The cyclic bicomplex

Let $C_{\bullet}$ be the cyclic module with
$$\begin{align*} d_i &\: C_n\to C_{n-1}, \\ t_n &\: C_n\to C_n. \end{align*}$$
Consider the following two-column bicomplex

$$\xymatrix{ \vdots \ar[d] & \vdots \ar[d] \\ C_2 \ar[d]_{b} ... ...C_1 \ar[d]_{-b'} \ar[l]_{1-t} \\ C_0 & C_0 \ar[l]_{1-t} \\ }$$

One checks that it has anticommuting squares, so it is indeed a bicomplex. It can be extended to the right using the map $N:=1+t+\hdots t^n\: C_n\to C_n$.

$$\xymatrix{ \vdots \ar[d] & \vdots \ar[d] & \vdots \ar[d] & \... ...} & C_0 \ar[l]_{N} & C_0 \ar[l]_{1-t} & \hdots \ar[l]_{N}\\ }$$

For example if $C_n=A\ox A^{\ox n}$ we have a cyclic bicomplex $C_{\bullet\bullet}(A)$ with $t$ being the cyclic operator, and $N=1+t+\hdots t^n$.





Whenever we have a sequence of complexes

$$K'_{\bullet} \mono K_{\bullet} \epi K''_{\bullet}$$

and we know that $K'_{\bullet}$ is acyclic, then the complexes $K_{\bullet}$ and $K''_{\bullet}$ are quasi-isomorphic. This allows us to quotient out the acyclic subcomplexes of a given complex when computing homology. But $(C_{\bullet}, -b')$ is not a subcomplex. We will get rid of one column at a time using

The cokernel of $(\id, -h\ga)$ is $(A'_{\bullet}, \dl)$. Applied infinitely many times to the cyclic bicomplex we end up with the total complex of the bicomplex $B_{\bullet}C_{\bullet}$

$$\xymatrix{ \vdots \ar[d] && \vdots \ar[d] && \vdots \ar[d] &... ... C_0 \ar[llu]_{B} && C_0 \ar[llu]_{B} && \hdots \ar[llu] \\ }$$

This is the normalized version of a bicomplex $C_{\bullet\bullet}$ used to define cyclic homology. Because the quasi-isomorphism in the lemma () we have

$$\rH_*(C_{\bullet}) = \rH_*(\Tot(B_{\bullet}C_{\bullet})).$$

We can rearrange the bicomplex $B_{\bullet}C_{\bullet}$ to obtain

$$\xymatrix{ \vdots \ar[d] & \vdots \ar[d] & \vdots \ar[d] & \... ...]_{B} & \\ C_1 \ar[d]_{b} & C_0 \ar[l]_{B} & &\\ C_0 & & & }$$

It is indeed a bicomplex, that is we have the identities

$$b^2=0,\quad B^2=0, \quad bB+Bb=0.$$

The morphism $B$ on the normalized complex $B_{\bullet} C_{\bullet}(A)$ is given explicitly by

$$B=(1-t)sN\: A\ox \Bar{A}^{\ox n} \to A\ox \Bar{A}^{\ox(n+1)},$$

$$(a_0, \hdots, a_n)\mapsto \sum_{i=0}^n (-1)^{in}(1, a_i, \hdots, a_n, a_0, \hdots, a_{n-1}).$$

In the non-normalized complex there are more terms, but they are trivial in the normalized complex.