4
$\begingroup$

Consider the Verma module $M(\lambda)=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)_{\mu}$. Denote its BGG dual by $M(\lambda)^{\vee}=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)^{\star}_{\mu}$ as in section $3.2$ of Humphreys' book.

When $\lambda$ is anti-dominant, $M(\lambda)$ is simple and $M(\lambda) \cong M(\lambda)^{\vee}$. My question is whether $$\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}?$$ I have doubts because on one hand, I feel that since $M(\lambda) \cong M(\lambda)^{\vee}$ we should have $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{\mu}$. On the other hand, I feel that the weights of $M(\lambda)^{\vee}$ should be negative of the weights of $M(\lambda)$. Hence, it might be true that $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}$.

Thank you for explaining.

Improve this question
$\endgroup$
4
  • $\begingroup$ It's useful to have some examples of Verma modules and their category $\mathcal{O}$ duals in mind (starting with rank $1$): the weight diagram is far from symmetric. Even when $\mu$ occurs as a weight of $M(\lambda)$ (with some multiplicity), it's usually not the case that $-\mu$ also occurs as a weight. (By the way, the possessive in English is awkward for names like mine ending in 's', but the simplest convention then is just to add an apostrophe.) $\endgroup$
    – Jim Humphreys
    Commented Jan 6, 2018 at 18:51
  • $\begingroup$ I corrected the English. If $\mu $ is a weight of $M(\lambda)$, then can $-\mu$ be a weight of BGG dual of $M(\lambda)$? Also, is my question on the dimension true? It will be very helpful if you explain a little bit. $\endgroup$
    – MathStudent
    Commented Jan 6, 2018 at 19:41
  • $\begingroup$ To be more explicit, the weights and multiplicities are the same in any Verma module and its BGG dual; only the composition factor picture is turned upside down. So your question has a negative answer. In rank 1 for example, where weights can be identified with integers, the antidominant Verma module $M(-1)$ equals the simple module $L(-1)$ and is then isomorphic to its BGG dual $M(-1)^\vee$; the weights (with multiplicity 1) are $-1,-3,-5, \dots$ in both cases. $\endgroup$
    – Jim Humphreys
    Commented Jan 6, 2018 at 21:33
  • $\begingroup$ To answer your newer question, it's quite possible (for weights which are not antidominant) that some weights of a Verma module have negatives which are also weights: again this is seen in rank 1, when you start with a nonnegative highest weight and get two composition factors. The top one is finite dimensional, with positive weights occurring along with their negatives as in the classical case. (Of course, $M(\mu)$ with $\mu = 0$ is an extreme example of this.) Dual Verma modules have the same weights and multiplicities. $\endgroup$
    – Jim Humphreys
    Commented Jan 6, 2018 at 21:44

1 Answer 1

Reset to default
4
$\begingroup$

If you compute the dual of a weight module $M$ using the usual notion of dual of a module over a Lie algebra, then you get a module $M^{\ast}$ such that $M^{\ast}_{-\mu}$ is the vector space dual of $M_{\mu}$ (and thus has the same dimension). This why you think the displayed equation is correct.

But that's not the notion of dual that anyone uses in category $\mathcal{O}$, because as you can see it doesn't send objects in category $\mathcal{O}$ to objects in category $\mathcal{O}$. Thus, $M^\vee$ is something different: it's the usual vector space dual (EDIT: not actually the full one; as the OP writes, it's the sum of the vector space duals of the individual weight spaces, which is a version of graded dual), with the action twisted by the Cartan involution $E_i\mapsto -F_i, F_i\mapsto -E_i, H_i\mapsto -H_i$ (for the classical groups, this is transpose and negate). Because of the negative sign in front of $H_i$, this flips the sign of weights, and $M^\vee_\mu$ is the vector space dual of $M_\mu$ (and thus the same dimension).

Improve this answer
$\endgroup$
4
  • $\begingroup$ @Ben: Your quick sketch of the BGG construction is partly out of focus, especially the usual vector space dual (too large when $M$ is infinite dimensional); their construction didn't use the Cartan involution, but rather a related anti-involution. See the remarks at the end of $\S4$ in mathnet.ru/php/… (the English translation is OK but has "complete" rather than "full subcategory" etc.). See also Gaitsgory's lectures math.harvard.edu/~gaitsgde/267y/index.html or my 2008 book. $\endgroup$
    – Jim Humphreys
    Commented Jan 7, 2018 at 14:21
  • 1
    $\begingroup$ @JimHumphreys Of course I'm well aware that the full dual is too large, but I didn't want to water down my point with topological issues (especially since the OP seemed to be straight on this point). I also am fairly sure that I'm doing exactly what BGG did (my Russian is a little rusty), but I broke it up into two steps of taking usual dual (which implicitly chooses the anti-involution of taking negative) and then twisting by Cartan, whereas BGG does it in one step composing these (to get an anti-involution which is the identity on the Cartan). $\endgroup$
    – Ben Webster ♦
    Commented Jan 7, 2018 at 18:57
  • $\begingroup$ I think this is a good description of what is going on, except I agree with Jim that calling it the usual vector space dual might lead to confusion (for someone other than the OP potentially reading this answer without thoroughly reading the question). I suppose a short term for what it really is would be "graded dual" (with the grading by the dual of the Cartan). $\endgroup$
    – Tobias Kildetoft
    Commented Jan 17, 2018 at 8:39
  • $\begingroup$ @TobiasKildetoft Since my not saying "graded dual" is apparently upsetting for people, I added a note. $\endgroup$
    – Ben Webster ♦
    Commented Jan 17, 2018 at 13:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .