The following proposition comes from Connes' paper in IHES. See the link Non-commutative differential geometry.
On page 109, Proposition 15. of Part II, he claims that
(1) The following equality defines a bilinear pairing between $K_1(\mathcal{A})$ and $H_{\lambda}^{2m-1}(\mathcal{A})$: $$\langle[u],[\varphi]\rangle=(2i\pi)^{-m}2^{-2m-1}\frac{1}{(m-1/2)\cdots1/2}(\varphi\#\mathrm{Tr})(u^{-1}-1,u-1,u^{-1}-1,\cdots,u-1)$$ where $\varphi\in Z_{\lambda}^{2m-1}(\mathcal{A})$ and $u\in\mathrm{GL}_k(\mathcal{A})$.
(2) One has $\langle[u],[S\varphi]\rangle=\langle[u],[\varphi]\rangle$.
Some notations:
Here, $\mathcal{A}$ is a unital complex algebra, $K_1(\mathcal{A})$ is the algebraic K-theory group of $\mathcal{A}$.
In the original text, it writes "between $K_1(\mathcal{A})$ and $H_{\lambda}^{\text{odd}}(\mathcal{A})$". I believe here is a typo.
The operation $S$ denotes the periodicity operator, defined on page 61, Part I (just before Theorem 1.) and also on page 106, Part II (before Lemma 11.). In short, $S\varphi$ is obtained from making the cup product of $\varphi$ and a fixed generator of $H_\lambda^2(\mathbb{C})\cong\mathbb{C}$.
$n=2m-1$ is an odd integer.
$\mathrm{Tr}$ is the usual trace on $M_k(\mathbb{C})$, the algebra of $m\times m$ complex matrices.
The question is: how to prove (2)? The same question has been raised in this post but with no answers. The comments there do not seem to really work well on this original version of pairing.
Connes has used a "normalizing" argument in proving (1):
Let $\widetilde{\mathcal{A}}$ denote the algebra obtained from $\mathcal{A}$ by adjoining a unit, whose elements are written as $(a,\lambda)$ and multiplication $(a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu)$ where $a,b\in\mathcal{A},\lambda,\mu\in\mathbb{C}$.
Let $\widetilde{\varphi}\in C_\lambda^n(\widetilde{\mathcal{A}})$ be defined by $$\widetilde{\varphi}((a^0,\lambda^0),\cdots,(a^n,\lambda^n))=\varphi(a^0,\cdots,a^n).$$ One can check that $\widetilde{\varphi}$ is a cocycle and $\widetilde{b\varphi}=b\widetilde{\varphi}$, where $b$ is the coboundary map of the complex $C_\lambda^*(\mathcal{A})$.
Such a $\widetilde{\varphi}$ is normalized in this sense: $$\widetilde{\varphi}(1_{\widetilde{\mathcal{A}}},\widetilde{a}^0,\cdots,\widetilde{a^{n-1}})=0,\quad\forall \widetilde{a^i}\in\widetilde{\mathcal{A}}.$$
However, it seems that the "normalization" argument may not work in proving (2) as expected.
One may expect that $S\widetilde{\varphi}=\widetilde{S\varphi}$, but this is not the case if I did not make mistakes in my computation: $$(S\widetilde{\varphi}-\widetilde{S\varphi})((a^0,\lambda^0),\cdots,(a^{n+2},\lambda^{n+2}))\\= \sum\limits_{j=0}^{n+1}\left(\lambda^{j}\lambda^{j+1}\varphi(a^0,\cdots,a^{j-1},a^{j+2},\cdots,a^{n+2})+\lambda^{j+1}\varphi(a^0,\cdots,a^{j-1},a^ja^{j+2},\cdots,a^{n+2})\right)-\lambda^{n+2}\lambda^0\varphi(a^1,\cdots,a^{n+1})-\lambda^0\varphi(a^{n+2}a^1,\cdots,a^{n+1}).$$ This seems not a coboundary. One may try the case $m=1$.
If one assumes that $\widetilde{S\varphi}=S\widetilde{\varphi}$, then one can easily obtain a simple identity like (2). (But the constant $c_m$ is not like the given form?)
There are some materials containing such a pairing besides Connes' paper.
In this paper page 24, the author used $d1=0$, which does not make sense (in universal differential graded algebra $\Omega(\mathcal{A})$). The computation he did is also strange. Why did he compute $(u^{-1}-1)du(u^{-1}-1)$? It should be $(u^{-1}-1)dud(u^{-1})$ even if $d1=0$.
In this paper by Khalkhali page 11, he stated a formula in Proposition 3.1, but the $\varphi$ he gave is automatically normalized. Also in page 13 just below the (26) formula, he said "Any cyclic cocycle can be represented by a normalized cocycle." but with no proof.
It is also mentioned somewhere that Quillen's paper Cyclic cohomology and algebra extensions contains some discussions on the pairing, but my university does not have access to this paper.
Another idea is to try the following steps:
The tilde operation $\varphi\mapsto\widetilde{\varphi},C_\lambda^n(\mathcal{A})\rightarrow C_\lambda^n(\widetilde{\mathcal{A}})$ is a cochain map. Moreover, it is a right inverse to $\iota^*:C_\lambda^n(\widetilde{\mathcal{A}})\rightarrow C_\lambda^n(\mathcal{A})$, where $\iota:\mathcal{A}\rightarrow\widetilde{\mathcal{A}}$ is the inclusion map $a\mapsto(a,0)$. Therefore, we may consider $H_\lambda^n(\mathcal{A})$ a subgroup of $H_\lambda^n(\widetilde{\mathcal{A}})$. All cohomology classes in this subgroup can be represented by normalized cocycles. Maybe the pairing should be described on $\widetilde{A}$.
