The following proposition comes from Connes' paper in IHES. See the link [Non-commutative differential geometry](http://www.numdam.org/item/PMIHES_1985__62__41_0.pdf).
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 $\langle[u],[S\varphi]\rangle=\langle[u],[\varphi]\rangle$.
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**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.
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The question is: how to prove **(2)**? The same question has been raised in [this post](https://mathoverflow.net/questions/293654/connes-chern-pairing-compatibility-with-periodicity-operator-in-the-odd-case) 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?)
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There are some materials containing such a pairing besides Connes' paper.
In [this paper](http://www.math.u-ryukyu.ac.jp/rmj/pdf/RyukyuMathJ-35-02.pdf) 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](https://arxiv.org/abs/1008.1212) 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.
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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 $\widetilde{A}$.