MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 down vote accepted

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 $B_{p,q}$.

share|cite|improve this answer
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
The problem I think is the meaning of words: what do you mean – Peter May Oct 12 '13 at 16:57
by a free resolution: in the presence of a differential on A you have to be careful. The modern approach is to ask for cofibrant approximations of DG modules over DGAs in an appropriate model structure. I can refer you to a brand new treatment: – Peter May Oct 12 '13 at 17:06
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.