# Two-sided bar construction

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.

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The grading on $B(A,A,A)$ is defined on line 18 of page 4 (Abbaspour calls it "degree"). –  M T Oct 12 '13 at 12:14
But then I don't understand how $(B(A,A,A),d)$ is a chain complex giving a free resolution of $(A,d_A)$ as an $A\otimes A^{op}$-module. Or is this not true and what is really going on is that $(B(A,A,A),d_1)$ is a chain complex (graded by wordlength on $T(s\bar{A})$) giving a free resolution of $(A,d_A)$ as an $A\otimes A^{op}$-module? –  Dave Oct 12 '13 at 12:28

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 $B_{p,q}$.
Thank you very much for this nice explanation. How does $(B=B(A,A,A),d)$ give rise to a free $A\otimes A^{op}$-resolution of $A$? Does one just consider $(B,(-1)^pd^v)$, where $B$ is now only graded by the homological degree? I would like to get to the Hochschild homology $Tor_*^{A\otimes A^{op}}(A,A)$ of $A$ via the resolution arising from the two-sided bar construction. –  Dave Oct 12 '13 at 14:46
Thank you, I wasn't aware of these problems. I was under the impression, that the case when $A$ is a differential, graded algebra would be similar to the case when $A$ is just an algebra. In this case the two-sided bar construction is a free resolution, but apparently the situation is much more complicated in case $A$ is graded, differential...which is a shame, since I now have no idea how to interpret Hochschild homology of a DG-Algebra. What does Hochschild homology of a DG-algebra "measures"? –  Dave Oct 12 '13 at 18:18
If $A$ is just an algebra, then $HH_*(A)$ "measures" how far $A$ is from being a flat $(A\otimes A^{op})$-module, right? –  Dave Oct 12 '13 at 18:19