On page 4 of this [paper](http://arxiv.org/pdf/1302.6534.pdf) 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.



Two-sided bar construction
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 $B\_{p,q}$. 


