Let $A$ be a unital algebra over $\mathbb{C}$. Let $C^n(A)$ be the space of all $n+1$-linear maps $f:A^{n+1} \to \mathbb{C}$ (to be called $n$-cochains). Define $b:C^n(A) \to C^{n+1}(A)$ by the formula $$(bf)(a_0,a_1,...,a_n,a_{n+1}) \\ :=\sum_{j=0}^n(-1)^jf(a_0,...,a_{j-1},a_ja_{j+1},a_{j+2},...,a_{n+1}+(-1)^{n+1}f(a_{n+1}a_0,a_1,...,a_n).$$ The cohomology of this complex is called Hochschild cohomology of $A$. Let us call the $n$-cochain $f$ normalized if $f(a_0,a_1,...,a_n)=0$ whether $a_i=1$ for some $i \geq 1$. One checks that the normalized cochains form a subcomplex of $(C^*(A),b)$.

How to prove that the inclusion of normalized cochains into all cochains induces an isomorphism in cohomology?

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    $\begingroup$ This should be explained in the book of Weibel or perhaps even in the book of Cartan--Eilenberg. Since you have been asking several questions about Hochschild cohomology, might I ask which books or notes you have access to, or are working from? $\endgroup$
    – Yemon Choi
    Jul 11, 2016 at 17:11
  • $\begingroup$ To address your actual question: one can either try to do this with an inductive construction of an explicit chain homotopy -- similar arguments can be found in parts of the literature on continuous Hochschild cohomology of Banach algebras -- or you can take a more conceptual approach, as follows. There is a "normalised" version of the 2-sided bar resolution of A by A-biprojective modules, and so using the comparison lemma for projective resolutions, the natural map from the un-normalised resolution to the normalised one will induced isomorphism $H^*_n(A,M)\to H^*(A,M)$ for all bimodules $M$ $\endgroup$
    – Yemon Choi
    Jul 11, 2016 at 17:15
  • $\begingroup$ Going back to my first comment: I strongly recommend investing in a copy of Loday's book, or at least getting a copy from a library (or whatever other method you prefer) $\endgroup$
    – Yemon Choi
    Jul 11, 2016 at 17:16
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    $\begingroup$ Thank You, I will try to look into these books. My motivation is the index formula in noncommutative geometry which states that the pairing between K-theory and K-homology, which is an index pairing can be computed by the pairing of Chern characters which are elements of cyclic cohomology and cyclic cohomology-but while Hochschild cohomology seems to be more standard than cyclic cohomology, I posed this question in the context of Hochschild complex. $\endgroup$
    – truebaran
    Jul 11, 2016 at 20:29
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    $\begingroup$ @darijgrinberg I think that if you take $a_0$ as the "coefficient module" (well, the coefficient module $M$ is $A'$, and so an $n$-cocycle with values in $A'$ is an $(n+1)$-linear map where the first argument comes from the coeff module) then the $i\geq 1$ is correct. If one wanted to "normalise in the module variable", so to speak, then I think this does work for cyclic cohomology -- but I may have misremembered this last comment $\endgroup$
    – Yemon Choi
    Jul 23, 2016 at 13:11

2 Answers 2


I would like someone to take a look for the following argument-maybe somebody find it useful. 1. Let us recall the so called bar resolution for an algebra $A$. It is the sequence of the form $$...P_n \stackrel{b'}{\longrightarrow} P_{n-1} \stackrel{b'}{\longrightarrow} ... \stackrel{b'}{\longrightarrow}P_0 \stackrel{m}{\longrightarrow} A$$ where $m$ is the multiplication map and where $P_n=A \otimes A^{op} \otimes A^{\otimes n}$ which is isomorphic to $A^{\otimes (n+2)}$ via the map $$a \otimes a' \otimes a_1 \otimes ... \otimes a_n \mapsto a \otimes a_1 \otimes ... \otimes a_n \otimes a'.$$ There is a one-to-one correspondence between $A-A$ bimodule structures and left $A \otimes A^{op}$ module structures and we view each $P_k$ as left $A \otimes A^{op}$ module. One can check that even if $A$ is noncommutative, operator $b'$ is always $A \otimes A^{op}$ linear (unlike operator $b$). Since each $P_n$ is of the form $A \otimes A^{op} \otimes (something)$ we conclude that each $P_n$ is free $A \otimes A^{op}$ module, therefore there are projective in particular. Exactness of this sequence is provided by the existence of an operator $s$ which acts as follows $$s(a_1 \otimes ... \otimes a_n):=1 \otimes a_1 \otimes ... \otimes a_n.$$ It is straighforward that $$b's+s'b=1$$ so the above complex (in the category of left $A \otimes A^{op}$ modules) is contractible so it is exact. This is how we get bar resolution (beeing a projective resolution for $A$). The general theory tells us that Hochschild homology $H_*(A,M)$ may be computed by tensoring this resolution with $M$.
2. Denote by $D_n$ the subspace of $A^{\otimes n}$ generated by the expressions $a_1\otimes a_2 \otimes ... \otimes a_n$ where some $a_i=1$ and let $\overline{A^{\otimes n}}$ denote the quotient $A^{\otimes n}/D_n$. We construct projective resolution $(\overline{P_n})_n$ for $A$ where $\overline{P_n}:=A \otimes A^{op} \otimes \overline{A^{\otimes n}}$. The fact that this is projective resolution is proved similarly as for bar resolution provided we can show that $b'$ and $s$ ,,pass to the quotient''---this can be directly and easily verified.
3. General homological algebra tells us that the projective resolution is unique up to chain homotopy: there is some subtlety---slightly more is true. If we take two projective resolutions $(P_n)_n$ and $(\overline{P_n})_n$ there is up to chain homotopy one and only one chain map $P_* \to \overline{P_*}$ and $\overline{P_*} \to P_*$. But there is natural chain map $P_* \to \overline{P_*}$ namely the projection map $p$. General theory provides us with the chain map going opposite way: $q:\overline{P_*} \to P_*$ and from uniqueness we have that $pq$ and $qp$ are chain homotopic to identities.
4. From this we get that the maps $1_M \otimes p: M \otimes_{A \otimes A^{op}} P_* \to M \otimes_{A \otimes A^{op}} \overline{P_*}$ and $1_M \otimes q: M \otimes_{A \otimes A^{op}} \overline{P_*} \to M \otimes_{A \otimes A^{op}} P_*$ when composed are chain homotopic to identities, therefore both of them are quasi-isomorphisms.
5. Let us recall the argument that the (standard) bar resolution computes Hochschild homology. We consider the map $$\alpha: M \otimes_{A \otimes A^{op}} A \otimes A^{op} \otimes A^{\otimes n} \to M \otimes A^{\otimes n},$$ $$ \alpha(x \otimes a \otimes a' \otimes a_1 \otimes ... \otimes a_n):=a'xa \otimes a_1 \otimes ... \otimes a_n $$ One shows that such $\alpha$ is an isomorphism (in particular it is the chain map: it intertwines $b'$ from the bar resolution with the standard Hochschild boundary $b$).
6. We can repeat this argument for $\overline{A^{\otimes n}}$ in place of $A^{\otimes n}$:

