Let $k$ be a commutative ring and $A$ a $k$-algebra. Denote by $C_{\bullet}(A)$ the Hochschild chain complex, with $C(A)_n = A^{\otimes n+1}$. Let $B$ denote Connes' operator
$$B_n \colon A^{\otimes n+1} \to A^{\otimes n+2}\, .$$
This induces a bicomplex $\mathcal{B}_{\bullet\bullet}(A)$ with $\mathcal{B}(A)_{pq} = A^{\otimes q-p+1}$ if $q \geq p$ and $0$ otherwise, where the vertical differential $b$ is the Hochschild differential and the horizontal differential is given by $B$. Define the cyclic homology of $A$ as
$$HC_{\bullet}(A) := H_n(\text{Tot}\, \mathcal{B}_{\bullet\bullet}(A)) \, .$$
This definition makes sense for any mixed complex $(M, b, B)$ (i.e. $M$ is a graded $k$-module together with a homological differential $b$ and a cohomological differential $B$ such that $bB + Bb = 0$). Any mixed complex can be identified with a dg $\Lambda$-module over the dg-algebra $\Lambda$ given by
$$\Lambda_0 = k \, , \; \Lambda_1 = k[\epsilon]/\epsilon^2 \, \; \Lambda_i = 0 \quad \text{for $i >1$}\, ,$$
where $\epsilon$ has degree $1$. An alternative definition of cyclic homology that works for any bicomplex $(M, b, B)$ is
$$HC_{\bullet}(M) = H^n(k \otimes_{\Lambda}^{\mathbf{L}} M) \, ,$$
where $-\otimes_{\Lambda}^{\mathbf{L}}-$ stands for the derived tensor product. My question is
Why are these two definitions equivalent?
If I wanted to compute $HC_{\bullet}(M)$ as in the second definition, I need a resolution of $k$ by dg $\Lambda$-modules, and then somehow compute the derived tensor product. I don't know how to proceed, probably because my comprehension of the "derived world" is very limited. Any help would be very much appreciated.
Best Answer
What you want is explained in this paper of Kassel, in Proposition 1.3.
A mixed module is a (non-negatively graded) dg-module $(M,b)$ for the algebra $\Lambda = k[t]/(t^2)$ where $t$ has degree $1$, so it comes equipped with an extra differential $B$ of degree $1$ that (anti)commutes with $b$.
The trivial module $k$, seen as a (dg) module over $\Lambda$ admits a very simple free resolution over $\Lambda$, where the maps are multiplication by $t$, where I use $s$ as a formal symbol to remember the degree where $\Lambda$ sits on:
$$C:\cdots\longrightarrow s^4\Lambda \longrightarrow s^3\Lambda \longrightarrow s^2\Lambda \longrightarrow s\Lambda \longrightarrow 0$$
If you tensor this with $M$ over $\Lambda$, you get a complex $C\otimes_\Lambda M$ representing the derived tensor product such that $$(C\otimes_\Lambda M)_n = \bigoplus_{i+j=n} s^iM_j = M_n\oplus M_{n-1} \oplus \cdots$$
and where we both have an internal differential $b$ that preserves this homological grading (and simply goes from $M_n$ to the next summand, etc) and the external differential $B$ going from one "row of $M_i$s" to the next one above. This is precisely what your cyclic bar (double) complex looks like:
$$\begin{matrix} \uparrow & & \uparrow & & \uparrow\\ M_2 &\to& M_1& \to &M_0 \\ \uparrow & &\uparrow \\ M_1 &\to &M_0 \\ \uparrow \\ M_0\end{matrix}$$
right? (You need to totalize at the end and look at diagonals, and this introduces the funny situation that as in Kassel's paper the total complex now goes $M_n\oplus M_{n-2}\oplus\cdots$.)