Let $A=k\mathbb \Pi$ be the group algebra of an abelian group $\Pi$ and let $B(A)=\bigoplus_{k=0}^\infty\,B^k(A)$ be the unnormalized bar complex of $A$ with generators $[a_0,\dots,a_k] \in B^k(A)=A^{\otimes (k+1)}$. Geometrically, we can think of these generators as labelled $k$-simplices. That is, vertices are labelled by $a_i$'s, edges are labelled by pairs $(a_i,a_j)$ for $i<j$ and so on.

On $B(A)$ we have a shuffle product $\ast_{p,q}:B^p(A)\otimes B^q(A) \to B^{p+q}(A)$.

My question is, do you have a reference for a geometric interpretation of the shuffle $[a_0,\dots,a_p]\ast[b_0,\dots,b_q]$ as a simplicial complex? For simplicity, we assume that there are no relations between the $a_i$'s and $b_j$'s.


1 Answer 1


Perhaps this is more naïve than you are looking for, but here is one interpretation: one representation of an $n$ simplex is the following

$$\Delta_n=\{(t_1,t_2,\dotsc,t_n)\mid 0\leq t_1\leq t_2\leq\dotsb\leq t_n\leq 1\}.$$

The product of two such simplices has a canonical decomposition into a disjoint union of subsimplices

$$\Delta_m\times\Delta_n=\bigsqcup \Delta_{m+n,\sigma}$$

where, if we use $s_i$ ($t_i$) for coordinates on $\Delta_m$ ($\Delta_n$), $\sigma$ represents a total order on the union $\{s_1,\ldots,s_m,t_1,\ldots,t_n\}$ compatible with the internal orders of each set of coordinates. In particular, $\sigma$ is a shuffle. So every term in a shuffle product corresponds to a subsimplex in this decomposition

Then, thinking of the elements of $A$ as functions on $[0,1]$ and of $[a_1,\dotsc,a_m]$ as an integral over $\Delta_m$, and similarly for the $b$ term, the product $[a_1,\dotsc,a_m]\star [b_1,\dotsc, b_n]$ represents an integral on the product of simplices, which is equal to the sum of the integrals over the subsimplices.

A good example here is Chen's iterated integrals, and the iterated integral representation of multiple zeta values, where this is used to show that the product of two iterated integrals is equal to a sum over shuffles.

  • 1
    $\begingroup$ Thx! This might be exactly the interpretation I need, I just need to think a bit more about this decomposition you mentioned before I accept :) $\endgroup$ May 11 at 12:35
  • $\begingroup$ Yup, it's perfect. Thx again! $\endgroup$ May 11 at 12:44
  • $\begingroup$ As a comment, in "Greg Friedman, Section 5 of: An elementary illustrated introduction to simplicial sets, Rocky Mountain J. Math. 42(2): 353-423 (2012)}" there is a nice exposition of this decomposition with illustrations $\endgroup$ May 12 at 7:57

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