The Rule of Three for commutation relations

Transcription

1 The Rule of Three for commutation relations Jonah Blasiak Drexel University joint work with Sergey Fomin (University of Michigan) May 2016

2 Symmetric functions The ring of symmetric functions Λ(x) = Λ(x 1,...,x N ) Q[x 1,...,x N ] consists of those polynomials which are fixed by the action of the symmetric group S N permuting the indices. Basic families of symmetric functions: Fact elementary symmetric functions e k (x) = e k (x 1,...,x N ) = Schur functions s λ (x) = s λ (x 1,...,x N ) N i 1 > >i k 1 x i1 x ik The e k (x) are algebraically independent and generate Λ(x). Hence Λ(x) = Q[e 1 (x),...,e N (x)].

3 Symmetric functions The ring of symmetric functions Λ(x) = Λ(x 1,...,x N ) Q[x 1,...,x N ] consists of those polynomials which are fixed by the action of the symmetric group S N permuting the indices. Basic families of symmetric functions: Fact elementary symmetric functions e k (x) = e k (x 1,...,x N ) = Schur functions s λ (x) = s λ (x 1,...,x N ) N i 1 > >i k 1 x i1 x ik The e k (x) are algebraically independent and generate Λ(x). Hence Λ(x) = Q[e 1 (x),...,e N (x)].

4 Noncommutative elementary symmetric functions U = Q u 1,...,u N = free associative ring generated by u = (u 1,...,u N ) We identify the monomials in U with words in the alphabet [N] and frequently write 312, cab, etc. for words/monomials. Definition (Noncommutative elementary symmetric functions) e k (u) = u i1 u i2 u ik N i 1 >i 2 > >i k 1 Equivalently, N x k e k (u) = (1+xu N ) (1+xu 1 ) U[x]. k=0

5 Noncommutative elementary symmetric functions U = Q u 1,...,u N = free associative ring generated by u = (u 1,...,u N ) We identify the monomials in U with words in the alphabet [N] and frequently write 312, cab, etc. for words/monomials. Definition (Noncommutative elementary symmetric functions) e k (u) = u i1 u i2 u ik N i 1 >i 2 > >i k 1 Equivalently, N x k e k (u) = (1+xu N ) (1+xu 1 ) U[x]. k=0

6 Noncommutative elementary symmetric functions U = Q u 1,...,u N = free associative ring generated by u = (u 1,...,u N ) We identify the monomials in U with words in the alphabet [N] and frequently write 312, cab, etc. for words/monomials. Definition (Noncommutative elementary symmetric functions) e k (u) = u i1 u i2 u ik N i 1 >i 2 > >i k 1 Equivalently, N x k e k (u) = (1+xu N ) (1+xu 1 ) U[x]. k=0

7 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 ))

8 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 ))

9 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3)

10 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) =

11 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) =

12 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) =

13 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) =

14 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) =

15 The plactic algebra The plactic algebra is the quotient of U by the relations bca = bac, cab = acb for a < b < c, bba = bab, baa = aba for a < b. Theorem (Lascoux-Schützenberger) The e k (u) pairwise commute in the plactic algebra. Example (e 3 (u 1,u 2,u 3 ) commutes with e 1 (u 1,u 2,u 3 )) 321(1+2+3) = = (1+2+3)321

16 Generalizations to other algebras Let R be the quotient of U by the relations bca = bac, cab = acb for a < b < c, Theorem (Fomin-Greene) cca = cac, caa = aca for c a > 1, bab+baa = bba+aba The e k (u) pairwise commute in R. Other variations For LLT polynomials (Lam 2005), For k-schur functions (Lam 2006), for b = a+1. The forgotten monoid (Novelli-Schilling 2008), For Schubert times Schur (Benedetti-Bergeron 2014), The switchboard ideal (B. Fomin 2015, A. N. Kirillov 2015).

17 Generalizations to other algebras Let R be the quotient of U by the relations bca = bac, cab = acb for a < b < c, Theorem (Fomin-Greene) cca = cac, caa = aca for c a > 1, bab+baa = bba+aba The e k (u) pairwise commute in R. Other variations For LLT polynomials (Lam 2005), For k-schur functions (Lam 2006), for b = a+1. The forgotten monoid (Novelli-Schilling 2008), For Schubert times Schur (Benedetti-Bergeron 2014), The switchboard ideal (B. Fomin 2015, A. N. Kirillov 2015).

