00:00:00 / 00:00:00

The one-sided cycle shuffles in the symmetric group algebra

By Darij Grinberg

Appears in collection : Combinatorics and Arithmetic for Physics : Special Days

We study a new family of elements in the group ring of a symmetric group – or, equivalently, a class of ways to shuffle a deck of cards. Fix a positive integer n. Consider the symmetric group S_n. For each 1 ≤ ℓ ≤ n, we define an element t_ℓ := cyc_ℓ + cyc{ℓ,ℓ+1} + cyc_{ℓ,ℓ+1,ℓ+2} + · · · + cyc_{ℓ,ℓ+1,...,n} of the group ring ${\mathbb R} [S_n]$, where cyc_{i₁,i₂,...,i_k} denotes the cycle that rotates through the given elements i₁, i₂ , . . . , i_k . We refer to these n elements t₁, t₂, . . . , t_n as the somewhere-to-below shuffles, since the standard interpretation of elements of ${\mathbb R} [S_n]$ in terms of card shuffling allows us to view them as shuffling operators. Note that t₁ is the well-known top-to-random shuffle studied by Diaconis, Fill, Pitman and others, whereas t_n = id. Similar families of elements of ${\mathbb R} [S_n]$ include the Young-Jucys-Murphy el- ements, the Reiner-Saliola-Welker elements, and the Diaconis-Fill-Pitman ele- ments. Unlike the latter three families, the somewhere-to-below shuffles t₁, t₂, . . . , t_n do not commute. However, they come close to commuting: There is a basis of ${\mathbb R} [S_n]$ on which they all act as upper-triangular matrices; thus, they generate an algebra whose semisimple quotient is commutative (which entails, in particular, that their commutators are nilpotent). This basis can in fact be constructed combinatorially, and bears several un- expected connections, most strikingly to the Fibonacci sequence. One of the consequences is that any ${\mathbb R}$-linear combination λ₁ t₁ + λ₂ t₂ + · · · + λ_n t_n (with λ₁ , λ₂ , . . . , λ_n ∈ {\mathbb R}) can be triangularized and its eigenvalues explicitly computed (along with their multiplicities); the number of distinct eigenvalues is at most the Fibonacci number f_{n+1}. If all these f_{n+1} eigenvalues are indeed distinct, then the matrix is diagonalizable. While we have been working over ${\mathbb R}$ for illustrative purposes, all our proofs hold over any commutative ring (or, for the diagonalizability claim, over any field). Several open questions remain (joint work with Nadia Lafrenière)

Information about the video

  • Date of recording 29/11/2022
  • Date of publication 30/11/2022
  • Institution IHES
  • Language English
  • Audience Researchers
  • Format MP4

Domain(s)

Last related questions on MathOverflow

You have to connect your Carmin.tv account with mathoverflow to add question

Ask a question on MathOverflow




Register

  • Bookmark videos
  • Add videos to see later &
    keep your browsing history
  • Comment with the scientific
    community
  • Get notification updates
    for your favorite subjects
Give feedback