I have a reductive group $G$, sitting inside $GL\_n$, everything over some algebraically closed field of characteristic 0. Let $\mu$ be a cocharacter (of a maximal torus in $G$) such that the induced filtration on the vector space is a 1-step filtration, with the filtered piece having dimension $n/2$ (I'm assuming that $n$ is even). Let $U$ be the subgroup of $G$ which acts trivially on the associated graded, and let $L$ be the centraliser of the cocharacter (in $G$). Is the adjoint action of $L$ on $Lie(U)$ irreducible? 


And does the same result hold for any cocharacter (without the added conditions on the filtration)?



Levi action on the unipotent radical
The adjoint action of $L$ on ${Lie}(U)$ is very rarely irreducible, but this does happen in your case provided that the derived subgroup of $G$ is simple. If $Lie(L)$ intersects properly with at least two simple ideals of $Lie(G)$ then intersecting ${Lie}(U)$ with those ideals we'll obtain some non-trivial proper $L$-submodules of $Lie(U)$.


So assume from now that the derived subgroup of $G$ is simple. One knows that the centraliser $L$ of $\mu$ in $G$ is a Levi subgroup of the parabolic subgroup $P\_G(\mu)=L\cdot U$ of $G$. Now $G\subset GL\_{n}$ by our general assumption and, moreover, we know that the unipotent radical, $\widetilde{U}$, of the parabolic subgroup $P\_{GL\_n}(\mu)=Z\_{GL\_n}(\mu)\cdot \widetilde{U}$ of $GL\_n$ is abelian (this can be checked directly by looking at $n\times n$ matrices). As $U\subseteq \widetilde{U}$ we see that $U$ is abelian, too. But then it is well known that $L$ acts irreducibly on ${Lie}(U)$, although all proofs of this fact seem to rely on the classification of simple algebraic groups
(there is a related discussion of irreducibility on this website prompted by a question of Jim Humphreys).


