1
$\begingroup$

I am reading a book that uses a lot of homological algebra and I am not familiar with the tools of it. To prove a part of a theorem I suspect the following must be true.

Let $X$ be an infinite $G$-set. Consider a chain complex $(C_{*}(X), \partial)$ where $C_n(X)$ is the free abelian group generated by $X^{n+1}$ and $\partial((x_0,\ldots,x_n))=\sum_i (-1)^{i} (x_0, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n)$.

If $C_n^{\neq}(X)$ is free abelian group generated by the set $\{(x_0, \ldots,x_n)~|~x_i \neq x_j \forall i \neq j\}$ and $C_n^D(X)$ is free abelian group generated by the set $\{(x_0, \ldots, x_n)~|~x_i =x_{i+1} ~\textrm{for some}~0 \leq i \leq n-1\}$, then is it true that for any $G$ module $M$, $H_n(C_n ^{\neq} (X) \otimes_{\mathbb{Z}[G]}M) \cong H_n(C_n^{nor}(X) \otimes_{\mathbb{Z}[G]} M)?$

Here $C_n^{nor}(X)= C_n(X)/C_n^{D}(X)$ and is called normalized complex.

$\endgroup$
5
  • $\begingroup$ Is the definition of $C_n^D(X)$ correct? If so, I don't understand how $C_*^{nor}(X)$ is a chain complex, as $\partial$ doesn't map $C_n^D(X)$ into $C_{n-1}^D(X)$. $\endgroup$ Commented Feb 11, 2020 at 10:44
  • $\begingroup$ Jeremy Rickard: Sorry, I have corrected it. $\endgroup$
    – eyp
    Commented Feb 11, 2020 at 10:49
  • $\begingroup$ I still don't think that there's a well-defined differential, since $\partial((x_0,\dots,x_n))$ is in $C_{n-1}^D(X)$ only if $x_i=x_{i+1}$ for at least two values of $i$. $\endgroup$ Commented Feb 11, 2020 at 10:52
  • 1
    $\begingroup$ Jeremy Rickard: It is well-defined. $$\partial((x_0, \ldots, x_{i-1},x_i,x_i,x_{i+2},\ldots, x_n)) = \sum_{j=0} ^{j=i-1} (-1)^j (x_0, \ldots, \hat{x_j}, \ldots, x_{i-1},x_i,x_i,x_{i+2}, \ldots,x_n) + \sum_{j= i+2} ^{j=n} (-1)^j (x_0, \ldots,x_i,x_i,x_{i+2},\ldots, \hat{x_j}, \ldots, x_n),$$ here by $\hat{x_j}$ I mean to remove it. Thus $\partial(C_n ^D(X))\subset C_{n-1} ^D (X)$. $\endgroup$
    – eyp
    Commented Feb 11, 2020 at 11:26
  • $\begingroup$ Ah, yes, OK! Sorry. $\endgroup$ Commented Feb 11, 2020 at 12:49

0

You must log in to answer this question.