I am interested in whether the transgression maps for group cohomology and group homology are related via a version of the universal coefficient theorem.

Let $G$ be a group, $H$ a normal subgroup of $G$ and let $A$ and $B$ finite rank free $\mathbb{Z}$-modules equipped with actions of $G$ and a $G$-equivariant perfect pairing

$$A \times B \rightarrow \mathbb{Z}.$$

The Lyndon-Hochschild-Serre spectral sequences in cohomology and homology give transgression maps:

$$d^2 : \operatorname{H}^1(H,B \otimes \mathbb{C}^{\times})^{G/H} \rightarrow \operatorname{H}^2(G/H, (B \otimes \mathbb{C}^{\times})^H),$$

$$d_2 : \operatorname{H}_2(G/H, A_H) \rightarrow \operatorname{H}_1(H, A)_{G/H}.$$

Assume that $H$ acts trivially on $A$ and $B$, and suppose that the order of $G/H$ is $m < \infty$.

Since $H$ acts trivially on $B$, the universal coefficients theorem gives a map

$$\operatorname{H}^1(H,B \otimes \mathbb{C}^{\times}) \rightarrow \operatorname{Hom}(\operatorname{H}_1(H, \mathbb{Z}), B \otimes \mathbb{C}^{\times}).$$

Since $A$ is torsion-free, $B \otimes \mathbb{C}^{\times} = \operatorname{Hom}(A, \mathbb{C}^{\times})$, so $$\operatorname{Hom}(\operatorname{H}_1(H,\mathbb{Z}),B \otimes \mathbb{C}^{\times})^{G/H} = \operatorname{Hom}(\operatorname{H}_1(H,\mathbb{Z}) \otimes A, \mathbb{C}^{\times})^{G/H} $$ $$= \operatorname{Hom}(\operatorname{H}_1(H, A), \mathbb{C}^{\times})^{G/H} = \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H}, \mathbb{C}^{\times}).$$

We thus get an isomorphism between the domain of $d^2$ and dual of the codomain of $d_2$. We can't use the universal coefficient theorem for the other pair since $G/H$ may not act trivially, but we can replace it as follows. Tate-Nakayama duality says that cup product induces a perfect pairing $$\hat{\operatorname{H}}^n(G/H, A) \times \hat{\operatorname{H}}^{-n}(G/H, B) \rightarrow \mathbb{Z} / m \mathbb{Z},$$ where $\hat{\operatorname{H}}$ denotes Tate cohomology.

Via the exponential sequence $$0 \rightarrow B \rightarrow B \otimes \mathbb{C} \rightarrow B \otimes \mathbb{C}^{\times} \rightarrow 0$$ one obtains an isomorphism

$$\operatorname{H}^2(G/H, (B \otimes \mathbb{C}^{\times})^H) = \operatorname{H}^2(G/H, B \otimes \mathbb{C}^{\times}) = \operatorname{H}^3(G/H, B).$$

Following this isomorphism by Tate-Nakayama duality, we finally get a partial diagram

\begin{array}{cccc} \operatorname{H}^1(H,B \otimes \mathbb{C}^{\times})^{G/H} &\xrightarrow{d^2}&\operatorname{H}^2(G/H, B \otimes \mathbb{C}^{\times}) \\ \downarrow & & \downarrow \\ \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H},\mathbb{C}^{\times}) & & \operatorname{Hom}(\operatorname{H}_2(G/H,A),\mathbb{C}^{\times}) \end{array}

Does adding $$d_2^{\vee} : \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H}, \mathbb{C}^{\times}) \rightarrow \operatorname{Hom}(\operatorname{H}_2(G/H, A_H), \mathbb{C}^{\times})$$ make the diagram commute? Does anyone know a reference for such a statement?

  • 1
    $\begingroup$ I don't really see how to define the vertical maps in your diagram, unless you assume that $A$ is trivial as a $G$-module? $\endgroup$
    – Mark Grant
    Feb 11 '13 at 16:50
  • $\begingroup$ Dear Mark, thank you for your comment, you are absolutely right. I have edited the question to include the assumption that $G$ acts trivially on $A$. $\endgroup$ Feb 11 '13 at 17:12
  • 1
    $\begingroup$ The codomain of $F$ is (trivial coefficients) $\operatorname{Hom}(H_2(G/H;\mathbb{Z}),A)$ while the codomain of $d_2^{\vee}$ is $\operatorname{Hom}(H_2(G/H,A^\vee), \mathbb{C}^{\times})$. So when you ask if $F=d_2^\vee$, does this mean $A=\mathbb{C}^{\times}$ and $A^\vee = \mathbb{Z}$ ? $\endgroup$
    – Ralph
    Feb 11 '13 at 22:51
  • $\begingroup$ Dear Ralph : note that $Hom(H_2(G/H,A^{\vee}), \mathbb{C}^{\times}) = Hom(H_2(G/H,\mathbb{Z}) \otimes A^{\vee}, \mathbb{C}^{\times})$, which by adjointness of $Hom$ and $\otimes$, equals $Hom(H_2(G/H, \mathbb{Z}), Hom(A^{\vee}, \mathbb{C}^{\times})) = Hom(H_2(G/H, \mathbb{Z}), A)$. This is all because $A$ is assumed to be torsion-free, implying that $H_2(G/H, A^{\vee}) = H_2(G/H, \mathbb{Z}) \otimes A^{\vee}$ (similar for $H_1$). $\endgroup$ Feb 12 '13 at 0:04
  • $\begingroup$ It seems that Ralph's answer to mathoverflow.net/questions/94424/… should be useful in obtaining a general statement. In fact, all that is missing is the precise relationship between the homology transgression $H_2(G/H;A)\to H_1(H;A)_{G/H}$ and the class in $H^2(G/H;H_1(H))$ which classifies the extension $H_1(H)\to G/H'\to G/H$. $\endgroup$
    – Mark Grant
    Feb 12 '13 at 7:19

