$\newcommand{\Tors}{{\rm Tors}} \newcommand{\tf}{{\rm\, t.f.}} \newcommand{\Gt}{{\Gamma\!,\,\Tors}} \newcommand{\Gtf}{{\Gamma\!,\tf}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\HH}{{\mathbb H}}$Let $A$ be an abelian group. We denote by $A_\Tors$ the torsion subgroup of $A$. We set $A_\tf=A/A_\Tors\,$, which is a torsion free group.

Let $\Gamma$ be a finite group, and let $M$ be a $\Gamma$-module, that is, an abelian group on which $\Gamma$ acts. We denote by $M_\Gamma$ the group of coinvariants of $\Gamma$ in $M$, that is, $$ M_\Gamma=M\, \big/\bigg\{\sum_{\gamma\in \Gamma}(\,{}^\gamma y_\gamma-y_\gamma\,) \ \big|\ y_\gamma\in M\bigg\}.$$ We write $M_\Gt:= (M_\Gamma)_\Tors$ (which is the torsion subgroup of $M_\Gamma$), $\ M_\Gtf=M_\Gamma/ M_\Gt\ $ (which is a torsion free group). We consider the functors: \begin{align*} &(\Gamma,M)\,\rightsquigarrow\, M_\Gt\,,\\ &(\Gamma,M)\,\rightsquigarrow\, M_\Gtf\otimes(\Q/\Z)=M_\Gamma\otimes (\Q/\Z). \end{align*}

Theorem. Let $\Gamma$ be a finite group, and let $$0\to M_1\xrightarrow{i} M_2\xrightarrow{j} M_3\to 0$$ be a short exact sequence of $\Gamma$-modules. Then there exists a natural homomorphism $$\delta\colon (M_3)_\Gt\to (M_1)_\Gamma\otimes (\Q/\Z)$$ such that the following sequence is exact: \begin{multline*} (M_1)_\Gt \xrightarrow{i_*} (M_2)_\Gt \xrightarrow{j_*} (M_3)_\Gt\xrightarrow{\delta}\\ (M_1)_\Gamma\otimes(\Q/\Z)\xrightarrow{i_*} (M_2)_\Gamma\otimes(\Q/\Z)\xrightarrow{j_*} (M_3)_\Gamma\otimes(\Q/\Z)\to 0. \end{multline*}

Question. Is this exact sequence known? If not, does this follow easily from some more general known exact sequence?

I do have a proof, I am asking for a reference!

Special cases. $\DeclareMathOperator\Gal{Gal}$Let $\Gamma=\Gal(E/F)$ be the Galois group of a finite Galois extension $E/F$ of nonarchimedean local fields. Assume that our $\Gamma$-modules $M_i$ for $i=1,2,3$ are finitely generated and torsion-free, and for each $i=1,2,3$, let $T_i$ be the corresponding algebraic $F$-torus, which splits over $E$, with cocharacter group $M_i$. We have a short exact sequence of $F$-tori $$ 1\to T_1\to T_2\to T_3\to 1$$ and the corresponding Galois cohomology exact sequence $$ H^1(F,T_1)\to H^1(F,T_2)\to H^1(F,T_3)\xrightarrow{\delta} H^2(F,T_1)\to H^2(F,T_2)\to H^2(F,T_3)\to 1, $$ where $H^n(F,T_i):=H^n(\Gal(F_s/F), M_i\otimes F_s^\times)$ is the cohomology of the profinite group $\Gal(F_s/F)$. This is our sequence in this special case.

