j1) Mn,j bn−1 jj=1=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n n+j bn det (ai,j)1≤i,j≤n X (−1) Mn,j = = an,j bn−1 bn−1 bn−1 j=1 | {z } =: cn,j nn (q2n; q)2 XX b det (a n) aa (c−1)n+jM n (qn; q2i,j)2 1≤i,j≤n n,jn,j n,j n,j n jj=1=1
1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
nn n+j XX (−1)n+j Mn,j 0 = aai,ji,j(c−n,j1) Mn,j bn−1 jj=1=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n n+j bn det (ai,j)1≤i,j≤n X (−1) Mn,j = = an,j bn−1 bn−1 bn−1 j=1 | {z } =: cn,j n+n (−1) Mn,n cn,n = bn−1 nn (q2n; q)2 XX b det (a n) aa (c−1)n+jM n (qn; q2i,j)2 1≤i,j≤n n,jn,j n,j n,j n jj=1=1
n+n (−1) Mn,n bn−1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
nn n+j XX (−1)n+j Mn,j 0 = aai,ji,j(c−n,j1) Mn,j bn−1 jj=1=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n n+j bn det (ai,j)1≤i,j≤n X (−1) Mn,j = = an,j bn−1 bn−1 bn−1 j=1 | {z } =: cn,j
cn,n = 1 nn (q2n; q)2 XX b det (a n) aa (c−1)n+jM n (qn; q2i,j)2 1≤i,j≤n n,jn,j n,j n,j n jj=1=1
n+n (−1) Mn,n bn−1
nn n+j XX (−1) Mn,j aai,ji,jcn,j bn−1 jj=1=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n n+j bn det (ai,j)1≤i,j≤n X (−1) Mn,j = = an,j bn−1 bn−1 bn−1 j=1 | {z } =: cn,j
cn,n = 1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
n X n+j 0 = ai,j (−1) Mn,j j=1 nn (q2n; q)2 XX b det (a n) aa (c−1)n+jM n (qn; q2i,j)2 1≤i,j≤n n,jn,j n,j n,j n jj=1=1
n+n (−1) Mn,n bn−1
nn XX n+j aai,ji,j(c−n,j1) Mn,j jj=1=1
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n n+j bn det (ai,j)1≤i,j≤n X (−1) Mn,j = = an,j bn−1 bn−1 bn−1 j=1 | {z } =: cn,j
cn,n = 1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
n n+j X (−1) Mn,j 0 = ai,j bn−1 j=1 | {z } = cn,j 2n 2 n n+j (q ; q)n X (−1)n+j Mn,j bn detn (a2i,j2)1≤i,j≤n an,j (−1) Mn,j (q ; q )n bn−1 j=1 | {z } =: cn,j n+n (−1) Mn,n bn−1
n n+j X (−1)n+j Mn,j ai,j (−1) Mn,j bn−1 j=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
n bn det (ai,j)1≤i,j≤n X = = a c b b n,j n,j n−1 n−1 j=1
cn,n = 1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
n X 0 = ai,jcn,j j=1 n n+j det (ai,j)1≤i,j≤n X (−1)n+j Mn,j bn det (ai,j)1≤i,j≤n an,j (−1) Mn,j bn−1 bn−1 j=1 | {z } =: cn,j n+n (−1) Mn,n bn−1
n n+j X (−1)n+j Mn,j ai,j (−1) Mn,j bn−1 j=1 | {z } = cn,j
Zeilberger’s holonomic ansatz
Laplace expansion w.r.t. the n-th row:
2n 2 n bn (q ; q) X = n = a c b (qn; q2)2 n,j n,j n−1 n j=1
cn,n = 1
Now copy the i-th row (1 ≤ i < n) into the n-th row:
n X 0 = ai,jcn,j j=1 Zeilberger’s holonomic ansatz
Problem: Given ai,j and bn 6= 0. Show det (ai,j)1≤i,j≤n = bn.
