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](https://books.google.fr/books?id=8GCP4Ng6risC&printsec=frontcover#v=onepage&q&f=false).


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.



Weights of BGG dual of Verma module
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). 