More generally, let again $\Gamma=\Gal(E/F)$ be the Galois group of a finite Galois extension $E/F$ of nonarchimedean local fields. Assume that our $\Gamma$-modules $M_i$ for $i=1,2,3$ are finitely generated, but now we do not assume that they are torsion-free. For each $i=1,2,3$, we choose a torsion-free resolution $$ 0\to M_i^{-1}\to M_i^0\to M_i\to 0.$$ We can choose these resolutions compatibly, so that we obtain a short exact sequence of complexes of torsion-free $\Gamma$-modules $$0\to (M_1^{-1}\to M_1^0)\to (M_2^{-1}\to M_2^0) \to (M_3^{-1}\to M_3^0)\to 0$$ and a short exact sequence of complexes of $F$-tori $$0\to (T_1^{-1}\to T_1^0)\to (T_2^{-1}\to T_2^0) \to (T_3^{-1}\to T_3^0)\to 0.$$ Write $T_i^\bullet$ for the complex of tori $(T_i^{-1}\to T_i^0)$. Then we know the hypercohomology groups $\HH^n(F,T_i^\bullet)$: $$ \HH^1(F,T_i^\bullet)\cong(M_i)_\Gt,\quad\ \HH^2(F,T_i^\bullet)\cong(M_i)_\Gamma\otimes(\Q/\Z);$$ see M. Borovoi, Abelian Galois cohomology of reductive groups, Memoirs AMS 132(626), 1998, Proposition 4.1. We have a hypercohomology exact sequence \begin{multline*} \HH^1(F,T_1^\bullet)\to \HH^1(F,T_2^\bullet)\to \HH^1(F,T_3^\bullet)\xrightarrow{\delta} \HH^2(F,T_1^\bullet)\to \HH^2(F,T_2^\bullet)\to \HH^2(F,T_3^\bullet)\to 1. \end{multline*} This is our sequence in this special case.

  • 4
    $\begingroup$ I'm not sure if this is a kind of answer that you're looking for or if I'm repeating the proof you already have, possibly in a different language. We can get this exact sequence from the spectral sequence corresponding to the derived functor of tensor product with $\mathbb{Q}/\mathbb{Z}$ applied to the long exact sequence of $\Gamma$-homology corresponding to the given SES. On the one hand this long exact sequence is an acyclic complex so the spectral sequence must converge to zero. $\endgroup$
    – SashaP
    Sep 4, 2022 at 16:26
  • 5
    $\begingroup$ On the other hand, the six groups appearing in the desired sequence appear on the 1st page of this spectral sequence and, since both $Tor_0(H_1(\Gamma,M_3),\mathbb{Q}/\mathbb{Z})$ and $Tor_0(H_1(\Gamma,M_2),\mathbb{Q}/\mathbb{Z})$ are zero (because these homology groups are annihilated by the order of $\Gamma$), no non-zero differentials on page 2 and onward can touch these 6 groups, which gives the desired sequence. $\endgroup$
    – SashaP
    Sep 4, 2022 at 16:26
  • 2
    $\begingroup$ @SashaP: Yes, this is exactly the kind of answer I am looking for. Please kindly post an answer with as much details as possible! $\endgroup$ Sep 4, 2022 at 16:56
  • 1
    $\begingroup$ @SashaP: The answer that I have is by direct checks on cocycles (four pages of calculations). The preprint will appear in arXiv on Wednesday September 7, and I have enough time to replace my direct checks by your nice proof. $\endgroup$ Sep 4, 2022 at 17:00
  • $\begingroup$ @MikhailBorovoi Perhaps the "direct check" is reproducing the proof of the Snake Lemma? $\endgroup$
    – Pedro
    Jan 9 at 12:21

2 Answers 2


$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me know if anything looks dubious.

Consider the long exact sequence $$\ldots\to H_1(\Gamma, M_2)\to H_1(\Gamma,M_3)\to M_{1,\Gamma}\to M_{2,\Gamma}\to M_{3,\Gamma}\to 0$$ which we will view as an (acyclic) complex and denote its terms by $C^n, n\leq 0$, for brevity, so that $C^0=M_{3,\Gamma},C^{-1}=M_{2,\Gamma}$ etc. Consider the spectral sequence associated with the derived functor of $-\otimes_{\bZ}\bQ/\bZ$ and the 'bête' filtration on the complex $\ldots C^1\to C^0$. I'll use the cohomological grading conventions so that the first page of the spectral sequence looks like $E_{1}^{i,j}=\Tor_{-j}^{\bZ}(C^{-i},\bQ/\bZ)$. Note that for an abelian group $M$ we have $\Tor^{\bZ}_0(M,\bQ/\bZ)=M\otimes\bQ/\bZ,\Tor^{\bZ}_1(M,\bQ/\bZ)=M_{\mathrm{Tors}}$, and $\Tor_{>1}(M,\bQ/\bZ)=0$, hence the $1$st page looks like this, having only $2$ potentially non-zero rows:

$$\begin{matrix}\dots & \color{red}{H_1(\Gamma,M_2)\otimes\bQ/\bZ} & \color{red}{H_1(\Gamma,M_3)\otimes\bQ/\bZ} & M_{1,\Gamma}\otimes\bQ/\bZ & M_{2,\Gamma}\otimes\bQ/\bZ & M_{3,\Gamma}\otimes\bQ/\bZ & \\ \dots & H_1(\Gamma,M_2)_{\mathrm{Tors}} & H_1(\Gamma,M_3)_{\Tors} & M_{1,\Gamma,\Tors} & M_{2,\Gamma,\Tors} & M_{3,\Gamma,\Tors}\end{matrix}$$

The differentials $d_{i,j}:E_1^{i,j}\to E^{i+1,j}_1$ are simply the maps induced by the maps in the above exact sequence. Since we started with an acyclic complex, the spectral sequence must converge to zero.