Method: “Pull out of the hat” a function cn,j and prove
cn,n = 1 (n ≥ 1), n X cn,jai,j = 0 (1 ≤ i < n), j=1 n X bn c a = (n ≥ 1). n,j n,j b j=1 n−1
Then det (ai,j)1≤i,j≤n = bn holds. No! Argue by induction on n.
Advocatus Diaboli
What if det (ai,j)1≤i,j≤m = 0 for some m???
Then cn,j is not uniquely deter- mined!
Proof is wrong! Advocatus Diaboli
What if det (ai,j)1≤i,j≤m = 0 for some m???
Then cn,j is not uniquely deter- mined!
Proof is wrong!
No! Argue by induction on n. Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values. But there is no reason why c should admit n such a recursive Example: The binomial coefficientn,j fn,k = k can be described by description. (n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 All linear combinations of shifts are again valid recurrences:¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: (n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 n+1 k ¯ k+n+1 k ¯ (k + 1)fn+1(,kq+1 −−(nq −)fkn+1+,k 1)f=n+1,k (q= 0 − q )fn,k (n + 1)f 2k+1− (nk−¯k + 1)f −n (n +k 1)¯ f = 0 n+1(q,k+1 − q )fn,k+1 =n+1,k (q − q )fn,kn,k+1
They form a left ideal in somef¯0,0 noncommutative= 1 operator algebra.
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! But there is no reason why c should admit n such a recursive Example: The binomial coefficientn,j fn,k = k can be described by description. (n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 All linear combinations of shifts are again valid recurrences:¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: (n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 n+1 k ¯ k+n+1 k ¯ (k + 1)fn+1(,kq+1 −−(nq −)fkn+1+,k 1)f=n+1,k (q= 0 − q )fn,k (n + 1)f 2k+1− (nk−¯k + 1)f −n (n +k 1)¯ f = 0 n+1(q,k+1 − q )fn,k+1 =n+1,k (q − q )fn,kn,k+1
They form a left ideal in somef¯0,0 noncommutative= 1 operator algebra.
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values. But there is no reason why cn,j should admit such a recursive description.
All linear combinations of shifts are again valid recurrences:¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: (n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 n+1 k ¯ k+n+1 k ¯ (k + 1)fn+1(,kq+1 −−(nq −)fkn+1+,k 1)f=n+1,k (q= 0 − q )fn,k (n + 1)f 2k+1− (nk−¯k + 1)f −n (n +k 1)¯ f = 0 n+1(q,k+1 − q )fn,k+1 =n+1,k (q − q )fn,kn,k+1
They form a left ideal in somef¯0,0 noncommutative= 1 operator algebra.
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values. n Example: The binomial coefficient fn,k = k can be described by
(n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 But there is no reason why cn,j should admit such a recursive description.
All linear combinations of shifts are again valid recurrences:
(n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 (k + 1)fn+1,k+1 − (n − k + 1)fn+1,k = 0 (n + 1)fn+1,k+1 − (n − k + 1)fn+1,k − (n + 1)fn,k+1 = 0 They form a left ideal in some noncommutative operator algebra.
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values. n Example: The binomial coefficient fn,k = k can be described by
(n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 ¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: n+1 k ¯ k+n+1 k ¯ (q − q )fn+1,k = (q − q )fn,k 2k+1 k ¯ n k ¯ (q − q )fn,k+1 = (q − q )fn,k
f¯0,0 = 1 But there is no reason why cn,j should admit such a recursive description.
¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: n+1 k ¯ k+n+1 k ¯ (q − q )fn+1,k = (q − q )fn,k 2k+1 k ¯ n k ¯ (q − q )fn,k+1 = (q − q )fn,k
f¯0,0 = 1
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values. n Example: The binomial coefficient fn,k = k can be described by
(n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 All linear combinations of shifts are again valid recurrences:
(n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 (k + 1)fn+1,k+1 − (n − k + 1)fn+1,k = 0 (n + 1)fn+1,k+1 − (n − k + 1)fn+1,k − (n + 1)fn,k+1 = 0 They form a left ideal in some noncommutative operator algebra. n Example: The binomial coefficient fn,k = k can be described by
(n − k + 1)fn+1,k = (n + 1)fn,k
(k + 1)fn,k+1 = (n − k)fn,k
f0,0 = 1 All linear combinations of shifts are again valid recurrences:¯ n Analogously, we get for the q-binomial coefficient fn,k = k q: (n − k)fn+1,k+1 − (n + 1)fn,k+1 = 0 n+1 k ¯ k+n+1 k ¯ (k + 1)fn+1(,kq+1 −−(nq −)fkn+1+,k 1)f=n+1,k (q= 0 − q )fn,k (n + 1)f 2k+1− (nk−¯k + 1)f −n (n +k 1)¯ f = 0 n+1(q,k+1 − q )fn,k+1 =n+1,k (q − q )fn,kn,k+1
They form a left ideal in somef¯0,0 noncommutative= 1 operator algebra.
Holonomic systems Of course, it is unlikely to get a closed-form description for cn,j! Instead we aim at some “suitable description”, viz. implicitly via linear recurrences (“holonomic system”) plus initial values.
But there is no reason why cn,j should admit such a recursive description. TheirThe staircasetotal Gr¨obnerbasis size of is the244MB has Gr¨obnerbasis: the (several form 1000 pages of paper)!
cn,j+4 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 Great! We found the+ certificatecn+2,j + for c then+2 determinant,j+1 evaluation! cn+1,j+3 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+2,j+2 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+3,j+1 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+4,j = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 j n where each is a polynomial in Q[q, q , q ] of total degree ≤ 100.
Guessing
Manuel Kauers guessed some recurrences for cn,j. The staircasetotal size of is the244MB Gr¨obnerbasis: (several 1000 pages of paper)!
Great! We found the certificate for the determinant evaluation!
Guessing
Manuel Kauers guessed some recurrences for cn,j. Their Gr¨obnerbasis has the form
cn,j+4 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+1,j+3 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+2,j+2 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+3,j+1 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+4,j = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 j n where each is a polynomial in Q[q, q , q ] of total degree ≤ 100. TheirThe total Gr¨obnerbasis size is 244MB has the (several form 1000 pages of paper)!
cn,j+4 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 Great! We found the+ certificatecn+2,j + for c then+2 determinant,j+1 evaluation! cn+1,j+3 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+2,j+2 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+3,j+1 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+4,j = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 j n where each is a polynomial in Q[q, q , q ] of total degree ≤ 100.
Guessing
Manuel Kauers guessed some recurrences for cn,j. The staircase of the Gr¨obnerbasis: TheirThe staircase Gr¨obnerbasis of the has Gr¨obnerbasis: the form
cn,j+4 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+1,j+3 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+2,j+2 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 cn+3,j+1 = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+1,j + cn+1,j+1 + cn+1,j+2 + cn+2,j + cn+2,j+1 + cn+3,j cn+4,j = cn,j + cn,j+1 + cn,j+2 + cn,j+3 + cn+2,j + cn+2,j+1 j n where each is a polynomial in Q[q, q , q ] of total degree ≤ 100.
Guessing
Manuel Kauers guessed some recurrences for cn,j. The total size is 244MB (several 1000 pages of paper)!
Great! We found the certificate for the determinant evaluation! Advocatus Diaboli
The guessed recurrences can be artifacts, that do not describe the true function cn,j!!! Artifacts?
The guessed recurrences are very unlikely to be artifacts for several reasons: • solutions of a dense overdetermined linear systems • many polynomial coefficients factor nicely
• recurrences produce correct values for cn,j that were not used for guessing Advocatus Diaboli Convincing, but not a proof!
