I paraphrase part of the wikipedia article on the Weyl character formula: Weyl character formula.

If $\pi$ is an irreducible finite-dimensional representation of a complex semisimple Lie algebra $\mathfrak{g}$ and $\mathfrak{h}$ is a choice of Cartan subalgebra of $\mathfrak{g}$, then the Weyl character formula states that the character $\operatorname{ch}_\pi$ of $\pi$ is given by

$$ \operatorname{ch}_\pi(H) = \frac{\sum_{w \in W} \epsilon(w) e^{w(\lambda+\rho)(H)}}{\prod_{\alpha \in \Delta^+}(e^{\alpha(H)/2} - e^{-\alpha(H)/2})},$$

where $W$ is the Weyl group, $H \in \mathfrak{h}$, $\epsilon(w)$ is the determinant of the action of $w \in W$ on the Cartan subalgebra $\mathfrak{h}$, $\Delta^+$ denotes the set of positive roots of $(\mathfrak{g}, \mathfrak{h})$, $\lambda$ denotes the highest weight of $\pi$ and $\rho$ is half the sum of all positive roots of $(\mathfrak{g}, \mathfrak{h})$ (i.e. half the sum of all the elements of $\Delta^+$).

If $\mathfrak{g} = \mathfrak{sl}(n)$, it is known that the denominator of the RHS of the Weyl character formula can be written as a determinant (actually a Vandermonde determinant). While it does seem counterproductive, since the numerator looks simple enough (well, to some extent), yet I am interested whether or not the numerator can also be written as a determinant, at least for $\mathfrak{g} = \mathfrak{sl}(n)$, though I am also interested in the other cases too. After all, the Weyl group in this special case is $S_n$, the symmetric group on $n$ elements, and an $n \times n$ determinant can be expanded as an alternating sum over $S_n$. So this does seem promising. I apologize if it turns out to be a trivial question perhaps (I currently have limited access to online journals etc.).