Richard Stanley – master connector

Stanley80 Conference

June 4, 2024

What do the following people have in common?

• Bill Walsh – legendary NFL coach

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What do the following people have in common?

• Bill Walsh – legendary NFL coach

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• Alice Waters – celebrity chef, owner of Chez Panisse

What do the following people have in common?

• Bill Walsh – legendary NFL coach

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• Alice Waters – celebrity chef, owner of Chez Panisse

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• Richard Stanley – father of modern algebraic combinatorics

SUPERBOSSES!�Leaders who spawn an extraordinary number of other leaders

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• Bill Walsh: between 1979 and 2015, 32 of the 72 head coaches in the SuperBowl were disciples of Walsh
• Alice Waters: trained dozens of award-winning chefs (eg., Judy Rodgers, Jeremiah Tower, Joyce Goldstein, Joanne Weir, David Lebovitz, …)
• Richard Stanley – educated and inspired decades of grad students, post-docs, assistant profs, colleagues … JUST LOOK AROUND THE ROOM!

RESEARCH QUESTION: What characteristics do superbosses have in common?

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Some superboss characteristics

• Superbosses have an uncanny ability to draw connections between seemingly unrelated areas of their business, often using analogue and metaphor as tools to make these connections.
• By making these connections, Superbosses bring to bear talent and expertise from unexpected places within their organizations to work on its most interesting and important issues.
• Superbosses are not afraid to assign complex problems to inexperienced protégés

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Lessons that I learned from Richard

• When looking to solve problems, cast a wide net mathematically.

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• “I wonder if there is a q-analogue …”.
• Analogues, that introduce additional parameters often deepen your understanding and the interest of results that you have achieved. Analogues usually lead to interesting places.

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• Although a Superboss, Richard has never been “bossy”. He knows that the most powerful form of inspiration is by example.

My first contact with Richard (circa 1980)

• I had computed the value of the Mobius function in (Π_N)^σ the lattice of partitions fixed by a permutation σ in S_N
• Richard contacted me with a connection:

“Using your computation, I have been able to show that the action of S_N on the top homology of the partition lattice, twisted by the sgn, is the induction of the linear representation on the N-cycle which assigns the generator the primitive N-th root of unity”

• This induced representation is equal to Lie_N which initiated my interest in Lie_N and its analogues.

Led me to a decades-long interest in �n-analogues Lie_N

[x _σ1, … , x _σn] = sgn(σ) [x₁, … , x_n]

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(*) [[x₁, … , x_n], x_n+1, … , x_2n-1] = ?

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• (*) should be equivalent to the Jacobi Identity in the case n = 2

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• Analogue of Lie_N will be the action of S_N on the multilinear part of the space generated by k different n-bracketings subject to the relation (*). Here N = k(n-1) +1.

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Lie n-algebras (H and Wachs, 1995)

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(**) [[x₁, … , x_n], x_n+1, … , x_2n-1] = Σ_S (-1)^L [[x_S], x_T]

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sum is over all n-subsets S of 2n-1 and T is the complement of S.

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We did this in the superalgebra setting:

Generating set split into even and odd generators X₀ ∪ X₁

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Bracket is graded-skew symmetric and changes parity.

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Lie n-algebras

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MOTIVATION: For each d, define the map ∂:∧^dL → ∧^d-nL by

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∂(a₁∧a₂∧…∧a_d) = Σ_S (-1)^M [a_S]∧a_t1∧…∧a_t(d-n)

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(the exterior product is “graded” exterior).

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This choice (**) of Jacobi Identity implies that ∂² = 0. So can define

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H_*(L) = ker(∂)/im(∂)

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This analogue leads to two elegant results

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Assume the generators of the Free Lie algebra are even. Then,

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1) The analogue of Lie_N is isomorphic to the top homology of the lattice of partitions of N in which each block size is congruent to 1 mod n.

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2) The homology of the Free Lie algebra can be derived in a simple way from the homology of the lattice of partitions of N in which each block size is congruent to 1 mod n.

(NOTE: For n>2, this homology is non-zero but the formula reduces to zero in the case that n=2.)

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Analogues lead to interesting places and surprising connections.

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A different analogue …

n-ary Filippov algebras

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(***) [[x₁, … , x_n], x_n+1, … , x_2n-1] = Σ_i [x₁, … x_i-1, [x_i, x_n+1, … , x_2n-1], x_i+1, … ,x_n]

MOTIVATION: comes up in Mathematical Physics (String Theory)

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1. The multilinear part of the space with k brackets is spanned by ”combs”, i.e., brackets of the form [[[x_A],x_B] … ],x_Z]
2. Let ρ_n,k denote the representation of S_N on the multilinear part the space spanned by k brackets. This is the analogue of Lie_N.
3. By 1), ρ_n,k is contained in the set of shapes with at most k columns.

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Decompositions of some ρ_n,k

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EASY TO SEE:

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ρ_2,k = Lie_k+1

ρ_n,1 = 1ⁿ

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Decompositions of some ρ_n,k

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Theorem: [Friedman, H, Stanley, Wachs]

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ρ_n,2 = 2^n-11

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Sketch of Proof: M = matrix with rows and columns indexed by n-subsets of 2n-1.

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M_U,V = coef of X _V in expansion of X_U via (***)

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[[x₁,…, x_n],x_n+1,…,x_2n-1] - Σ_i [x₁,…,x_i-1,[x_i,x_n+1,…,x_2n-1],x_i+1,…,x_n] = 0

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1 if U = V

M_U,V = (-1)^d if U ∩ V = {d}

0 otherwise

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• M acts on the S_2n-1 module 1ⁿ⊗1^n-1. We show that M is the scalar

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1 + (n-i)(-1)^(n-i)

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on the irreducible 2ⁱ1^(2n-1-2i).

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• We construct a vector in 2ⁱ1^(2n-1-2i) via Young symmetrizers. Calculate the coefficient of a basis element before and after action by M. Then use that M must be a scalar on the irreducible.

Stabilization and the case k = 4

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ρ_2,4 = 41 + 32 + 31² + 2²1 + 21³

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ρ_3,4 = (4²1 + 432 + 431² + 42²1 + 421³) + (3²1³ + 32³)

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ρ_4,4 = 4³1 + 4²32 + 4²31² + 4²2²1 + 4²21³ + 43²1³ +432³

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Theorem (Friedman, H, Wachs)

1) (Inheritance) For every n,k:

ρ_n+1,k = k ρ_n,k + β_n,k

where “k ρ_n,k” means that you take every shape in the decomposition of ρ_n,k and add a part of size k to the top row and where β_n,k contains shapes all of which have fewer than k columns.

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2) (Stability) For n ≥ k, β_n,k = 0. In other words,

ρ_n+1,k = k ρ_n,k

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OPEN QUESTIONS:

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• Are there other contexts in which the “k ρ_n,k “ construct appears?
• What are the β_n,k for n<k?
• Known for n = 2,3,4.
• Evidence suggests that they may have to do with the types of bracketings.

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CONCLUSION: Happy Birthday Richard! And thank you for all your teachings amongst which are:

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Cast a wide net mathematically when looking for connections.

Analogues often lead you to magical places.