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So rank $1$ local Langlands is special in as that it is given by the Artin map $$\text{GL}_1(K)\to G_K^{ab},$$ whereas in the higher rank (to the best of my knowledge) there doesn't exist a map $$\text{GL}_n(K)\to G_K$$ which realizes the rank $n$ local Langlands. In fact, I think I've read that there exists a good reason that no such map can exist. I can't find a source claiming this though, so here I am.

Is there a reason that no "higher" Artin map can exist?

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  • $\begingroup$ Even for $n=1$, the Artin map is not a map from $\text{GL}_1(K)$ to $G_K$, the abelianization is important. $\endgroup$
    – Antoine Labelle
    Commented Aug 23, 2022 at 16:51
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    $\begingroup$ @AntoineLabelle Can one prove that the Artin map can't be lifted from $G_K^{\mathrm{ab}}$ to $G_K$? Clearly such a lift won't be canonical. $\endgroup$
    – David Loeffler
    Commented Aug 24, 2022 at 5:16

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There is no way of reformulating local Langlands for $n > 1$ in terms of such a map.

Local Langlands is a bijection between irreducible smooth representations of $\operatorname{GL}_n(K)$, and $n$-dimensional Frobenius-semisimple Weil–Deligne representations. However, if $n > 1$ then an irreducible smooth $\operatorname{GL}_n(K)$-rep will almost always be infinite-dimensional, so there's no way of getting it by composing a Galois representation with a map from $\operatorname{GL}_n(K) \to G_K$. (Moreover, the "irreducibility" constraint also trips you up: there are lots of $n$-dimensional Weil–Deligne reps which are not irreducible, but they still correspond to something irreducible on the $\mathrm{GL}_n$ side, so they couldn't be given by composing a reducible representation with a map.)

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