relation between lowest and highest weight of representation of $\mathfrak{sl}_n $ Having a finite-dimensional highest weight representation $L(\Lambda)$ of $\mathfrak{sl}_n,$ where $\Lambda=\lambda_1 w_1+...+\lambda_{n-1}w_{n-1}$ and $w_1,..,w_{n-1}$ are fundamental weights, is it true that its lowest weight space has the weight $\mu=-\lambda_{n-1} w_1-...-\lambda_1 w_{n-1}$? I saw that happening in a couple of examples.
 A: Edit: Sorry, in an earlier answer I misread your post.
This is indeed true, assuming that the fundamental weights $w_1, ..., w_{n-1}$ are ordered according to the standard ordering of the simple roots in the Dynkin diagram. Namely, in general, the lowest weight of a highest weight representation $L(\Lambda)$ is the negative of the highest weight of the dual representation $L(\Lambda)^*$; and this, in turn, is given by $-w_0(\Lambda)$ where $w_0$ is the longest element of the Weyl group.
See Relationship between highest and lowest weights for a simple complex Lie algebra, finding highest weight of dual of a representation of a semisimple lie algebra, Highest weight of dual representation of $\mathfrak{sl}_3$, Irreducible Dual Representation.
The upshot is: The lowest weight of $L(\Lambda)$ is $w_0(\Lambda)$. Now in a root system of type $A_{n-1}$, $w_0$ is given on simple roots in the usual ordering by $w_0(\alpha_i) = -\alpha_{n-i-1}$, and is easily seen to do the same on the fundamental weights, i.e. $w_0(w_i) =-w_{n-1-i}$. Since its action is linear, this explains your formula.
