# Why is the Langlands dual group always taken over $\mathbb{C}$?

Whenever I read a statement of the Langlands conjectures for a reductive group $$G$$, they are formulated in terms of the Langlands dual group, which is essentially the reductive group over $$\mathbb{C}$$ whose root datum is dual to that of $$G$$.

Instead of $$\mathbb{C}$$ we could use any separably closed field to obtain a "dual group”. Is there a compelling reason why the conjectures are not formulated in this generality?

• Many of the papers by Adler and Lansky, for example Lifting representations of finite reductive groups I, deal with dual groups over the 'original' field (which is (probably) not separably closed). Jul 15 '19 at 2:18
• Also, when dealing with the classification of representations of a finite reductive group the dual is taken over the same base field. See for example Digne and Michel's book. Jul 15 '19 at 11:05

The Satake isomorphism gives a relationship between the convolution algebra of $$F$$-valued functions on $$G (\mathcal O) \backslash G(K) / G(\mathcal O)$$ and the ring of conjugacy-invariant polynomial functions on the dual group over $$F$$. This is one example of where you might want the field of definition of your dual group to equal the field that your automorphic representations are defined over.
Outside the $$p$$-adic Langlands programs, and some steps when you are proving cases of the Langlands correspondence by $$\ell$$-adic cohomology, it is usually most convenient to have the automorphic representations with coefficient field $$\mathbb C$$ (so you can do analysis).
So then it's natural to have the Langlands dual group defined over $$\mathbb C$$ as well.
• Certainly what you say is true, but a naive person could also wonder why, for example, $p$-adic principal series are repns of $p$-adic groups on complex vector spaces... which might be construed as reducing the imperative to have "the Langlands dual" be a complex group despite acting on complex vector spaces, etc. (Not that I have any much better rationale, myself...!) Jul 15 '19 at 0:43
• @paulgarrett Well in local Langlands, it may be most natural to compare representations of $p$-adic groups into $\ell$-adic vector spaces and $\ell$-adic representations of the Galois groups of $p$-adic fields. If one agrees that Galois representations of this type are interesting, it's natural for there to be one $\ell$ and one $p$ on the others side. Then because the $\ell$-adic topology isn't relevant on the representation side, we can replace it with another algebraically closed fields. Then one has the $p$-adic Langlands program also, of course. Jul 15 '19 at 1:01