We have the following general result:

Lemma: Let $K$ be a filtered complex of ablian groups and let $I$ be an ablian group. Filter the cocomplex $K^\ast := \text{Hom}(K,I)$ by the dual filtration (def. in proof). Then there is a homomorphism $\phi_r: E_r(K^\ast) \to E^r(K)^\ast$ of abelian groups such that the following diagram commutes: $$\begin{array}{ccc} E_r^{ij} & \xrightarrow{d_r} & E_r^{i+r,j-r+1} \newline {\scriptstyle\phi_r}\downarrow\;\; & & \;\;\downarrow{\scriptstyle\phi_r} \newline (E^r_{ij})^\ast & \xrightarrow[(d^r)^\ast]{} & (E^r_{i+r,j-r+1})^\ast \end{array}$$ If $I$ is injective, then $(\phi_r)_{r\ge 0}$ is an isomorphism of spectral sequences.

As a corollary we obtain:

Let $A$ be a $G$-modules, $I$ an injective abelian group (with trivial $G$-action), and denote the homology LHS spectral sequence with coefficients in $A$ by $E^r(A)$ (and respectively for cohomology). Then there is an isomorphism $\phi_r: E_r(A^\ast) \to E^r(A)^\ast$ of spectral sequences.

Proof: Let $P$ resp. $Q$ be projective resolutions of $\mathbb{Z}$ over $G$ resp. $G/H$ and put $X=Q \otimes P$. Then the cohomology LHS spectral sequence is the spectral sequence of of the filtration of the cocomplex $$\text{Hom}_G(X,A^\ast)=\text{Hom}_G(X,\text{Hom}(A,I))\cong \text{Hom}_G(X\otimes A,I) \cong \text{Hom}(X\otimes_GA,I)$$ (the last iso needs $I$ to be trivial) and the homology LHS spectral sequence is the spectral sequence of the filtration of $X\otimes_G A$. Now the results follows from the lemma. qed.

Remark: If $I$ is injective, then the lemma can be proved easier by using exact couples (in the same manner as in Tyler Lawson's answer to the question linked by Mark Grant). But if there are additional structures in the spectral sequences like products, Steenrod power operations, etc. that one wants to compare, then it's helpful to have an explicit map $\phi_r$ as defined below. For example, this way one can show that $\phi_2^{2,0}: H^2(G/H,(A^\ast)^H)\to H_2(G/H,A_H)^\ast$ is induced by $$\text{Hom}_{G/H}(Q_2,(A^\ast)^H) \to (Q_2 \otimes_{G/H}A_H)^\ast,\; f \mapsto (x \otimes \bar{a} \mapsto f(x)(a)).$$

Proof of the lemma: Let $C := K^\ast$ with differential $\delta = d^\ast$. The dual filtration is defined by $F^rC^i=\lbrace f \in \text{Hom}(K_i,I)\mid f|F_{r-1}K_i=0\rbrace$. As usual set $$Z^r_{ij}=\lbrace x \in F_iK_{i+j}\mid dx \in F_{i-r}K_{i+j-1}\rbrace\;,\quad B^r_{ij}=dZ_{i+r-1,j-r+2}^{r-1}+Z^{r-1}_{i-1,j+1}$$

$$Z_r^{ij}=\lbrace f \in F^iC^{i+j}\mid \delta f \in F^{i+r}C^{i+j+1}\;,\quad B_r^{ij}=\delta Z_{r-1}^{i-r+1,j+r-2}+Z^{i+1,j-1}_{r-1}$$ It's straightforward to show that $Z_r^{ij} \to \text{Hom}(Z^r_{ij},I),\;f \mapsto f|Z^r_{ij}$ induces a hom. $$\phi_r: E_r^{ij}=Z_r^{ij}/B_r^{ij} \to \text{Hom}(Z^r_{ij}/B^r_{ij},I)=(E^r_{ij})^\ast$$ that makes the diagram in the lemma commute.

Now suppose $I$ is injective. We want to show that $\phi_r$ is bijective. In case $r=0$, $E^0_{ij}=F_iK_{i+j}/F_{i-1}K_{i+j}, E_0^{ij}=F^iC^{i+j}/F^{i+1}C^{i+j}$ and bijectivity is easy to establish. Assume $\phi_r$ is bijective. Then, by the diagram above, $H(\phi_r)$ is bijective and the bijectivity of $\phi_{r+1}$ follows from the commutativity of the diagram below:

$$\begin{array}{ccc} E_{r+1}^{ij} & \xrightarrow{} & \xrightarrow[\alpha]{\sim} & H(E_r^{ij}) \newline {\scriptstyle\phi_{r+1}}\downarrow\;\; & & & {\scriptstyle\cong}\downarrow {\scriptstyle H(\phi_r)} \newline (E_{ij}^{r+1})^\ast & \xleftarrow[\beta^\ast]{\sim} H(E^r_{ij})^\ast & \xleftarrow[\gamma]{\sim} & H( (E^r_{ij})^\ast) \end{array}$$ Here $\alpha$ is induced by $F_{r+1}^{ij}\hookrightarrow F_r^{ij}$, $\beta$ by $F_{ij}^{r+1}\hookrightarrow F^r_{ij}$ and $\gamma$ (which reflects the fact that the ,exact functor $\text{Hom}(-,I)$ commutes with homology) is given by $\bar{f} +\text{im}\ d_r \mapsto (\bar{x} + \text{im}\; d^r \mapsto f(x)\;)$. qed.


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.