Let $(V,r,N)$ be an irreducible Weil-Deligne representation. By irreducibility, $N=0$ automatically. Is it true that the Frobenius element acts semisimply?
This is not true if we only assume that the representation is indecomposable. The tate module of an elliptic curve that has semistable but not good reduction is such an example.
Definitions:
Let $F$ be a local field, $\Gamma_F$ it's Galois group and $I_F$ the Inertia subgroup. Recall that there is an exact sequence: $$0 \to I_F \to \Gamma_F\to \hat {\mathbb Z} \to 0.$$ The Weil group $W_F \subset \Gamma_F$ is the inverse image of $\mathbb Z\subset \hat{\mathbb Z}$. A Weil-Deligne representation is a representation $r$ of $W_F$ on a $\mathbb C$ vector space $V$ so that $r(I_F)$ is finite and some technical condition on $N$ that we ignore here because it turns out be zero in the irreducible case anyway.