10
$\begingroup$

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.).

$\endgroup$
5
  • 8
    $\begingroup$ There is the bialternant formula for Schur functions, I think that's what you want. $\endgroup$ Aug 8 at 1:19
  • 5
    $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ Aug 8 at 1:20
  • $\begingroup$ Thank you so much @Sam Hopkins! Could you post it as an answer please? $\endgroup$
    – Malkoun
    Aug 8 at 1:21
  • $\begingroup$ Every number is equal to the Vandermonde determinant of size 2. $\endgroup$
    – markvs
    Aug 8 at 1:28
  • $\begingroup$ @MarkSapir, fair enough, I was not a 100% precise, but Sam Hopkins understood the kind of formula I was looking for... $\endgroup$
    – Malkoun
    Aug 8 at 1:35
14
$\begingroup$

The classical definition of the Schur polynomials, which considerably predates the Weyl character formula, is as a ratio of two determinants (a so-called "bialternant"): see, e.g., https://en.wikipedia.org/wiki/Schur_polynomial#Definition_(Jacobi's_bialternant_formula).

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 ).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.