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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 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 $Z(M)^0$?

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    $\begingroup$ It's probably best to spell out the meaning of the Z operator (which could refer to either the center or the centraiizer but here must mean the center). Aside from that, "its Levi" is inaccurate, since there are many conjugate Levi factors (as there are many conjugate Borel subgroups). $\endgroup$
    – Jim Humphreys
    Commented Dec 13, 2018 at 21:45
  • $\begingroup$ @JimHumphreys, in this case, $\mathrm C_G(M) = \mathrm Z(M)$, so probably it's not an ambiguity that requires much attention. $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 22:19
  • $\begingroup$ (@JimHumphreys, also, just as one may speak of "the set of simple roots" after having chosen a 'Borus', surely one may also speak of "the Levi subgroup" after having chosen a torus? @‍S.D. must have meant that $P$ contains $T$.) $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 22:30
  • $\begingroup$ @LSpice thanks. Yes I meant P contains T. And Z is the center of M. $\endgroup$
    – user100841
    Commented Dec 14, 2018 at 7:27

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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.

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    $\begingroup$ Thanks for your answer but the weight lattice is not the same as the space of characters, to begin with. $\endgroup$
    – user100841
    Commented Dec 13, 2018 at 8:44
  • $\begingroup$ Hmm, you are right; so I should have passed initially to the adjoint quotient of $G$, in which case, by the same argument, the problem becomes: if $G$ is adjoint, then does $M$ have connected centre? (Of course this destroys my counterexample.) I think that this is still false, but I will consider more. $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 12:29
  • $\begingroup$ OK, I think that I have fixed the issue. I apologise for rejecting your initial edit, by the way; I thought that you were just deleting my re-phrasing of the problem, and didn't realise that your goal was to harmonise notation. Now I can't undo my reject vote. I'll stick with $\mathrm X^*$, but I have added a note in order hopefully to avoid confusion. $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 13:54
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    $\begingroup$ I don't see this statement on p. 8 of the linked paper. The closest that I see is the claim that $\smash{\check\Lambda}_{Z_0(M)}^{\mathbb Q} \to \smash{\check\Lambda}_{G, P}^{\mathbb Q}$ is an isomorphism. If that's the statement you mean, it is different in many ways from what you said. A probably minor point: the paper's definition of $\Lambda_{G, P}$ is different from yours. (It is a sublattice of $\Lambda_G$, not a quotient.) $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 18:58
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    $\begingroup$ In fact I'm not quite sure how to make sense of $\check\Lambda_{Z_0(M)}$; according to the text it seems it should be a coweight lattice, but the coweight lattice of a torus is trivial. Anyway, tensoring with $\mathbb Q$ kills all my finite-index worries in the semisimple case; those are just issues with integer lattices, not rational vector spaces. $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 19:02

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