What is the correct definition of the (derived) tensor product over a dg-algebra?

Question

Let $A_\bullet$ be a dg-algebra over a field $k$. Let $M_\bullet$ (resp. $N_\bullet$) be a right (resp. left) $A_\bullet$-module. There is then a notion of the derived tensor product: $$M_\bullet\otimes^L_{A_\bullet}N_\bullet=\bigoplus_{p\geq 0}M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet$$ (where $\otimes$ is $\otimes_k$) with differential given by the internal differential on each direct summand plus an alternating sum of all possible maps multiplying adjacent elements (decreasing $p$ by one). We can rewrite this as: $$(M_\bullet\otimes^L_{A_\bullet}N_\bullet)_j=\bigoplus_{p\geq 0}(M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet)_{j-p}$$ But now (based on my limited understanding) it is not clear to me why we shouldn't instead take direct product instead of direct sum, and define: $$(M_\bullet\otimes^L_{A_\bullet}N_\bullet)_j=\prod_{p\geq 0}(M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet)_{j-p}$$ Note that with this definition the differential still makes sense.

The definition with the direct sum seems to be standard. On the other hand, I have encountered a context where I want to write down a certain element in $M_\bullet\otimes^L_{A_\bullet}N_\bullet$, but it lies in the direct product (and not in the direct sum). Hence my question is:

Which of the two definitions of $M_\bullet\otimes^L_{A_\bullet}N_\bullet$ given above is correct, and why?

Note that when $A_\bullet$ is concentrated in nonnegative degrees, taking direct product instead of direct sum makes no difference, since for fixed $j$, the factors on the right become trivial for sufficiently large $p$. Hence this question is only non-vacuous when $A_\bullet$ is nontrivial in negative degrees.

Answer 1

Omitting the stars, what you have written down is a two-sided bar construction $B(M,A,N)$, unreduced since you have not assumed that $A$ is augmented. Note that in degree $n$ it is the direct sum over $p+q=n$ of the elements of the $p$th term in degree $q$. One conceptual way of thinking about it is that that we can write $B(M,A,N) = M\otimes_A B(A,A,N)$. Depending on taste and background, one can view $B(A,A,N)$ as a kind of projective resolution of $N$, a point of view that goes back to Cartan and Eilenberg, or one can view $B(A,A,N)$ as a cofibrant approximation of $N$ in model theoretic terms. For a recent exposition that explains and compares these points of view (allowing $k$ to be a general commutative ring) see http://arxiv.org/pdf/1310.1159.pdf. For either point of view, you want sums and not products.

Answer 2

In the case when $M=N=A$, your question reduces to the comparison between the usual direct sum Hochschild chain complex versus the direct product chain complex. It is important to use the direct sum chain complex in proving that Hochschild homology is a homotopy invariant of $A$. To see this, observe that the Hochschild differential decomposes to $d_1+d_2$ with $d_i$ induced by $m_i$ of $A$. In the case of direct sum, we can start with $d_1$ to compute Hochschild homology using spectral sequence method, which would yield the same complex if $A$ and $B$ are quasi-isomorphic. This argument fails in the direct product case.