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Let $G$ be a locally profinite group, and $(\pi, V)$ be a representation of $G$ over a complex vector space $V$. Then the representation is called smooth if, for every $v$ in $V$, there is a compact open subgroup $K$ of $G$ (depending on $v$) such that $\pi(g)v=v$ for all $g$ in $K$.

Let $V^K = \{v \in V \;|\, \pi(g)v=v \quad \forall g \in K \}$ be the space of $\pi(K)$-fixed vectors. The smooth representation is called admissible if the space $V^K$ is finite dimensional for each compact open $K$ of $G$.

I believe the above are standard definitions. My confusion stems from a passage on pg 6 of the paper http://www.math.harvard.edu/~gross/preprints/AdjointGamma5.pdf :

A representation ρ : W → GL(V ) on a finite dimensional complex vector space V is called admissible if ker ρ is open and ρ(Fr) is semisimple.

Here $k$ is a local, nonarchimedean field of characteristic zero, and $W$ is the Weil group of $k$, so $W$ is locally profinite, and Fr is the geometric Frobenius.

Here is my question:

So now if I try to apply the earlier definitions of "admissible", I see that since $V$ is already finite dimensional, I need only show that the representation is smooth? Now it seems that having the kernel of $\rho$ be open is sufficient to get smoothness. So why is the semisimplicity of $\rho(Fr)$ necessary for smoothness and/or admissibility?

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1 Answer 1

up vote 3 down vote accepted

"Admissible" is used too often in mathematics, and it means roughyly "good", "relevant to our situation". Your first definition of admissibility is most of the time used when $G$ is the $K$-points of a connected, reductive group over $K$ a local non-archimedean field (even though the definition makes sense for a broader class of groups). These representations are almost always infinite-dimensional (that is why we introduce the notions of smooth and admissible, "abstract" infinite-dim. reps are too wild).

A representation of a Galois-type group (like $\rho$) is very different. From a naive point of view, these two kinds of reps are unrelated (of course from what you write I guess you are aware of the Langlands programme, whose goal is precisely to relate the two). So the second definition of "admissible" is not equivalent to the first one (and it is not intended to be, it's just poor terminology).

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Thanks. That's helpful. –  John M Sep 3 '11 at 23:47

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