18 The Rule of Three For S {1,...,N}, denote e k (u S ) = i 1 >i 2 > >i k i 1,...,i k S u i1 u i2 u ik. Theorem (A. N. Kirillov, B. Fomin) For a quotient R of U, the following are equivalent: e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S of size 3; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S. Equivalent formulation The elements e k (u S )e l (u S ) e l (u S )e k (u S ) for S 4 lie in the two-sided ideal generated by { e k (u S )e l (u S ) e l (u S )e k (u S ) } k,l, S 3

19 The Rule of Three For S {1,...,N}, denote e k (u S ) = i 1 >i 2 > >i k i 1,...,i k S u i1 u i2 u ik. Theorem (A. N. Kirillov, B. Fomin) For a quotient R of U, the following are equivalent: e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S of size 3; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S. Equivalent formulation The elements e k (u S )e l (u S ) e l (u S )e k (u S ) for S 4 lie in the two-sided ideal generated by { e k (u S )e l (u S ) e l (u S )e k (u S ) } k,l, S 3

20 The Rule of Three For S {1,...,N}, denote e k (u S ) = i 1 >i 2 > >i k i 1,...,i k S u i1 u i2 u ik. Theorem (A. N. Kirillov, B. Fomin) For a quotient R of U, the following are equivalent: e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S of size 3; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R, for all k, l, and S. Equivalent formulation The elements e k (u S )e l (u S ) e l (u S )e k (u S ) for S 4 lie in the two-sided ideal generated by { e k (u S )e l (u S ) e l (u S )e k (u S ) } k,l, S 3

21 Example The relation e 1 (u S )e 2 (u S ) = e 2 (u S )e 1 (u S ) for S = 4 follows from the relations of this type for S 3. Example e 1 (a,b,c,d)e 2 (a,b,c,d) e 2 (a,b,c,d)e 1 (a,b,c,d) = [a+b+c+d,ba+ca+cb+da+db+dc] = [e 1 (a,b,c),e 2 (a,b,c)]+[e 1 (a,b,d),e 2 (a,b,d)] +[e 1 (a,c,d),e 2 (a,c,d)]+[e 1 (b,c,d),e 2 (b,c,d)] [e 1 (a,b),e 2 (a,b)] [e 1 (a,c),e 2 (a,c)] [e 1 (a,d),e 2 (a,d)] [e 1 (b,c),e 2 (b,c)] [e 1 (b,d),e 2 (b,d)] [e 1 (c,d),e 2 (c,d)]

22 Example The relation e 1 (u S )e 2 (u S ) = e 2 (u S )e 1 (u S ) for S = 4 follows from the relations of this type for S 3. Example e 1 (a,b,c,d)e 2 (a,b,c,d) e 2 (a,b,c,d)e 1 (a,b,c,d) = [a+b+c+d,ba+ca+cb+da+db+dc] = [e 1 (a,b,c),e 2 (a,b,c)]+[e 1 (a,b,d),e 2 (a,b,d)] +[e 1 (a,c,d),e 2 (a,c,d)]+[e 1 (b,c,d),e 2 (b,c,d)] [e 1 (a,b),e 2 (a,b)] [e 1 (a,c),e 2 (a,c)] [e 1 (a,d),e 2 (a,d)] [e 1 (b,c),e 2 (b,c)] [e 1 (b,d),e 2 (b,d)] [e 1 (c,d),e 2 (c,d)]

23 Example The relation e 1 (u S )e 2 (u S ) = e 2 (u S )e 1 (u S ) for S = 4 follows from the relations of this type for S 3. Example e 1 (a,b,c,d)e 2 (a,b,c,d) e 2 (a,b,c,d)e 1 (a,b,c,d) = [a+b+c+d,ba+ca+cb+da+db+dc] = [e 1 (a,b,c),e 2 (a,b,c)]+[e 1 (a,b,d),e 2 (a,b,d)] +[e 1 (a,c,d),e 2 (a,c,d)]+[e 1 (b,c,d),e 2 (b,c,d)] [e 1 (a,b),e 2 (a,b)] [e 1 (a,c),e 2 (a,c)] [e 1 (a,d),e 2 (a,d)] [e 1 (b,c),e 2 (b,c)] [e 1 (b,d),e 2 (b,d)] [e 1 (c,d),e 2 (c,d)]

24 Original motivation: positivity Positivity problems in algebraic combinatorics ask to find positive combinatorial formulae for nonnegative quantities arising in geometry and representation theory. Key example of a positive combinatorial formula The Littlewood-Richardson Rule is a positive combinatorial formula for Littlewood-Richardson coefficients, the decomposition multiplicities of a tensor product of irreducible representations of GL n. Two important unsolved positivity problems Find a positive combinatorial formula for Kronecker coefficients: decomposition multiplicities of a tensor product of irreducible representations of the symmetric group. Kostka-Macdonald coefficients: coefficients in the Schur expansion of a transformed Macdonald polynomial.

