# Can the numerator in Weyl's character formula be written as a determinant?

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.).

• There is the bialternant formula for Schur functions, I think that's what you want. Aug 8 at 1:19
• Aug 8 at 1:20
• Thank you so much @Sam Hopkins! Could you post it as an answer please? Aug 8 at 1:21
• Every number is equal to the Vandermonde determinant of size 2. Aug 8 at 1:28
• @MarkSapir, fair enough, I was not a 100% precise, but Sam Hopkins understood the kind of formula I was looking for... Aug 8 at 1:35

Of course the Schur polynomials are $$\mathfrak{sl}_n$$ characters. There are similar things in other types too, if you are interested in that (for example, see Proposition 1.1 in Okada's "Applications of Minor Summation Formulas to Rectangular-Shaped Representations of Classical Groups", https://doi.org/10.1006/jabr.1997.7408 ).