Two-sided bar construction

Question

On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction $$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ of a differential graded algebra $(A,d_A)$ (over a field). The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$. As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$?

Thanks to anyone who can shed some light on the two-sided bar construction.

Answer 1

I haven't looked at Abbaspour's paper, but here is what is going on, in a bit greater generality. Let $N$ be a right, $M$ a left DG $A$-module. Then $B=B(N,A,M)$ is defined and it is bigraded. The grading with differentials that raise degree, which you apparently have in mind, is a bit awkward, so regrade by $A_n = A^{-n}$ and similarly for $M$ and $N$. Then we have $B_{p,q}$, where $p$ is the homological degree (via $N\otimes \bar{A}^{p}\otimes M$ and $q$ is the internal degree (add up the degrees of $n$, the $a_i$, and $m$ of an element $n[a_1,\cdots,a_p]m$). There is a horizontal (or internal) differential $d^h\colon B_{p,q} \to B_{p,q-1}$ given by the differentials on $N$, $A$, and $M$ and there is a vertical (or homological) differential $d^v: B_{p,q}\to B_{p-1,q}$.
These commute. Now regrade by total degree, $B_n = \sum_{p+q=n} B_{p,q}$. Then the differential is given by $d = d^h + (-1)^p d^v$ on the summand $B_{p,q}$.