And even if the recursive de- scription of cn,j is correct, this wouldn’t prove anything yet!!!
Show:
cn,n = 1 n X cn,jai,j = 0 j=1 n X bn c a = n,j n,j b j=1 n−1 • We find an element in the annihilating ideal of cn,j of the form
pvcn+v,j+v = pv−1cn+v−1,j+v−1 + ··· + p1cn+1,j+1 + p0cn,j
j n with v ∈ N and pi ∈ Q[q, q , q ]. • Substituting j → n yields a recurrence for the diagonal sequence cn,n.
• Show that the corresponding operator factors into P1P2 where P2 corresponds to cn+1,n+1 = cn,n.
• Show that c1,1 = ··· = cv,v = 1. −→ works with v = 7.
The first identity
Prove the identities using the recursive description of cn,j.
How to prove cn,n = 1 for all n ≥ 1? The first identity
Prove the identities using the recursive description of cn,j.
How to prove cn,n = 1 for all n ≥ 1?
• We find an element in the annihilating ideal of cn,j of the form
pvcn+v,j+v = pv−1cn+v−1,j+v−1 + ··· + p1cn+1,j+1 + p0cn,j
j n with v ∈ N and pi ∈ Q[q, q , q ]. • Substituting j → n yields a recurrence for the diagonal sequence cn,n.
• Show that the corresponding operator factors into P1P2 where P2 corresponds to cn+1,n+1 = cn,n.
• Show that c1,1 = ··· = cv,v = 1. −→ works with v = 7. n For n ≥ 7 we have p7(q, q ) 6= 0.
Advocatus Diaboli
The leading coefficient p7 could have singularities!!! Advocatus Diaboli
The leading coefficient p7 could have singularities!!!
n For n ≥ 7 we have p7(q, q ) 6= 0. The third identity Recall: n X bn c a = n,j n,j b j=1 n−1 with ! i+j−1 i + j − 2 i + j − 1 i ai,j = q + q +(1+q )δi,j −δi,j+1. i − 1 q i q gives n X bn (1 + qn) − c + c0 = n,n−1 n,j b j=1 n−1 with ! 0 n+j−1 n + j − 2 n + j − 1 cn,j = q + q cn,j n − 1 q n q • Closure properties yield a recurrence for the left-hand side. • Recurrence for right-hand side is a right factor. • Compare finitely many initial values (again v = 7).
The third identity
n X bn How to prove (1 + qn) − c + c0 = ? n,n−1 n,j b j=1 n−1 0 • Compute an annihilating ideal for cn,j via closure properties. • Find a relation in this ideal of the form
0 0 0 pvcn+v,j + ··· + p1cn+1,j + p0cn,j = tn,j+1 − tn,j
n where the pv, . . . , p0 are rational functions in Q(q, q ) and tn,j j n 0 is a Q(q, q , q )-linear combination of certain shifts of cn,j. • Creative telescoping yields a recurrence for the sum. The third identity
n X bn How to prove (1 + qn) − c + c0 = ? n,n−1 n,j b j=1 n−1 0 • Compute an annihilating ideal for cn,j via closure properties. • Find a relation in this ideal of the form
0 0 0 pvcn+v,j + ··· + p1cn+1,j + p0cn,j = tn,j+1 − tn,j
n where the pv, . . . , p0 are rational functions in Q(q, q ) and tn,j j n 0 is a Q(q, q , q )-linear combination of certain shifts of cn,j. • Creative telescoping yields a recurrence for the sum. • Closure properties yield a recurrence for the left-hand side. • Recurrence for right-hand side is a right factor. • Compare finitely many initial values (again v = 7). n CK’sChyzak’sTakayama’sZeilberger’s rationalpolynomial algorithm: algorithm: slow ansatz: ansatz: algorithm:ansatzaansatzrefine faster witheliminate variant with (e.g.unknownwhich with is alsop Gr¨obneri( basedq, q ) n j j andonbases) elimination.rk( theq, q variable, q ). Leadsq . to a coupled first-order parametrized PLL n j l n j l=0 rk,l(q, q )(q ) linearEstimate:Input system recurrences52428800r of(q,q-difference q haven, daysq )j j-degrees = equations.X betweenn 24j l and 30 (in the rk (q, q , q ) = r (q,n q )(jq ) . Estimate:q = 1 case).∞ After?k 48h, this wasd reducedkk,l(q, q , toq ) 23. l=0 Estimate: 1677721600 days where the denominators d can be “guessed” by looking at k n theLeads leading to a linear coefficients system of over theQ Gr¨obnerbasis.(q, q ). We used this ansatz for proving TSPPn (took about 40 days). Leads to a linear system over Q(q, q ) with 377 unknowns. Estimate: 4000 days