Now, since $H_i(\Gamma, M)$ is annihilated by $|\Gamma|$ for all $\Gamma$-modules $M$, the two entries that are highlighted in red are in fact zero (and all the hidden entries in the top row are zero likewise). This implies that the only possibly non-trivial differential on the second page is $$E_{2}^{-2,0}=\ker (E_1^{-2,0}\to E_1^{-1,0})=\ker (M_{1,\Gamma}\otimes\bQ/\bZ \xrightarrow{i_*} M_{2,\Gamma}\otimes\bQ/\bZ )\to E_2^{0,-1}=\mathrm{coker} (M_{2,\Gamma,\Tors} \xrightarrow{j_*} M_{3,\Gamma,\Tors})$$ This differential must be an isomorphism, while all other $E_2^{i,j}$ must already be zero as otherwise something would survive to the abutment $E_3^{i,j}=E_{\infty}^{i,j}$ of the spectral sequence. But this is exactly saying that the $6$-term sequence in question is exact, with $\delta$ being defined as the inverse to the differential $E_2^{-2,0}\to E_2^{0,-1}$.

  • 1
    $\begingroup$ Many thanks! It seems to be clear. Now I must learn the terms, starting with ${\rm Tor}^{\Bbb Z}_{-j}$. OK, I have good books and someone to ask questions. $\endgroup$ Sep 4, 2022 at 18:35

$ \newcommand{\G}{\Gamma} \newcommand{\rsa}{\rightsquigarrow} \newcommand{\Z}{{\mathbb Z}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\Lam}{\Lambda} \newcommand{\Tor}{{\rm Tor}} \newcommand{\Gt}{{\Gamma, {\rm Tors}}} \newcommand{\lra}{\longrightarrow} \newcommand{\li}{{\overset{i_*}{\lra}}} \newcommand{\lj}{{\overset{j_*}{\lra}}} \newcommand{\isoto}{{\overset\sim\lra}} $We prove more:

Theorem 1. A finite group $\G$ and a short exact sequence of $\G$-modules \begin{equation}\label{e:1-2-3}\tag{1} 0\to B_1\overset i\lra B_2\overset j\lra B_3\to 0 \end{equation} give rise to a long exact sequence \begin{multline}\label{e:1-2-3-long}\tag{2} \to H_1(\G,B_1)\li H_1(\G,B_2) \lj H_1(\G,B_3)\overset{\delta_1}\lra\\ (B_1)_\Gt \li (B_2)_\Gt \lj (B_3)_\Gt\overset{\delta_0}\lra\\ \Q/\Z\otimes_\Z (B_1)_\G \li \Q/\Z\otimes_\Z(B_2)_\G \lj \Q/\Z\otimes_\Z(B_3)_\G\to 0 \end{multline} depending functorially on $\G$ and on the sequence \eqref{e:1-2-3}.

Proof of Theorem 1 due to Vladimir Hinich (private communication).

The functor from the category $\G$-modules to the category of abelian groups $$B\rsa \Q/\Z\otimes_\Z B_\G$$ is the same as $$B\rsa\Q/\Z\otimes_\Lam\! B$$ where $\Lam=\Z[\G]$ is the group ring of $\G$. From the short exact sequence of $\G$-modules \eqref{e:1-2-3}, we obtain a long exact sequence \begin{multline*} \to\Tor_2^\Lam(\Q/\Z,B_1) \li \Tor_2^\Lam(\Q/\Z,B_2) \lj \Tor_2^\Lam(\Q/\Z,B_3)\overset{\delta_1}\lra\\ \Tor_1^\Lam(\Q/\Z,B_1) \li \Tor_1^\Lam(\Q/\Z,B_2) \lj \Tor_1^\Lam(\Q/\Z,B_3)\overset{\delta_0}\lra\\ \Q/\Z\otimes_\Lam\! B_1\li \Q/\Z\otimes_\Lam\! B_2\lj \Q/\Z\otimes_\Lam\! B_3\to 0 \end{multline*} depending functorially on $\G$ and on the short exact sequence \eqref{e:1-2-3}. Now Theorem 1 follows from the next proposition.

