In several of Langlands' papers: Representations of Abelian Algebraic Groups, On Artin's L-functions, On the Functional Equation of Artin's L-functions, Langlands takes a finite Galois extension $K/F$ of local fields and considers an exact sequence
$$1 \rightarrow K^{\ast} \rightarrow W_{K/F} \rightarrow \operatorname{Gal}(E/F) \rightarrow 1.$$
Here $W_{K/F}$ is the group $W_F/W_K^c$, where $W_K \subset W_F$ are the Weil groups of $K$ and $F$, and $W_K^c$ is the closure of the derived group of $W_K$. To define this exact sequence, one considers the reciprocity law isomorphism $K^{\ast} \rightarrow W_K^{\operatorname{ab}} = W_K/W_K^c$ and composes it with the map $W_K/W_K^c \rightarrow W_F/W_K^c$.
For example, here is from Langlands' short paper On Artin's L-functions.
My question is, what is the reciprocity law isomorphism $K^{\ast} \rightarrow W_K^{\operatorname{ab}}$ Langlands is using here? There are two: the classical one, which sends a uniformizer to an arithmetic Frobenius, and the "modern" one, which sends a uniformizer to a geometric (inverse) Frobenius.
I can't seem to find in these papers which convention Langlands is using. I originally assumed he was following the modern approach, but now I am not so sure. When he mentions "the Frobenius substitution" in his paper on abelian algebraic groups, I can't tell whether he is talking about the arithmetic Frobenius or the geometric one.
Does anyone who has read Langlands' early papers know which one he means?
