A map from the coinvariants of the dual to the dual of the invariants for a G-module

Question

Suppose $G$ is a group and $X$ is a $\mathbb{Z}[G]$-module. Recall that the augmentation ideal $I \subset \mathbb{Z}[G]$ is generated by elements of the form $g - 1$ for $g \in G$, the coinvariants are defined as $X_G := X / IX$, and the invariants are given by $X^G := \{x \in X : gx = x, \forall g \in G\}$.

If $G$ is finite then the trace map $X_G \to X^G$ given by $x \mapsto \sum_{g\in G} gx$ is used in Tate cohomology, but this map doesn't make sense in the case that $G$ is infinite.

Suppose instead that we take a $\mathbb{Z}$-module $A$ and set $X^* := \operatorname{Hom}_{\mathbb{Z}}(X, A)$. Then there is a well-defined map \begin{align*} \rho:(X^*)_G &\to (X^G)^* \\ [f] &\mapsto f|_{X^G} \end{align*}

even when $G$ is infinite. Is there a nice description of the kernel and cokernel of $\rho$?

Answer 1

I will answer the question in the case that $G$ is cyclic or pro-cyclic. Write $t$ for a (topological) generator of $G$. Then we can describe $X^G$ and $X_G$ as the kernel and cokernel of $t-1$: $$ 0 \to X^G \to X \xrightarrow{t-1} X \to X_G \to 0. $$ Write $Y = (t-1)X$ for the augmentation ideal. We can split into two short exact sequences \begin{align*} &0 \to X^G \to X \to Y \to 0 \\ &0 \to Y \to X \to X_G \to 0. \end{align*} Note that the map $X \to Y$ is given by the action of $t-1$, while the map $Y \to X$ is just the inclusion. Dualizing, we get sequences \begin{align*} &0 \to Y^* \xrightarrow{t-1} X^* \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)\\ &0 \to (X_G)^* \to X^* \longrightarrow Y^* \to \operatorname{Ext}^1_{\mathbb{Z}}(X_G, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A). \end{align*} Since the cokernel of $Y^* \xrightarrow{t-1} X^*$ is $(X^*)_G$ by definition, we get a sequence $$0 \to (X^*)_G \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A).$$ The first map in this sequence is $\rho$, so $\rho$ is injective in this case, with cokernel equal to the kernel of $\operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)$.

If $A$ is divisible as an abelian group ($A = \mathbb{C}^\times$ for example), then $\operatorname{Ext}^1_{\mathbb{Z}}(-, A)$ vanishes and $\rho$ is an isomorphism. In this case we also get an isomorphism $(X_G)^* \to (X^*)^G$.