25 Original motivation: positivity Positivity problems in algebraic combinatorics ask to find positive combinatorial formulae for nonnegative quantities arising in geometry and representation theory. Key example of a positive combinatorial formula The Littlewood-Richardson Rule is a positive combinatorial formula for Littlewood-Richardson coefficients, the decomposition multiplicities of a tensor product of irreducible representations of GL n. Two important unsolved positivity problems Find a positive combinatorial formula for Kronecker coefficients: decomposition multiplicities of a tensor product of irreducible representations of the symmetric group. Kostka-Macdonald coefficients: coefficients in the Schur expansion of a transformed Macdonald polynomial.

26 Original motivation: positivity Positivity problems in algebraic combinatorics ask to find positive combinatorial formulae for nonnegative quantities arising in geometry and representation theory. Key example of a positive combinatorial formula The Littlewood-Richardson Rule is a positive combinatorial formula for Littlewood-Richardson coefficients, the decomposition multiplicities of a tensor product of irreducible representations of GL n. Two important unsolved positivity problems Find a positive combinatorial formula for Kronecker coefficients: decomposition multiplicities of a tensor product of irreducible representations of the symmetric group. Kostka-Macdonald coefficients: coefficients in the Schur expansion of a transformed Macdonald polynomial.

27 Noncommutative Schur functions Noncommutative Schur functions are a powerful tool for solving positivity problems. They have led to positive combinatorial formulae for the Schur expansion of Stanley symmetric functions and stable Grothendieck polynomials. the Schur expansion of LLT polynomials indexed by 3-tuples of skew shapes and transformed Macdonald polynomials indexed by shapes with 3 columns. Kronecker coefficients for one hook shape and two arbitrary shapes. The first step to apply this tool is to identify a quotient of U in which the e k (u) commute.

28 Noncommutative Schur functions Noncommutative Schur functions are a powerful tool for solving positivity problems. They have led to positive combinatorial formulae for the Schur expansion of Stanley symmetric functions and stable Grothendieck polynomials. the Schur expansion of LLT polynomials indexed by 3-tuples of skew shapes and transformed Macdonald polynomials indexed by shapes with 3 columns. Kronecker coefficients for one hook shape and two arbitrary shapes. The first step to apply this tool is to identify a quotient of U in which the e k (u) commute.

29 The Rule of Three for two sets of variables For S [N], let e k (v S ) denote the elementary symmetric function in the noncommuting variables {v i i S}. Theorem (B. Fomin) For a quotient R of Q u 1,...,u N,v 1,...,v N, the following are equivalent: e k (u S )e l (v S ) = e l (v S )e k (u S ) in R, for all k, l, and S of size 3; e k (u S )e l (v S ) = e l (v S )e k (u S ) in R, for all k, l, and S.

30 Reformulation using generating functions Let g i = 1+xu i and h i = 1+yv i for i = 1,...,N. For S [N], let g S denote the descending product of g i, i S. Define h S similarly. For example, g [N] = g N g N 1 g 1. g S = N k=0 xk e k (u S ) h S = N k=0 yk e k (v S ) Fact For a quotient R of Q u 1,...,u N,v 1,...,v N, the following are equivalent e k (u S )e l (v S ) = e l (v S )e k (u S ) in R for all k, l; g S h S = h S g S in R[[x,y]].

31 Reformulation using generating functions Let g i = 1+xu i and h i = 1+yv i for i = 1,...,N. For S [N], let g S denote the descending product of g i, i S. Define h S similarly. For example, g [N] = g N g N 1 g 1. g S = N k=0 xk e k (u S ) h S = N k=0 yk e k (v S ) Fact For a quotient R of Q u 1,...,u N,v 1,...,v N, the following are equivalent e k (u S )e l (v S ) = e l (v S )e k (u S ) in R for all k, l; g S h S = h S g S in R[[x,y]].