a) $\alpha$ indeed passes to quotients---denote this map by $\overline{\alpha}$;

b) $b$ is well defined as a mapping $M \otimes \overline{A^{\otimes n}} \to M \otimes \overline{A^{\otimes (n-1)}}$.

The above statements may be verified easily. So we proved that our normalized projective resolution computes the homology of the normalized complex. But being a projective resolution for $A$ it must also compute ordinary Hochschild homology. Therefore we get that the homology of the normalized complex is isomorphic to Hochschild homology.
7. We want slightly more: we want our projection $M \otimes A^{\otimes n} \to M \otimes \overline{A^{\otimes}}$ to induce the isomorphism in homology. But this map is the composition $\overline{\alpha} \circ (1_M \otimes p) \circ \alpha^{-1}$---the composition of isomorphism, quasi isomorphism and isomorphism, therefore it is quasi isomorphism.
8. For co homology the argument is similar modulo the following:
a) normalized cochains (i.e. maps defined on $\overline{A^{\otimes n}}$) may be viewed as ordinary cochains, i.e. maps on $A^{\otimes n}$
b) applying $Hom(-,M)$ functor to the quotient map gives the inclusion.
So the inclusion of normalized complex into the whole Hochschild complex induces isomorphism in cohomology.
9. We would like to have analogous result for cyclic theory. Since cyclic homology (or cohomology) cannot be computed using bar resolution, we cannot repeat everything for cyclic theory. However we can start defining cyclic theory using normalized complexes and develop the theory in the context of normalized complexes, until we arrive at another, normalized version of Connes $IBS$ sequence. Then we have a commutative diagram where the first row is normalized IBS sequence, the second is ordinary IBS sequence and vertical maps comes from the inclusion. Using the inductive argument (our rows starts with some zeros, which allows us to start the induction) and five lemma we obtain that the inclusion of cyclic cochains into all cochains induces the isomorphism in cohomology.

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    $\begingroup$ Disclaimer: apparently this argument does not work. In order to obtain $IBS$ long exact sequence one uses the short exact sequence where the complex $(C*,b')$ appears. However $b'$ does not preserves normalized cochains therefore we cannot arque like in the ordinary case of arbitrary cochains. I have no idea how to fix this problem and I would be very interested in seeing a solution. $\endgroup$
    – truebaran
    Sep 16, 2016 at 22:08

It makes sense to give a precise reference:

It is proved in page 13 (and 46) of Loday's book in detail. The original reference may be Loday&Quillen's paper, where they proved it in page 8 and drawn the double complex quite explicitly. I am not sure if their results were proved earlier by others.

In my humble view, the whole book by Loday can be viewed as his expository service to the community of their paper. So it makes sense to read the book together with the paper. If you are into index theory, recent work by Victor Nistor and his students may be of interest to you.

  • $\begingroup$ I agree with you, however I'm curious whether the above argument can be repaired to work in the setting of ordinary complexes instead of double complexes. This question is inspired by the Chapter 3 in M. Khalkhali book "Basic Noncommutative Geometry". $\endgroup$
    – truebaran
    May 29, 2017 at 22:00
  • $\begingroup$ There is nothing wrong in their answer. So I do not understand. I do think double complex (or the more general spectral sequence in Loday's book) is a natural answer. $\endgroup$ May 29, 2017 at 22:19

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