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 diagram? Does anyone know a reference for such a statement?

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?

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

Let $G$ be a group, $H$ a normal subgroup of $G$ and let $A$ and $B$ be $G$-modules. The Lyndon-Hochschild-Serre spectral sequences in cohomology and homology give transgression maps:

$$d^2 : H^1(H,A)^{G/H} \rightarrow H^2(G/H, A^H),$$

$$d_2 : H_2(G/H, B_H) \rightarrow H_1(H, B)_{G/H}.$$

Assume that $G$ acts trivially on $A$. Via the universal coefficients theorem, applied functorially to $d^2$, we get a commutative diagram

\begin{array}{cccc} H^1(H,A)^{G/H} &\xrightarrow{d^2}&H^2(G/H, A^H)\\ \downarrow & & \downarrow \\ \operatorname{Hom}(H_1(H;\mathbb{Z}),A)^{G/H} & \xrightarrow{F} & \operatorname{Hom}(H_2(G/H;\mathbb{Z}),A^H) \end{array}

Suppose that $A$ is torsion free and $B = A^\vee := \operatorname{Hom}(A, \mathbb{C}^\times)$.

• Is $F$ equal to $$d_2^{\vee} : \operatorname{Hom}(H_1(H,A^{\vee})_{G/H}, \mathbb{C}^{\times}) \rightarrow \operatorname{Hom}(H_2(G/H, (A^{\vee})_H), \mathbb{C}^{\times})?$$
• Does anyone know a reference for such a statement, and whether it generalizes to the case that $A$ is not torsion-free (used in identifying $H_1(H, A^\vee)$ with $H_1(H, \mathbb{Z}) \otimes A^\vee$)?

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 diagram? Does anyone know a reference for such a statement?