32 Reformulation using generating functions Let g i = 1+xu i and h i = 1+yv i for i = 1,...,N. For S [N], let g S denote the descending product of g i, i S. Define h S similarly. For example, g [N] = g N g N 1 g 1. g S = N k=0 xk e k (u S ) h S = N k=0 yk e k (v S ) Fact For a quotient R of Q u 1,...,u N,v 1,...,v N, the following are equivalent e k (u S )e l (v S ) = e l (v S )e k (u S ) in R for all k, l; g S h S = h S g S in R[[x,y]].

33 Rule of Three for power series Conjecture (B. Fomin) Let R be a quotient ring of Q u 1,...,u N,v 1,...,v N. Let g 1,...,g N, h 1,...,h N R[[x,y]] be power series of the form g i = k α ik(u i x) k, h i = k β ik(v i y) k, satisfying α i0 = β i0 = 1 and α i1 β i1 0. Then the following are equivalent: g S h S = h S g S for all S of size 3; g S h S = h S g S for all S. Note: the previous theorem is the special case g i = 1+xu i, h i = 1+yv i.

34 Rule of Three for power series Conjecture (B. Fomin) Let R be a quotient ring of Q u 1,...,u N,v 1,...,v N. Let g 1,...,g N, h 1,...,h N R[[x,y]] be power series of the form g i = k α ik(u i x) k, h i = k β ik(v i y) k, satisfying α i0 = β i0 = 1 and α i1 β i1 0. Then the following are equivalent: g S h S = h S g S for all S of size 3; g S h S = h S g S for all S. Note: the previous theorem is the special case g i = 1+xu i, h i = 1+yv i.

35 van Kampen diagrams Given a presentation of a group G, a van Kampen diagram is a planar finite cell complex, connected and simply connected, with a specific embedding in R 2, and 1-cells labeled by generators, 2-cells labeled by relations. Let G be the group with generators a,b,c and relations: aba 1 b 1 = id aca 1 c 1 = id b c b a a a a b c b

36 van Kampen diagrams Given a presentation of a group G, a van Kampen diagram is a planar finite cell complex, connected and simply connected, with a specific embedding in R 2, and 1-cells labeled by generators, 2-cells labeled by relations. Let G be the group with generators a,b,c and relations: aba 1 b 1 = id aca 1 c 1 = id b c b a a a a b c b

37 van Kampen s Lemma van Kampen s Lemma A word w in the generators and their inverses is equal to the identity if and only if there exists a van Kampen diagram whose boundary is labeled by w. The given relations imply the boundary relation abcba 1 b 1 c 1 b 1 = id. The boundary relation can also be read as abcb = bcba. b c b a a a a b c b

38 van Kampen s Lemma van Kampen s Lemma A word w in the generators and their inverses is equal to the identity if and only if there exists a van Kampen diagram whose boundary is labeled by w. The given relations imply the boundary relation abcba 1 b 1 c 1 b 1 = id. The boundary relation can also be read as abcb = bcba. b c b a a a a b c b

39 Simplified purely group-theoretic statement Theorem Let g 1,...,g N,h 1,...,h N be elements of a group G satisfying the following relations: g b h b = h b g b 1 < b < N, g a+1 g a h a+1 h a = h a+1 h a g a+1 g a 1 a < N, g a h c = h c g a 1 a,c N with a c 2. Then the following relation also holds in G: g N g N 1 g 1 h N h N 1 h 1 = h N h N 1 h 1 g N g N 1 g 1. Corollary The e k (u) pairwise commute in the nilcoxeter algebra.

40 Simplified purely group-theoretic statement Theorem Let g 1,...,g N,h 1,...,h N be elements of a group G satisfying the following relations: g b h b = h b g b 1 < b < N, g a+1 g a h a+1 h a = h a+1 h a g a+1 g a 1 a < N, g a h c = h c g a 1 a,c N with a c 2. Then the following relation also holds in G: g N g N 1 g 1 h N h N 1 h 1 = h N h N 1 h 1 g N g N 1 g 1. Corollary The e k (u) pairwise commute in the nilcoxeter algebra.

