Let $\frak{g}$ be a complex simple Lie algebra and $V$ a finite dimensional representation $V$ of $\frak{g}$. Moreover, let $\frak{h}$ be a choice of Cartan subalgebra of $\frak{g}$.

A highest weight vector in $V$, unique up to scalar multiple, is a vector satisfying $E \triangleright v = 0$, for all positive roots $E$. Explicitly, we know that $V$ determined by the weight $(\lambda_1, \dots, \lambda_r)$ of its highest weight vector. (More explicitly, $v$ is a common eigenvector for the elements of $\frak{h}$, and the associated set eigenvalues of $v$ can be presented as $(\lambda_1, \dots, \lambda_r)$, with respect to a choice of basis of $\frak{h}$.)

Similarly one can define lowest weight vectors in terms of negative weights. Do these uniquely identify the representations of $\frak{g}$, and can one express $(\mu_1, \dots, \mu_r)$, the weight of a lowest vector, in terms of the highest weight $(\lambda_1, \dots, \lambda_r)$?

My naive guess would be that $$ (\mu_1, \dots, \mu_r) = (-\lambda_1, \dots, -\lambda_r). $$ Is this too naive?

  • 2
    $\begingroup$ The Weyl group $W$ acts on the weight lattice and, if $\lambda$ is the highest weight, then $w_0.\lambda$ is the lowest weight, where $w_0$ is the longest element. $\endgroup$
    – David Hill
    Mar 5 '18 at 16:08

Lowest weights can be used as an alternative to highest weights without problems. Indeed, the concept of a highest weight uses a choice of a system of positive roots $\Delta^+\subset\Delta$. If you instead use $-\Delta^+$ as a system of positive roots (which is a legitimate choice), then you get a highest weight that is the negative of the lowest weight in the original convention. The change of positive system can be implemented by the action of an element of the Weyl group and moving from $\Delta^+$ to $-\Delta^+$ is implemented by the longest element $w_0$ in the Weyl group. This explains the comment by @DavidHill that in terms of the highest weight $\lambda$, the lowest weight is given by $w_0(\lambda)$.

You can also see directly that the idea that the lowest weight would be $-\lambda$ is too naive. The point is that if you use a weight basis for the representation $V$, then the dual basis of $V^*$ consists of weight vectors whose weight are exactly the negatives of the weights of the basis elements in $V$. In particular, $-\lambda$ is the lowest weight of $V^*$ (which in general is not isomorphic to $V$). This gives an alternative description of the lowest weight of $V$ as the negative of the highest weight of $V^*$.

  • 1
    $\begingroup$ I see, so I was failing to keep in mind that $V$ and $V^*$ are not always isomorphic. Thanks for clarifying this! $\endgroup$ Mar 7 '18 at 18:15

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