Let $G$ be a reductive group over an algebraically closed field $k$. Let $T$ be a maximal torus, $B$ be a Borel subgroup and $I\_G$ is the set of simple roots. Let $P$ be a standard parabolic subgroup, $M$ be its Levi containing $T$ and let $I\_M$ be the set of simple roots of $M$ (with the natural choice of Borel of $M$). Let $\Lambda\_G$ be the weight lattice of G. We define $\Lambda\_{G,P}:=\frac{\Lambda\_G}{\text{span of $\alpha\_i$, $i\in I\_M$}}$.



> 
> Is $\Lambda\_{G,P}=X(Z(M)^0)$, where $Z(M)^0$ is the component of $Z(M)$ (
>  the center of $M$) containing the identity and $X(Z(M)^0)$ is the character group of the
>  torus $Z(M)^0$?
> 
> 
> 



Relative weight lattice
I'll use $\mathrm X^\*$ instead of $X$ for character lattices, since I can never remember which is which in the $X$/$Y$ notation. I have also updated this answer from its original wrong formulation to a hopefully correct one.


$\DeclareMathOperator\srank{srank}$Note that $\Lambda\_{G, P}$ is a lattice of rank $\srank(G) - \srank(M)$, where $\srank$ stands for the semisimple rank.


Exactly as written, the answer is 'no'; for example, if $G = M$ is a non-trivial torus, then $\Lambda\_{G, P}$ is trivial but $\mathrm X^\*(\mathrm Z(M)^\circ) = \mathrm X^\*(G)$ is not.


If $G$ is semisimple, then $\srank(G) - \srank(M) = \dim(\mathrm Z(M)^\circ)$, so that $\Lambda\_{G, P}$ and $\mathrm X^\*(\mathrm Z(M)^\circ)$ are lattices of the same rank, hence abstractly isomorphic. However, there is a natural map $\Lambda\_{G, P} \to \mathrm X^\*(\mathrm Z(M)^\circ)$ given by restriction, and it need not be an isomorphism; its image may have finite index. Consider the case $G = \mathrm{SL}\_2$ and $M = T$.


$\DeclareMathOperator\Span{Span}$If $G$ is adjoint, then $\Lambda\_G = \mathrm X^\*(T)$ and $\Span\_{\mathbb Z} \{\alpha : \alpha \in I(M, B \cap M, T)\}$ is the annihilator of $\mathrm Z(M)$ in $\mathrm X^\*(T)$, so that the restriction map $\Lambda\_{G, P} \to \mathrm X^\*(\mathrm Z(M))$ is an isomorphism. Since $\Lambda\_{G, P}$ is torsion free, so is $\mathrm X^\*(\mathrm Z(M))$, which means that $\mathrm Z(M)$ is connected, and hence we have finally that $\Lambda\_{G, P} \to \mathrm X^\*(\mathrm Z(M)) = \mathrm X^\*(\mathrm Z(M)^\circ)$ is an isomorphism in this case.