41 The van Kampen diagram proving the Theorem g b h b = h b g b 1 < b < N g 1 h 6 g 2 h 5 g 3 h 4 g 4 h 3 g 5 h 6 g 4 h 5 g 3 h 4 g 2 h 3 g 1 h 2 g 6 h 5 g 5 h 4 g 4 h 3 g 3 h 2 g 2 h 1 h 6 g 5 h 5 g 4 h 4 g 3 h 3 g 2 h 2 g 1 h 5 g 6 h 4 g 5 h 3 g 4 h 2 g 3 h 1 g 2 h 4 g 3 h 3 g 4 h 2 g 5 h 1 g 6

42 The van Kampen diagram proving the Theorem g a+1g ah a+1h a = h a+1h ag a+1g a g 1 h 6 g 2 h 5 g 3 h 4 g 4 h 3 g 5 h 6 g 4 h 5 g 3 h 4 g 2 h 3 g 1 h 2 g 6 h 5 g 5 h 4 g 4 h 3 g 3 h 2 g 2 h 1 h 6 g 5 h 5 g 4 h 4 g 3 h 3 g 2 h 2 g 1 h 5 g 6 h 4 g 5 h 3 g 4 h 2 g 3 h 1 g 2 h 4 g 3 h 3 g 4 h 2 g 5 h 1 g 6

43 The van Kampen diagram proving the Theorem g ah c = h cg a for a c 2 g 1 h 6 g 2 h 5 g 3 h 4 g 4 h 3 g 5 h 6 g 4 h 5 g 3 h 4 g 2 h 3 g 1 h 2 g 6 h 5 g 5 h 4 g 4 h 3 g 3 h 2 g 2 h 1 h 6 g 5 h 5 g 4 h 4 g 3 h 3 g 2 h 2 g 1 h 5 g 6 h 4 g 5 h 3 g 4 h 2 g 3 h 1 g 2 h 4 g 3 h 3 g 4 h 2 g 5 h 1 g 6

44 The Rule of Three for power series Special case of the main conjecture for one set of variables: Theorem (B. Fomin) Let R be a quotient of U. Let g 1,...,g N,h 1,...,h N U[[x,y]] be power series of the form g i = k α ik(u i x) k, h i = k β ik(u i y) k, satisfying α i0 = β i0 = 1 and α i1 β i1 0. The following are equivalent: g S h S = h S g S in R[[x,y]] for all S of size 3; g S h S = h S g S in R[[x,y]] for all S. Note: R[[x,y]] is the ring of formal power series with coefficients in R. g S denotes the descending product of the g i, i S.

45 Proof (Step 1) The proof uses a lemma and theorem which are entirely group-theoretic, and a Lagrange inversion argument. Lemma Let g 1,g 2,g 3,h 1,h 2,h 3 be elements of a group G satisfying the following relations: g a h a = h a g a for all a, g b g a h b h a = h b h a g b g a for all a < b. Then the following are equivalent: h 1 2 g 2(g 1 3 h 1g 3 h 1 1 ) = (g 1 3 h 1g 3 h 1 1 )h 1 2 g 2, g 3 g 2 g 1 h 3 h 2 h 1 = h 3 h 2 h 1 g 3 g 2 g 1.

46 The van Kampen diagram proving the lemma g 1 h 3 g 3 h 1 g 2 h 3 h 3 g 1 g 1 g 3 h 1 g 3 h 1 h 2 h 2 g 2 g 2 g 3 g 2 h 2 h 2 g 3 h 1 g 3 h 1 h 1 h 3 h 2 g 2 g 1

47 Proof (Step 2) Proposition Let R be a quotient of U. Suppose g i = k α ik(u i x) k, h i = k β ik(u i y) k R[[x,y]], satisfying α i0 = β i0 = 1 and α i1 β i1 0. Then if g i h i commutes with some z R[[x,y]], then g i and h i also commute with z. This follows from a Lagrange inversion argument.

48 Proof (Step 3) Theorem Let g 1,...,g N,h 1,...,h N be elements of a group G satisfying the following relations: g a h a = h a g a for all a, g b g a h b h a = h b h a g b g a for all a < b, g b (g 1 c h a g c h 1 a ) = (g 1 c h a g c h 1 a )g b for all a < b < c, h 1 b (g 1 c h a g c ha 1 ) = (gc 1 h a g c ha 1 )h 1 b for all a < b < c. Then the following relation also holds in G: g N g N 1 g 1 h N h N 1 h 1 = h N h N 1 h 1 g N g N 1 g 1.