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j? n CK’sChyzak’sTakayama’s rationalpolynomial algorithm: algorithm: ansatz: ansatz:ansatzaansatzrefine faster with variant with unknownwhich is alsopi( basedq, q ) n j andon elimination.rk(q, q , q ). Leads to a coupled first-order parametrized PLL n j l n j l=0 rk,l(q, q )(q ) linearEstimate: system52428800r of(q,q-difference q n, daysq )j = equations.X n j l rk (q, q , q ) = r (q,n q )(jq ) . Estimate: ∞?k dkk,l(q, q , q ) l=0 where the denominators d can be “guessed” by looking at k n theLeads leading to a linear coefficients system of over theQ Gr¨obnerbasis.(q, q ). We used this ansatz for proving TSPPn (took about 40 days). Leads to a linear system over Q(q, q ) with 377 unknowns. Estimate: 4000 days
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j?
Zeilberger’s slow algorithm: eliminate (e.g. with Gr¨obner bases) the variable qj. Input recurrences have j-degrees between 24 and 30 (in the q = 1 case). After 48h, this was reduced to 23. Estimate: 1677721600 days n CK’sChyzak’sZeilberger’s rationalpolynomial algorithm: slow ansatz: ansatz: algorithm:ansatzansatzrefine witheliminate with (e.g.unknown withp Gr¨obneri(q, q ) n j j andbases)rk( theq, q variable, q ). Leadsq . to a coupled first-order parametrized PLL n j l n j l=0 rk,l(q, q )(q ) linearInput system recurrencesr of(q,q-difference q haven, q )j j-degrees = equations.X betweenn 24j l and 30 (in the rk (q, q , q ) = r (q,n q )(jq ) . Estimate:q = 1 case).∞ After?k 48h, this wasd reducedkk,l(q, q , toq ) 23. l=0 Estimate: 1677721600 days where the denominators d can be “guessed” by looking at k n theLeads leading to a linear coefficients system of over theQ Gr¨obnerbasis.(q, q ). We used this ansatz for proving TSPPn (took about 40 days). Leads to a linear system over Q(q, q ) with 377 unknowns. Estimate: 4000 days
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j?
Takayama’s algorithm: a faster variant which is also based on elimination. Estimate: 52428800 days CK’sTakayama’sZeilberger’s rationalpolynomial algorithm: slow ansatz: ansatz: algorithm:ansatzarefine faster witheliminate variant (e.g.which with is also Gr¨obner based onbases) elimination. the variable qj. PLL n j l n j l=0 rk,l(q, q )(q ) Estimate:Input recurrences52428800r (q, q haven, daysq )j j-degrees = X betweenn 24j l and 30 (in the rk (q, q , q ) = r (q,n q )(jq ) . q = 1 case). Afterk 48h, this wasd reducedkk,l(q, q , toq ) 23. l=0 Estimate: 1677721600 days where the denominators d can be “guessed” by looking at k n theLeads leading to a linear coefficients system of over theQ Gr¨obnerbasis.(q, q ). We used this ansatz for proving TSPPn (took about 40 days). Leads to a linear system over Q(q, q ) with 377 unknowns. Estimate: 4000 days
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j?
