Let $A$ be a complex unital algebra. Consider the cyclic (cohomological) bicomplex $\mathcal{C}(A)$. This is a bicomplex
where in the $p$-th row one has $C^p(A)$ (the space of all $p+1$-linear forms) and the horizontal
differentials are $1-\lambda$ and $N$ alternately (here $\lambda$
is the cyclic operator and $N=1+\lambda+...+\lambda^n$) and the even columns looks like
$$C^0(A) \stackrel{b}{\longrightarrow} C^{1}(A) \stackrel{b}{\longrightarrow} ...$$
while the odd columns have a form
$$C^0(A) \stackrel{-b'}{\longrightarrow} C^{1}(A) \stackrel{-b'}{\longrightarrow} ...$$
Let us also consider the so called $(b,B)$ bicomplex to be denoted $\mathcal{B}(A)$: the zeroth row is $C^0(A)$ the first is $C^1(A) \stackrel{B}{\longrightarrow} C^0(A)$ the second is $C^2(A) \stackrel{B}{\longrightarrow} C^1(A) \stackrel{B}{\longrightarrow} C^0(A)$ and so on while the zeroth column is
$$ C^0(A) \stackrel{b}{\longrightarrow} C^1(A) \stackrel{b}{\longrightarrow} C^2(A) \stackrel{b}{\longrightarrow} ...$$
the first is
$$0 \rightarrow C^0(A) \stackrel{b}{\longrightarrow} C^1(A) \stackrel{b}{\longrightarrow} C^2(A) \stackrel{b}{\longrightarrow} ...$$
the second is
$$ 0 \rightarrow 0 \rightarrow C^0(A) \stackrel{b}{\longrightarrow} C^1(A) \stackrel{b}{\longrightarrow} ...$$ and so on. Here $B=Ns(1-\lambda)$ where $s:C^n(A) \to C^{n-1}(A)$ is defined via $(s \varphi)(a_0,...,a_{n-1})=(-1)^{n-1}\varphi(a_0,...,a_{n-1},1)$. The claim is that total complexes of $\mathcal{C}(A)$ and $\mathcal{B}(A)$ are homotopy equivalent. Here is the sketch of the proof based upon the book of Masoud Khalkhali Basic Noncommutative Geometry:
- consider the maps $$I: Tot \mathcal{B}(A) \to Tot \mathcal{C}(A), \quad I=id+Ns$$ and
$$J:Tot \mathcal{C}(A) \to Tot \mathcal{B}(A), \quad J=id+sN$$
-one checks that $I,J$ are chains maps
-consider operators
$$g: Tot\mathcal{B}(A) \to Tot \mathcal{B}(A), \quad g=Ns^2B_0$$
$$h: Tot\mathcal{C}(A) \to Tot \mathcal{C}(A), \quad h=s$$
where $B_0=s(1-\lambda)$.
-one checks that
$$I \circ J=id+h \delta+\delta h,$$
$$J \circ I=id+g \delta'+\delta'g$$
where $\delta,\delta'$ are differential of $Tot\mathcal{C}(A)$ and $Tot\mathcal{B}(A)$ respectively.
I don't see how to interpret the formulas for $I$ and $J$: moreover I have a problem in understanding how precisely the homotopies $g,h$ acts: $h$ lowers the degree by one which is fine but $g$ lowers the degree of the top component by $3$ so I'm also afraid that something is wrong with this formula.
My interpretation of maps $I,J$ was the following. The $n$-th term of $Tot\mathcal{C}(A)$ is $C^n \oplus C^{n-1} \oplus C^{n-2} \oplus ... \oplus C^1\oplus C^0$ ($C^k:=C^k(A)$) while for $(b,B)$ bicomplex we have $(Tot\mathcal{B}(A))^n=C^n \oplus C^{n-2} \oplus ... \oplus C^{\varepsilon}$ where $\varepsilon \in \{0,1\}$ for $n$ even and odd respectively. Then I interpreted $I$ as something acting like $$I(\varphi_n,\varphi_{n-2}, ..., \varphi_2,\varphi_0):=(\varphi_n,Ns \varphi_n,\varphi_{n-2}, Ns \varphi_{n-2},...,\varphi_2,Ns\varphi_2,\varphi_0)$$ for $n$ even and $$I(\varphi_n,\varphi_{n-2}, ..., \varphi_3,\varphi_1):=(\varphi_n,Ns \varphi_n,\varphi_{n-2}, Ns \varphi_{n-2},...,\varphi_1,Ns\varphi_1)$$ for n-odd while $$J(\varphi_n,\varphi_{n-1},...,\varphi_1,\varphi_0):=(\varphi_n,sN\varphi_{n-1}+\varphi_{n-2},sN\varphi_{n-3}+\varphi_{n-4},...,sN\varphi_3+\varphi_2,sN\varphi_1+\varphi_0)$$ for even $n$ and $$J(\varphi_n,\varphi_{n-1},...,\varphi_1,\varphi_0):=(\varphi_n,sN\varphi_{n-1}+\varphi_{n-2},sN\varphi_{n-3}+\varphi_{n-4},...,sN\varphi_4+\varphi_3,sN\varphi_2+\varphi_1)$$ for $n$ odd.
Additionally the differential $\delta$ acts as follows: $$\delta(\varphi_n,\varphi_{n-1},...,\varphi_1,\varphi_0)=$$ $$=(b\varphi_n,(1-\lambda)\varphi_n-b'\varphi_{n-1},N\varphi\varphi_{n-1}+b\varphi_{n-2},(1-\lambda)\varphi_{n-2}-b'\varphi_{n-3},N\varphi_{n-3}+b\varphi_{n-4},...$$ $$...,N\varphi_1+b\varphi_0,(1-\lambda)\varphi_0)$$ for $n$ even and $$\delta(\varphi_n,\varphi_{n-1},...,\varphi_1,\varphi_0)=$$ $$=(b\varphi_n,(1-\lambda)\varphi_n-b'\varphi_{n-1},N\varphi\varphi_{n-1}+b\varphi_{n-2},(1-\lambda)\varphi_{n-2}-b'\varphi_{n-3},N\varphi_{n-3}+b\varphi_{n-4},...$$ $$...,(1-\lambda)\varphi_1-b'\varphi_0,N\varphi_0)$$ for $n$ odd. Working with these formulas I didn't get that $I \circ J=id+h\delta+\delta h$. Comparing the terms for example for $C^0$ I got that $s(1-\lambda)\varphi_0$ should be zero. Other relations which looks to be false are for example $Ns\varphi_3+sb\varphi_2+bs\varphi_2=0$. I'm sure that I've messed something with these formulas. Since this is rather tedious task to do every computation
I would like to know how to interpret formulas for $I,J$ and $g,h$ in order to finish my calculations.
I would appreciate any help.