49 The van Kampen diagram proving the theorem g a h d g d h a g b h d h d g a g a g d h a g d h a h c h c g b g d h a g d h a g b g c h b g c g d h a g d h a g c h b g b h c g d h a g d h a h c g d gc h b g d h a g d h a h b h a h d h c h b g c g b g a

50 Computations Definition Let g 1,...,g N,h 1,...,h N be invertible elements of the ring Q u 1,...,u M,v 1,...,v M [[x,y]]. The Rule of Three holds (for g 1,...,g N,h 1,...,h N ) if for any quotient R of Q u 1,...,u M,v 1,...,v M, the following are equivalent: g S h S = h S g S in R[[x,y]] for all S of size 3; g S h S = h S g S in R[[x,y]] for all S. Problem Determine natural conditions on the g i,h i which ensure that the Rule of Three holds.

51 Computations Rule of Three fails: g 4 = (1+xu 4 ) h 4 = (1+xv 4 ) g 3 = (1+xu 3 ) h 3 = (1+xv 3 ) g 2 = (1+xu 2 ) h 2 = (1+xv 2 ) g 1 = (1+xu 1 ) h 1 = (1+xv 1 ) Corollary The purely group-theoretic version of the Rule of Three is false: there exists a group G with elements g 1,...,g 4,h 1,...,h 4 such that g S h S = h S g S for all S of size 3, but g S h S h S g S for S = {1,2,3,4}.

52 Computations Rule of Three fails: g 4 = (1+xu 4 ) h 4 = (1+xv 4 ) g 3 = (1+xu 3 ) h 3 = (1+xv 3 ) g 2 = (1+xu 2 ) h 2 = (1+xv 2 ) g 1 = (1+xu 1 ) h 1 = (1+xv 1 ) Corollary The purely group-theoretic version of the Rule of Three is false: there exists a group G with elements g 1,...,g 4,h 1,...,h 4 such that g S h S = h S g S for all S of size 3, but g S h S h S g S for S = {1,2,3,4}.

53 Computations Rule of Three fails: g 4 = (1+xu 8 )(1+xu 7 ) h 4 = (1+yu 8 )(1+yu 7 ) g 3 = (1+xu 6 )(1+xu 5 ) h 3 = (1+yu 6 )(1+yu 5 ) g 2 = (1+xu 4 )(1+xu 3 ) h 2 = (1+yu 4 )(1+yu 3 ) g 1 = (1+xu 2 )(1+xu 1 ) h 1 = (1+yu 2 )(1+yu 1 ) Conjecturally, the Rule of Three holds: g 4 = (1+xu 8 )(1+xu 7 ) h 4 = (1+yu 8 )(1+yu 7 ) g 3 = (1+xu 6 )(1+xu 5 ) h 3 = (1+yu 6 )(1+yu 5 ) g 2 = (1+xu 4 )(1+xu 3 ) h 2 = (1+yu 4 )(1+yu 3 ) g 1 = (1+xu 2 ) h 1 = (1+yu 2 )

54 Consider g i,h i of the following form: Computations g i = (1+z i u i ), h i = (1+z iv i ), for z i,z i {x,y}. A pattern is a sequence z 1 z 2 z 3 z 4 z 1 z 2 z 3 z 4 {x,y}8. Rule of Three fails for the following 34 patterns: xxxx xxxx yyyy yyyy xxyy xxyy yyxx yyxx xyxy xyxy yxyx yxyx xyyx xyyx yxxy xyyx xyyx yxxy yxxy yxxy xyyx yyyy yxxy xxxx xxxx yxxy yyyy xyyx xyyy xyyy yxxx yxxx xxxy xxxy yyyx yyyx xxyx xxyx yyxy yyxy xxyx yxyy yyxy xyxx xyxx yyxy yxyy xxyx xyxx xyxx yxyy yxyy xyyx yxyy yxxy xyxx xyyx yyxy yxxy xxyx xyxx yxxy yxyy xyyx xxyx yxxy yyxy xyyx Rule of Three holds for the other patterns.

55 Consider g i,h i of the following form: Computations g i = (1+z i u i ), h i = (1+z iv i ), for z i,z i {x,y}. A pattern is a sequence z 1 z 2 z 3 z 4 z 1 z 2 z 3 z 4 {x,y}8. Rule of Three fails for the following 34 patterns: xxxx xxxx yyyy yyyy xxyy xxyy yyxx yyxx xyxy xyxy yxyx yxyx xyyx xyyx yxxy xyyx xyyx yxxy yxxy yxxy xyyx yyyy yxxy xxxx xxxx yxxy yyyy xyyx xyyy xyyy yxxx yxxx xxxy xxxy yyyx yyyx xxyx xxyx yyxy yyxy xxyx yxyy yyxy xyxx xyxx yyxy yxyy xxyx xyxx xyxx yxyy yxyy xyyx yxyy yxxy xyxx xyyx yyxy yxxy xxyx xyxx yxxy yxyy xyyx xxyx yxxy yyxy xyyx Rule of Three holds for the other patterns.

