Suppose $(\pi, V)$ is a continuous representation of a real reductive group $G$, then the Lie algebra of $G$ act on the smooth vectors of $V$, and the $(g,K)$-modules are built from here.

It seems that we don't discuss the Lie algebra action on the representation space in p-adic group case. Is there any simple reason for that missing analog?


If one is doing smooth representations of the $p$-adic group $G$ on complex vector spaces, then the action is locally constant (small open subgroups act trivially on vectors), and so any interpretation of what it means to differentiate the action will give you the trivial action. (Note also the the Lie algebra has $p$-adic coefficients, while the representation $V$ has complex coefficients, so there is not much scope for a non-trivial action!).

On the other hand, if you mean not-necessarily-smooth actions of $G$ on a $p$-adic vector space, as one has in the theory developed by Schneider and Teitelbaum that is related to the $p$-adic Langlands program, then one can and does differentiate the action (provided it is locally analytic), and one does obtain interesting Lie algebra actions.

  • $\begingroup$ Thanks Matt. Is there a relation between smooth representations over complex vector spaces and those over p-adic vector spaces? $\endgroup$ – user1832 Apr 8 '10 at 0:25
  • $\begingroup$ Yes. Firstly, one can always choose an isomorphism $\mathbb C \cong \overline{\mathbb Q}_p$ and so turn a smooth representation over $\mathbb C$ into one over (some extension of) the $p$-adics; this is the analog in representation theory of regarding a complex Hecke eigenform as a $p$-adic Hecke eigenform, which one often does in arithmetic contexts. These smooth $p$-adic reps. then sit in the larger world of locally analytic $p$-adic reps. To get a feeling for this world, and why one cares about it, you could look at some of the recent expository literature on $p$-adic Langlands. $\endgroup$ – Emerton Apr 8 '10 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.