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Proof of the Q-TSPP Conjecture

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Proof of the Q-TSPP Conjecture 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 2 N with finite sum P jπj = 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 2 N with finite sum P jπj = 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 2 N with finite sum P jπj = 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 2 N with finite sum P jπj = 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 2 N with finite sum P jπj = 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 2 N with finite sum P jπj = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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: qjπ=S3j = 1 − qi+j+k−2 π2T (n) 1≤i≤j≤k≤n Y i + j + k − 1 Stembridge's theorem: jT (n)j = 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.
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