# Why does the type-A subdivision algebra look like the Rota-Baxter algebra axiom?

Let $\mathbf{k}$ be a commutative ring, and $\beta$ an element of $\mathbf{k}$. Fix a positive integer $n$, and set $\left[n\right] = \left\{1,2,\ldots,n\right\}$.

The $n$-th type-A subdivision algebra over $\mathbf{k}$ for parameter $\beta$ is the commutative $\mathbf{k}$-algebra $B$ with

• generators $x_{i,j}$ indexed by all the $n\left( n-1\right) /2$ pairs $\left( i,j\right)$ of integers satisfying $1\leq i<j\leq n$;

• relations $x_{i,j}x_{j,k}=x_{i,k}\left( x_{i,j}+x_{j,k}+\beta\right)$ for all $\left( i,j,k\right) \in\left[ n\right] ^{3}$ satisfying $i<j<k$.

Alternatively, we can define $B$ in a more symmetric fashion: Namely, $B$ is the commutative $\mathbf{k}$-algebra with

• generators $x_{i,j}$ indexed by all the $n\left( n-1\right)$ pairs $\left( i,j\right)$ of distinct integers in $\left[n\right]$;

• relations $x_{i,j}+x_{j,i}=-\beta$ whenever $i\neq j$, as well as $x_{i,j}x_{j,k}+x_{j,k}x_{k,i}+x_{k,i}x_{i,j}+\beta\left( x_{i,j} +x_{j,k}+x_{k,i}\right) +\beta^{2}=0$ whenever $i,j,k$ are distinct elements of $\left[n\right]$.

The $\mathbf{k}$-algebra $B$ has appeared in various contexts. It was originally introduced by Karola Mészáros as the abelianization of Anatol Kirillov's quasi-classical Yang-Baxter algebra. It is a deformation of the Orlik-Terao algebra of the braid arrangement of type $A_{n-1}$ (with the case $\beta=0$ corresponding to the Orlik-Terao algebra). It is probably isomorphic to a $\mathbf{k}$-subalgebra of the localization of the polynomial ring $\mathbf{k}\left[ q_{1},q_{2} ,\ldots,q_{n}\right]$ at the multiplicative subset generated by the differences $q_{i}-q_{j}$ for $i<j$ (here I say "probably" because I can only show this for $\beta=0$, in which case it is isomorphic to the $\mathbf{k}$-subalgebra generated by all $\dfrac{1}{q_{i}-q_{j}}$). It comes up in the computation of volumes of flow polytopes and evaluations of Grothendieck polynomials. See my recent preprint arXiv:1704.00839 for more on it.

On the other hand, recall that a Rota-Baxter algebra of weight $\beta$ means a $\mathbf{k}$-algebra $R$ equipped with a $\mathbf{k}$-linear map $P:R\rightarrow R$ (called its Rota-Baxter operator) that satisfies \begin{equation} P\left( a\right) P\left( b\right) =P\left( P\left( a\right) b\right) +P\left( aP\left( b\right) \right) +\beta P\left( ab\right) \label{eq.rota-baxter.def} \tag{1} \end{equation} for all $a,b\in R$. (Some authors, like those of the Wikipedia page, prefer to put the $\beta P\left( ab\right)$ addend on the left instead of the right hand side, but this just boils down to replacing $\beta$ by $-\beta$.)

The axiom \eqref{eq.rota-baxter.def} of the Rota-Baxter algebra is uncannily similar to the relations \begin{equation} x_{i,j}x_{j,k}=x_{i,k}\left( x_{i,j}+x_{j,k}+\beta\right) \label{eq.relB} \tag{2} \end{equation} of the algebra $B$. Indeed, represent each monomial in the indeterminates $x_{i,j}$ as a multigraph on the vertex set $\left\{ 1,2,\ldots,n\right\}$, where each indeterminate $x_{i,j}$ appearing in the monomial contributes an edge $ij$ to the multigraph. Then, \eqref{eq.relB} can be visually rewritten as (where all vertices other than $i,j,k$ are omitted). Now, imagine writing an "$a$" between the $i$ and the $j$, and writing a "$b$" between the $j$ and the $k$, and interpreting each edge as a signal to apply $P$ to whatever stands under the edge. The above equality thus becomes which is precisely \eqref{eq.rota-baxter.def}.

Question. Can this resemblance be turned into anything concrete (e.g., an action of $B$ on Rota-Baxter algebras?). Barring that, can we define "Rota-Baxter algebras of other types"?

• Don't you also want to impose (in the second more symmetric presentation) relations $x_{ij}x_{ji}=0$, or something like this? – Vladimir Dotsenko Nov 20 '17 at 10:10
• @VladimirDotsenko: Nope. Maybe I should have said that I mean the "big" Orlik-Terao algebra, not the "small" one (i.e., my variables aren't nilpotent or anything like that). – darij grinberg Nov 20 '17 at 10:17

The reason for my question in comments was as follows. In some cases (Gerstenhaber algebras, Poisson algebras, etc.) an operad describing a certain algebraic structure is an operad in the category of cocommutative coalgebras (some people refer to that as a Hopf operad). Alternatively, one can think about a collection of commutative associative algebras that form a cooperad. For your algebras, the most obvious candidate for cooperad structure does lead to an honest cooperad structure only for $x_{ij}x_{ji}=0$. But in this case this actually leads to an operad of associative algebras presented via unusual generators, see https://arxiv.org/abs/math/0412206 for instance.
• Interesting! I've thought about operads (my original idea was that elements of $B$ should give rise to maps $R^{\otimes n} \to R$ for any Rota-Baxter algebra $R$; but this direction has disappointed me) but not about cooperads. – darij grinberg Nov 20 '17 at 17:20