$\newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}}$ In class field theory, one typically identifies many groups such as $$\begin{align*} &H^{-2}(G, \Z) \cong H_1(G, \Z) \cong H_0(G,I_G) \cong I_G/I_G^2 \cong G^{ab}\\ &H^1(G, \Q/\Z) \cong \text{Hom}(G, \Q/\Z)\\ &H^{-1}(G, \Q/\Z) \cong \frac{1}{n}\Z/\Z \text{ where } n = |G| \end{align*}$$ (All cohomology groups are Tate cohomology. $I_G$ is the augmentation ideal.)
My problem is how does one know how to manipulate them. For instance, the cup product $$H^{-2}(G, \Z) \otimes H^1(H, \Q/\Z) \rightarrow H^{-1}(G, \Q/\Z)$$ naturally induces $$G^{ab} \otimes \text{Hom}(G, \Q/\Z) \rightarrow \frac{1}{n}\Z/\Z$$ via the above identification.
The most reasonable/natural expectation of the latter map should be to send $$s, \chi \mapsto \chi(s)$$ and I know this is true by establishing all inverses $$\begin{align*} G^{ab} &\rightarrow H^{-2}(G, \Z)\\ \text{Hom}(G, \Q/\Z) &\rightarrow H^1(G, \Q/\Z) \end{align*}$$ in term of the cochain complex for Tate cohomology and makes use of the explicit definition of cup product in Cassels and Frohlich. [I might get the first map wrong. Tracing the chain of isomorphism backward, it should be $$\begin{align*} g \in G^{ab} &\mapsto \text{class of } g - 1 \text{ in } I_G/I_G^2\\ &\mapsto \text{class of } 1 \otimes (g - 1) \in \Z[G] \otimes_G I_G \text{ in } H_0(G, I_G)\\ &\mapsto \text{class of } (g, 1) \otimes 1 \in \Z[G^2] \otimes_G \Z \text{ in } H_1(G, \Z) \\ & \qquad \text{ via explicit computation of the connecting hom}\\ &\mapsto \text{class of } \Z[G^2]^* \rightarrow \Z; f \mapsto f(g,1) \text{ in } H^{-2}(G, \Z) \end{align*}$$ while the second one is much easier $$\chi \in \text{Hom}(G, \Q/\Z) \mapsto [\text{class of cochain } \Z[G^2] \rightarrow \Q/\Z : (g_0, g_1) \mapsto \chi(g_0^{-1} g_1)].$$ If possible, please help me check these as well. Some instances of 1 are integers while other should be the identity $1_G$ of $G$.]
I have seen other instances such as $$\chi(\phi_{L/K}(\alpha)) = \text{inv}_K(\alpha \cup \delta\chi)$$ (Cassels and Frohlich, page 140, Prop. 1) but the given proof in no way illustrates how one knows all those correspondences. For instance, it is stated that
Moreover, the identification between $H^{-2}(G, \Z)$ and $G^{ab}$ has been so made in order to ensure that $s \cup \chi = \chi(s)$.
So my questions are:
- How does one make sense of what to do when dealing with these implicit identification between cohomology groups? In particular, is there a way to know that the induced map $G^{ab} \otimes \text{Hom}(G, \Q/\Z) \rightarrow \frac{1}{n}\Z/\Z$ that I mentioned above must be $s \cup \chi = \chi(s)$ without doing all those explicit computation?
- Is there an intuitive explanation of the cup product? In particular, how does one expect to prove 2-periodicity using cup product in the first place; given its complexity?