56 Further results Theorem (B. Fomin) The commutation relations e k (u S )e l (v S ) = e l (v S )e k (u S ) hold for all k,l, and S [N] if and only if the following relations hold: e 1 (u S )e 1 (v S ) = e 1 (v S )e 1 (u S ) for 1 S 2, e 2 (u S )e 1 (v S ) = e 1 (v S )e 2 (u S ) for 2 S 3, e 1 (u S )e 2 (v S ) = e 2 (v S )e 1 (u S ) for 2 S 3, e 3 (u S )e 1 (v S ) = e 1 (v S )e 3 (u S ) for S = 3, e 1 (u S )e 3 (v S ) = e 3 (v S )e 1 (u S ) for S = 3. Amazingly, the relations e 2 (u S )e 2 (v S ) = e 2 (v S )e 2 (u S ) for S of size 2 and 3 and the relations e 2 (u S )e 3 (v S ) = e 3 (v S )e 2 (u S ) for S = 3 follow from the above relations.

57 Further results Theorem (B. Fomin) The commutation relations e k (u S )e l (v S ) = e l (v S )e k (u S ) hold for all k,l, and S [N] if and only if the following relations hold: e 1 (u S )e 1 (v S ) = e 1 (v S )e 1 (u S ) for 1 S 2, e 2 (u S )e 1 (v S ) = e 1 (v S )e 2 (u S ) for 2 S 3, e 1 (u S )e 2 (v S ) = e 2 (v S )e 1 (u S ) for 2 S 3, e 3 (u S )e 1 (v S ) = e 1 (v S )e 3 (u S ) for S = 3, e 1 (u S )e 3 (v S ) = e 3 (v S )e 1 (u S ) for S = 3. Amazingly, the relations e 2 (u S )e 2 (v S ) = e 2 (v S )e 2 (u S ) for S of size 2 and 3 and the relations e 2 (u S )e 3 (v S ) = e 3 (v S )e 2 (u S ) for S = 3 follow from the above relations.

58 Further results Let g 1,...,g N,h 1,...,h N U[[x,y]] have the following form: g i = (1+xu i ) and h i = (1+yu i ), or g i = (1 xu i ) 1 and h i = (1 yu i ) 1. Definition Define the noncommutative super elementary symmetric functions e k (u) by g N g 1 = N k=0 xk e k (u). By the main theorem, the Rule of Three holds for this choice of g i,h i. The following stronger result holds: Theorem (B. Fomin) For a quotient R of U, the following are equivalent: e 1 (u S )e k (u S ) = e k (u S )e 1 (u S ) in R for S 3, k 2; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R for all k,l, and S.

59 Further results Let g 1,...,g N,h 1,...,h N U[[x,y]] have the following form: g i = (1+xu i ) and h i = (1+yu i ), or g i = (1 xu i ) 1 and h i = (1 yu i ) 1. Definition Define the noncommutative super elementary symmetric functions e k (u) by g N g 1 = N k=0 xk e k (u). By the main theorem, the Rule of Three holds for this choice of g i,h i. The following stronger result holds: Theorem (B. Fomin) For a quotient R of U, the following are equivalent: e 1 (u S )e k (u S ) = e k (u S )e 1 (u S ) in R for S 3, k 2; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R for all k,l, and S.

60 Further results Let g 1,...,g N,h 1,...,h N U[[x,y]] have the following form: g i = (1+xu i ) and h i = (1+yu i ), or g i = (1 xu i ) 1 and h i = (1 yu i ) 1. Definition Define the noncommutative super elementary symmetric functions e k (u) by g N g 1 = N k=0 xk e k (u). By the main theorem, the Rule of Three holds for this choice of g i,h i. The following stronger result holds: Theorem (B. Fomin) For a quotient R of U, the following are equivalent: e 1 (u S )e k (u S ) = e k (u S )e 1 (u S ) in R for S 3, k 2; e k (u S )e l (u S ) = e l (u S )e k (u S ) in R for all k,l, and S.