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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 parabolic subgroup, $M$ be its Levi 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 $Z(M)$ containing the identity and $X(Z(M)^0)$ is the character group of the torus $Z(M)^0$?

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 $Z(M)^0$?

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 parabolic subgroup, $M$ be its Levi 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 $\alpha_i$, $i\in I_M$}}$.

My question: Is $\Lambda_{G,P}=X(Z(M)^0)$, where $Z(M)^0$ is the component of $Z(M)$ containing the identity and $X(Z(M)^0)$ is the character group of the torus $Z(M)^0$?

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 parabolic subgroup, $M$ be its Levi 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 $Z(M)$ containing the identity and $X(Z(M)^0)$ is the character group of the torus $Z(M)^0$?

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 parabolic subgroup, $M$ be its Levi and let $I_M$ be the set of simple roots of $M$ (with the natural choice of Borel of $M$). Let $\wedge_G$ be the weight lattice of G. We define $\wedge_{G,P}:=\frac{\wedge_G}{\text{span of} \alpha_i,i\in I_M}$.

My question: Is $\wedge_{G,P}=X(Z(M)^0)?$, where $Z(M)^0$ is the component of $Z(M)$ containing the identity and $X(Z(M)^0)$ is the character group of the torus $Z(M)^0$.

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 parabolic subgroup, $M$ be its Levi 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 $\alpha_i$, $i\in I_M$}}$.

My question: Is $\Lambda_{G,P}=X(Z(M)^0)$, where $Z(M)^0$ is the component of $Z(M)$ containing the identity and $X(Z(M)^0)$ is the character group of the torus $Z(M)^0$?