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Homomesies Lurking in the Twelvefold Way

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Homomesies Lurking in the Twelvefold Way Homomesies Lurking in the Twelvefold Way Tom Roby (UConn) Describing joint research with Michael Joseph & Michael LaCroix Special Session on Enumerative Combinatorics AMS Central Sectional Meeting University of St. Thomas Minneapolis, MN USA 28 October 2016 (Friday) Slides for this talk are available online (or will be soon) at http://www.math.uconn.edu/~troby/research.html Homomesies Lurking in the Twelvefold Way Tom Roby (UConn) Describing joint research with Michael Joseph & Michael LaCroix Special Session on Enumerative Combinatorics AMS Central Sectional Meeting University of St. Thomas Minneapolis, MN USA 28 October 2016 (Friday) Slides for this talk are available online (or will be soon) at http://www.math.uconn.edu/~troby/research.html Abstract Abstract: Given a group acting on a finite set of combinatorial objects, one can often find natural statistics on these objects which are homomesic, i.e., over each orbit of the action, the average value of the statistic is the same. Since the notion was codified a few years ago, homomesic statistics have been uncovered in a wide variety of situations within dynamical algebraic combinatorics. We discuss several examples lurking in Rota's Twelvefold Way related to actions on injections, surjections (joint work with Michael Joseph), and bijections/permutations (joint work with Michael LaCroix) of finite sets. Acknowledgments This seminar talk discusses joint work with Michael Joseph and Michael La Croix. Thanks to James Propp for suggesting the study of whirling and of Foatic actions, as well as earlier collaborations on the homomesy phenomenon. Please feel free to interrupt with questions or comments. Outline Actions, orbits, and homomesy; The Twelvefold Way; Foatic actions on Sn. Whirling injections and surjections; Subset Rotation Among the most basic objects counted by the Twelvefold Way are subsets, counted by binomial coefficients. Set [n] := 1; 2;::: n . f g [n] = k-element subsets of [n] . k f g [7] For example, 3 consists of 35 subsets (dropping braces & commas): 123, 124, 125, 126, 127, 134, 135, 136, 137, 145, 146, 147, 156, 157, 167, 234,. [n] The map that adds 1 to each element mod n acts on k . 156 267 137 124 235 346 457 156 7! 7! 7! 7! 7! 7! 7! It's easy to see that the cardinality of each orbit of this action is a divisor of n. EG: n = 4; k = 2 gives us two orbits: 0011 0101 1001 1010 1100 0101 0110 0011 Rotation of bit-strings We find another unexpected structure in the orbits if we consider [n] S = k , as length n binary strings with k 1's as we cyclically shift them. τ:= CR : S S by b = b1b2 bn bnb1b2 bn 1, and ! ··· 7! ··· − count f (b) = #inversions(b) = # i < j : bi > bj . f g 1001 1010 1100 0101 0110 0011 Rotation of bit-strings We find another unexpected structure in the orbits if we consider [n] S = k , as length n binary strings with k 1's as we cyclically shift them. τ:= CR : S S by b = b1b2 bn bnb1b2 bn 1, and ! ··· 7! ··· − count f (b) = #inversions(b) = # i < j : bi > bj . f g EG: n = 4; k = 2 gives us two orbits: 0011 0101 Rotation of bit-strings We find another unexpected structure in the orbits if we consider [n] S = k , as length n binary strings with k 1's as we cyclically shift them. τ:= CR : S S by b = b1b2 bn bnb1b2 bn 1, and ! ··· 7! ··· − count f (b) = #inversions(b) = # i < j : bi > bj . f g EG: n = 4; k = 2 gives us two orbits: 0011 0101 1001 1010 1100 0101 0110 0011 Rotation of bit-strings We find another unexpected structure in the orbits if we consider [n] S = k , as length n binary strings with k 1's as we cyclically shift them. τ:= CR : S S by b = b1b2 bn bnb1b2 bn 1, and ! ··· 7! ··· − count f (b) = #inversions(b) = # i < j : bi > bj . f g EG: n = 4; k = 2 gives us two orbits: 0011 0101 1001 2 1010 3 1100 7! 4 0101 7! 1 0110 7! 2 7! 0011 7! 0 7! Rotation of bit-strings We find another unexpected structure in the orbits if we consider [n] S = k , as length n binary strings with k 1's as we cyclically shift them. τ:= CR : S S by b = b1b2 bn bnb1b2 bn 1, and ! ··· 7! ··· − count f (b) = #inversions(b) = # i < j : bi > bj . f g EG: n = 4; k = 2 gives us two orbits: 0011 0101 1001 2 1010 3 1100 7! 4 0101 7! 1 7! 7!4 0110 2 AVG = 2 = 2 0011 7! 0 7!8 AVG= 4 = 2 100001 100010 100100 110000 010001 010010 011000 101000 001001 001100 010100 000110 001010 000011 000101 We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 100001 100010 100100 110000 010001 010010 011000 101000 001001 001100 010100 000110 001010 000011 000101 We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 100001 100010 100100 110000 010001 010010 011000 101000 001001 001100 010100 000110 001010 000011 000101 We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 100001 4 100010 5 100100 6 110000 7! 8 010001 7! 3 010010 7! 4 011000 7! 6 101000 7! 7 001001 7! 2 001100 7! 4 010100 7! 5 7! 000110 7! 2 001010 7! 3 000011 7! 0 000101 7! 1 7! 7! We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 100001 4 100010 5 100100 6 110000 7! 8 010001 7! 3 010010 7! 4 011000 7! 6 101000 7! 7 001001 7! 2 001100 7! 4 010100 7! 5 7! 000110 7! 2 001010 7! 3 000011 7! 0 000101 7! 1 247! 247! 12 AVG= 6 = 4 AVG= 6 = 4 AVG= 3 = 4 We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P More rotation EG: n = 6; k = 2 gives us three orbits: 000011 000101 001001 100001 4 100010 5 100100 6 110000 7! 8 010001 7! 3 010010 7! 4 011000 7! 6 101000 7! 7 001001 7! 2 001100 7! 4 010100 7! 5 7! 000110 7! 2 001010 7! 3 000011 7! 0 000101 7! 1 247! 247! 12 AVG= 6 = 4 AVG= 6 = 4 AVG= 3 = 4 Note that in each of the three orbits average of the statistic f is the same. We know two simple ways to prove this: one can show pictorially that the value of the sum doesn't change when you mutate b (replacing a 01 somewhere in b by 10 or vice versa), or one can write the number of inversions in b as i<j bi (1 bj ) and then perform algebraic manipulations. − P Equivalently: the average of f over each τ-orbit is the same as the average over the entire set S:O 1 1 f (s) = f (s): # #S s s S O X2O X2 So we can compute what the average should be before checking whether a statistic is homomesic. Main Definition: Homomesic statistics Definition ([PrRo15]) Given an (invertible) action τ on a finite set of objects S, call a statistic f : S C homomesic with respect to (S; τ) iff the average of f over! each τ-orbit is the same constant c for all , 1 O O i.e., f (s) = c does not depend on the choice of . # O O s (Call f c-mesicX2O for short.) Main Definition: Homomesic statistics Definition ([PrRo15]) Given an (invertible) action τ on a finite set of objects S, call a statistic f : S C homomesic with respect to (S; τ) iff the average of f over! each τ-orbit is the same constant c for all , 1 O O i.e., f (s) = c does not depend on the choice of .
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