Proposition 2. For a finite group $\G$, write $\Lam=\Z[\G]$. Then for any $\G$-module $B$ there are canonical and functorial isomorphisms $$\Tor_1^\Lam (\Q/\Z,B)\isoto B_\Gt\,,\quad\ \Tor_m^\Lam(\Q/\Z, B)\isoto H_{m-1}(\G,B)\,\text{ for }\,m\ge 2.$$

Proof. Consider the short exact sequence $$0\to \Z\to \Q\to\Q/\Z\to 0$$ regarded as a short exact sequence of $\G$-modules with trivial action of $\G$. Tensoring with $B$, we obtain a long exact sequence \begin{multline} \label{e:long-tensor}\tag{3} \dots\to\Tor_2^\Lam(\Z,B)\to\Tor_2^\Lam(\Q,B)\to \Tor_2^\Lam(\Q/\Z,B)\to\\ \Tor_1^\Lam(\Z,B)\to\Tor_1^\Lam(\Q,B)\to \Tor_1^\Lam(\Q/\Z,B)\to\\ \Z\otimes_\Lam\! B \to \Q\otimes_\Lam\! B \to \Q/\Z\otimes_\Lam\! B\to 0 \end{multline} where $\Tor^\Lam_m(\Z,B)\cong H_m(\G,B)$ for $m\ge 1$. We have canonical isomorphisms $$\Z\otimes_\Lam\! B=B_\G\quad\ \text{and}\quad\ \ker\big[\Z\otimes_\Lam\! B\to \Q\otimes_\Lam\! B\big]= B_\Gt\,.$$ By Lemma 3 below, we have $\Tor_m^\Lam(\Q,B)=0$ for $m\ge 1$, and the proposition follows from \eqref{e:long-tensor}.

Lemma 3. For a finite group $\G$ and any $\G$-module $B$, we have $\Tor_m^\Lam (\Q,B)=0$ for all $m\ge 1$.

Proof. Let $$P_\bullet:\quad\dots\to P_2\to P_1\to P_0\to\Z\to 0$$ be a $\Lam$-free resolution of the trivial $\G$-module $\Z$, for example, the standard complex; see Atiyah and Wall [1], Section 2. Tensoring with $\Q$ over $\Z$, we obtain a flat resolution of $\Q$ $$\dots\to \Q\otimes_\Z P_2\to \Q\otimes_\Z P_1\to \Q\otimes_\Z P_0\to\Q\to 0.$$ Tensoring with $B$ over $\Lam=\Z[\G]$, we obtain the complex $(\Q\otimes_\Z P_\bullet)\otimes_\Lam B\,$: \begin{equation}\label{e:QPB}\tag{4} \to (\Q\otimes_\Z P_2)\otimes_\Lam B\to (\Q\otimes_\Z P_1)\otimes_\Lam B \to (\Q\otimes_\Z P_0)\otimes_\Lam B\to\Q\otimes_\Lam B\to 0. \end{equation} By definition, $\Tor_m^\Lam(\Q,B)$ is the $m$-th homology group of this complex.

However, we can obtain the complex \eqref{e:QPB} from $P_\bullet$ by tensoring first with $B$ over $\Lam$, and after that with $\Q$ over $\Z$: $$\Q\otimes_\Z\big(P_\bullet\otimes _\Lam B\big)\,\cong\, (\Q\otimes_\Z P_\bullet)\otimes_\Lam B.$$ Since $\Q$ is a flat $\Z$-module, we obtain canonical isomorphisms $$\Tor_m^\Lam(\Q,B)\cong \Q\otimes_\Z \Tor_m^\Lam(\Z,B)=\Q\otimes_\Z H_m(\G,B).$$

Now, since the group $\G$ is finite, the abelian group $H_m(\G,B)$ is killed by multiplication by $\#\G$; see, for instance, Atiyah and Wall [1], Section 6, Corollary 1 of Proposition 8. It follows that $\Q\otimes_\Z H_m(\G,B)=0$. Thus $\Tor_m^\Lam (\Q,B)=0$, which completes the proofs of Lemma 3, Proposition 2, and Theorem 1.


[1] M.F. Atiyah and C.T.C. Wall, Cohomology of groups, in: Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), 94–115. Thompson, Washington, D.C., 1967.


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