Let $A$ be a unital algebra over a field $K$, $C^n(A)$ a space of all $n+1$ linear maps into scalar field $k$ (I'm interested in case $k=\mathbb{C}$) and $$(bf)(a_0,...,a_{n+1})=\sum_{i=0}^n(-1)^if(a_0,...,a_ia_{i+1},...,a_{n+1})+(-1)^{n+1}f(a_{n+1}a_0,a_1,...,a_n)$$ for $f \in C^n(A)$. One checks that $b^2=0$ so we get a cochain complex $$C^0(A) \stackrel{b}{\longrightarrow} C^1(A) \stackrel{b}{\longrightarrow} C^2(A) \stackrel{b}{\longrightarrow} ... $$ The cohomology of this complex is the Hochschild cohomology. We further introduce the cyclic operator $\lambda:C^n(A) \to C^n(A)$ where $\lambda f (a_0,a...,a_n)=(-1)^nf(a_n,a_0,a_1,...,a_{n-1})$. Then we put $C^n_{\lambda}(A):=ker(1-\lambda)$. Remarkably we have $bC^n_{\lambda}(A) \subset C^{n+1}_{\lambda}(A)$ hence we have a subcomplex. The cohomology of this complex is the cyclic cohomology. Suppose that $f: (C^*(B),b) \to (C^*(A),b)$ be a map of Hochschild complexes commuting with the cyclic operator. Then it induces two morphisms: at the level of Hochschild cohomology and at the level of cyclic cohomology. My question is:

Why is it true, that if the induced maps for Hochschild cohomology are isomorphisms then all induced maps in cyclic cohomology are isomorphisms as well?

The idea is to use so called $IBS$ long exact sequence of the form: $$... \to HC^n(A) \stackrel{I}{\rightarrow} HH^n(A) \stackrel{B}{\rightarrow} HC^{n-1}(A) \stackrel{S}{\rightarrow} HC^{n+1}(A) \to ... .$$ (Here $HC$ is cyclic cohomology, $HH$ is Hochschild cohomology, $I$ comes from the inclusion of complexes, there are also exact formulas for $B$ and $S$ but I don't know whether we have to use those explicit formulas)). I wonder is it enough to use some general cohomological lemma (something like five lemma) in order to conclude that the maps for cyclic cohomology are isomorphisms. The problem is that if we write another IBS-sequence for an algebra $B$ then one vertical arrow is isomorphism, after that we have two arrows for which we don't know whether are they isomorphisms and then another isomorphism: so I don't see how to apply five lemma. How to handle this problem?


1 Answer 1


This was something that used to puzzle me when I was first learning this stuff. Refreshing my memory just now, I think that the trick is to use the fact that the Connes–Tsygan sequence has some zero entries in low degrees. It starts $$ 0 \to HC^0(A) \to HH^0(A) \to 0 \to HC^1(A) \to HH^1(A) \to HC^0(A) \to HC^2(A) \to \dots $$ So, in the context of the original question, if the induced maps $HH^n(B)\to HH^n(A)$ are isomorphisms in all degrees, the maps $HC^0(B)\to HC^0(A)$ and $HC^1(B)\to HC^1(A)$ are iso.

Now we can set up an inductive argument: suppose we know that $HC^{k-1}(B)\to HC^{k-1}(A)$ and $HC^k(B)\to HC^k(A)$ are both iso for some $k\geq 1$. Then by considering $$ HH^k(B)\to HC^{k-1}(B) \to HC^{k+1}(B) \to HH^{k+1}(B)\to HC^k(B) $$ $$ HH^k(A) \to HC^{k-1}(A) \to HC^{k+1}(A) \to HH^{k+1}(A)\to HC^k(A) $$ and applying the five lemma, we deduce that $HC^{k+1}(B)\to HC^{k+1}(A)$ is iso. So by induction, the induced maps between cyclic cohomology groups are iso in all degrees.


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