TITLE: Why is the Langlands dual group always taken over $\mathbb{C}$?
QUESTION [9 upvotes]: 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?

REPLY [8 votes]: 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.