# Langlands paper on representations of abelian algebraic groups

I have been working through Langlands paper Representations of Abelian Algebraic Groups, and I can't understand why one of his maps is obvious and how it helps. First I'll give the notation

Take a algebraic torus over $F$ that splits over a Galois extension $K$ of $F$, then the torus corresponds to a lattice $L$ on which $G=Gal(K/F)$ acts. Then let $\widehat{L}=Hom(L,\mathbb{Z})$. Then we define $\widehat{T}=Hom(\widehat{L},\mathbb{C}^*)$, and we define the Weil group as the extension $$0 \longrightarrow C_K \longrightarrow W_{K/F} \longrightarrow Gal(K/F) \longrightarrow 0.$$ ($C_{K}$ is the idele class group) and finally let $N = \sum_{g \in G} g$

On page 13 of his paper hes shown that there is an isomorphism $$H^{1}(W_{K/F},\widehat{T}) \longrightarrow Hom(H_{1}(W_{K/F},\widehat{L}),\mathbb{C}^{*}).$$ And now he wants to show that that image of a continuous cocycle will be continuous. To do this he proceeds as follows. He defines $U_{K}$ as the elements on norm 1 in $C_K$ and constructs the exact sequence $$1 \longrightarrow U_K \longrightarrow C_K \longrightarrow M_K \longrightarrow 1,$$ $M_K$ being $\mathbb{Z}$ or $\mathbb{R}$. He then uses $L$ to make the sequence $$0 \longrightarrow Hom(L,U_K) \overset{\lambda}\longrightarrow Hom(L,C_K) \overset{\mu}\longrightarrow Hom(L,M_K) \longrightarrow 0$$. And now this is where I get lost, he claims there is an obvious map from $$N(Hom(L,C_K)) \cap Hom(L,U_K)/N(Hom(L,U_K))$$ to $$\hat{H}^{-1}(G,Hom(L,M_K))/ \mu \hat{H}^{-1}(G,Hom(L,C_K)).$$ Here the $\hat{H}$ means Tate groups.

Is this map obvious? and why is the second group finite? which he claims.

Thank you

• The topology on the $H_1$ must be defined for the motivating problem to make sense; there is content since cycles in $H_1$ are functions satisfying a "vanish almost everywhere" condition (unlike cocycles). We topologize $H_1(C_K,\widehat{L})$ as a finite power of $C_K$, and (with a fair amount of work!) on pp. 4-13 Langlands shows its Gal$(K/F)$-invariants are identified with $H_1(W,\widehat{L})$ via restriction. That defines the topology on the latter, as Langlands says back atop p.4 (!). For clarity in the MO question it would be good to remark upon it in some way (at least briefly). Commented Nov 24, 2012 at 14:22
• There are several expositions of Langlands paper in the literature, for example, Labesse, J.-P. Cohomologie, $L$-groupes et fonctorialité. (French) [Cohomology, $L$-groups and functoriality] Compositio Math. 55 (1985), no. 2, 163--184.
– anon
Commented Nov 24, 2012 at 19:31

## 1 Answer

The map is induced by a map $N(Hom(L, C_{K})) \cap Hom(L, U_{K})$ to $\hat{H}^{-1}(G, Hom(L, M_{K}))$, which is obtained as follows. If $z$ is in $Hom(L, U_{K})$ and $z = Nx$, where $x \in Hom(L, C_{K})$, where should we send $z$? Well, $N(\mu(x)) = 0$. Therefore $\mu(x)$ maps to something in $\hat{H}^{-1}(G, Hom(L, M_{K}))$ by definition of the latter group. And that element is where we want to send $z$.

For the finiteness question, try writing down the long exact sequence in Tate cohomology for the short exact sequence $0 \rightarrow U_{K} \rightarrow C_{K} \rightarrow M_{K} \rightarrow 0$.

• Thank you, this helps, I can see why its sort of obvious. As for the finiteness ill write it out and see if I can see it. Commented Nov 24, 2012 at 18:15
• Actually I'm still a little lost. Why is $N(\mu(x))=0$?. I'm guessing that $U_K$ are elements such that $|u|=1$ where $||$ is the product of all the valuations. But if so then wouldnt $\mu(x)=|| \circ x$? And if so I cant see why its norm would be zero. I feel I'm getting confused by the multiple meanings of norm. Commented Nov 25, 2012 at 0:56
• $N(\mu(x)) = \mu(N(x)) = 0$ because $N(x) = z$ is in $Hom(L, U_{K})$ and $0 \rightarrow Hom(L,U_{K}) \rightarrow Hom(L, C_{K}) \rightarrow Hom(L, M_{K}) \rightarrow 0$ is exact. Commented Nov 25, 2012 at 6:24
• Yes of course I'm so silly. Thank you Commented Nov 25, 2012 at 11:21