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While reading the book 'Langlands correspondence for loop groups', I came across the definition of the Weil group $W_F$ and the Weil-Deligne group $W'_F = W_F \ltimes \mathbb{C}$ with action $$\sigma x\sigma^-1 = ||\sigma||x, \sigma \in W_F,x\in \mathbb{C}.$$ In it, they give the definition of an $n$-dimensional complex representation of $W'_F$:

"An $n$-dimensional complex representation of $W'_F$ is by definition a homomorphism $\rho': W'_F \rightarrow GL_n(\mathbb{C})$, which may be described as a pair $(\rho,u)$, where $\rho$ is an $n$-dimensional representation of $W_F$, $u \in \mathfrak{gl}_n (\mathbb{C})$, and we have $\rho(\sigma)u\rho(\sigma) = ||\sigma||u$."

Now, I understand what $\rho$ means and why you define that action of $\rho(\sigma)$ on $\rho'(\mathbb{C})$, but I don't understand the meaning of $u \in \mathfrak{gl}_n (\mathbb{C})$, shouldn't that be $u \in GL_n(\mathbb{C})$ ?

For a reference of the book:, page 4.

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up vote 4 down vote accepted

In this formulation (which is completely standard), the image of $1\in \mathbb{C}$ under $\rho'$ is not, as you might expect, $u$, but rather $\text{exp}(u)$, which is indeed in $GL_n(\mathbb{C})$. $u$ itself is nilpotent, and is usually called the monodromy operator attached to the Weil-Deligne representation $\rho'$. You should check that the claimed relation $\rho(\sigma)u\rho(\sigma)^{-1}=||\sigma||u$ then holds (you forgot the inverse).

A very nice reference for this is Tate's article "Number theoretic background" in "Automorphic Forms, Representations, and L-functions", Proceedings of Symposia in Pure Mathematics Volume 33, Part 2.

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Thank you very much, but does $u$ have to be nilpotent ? In the book I mentioned, they make a distinction, with nilpotent (and $\rho$ continous) only necessary for admissible representations. – KevinDL Oct 17 '11 at 12:38
@KevinDL I guess, conventions might vary there slightly. I have only heard the word "admissible" being used on the reductive side, but it does make sense to use it here, since the Frobenius semi-simple Weil-Deligne representations with nilpotent $u$ conjecturally correspond to admissible irreducible representations of $GL_n$. Tate on the other hand just defines a Weil-Deligne representation to come with a nilpotent monodromy operator. – Alex B. Oct 17 '11 at 13:19
Ok, thank you very much for your answer. – KevinDL Oct 17 '11 at 13:23

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