2
$\begingroup$

This question is a continuation of the discussion Normalization of Hochschild cocycles but this time in the cyclic context. I would like to ask whether the following is true:

The inclusion of normalized cyclic cochains into all cyclic cochains induces an isomorphism in (cyclic) cohomology.

In the discussion linked above I gave the argument which however turns out to be invalid, as indicated in the comment. Let me summarize what I have figured out already after reading carefully Loday's book: let $C^k=C^k(A)$ (where $A$ is an algebra) be a space of $k+1$-linear maps $\varphi: A^{k+1} \to \mathbb{C}$. Since drawing diagrams here is somehow problematic I will try to briefly describe underlying complexes:

1. The cyclic bicomplex $\mathcal{C}$ has in zeroth row $C^0$, in the first row $C^1$ and so on: vertical differentials are $b$ and $-b'$ (for even/odd columns) while horizontal differentials are $1-\lambda$ and $N$ where $\lambda$ is cyclic operator and $N=1+\lambda+...+\lambda^k$.

2. $(b,B)$-bicomplex $\mathcal{B}$ has in the zeroth row: $C^0,0,0,...$ in the first $C^1,C^0,0,0,...$ in the second $C^2,C^1,C^0,0,...$. Vertical differential is $b$ and horizontal differential is $B=Ns(1-\lambda)$ where for $\varphi \in C^n$ we define $$(s\varphi)(a_0,...,a_{n-1})=(-1)^{n-1}\varphi(1,a_0,...,a_{n-1})$$

3. Normalized version of $\mathcal{B}$ bicomplex, denoted by $\overline{\mathcal{B}}$ where all cochains are normalized, meaning that $\varphi (a_0,a_1,...,a_n)=0$ whenever $a_k=1$ for some $k>0$ ($0$-entry is excluded!). Note that $b$ as well $B$ preserves normalization (unlike $b'$ which prevents us from considering normalized version of the cyclic bicomplex)

4. Finally we have ordinary cyclic complex $C^*_{\lambda}$ with differential $b$ together with its normalization $\overline{C^*_{\lambda}}$

Bicomplexes give rise to cohomology via totalization of complexes.

Theorem 1. The inclusion $C^*_{\lambda} \to Tot \ \mathcal{C}$, $\varphi \mapsto (0,0,...,0,\varphi)$ is a quasi-isomorphism.

Theorem 2. The map $(id,Ns):Tot \ \mathcal{C} \to Tot \ \mathcal{B}$ is also a quasiisomorphism.

Luckily, even the second map is not an inclusion, if we look at the corresponding degrees of cochains, we can see that the composition $C^*_{\lambda} \to Tot \ \mathcal{C} \to Tot \ \mathcal{B}$ is also an inclusion and is clearly a quasi-isomorphism.

Finally we have that the inclusion $Tot \ \overline{\mathcal{B}} \to Tot \ \mathcal{B}$ is a quasi-isomorphism.

If we could somehow show that the inclusion $\overline{C^*_{\lambda}} \to Tot \overline{\mathcal{B}}$ is a quasi-isomorphism we would obtain that the inclusion $\overline{C^*_{\lambda}} \to C^*_{\lambda}$ is also a quasi-iso. However I'm not so sure about this fact: while showing that the inclusion $C^*_{\lambda} \to Tot \ \mathcal{B}$ is quasi-iso we have after all used the auxiliary complex, namely cyclic bicomplex (which is absent in the normalized setting). Therefore I'm stuck and would be very grateful for any help.

$\endgroup$
2
+100
$\begingroup$

If I'm not mistaken this is just the dual version of what Loday and Quillen proves in Proposition 4.4 (or Proposition 2.2.14 in Loday's book, with the same proof). And sorry for nitpicking, but it's not exactly a quasi-iso, more precisely you get an exact sequence $$ H^*(\bar{C}^*_\lambda(A)) \to HC^*(A) \to HC^*(\mathbb{C}) \to H^{*+1}(\bar{C}^*_\lambda(A)). $$

$\endgroup$
  • $\begingroup$ Are you sure that it answers my question? Proposition 2.2.14 deals with reduced (co)homology. I don;t see where the complex $\overline{C^*_{\lambda}}$ enters in these considerations $\endgroup$ – truebaran Aug 17 '17 at 21:34
  • 1
    $\begingroup$ You say $\bar{C}^*_\lambda$ is the space of cyclically invariant linear functionals $\phi$ such that $\phi(a_0, \ldots, \phi_n) = 0$ whenever $a_i = 1$ for some $i$. Loday(–Quillen) says $\bar{C}_*^\lambda$ is the quotient of $A^{\otimes n+1}$ by the image of $1-t$ and the subspace generated by $a_0 \otimes \cdots \otimes a_n$ with $a_i =1$ for some $i$. So $\bar{C}^*_\lambda$ is the dual complex of $\bar{C}_*^\lambda$. And Loday relates $\overline{HC}_*(A) = \bar{H}^\lambda_*(A)$ to usual $HC_*(A)$ in the sequence (2.2.13.1), which is (the dual of) what I wrote above. $\endgroup$ – Makoto Yamashita Aug 21 '17 at 2:23
  • $\begingroup$ I'm a bit embarassed since I've asked this question some time ago, offered bounty on it and accepted the answer. At that time I only wanted to know whether the statement in my question is correct. However recently I went through the proof in Lodays book and I must admit that I don't understand the last 3 lines of the proof. Did you go through the proof in Loday's book? If so, are those last 3 lines clear for you? After taking the quotient bicomplex (passing from the filtration to gradation) I'm getting different bicomplex: in particular I don't have an idea from where the maps $1-t$ and $N$ $\endgroup$ – truebaran Feb 5 '18 at 1:15
  • $\begingroup$ ... came from (since our original bicomplex involves only $b$ and $B$). Note also that probably there is a typo in the definition of filtration of $\mathcal{B}$: instead of $k \otimes A^{\otimes(n+1)}$ there should be probably $k \otimes \overline{A}^{\otimes(n+1)}$. After passing to gradation I'm getting bicomplex with $A$ and $\overline{A}^{\otimes(n+1)}$ in each row. $\endgroup$ – truebaran Feb 5 '18 at 1:18
  • $\begingroup$ Yes, I think that's typo (Loday-Quillen has correct formula btw). As for how those maps appear, $N$ appears from $B$ when you identify $k \otimes \bar{A}^{\otimes n+1}$ (from "diagonal" of $F_n$) at the codomain with $\bar{A}^{\otimes n+1}$, and $(A \otimes A^{\otimes n}) / (k \otimes A^{\otimes n})$ (from "below diagonal" of $F_n$ modulo "diagonal" of $F_{n-1}$) at the domain again with $\bar{A}^{\otimes n+1}$. Similarly, $1-t$ is induced by $b$; if you compute the image of $b(1,a_1,\ldots,a_{n+1})$ in $(A \otimes A^{\otimes n}) / (k \otimes A^{\otimes n})$, most of the terms become trivial. $\endgroup$ – Makoto Yamashita Feb 13 '18 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.