Lie algebra case Let \mathfrak{g} be a finite-dimensional semisimple
complex Lie algebra with
Cartan subalgebra \mathfrak{h}. Let R be the associated
root system. We then say that an element \lambda\in\mathfrak h^* is
integral if :2\frac{\langle\lambda,\alpha\rangle}{\langle\alpha,\alpha\rangle} is an integer for each root \alpha. Next, we choose a set R^+ of positive roots and we say that an element \lambda\in\mathfrak h^* is
dominant if \langle\lambda,\alpha\rangle\geq 0 for all \alpha\in R^+. An element \lambda\in\mathfrak h^* is
dominant integral if it is both dominant and integral. Finally, if \lambda and \mu are in \mathfrak h^*, we say that \lambda is
higher than \mu if \lambda-\mu is expressible as a linear combination of positive roots with non-negative real coefficients. A
weight \lambda of a representation V of \mathfrak g is then called a
highest weight if \lambda is higher than every other weight \mu of V. The theorem of the highest weight then states: if :\langle\lambda,H\rangle is an integer whenever :e^{2\pi H}=I where I is the identity element of K. Every analytically integral element is integral in the Lie algebra sense, but there may be integral elements in the Lie algebra sense that are not analytically integral. This distinction reflects the fact that if K is not simply connected, there may be representations of \mathfrak g that do not come from representations of K. On the other hand, if K is simply connected, the notions of "integral" and "analytically integral" coincide. is then the same as in the Lie algebra case, except that "integral" is replaced by "analytically integral." == Proofs ==