n Chyzak’s algorithm: ansatz with unknown pi(q, q ) n j and rk(q, q , q ). Leads to a coupled first-order parametrized linear system of q-difference equations. Estimate: ∞? n CK’sChyzak’sTakayama’sZeilberger’s rational algorithm: algorithm: slow ansatz: algorithm:ansatzaansatz faster witheliminate variant with (e.g.unknownwhich with is alsop Gr¨obneri( basedq, q ) andonbases) elimination.r ( theq, q variablen, qj). Leadsqj. to a coupled first-order parametrized k PL n j l linearEstimate:Input system recurrences52428800 of q-difference haven daysj j-degrees equations.l=0 betweenrk,l(q, q )( 24q and) 30 (in the rk(q, q , q ) = n j Estimate:q = 1 case).∞ After? 48h, this wasd reducedk(q, q , toq ) 23. Estimate: 1677721600 days where the denominators dk can be “guessed” by looking at the leading coefficients of the Gr¨obnerbasis. n Leads to a linear system over Q(q, q ) with 377 unknowns.
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j?
CK’s polynomial ansatz: refine
L n j X n j l rk(q, q , q ) = rk,l(q, q )(q ) . l=0
n Leads to a linear system over Q(q, q ). We used this ansatz for proving TSPP (took about 40 days). Estimate: 4000 days n CK’sChyzak’sTakayama’sZeilberger’s polynomial algorithm: algorithm: slow ansatz: algorithm:aansatzrefine fastereliminate variant with (e.g.unknownwhich with is alsop Gr¨obneri( basedq, q ) n j j andonbases) elimination.rk( theq, q variable, q ). Leadsq . to a coupled first-order parametrized L linearEstimate:Input system recurrences52428800 of q-difference have days j-degrees equations.X between 24 and 30 (in the r (q, qn, qj) = r (q, qn)(qj)l. Estimate:q = 1 case).∞ After?k 48h, this was reducedk,l to 23. l=0 Estimate: 1677721600 days n Leads to a linear system over Q(q, q ). We used this ansatz for proving TSPP (took about 40 days). Estimate: 4000 days
A computational challenge
How to find the certificate
n 0 n 0 pv(q, q )cn+v,j + ··· + p0(q, q )cn,j = tn,j+1 − tn,j
n j 0 n j 0 where tn,j = r1(q, q , q )cn+3,j+2 + ··· + r10(q, q , q )cn,j?
CK’s rational ansatz: ansatz with
PL n j l n j l=0 rk,l(q, q )(q ) rk(q, q , q ) = n j dk(q, q , q ) where the denominators dk can be “guessed” by looking at the leading coefficients of the Gr¨obnerbasis. n Leads to a linear system over Q(q, q ) with 377 unknowns. We use homomorphic images (modular computations): • do all computations modulo some prime • plug in concrete integral values for q and qn • requires special modular GB reduction • fixing q and varying qn (and vice versa) allows to estimate the necessary interpolation points: • 1167 interpolation points for q • 363 interpolation points for qn • each case takes about a minute (GB reduction, linear solving, for sufficiently many primes) • estimated computation time: 1167 · 363 · 60s = 294 days!
A computational challenge Even generating this linear system (reducing the ansatz with the Gr¨obnerbasis) would already consume too much memory! • 1167 interpolation points for q • 363 interpolation points for qn • each case takes about a minute (GB reduction, linear solving, for sufficiently many primes) • estimated computation time: 1167 · 363 · 60s = 294 days!
A computational challenge Even generating this linear system (reducing the ansatz with the Gr¨obnerbasis) would already consume too much memory!
We use homomorphic images (modular computations): • do all computations modulo some prime • plug in concrete integral values for q and qn • requires special modular GB reduction • fixing q and varying qn (and vice versa) allows to estimate the necessary interpolation points: A computational challenge Even generating this linear system (reducing the ansatz with the Gr¨obnerbasis) would already consume too much memory!
We use homomorphic images (modular computations): • do all computations modulo some prime • plug in concrete integral values for q and qn • requires special modular GB reduction • fixing q and varying qn (and vice versa) allows to estimate the necessary interpolation points: • 1167 interpolation points for q • 363 interpolation points for qn • each case takes about a minute (GB reduction, linear solving, for sufficiently many primes) • estimated computation time: 1167 · 363 · 60s = 294 days! Further improvements
• efficient Gr¨obnerbasis reduction procedure
• combination of arithmetic in Q and in Zp • rewrite polynomials that have to be evaluated in compact form • build matrices efficiently • discard redundant equations • compute parallel • normalize w.r.t. a certain component in order to minimize the number of interpolation points • guess small factors in the components of the solution Guess small factors
Compute, for example, • with q = 19 • modulo the prime 2147483629
Assume we obtain as solution (after factoring):
(qn + 19)(qn + 2147483628)(q2n + 381qn + 2147483610) ...
Presumably the true solution (for symbolic q and over Q) is
(qn + q)(qn − 1)(q2n + (q2 + q + 1)qn − q) ...
Many such small factors can be guessed from modular results!
All these optimizations reduced the actual computation to 35 days. The final result was reduced once again with the Gr¨obner basis (non-modular) and yielded 0.
Advocatus Diaboli The polynomial degrees of the solution are not known: not enough interpolation points???
The size of the integer co- efficients is not known: not enough primes???
The guessed small factors can be wrong!!! Advocatus Diaboli The polynomial degrees of the solution are not known: not enough interpolation points???
The size of the integer co- efficients is not known: not enough primes???
The guessed small factors can be wrong!!!
The final result was reduced once again with the Gr¨obner basis (non-modular) and yielded 0. Its principal part confirms the conjectured evaluation
2 Y 1 − qi+j+k−1 b = . n 1 − qi+j+k−2 1≤i≤j≤k≤n
Result The certificate for the third identity has a size of 7 Gigabytes. Result The certificate for the third identity has a size of 7 Gigabytes.
Its principal part confirms the conjectured evaluation
2 Y 1 − qi+j+k−1 b = . n 1 − qi+j+k−2 1≤i≤j≤k≤n The second identity
n X cn,jai,j = 0 (1 ≤ i < n). j=1 Strategy similar as before, but one variable more. This means: i n • linear system over Q(q, q , q ) • (at least) two creative telescoping relations are necessary There are only finitely many which can be checked sepa- rately.
Advocatus Diaboli
Zeros in the denominators of the delta part???
Singularities in the leading coef- ficients of the principal parts??? Advocatus Diaboli
Zeros in the denominators of the delta part???
Singularities in the leading coef- ficients of the principal parts???
There are only finitely many which can be checked sepa- rately. Quod erat demonstrandum.
Theorem (KKZ). Let π/S3 denote the set of orbits of a plane partition π under the action of the symmetric group S3. Then the orbit-counting generating function is given by
X Y 1 − qi+j+k−1 q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n where T (n) denotes the set of totally symmetric plane partitions with largest part at most n. No thanks to • the cleaning professional who unplugged and messed up my computations!
Acknowledgements
Thanks to • the RISC sysadmins for providing the necessary resources, • Victor Moll for hosting me in New Orleans, • Manuel Kauers for guessing and advising, • Doron Zeilberger for offering a noble prize! Acknowledgements
Thanks to • the RISC sysadmins for providing the necessary resources, • Victor Moll for hosting me in New Orleans, • Manuel Kauers for guessing and advising, • Doron Zeilberger for offering a noble prize! No thanks to • the cleaning professional who unplugged and messed up my computations!