Fix a partition $\lambda$ of $n$ of length $l$. Consider a Levi subgroup $L_{\lambda}$ in $GL_{n}(\mathbb{C})$. There is the decomposition $L_{\lambda} = \prod_{i} GL_{\lambda_{i}}$. Now the corresponding Levi in $SL_{n}$, $L_{SL_{n}, \lambda}$ is given by $SL_{n} \cap L_{\lambda} $. So these are block matrices with overall determinant $1$.
My question is about the centre $Z(L_{SL_{n}, \lambda})$. Note that this will be given by block matrices of the form $diag(\gamma_{1} I_{\lambda_{1}}, \gamma_{2} I_{\lambda_{2}}, \dots, \gamma_{l} I_{\lambda_{l}})$ for some $\gamma_{i} \in \mathbb{C}^{*}$ satisfying the condition $\prod \gamma^{\lambda_{i}}_{i} = 1$.
I want to prove that $Z(L_{SL_{n}, \lambda}) = (\mathbb{C}^{*})^{l-1}$ when $n$ is prime.
Note that $Z(L_{\lambda}) = (\mathbb{C}^{*})^{l}$ for any $n$. In small cases I can explicitly define some maps, but they do not seem to work for all cases. For example, let $n = 5$, and $\lambda = (3,2)$. Then we can define a map $\mathbb{C}^{*} \to Z(L_{SL_{5}, \lambda})$ by $\gamma \mapsto diag(\gamma^{-2} I_{3} , \gamma^{3} I_{2})$, which is an isomorphism.
For $n$ non prime I believe the center can in general be disconnected. For example for $n=4$ $\lambda = (2,2)$ I think $Z(L_{SL_{4}, \lambda}) = \mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{C}^{*}$.