1
$\begingroup$

This question is a follow up to this question I posted on Math.SE. I will make this question self-contained, though.

In a certain point on the paper The Geometry of Rewriting Systems, Kenneth Brown defines a simplicial set $EM$ associated to a monoid $M$:

The $n$-simplices are $(n+1)$-tuples of elements of $M$, a typical such $(n+1)$-tuple being written in the form $(m_1,\dots,m_n)m$. If $m=1$, we suppress it from the notation and simply write $(m_1,\dots,m_n)$. The face operators in $EM$ are given by

$$d_i(m_1,\dots,m_n)m = \begin{cases} (m_2,\dots,m_n)m \mbox{ if } i=0 \\ (m_1,\dots,m_im_{i+1},\dots,m_n)m \mbox{ if } 0 < i < n \\ (m_1,\dots,m_{n-1})m_nm \mbox{ if } i=n, \end{cases}$$

and the degeneracy operators are given by

$$s_i(m_1,\dots,m_n)m = (m_1,\dots,m_i,1,\dots,m_{i+1},\dots,m_n)m$$.

There is an obvious right action of $M$ on $EM$ by simplicial maps, where the action of $m$ is given by $(m_1,\dots,m_n)m' \mapsto (m_1,\dots,m_n)m'm$. This action makes the normalized chain complex $C=C_*(EM)$ a complex of free right $\mathbb{Z}[M]$-modules, and, in fact, it is precisely the normalized bar resolution of $\mathbb{Z}$ over $\mathbb{Z}[M]$.

The answer in my Math.SE question suggested the Moore complex for simplicial monoids. However, $EM$ is not by itself a simplicial monoid (the function $(m_1,m_2) \mapsto m_1m_2$ is not a monoid homomorphism between $M^2$ and $M$ unless $M$ is commutative). One might want to consider the free simplicial monoid generated by $EM$ (which makes sense, since this complex is supposed to coincide with the normalized bar resolution of $M$), but there's a problem: the Moore complex associated to it is trivial (all the objects are $0$)!

When I looked up for references on normalized chain complexes associated to certain structures, all I could find was the case of (associative, unital) algebras over fields. Well, the action of $M$ over $EM$ gives you some sort of simplicial "module" (if you define a module as an action of a monoidal object over an object on a monoidal category), which would allow us to consider for each module of $n$-simplices its free (associative unital) algebra, and from that possibly an associated normalized complex. However, in this case, we would have one normalized complex for each module of $n$-simplices. Which one would coincide with the bar resolution of $M$? Well, I'm pretty sure this isn't the correct way to think about this. What is $C_*(EM)$?

$\endgroup$
3
  • 3
    $\begingroup$ Isn't it just the normalized chain complex of the simplicial set $EM$? That is, in degree $n$ you take the free abelian group on $EM_n$ modulo the images of the degeneracies? In other words, you just kill any word containing a $1$. (Note that I am not saying anything different than what John Palmieri said in his answer.) $\endgroup$ Apr 7, 2019 at 13:01
  • $\begingroup$ I didn't think about this possibility...For some reason I only thought of adapting Moore complexes for simplicial monoids. In this case, how exactly does the complex coincides with the bar resolution? By forgetting the $\mathbb{Z}[M]$ action on the $\mathbb{Z}[M]-$modules? $\endgroup$ Apr 7, 2019 at 13:35
  • 1
    $\begingroup$ There is no forgetting to be done, $C_*(EM)$ is a chain complex of $\mathbb{Z}[M]$-modules, and this chain complex is isomorphic to the normalized bar resolution. Just write down the definition of both and you will see! $\endgroup$ Apr 7, 2019 at 15:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.