From the Weyl character formula says that the irreducible representation of Lie algebra $\mathfrak{g}$ with highest weight $\lambda$ has character $$\frac{1}{\prod(e^{\alpha/2}-e^{-\alpha/2})}\sum_{w\in W} (-1)^w e^{w(\lambda+\rho)}$$ where the product is over the positive roots and $\rho$ is half of the sum of the positive roots. The trivial (!) representation with $\mathfrak{g}=\mathfrak{sl}_n$ is the Vandermonde identity $$\det \begin{pmatrix} x_1&x_2&\cdots&x_n\\ x_1^2&x_2^2&\cdots&x_n^2\\ \vdots&\vdots&&\vdots\\ x_1^n&x_2^n&\cdots&x_n^n \end{pmatrix}=\prod_{i<j}(x_j-x_i)$$

$$\text{}$$ I've heard* that the Jacobi Triple product identity $$\prod \left( 1 - x^{2n}\right) \left( 1 + x^{2n-1} y^2\right) \left( 1 +x^{2n-1}y^{-2}\right) \ = \ \sum x^{n^2} y^{2n}$$ is a trivial-representation case of a `Kac-Moody' version of the Weyl character formula.

If this is true, what does this general character formula say, and what other classical identities like this does it imply? Is there geometric/representation theoretic content to these results?

$$\text{}$$ $$\text{}$$ *In page 221 of 'Moonshine, Beyond the Monster'' (which gives the Kac-Moody character formula, but not its relation to such identities).

  • $\begingroup$ Note to younger self: one way to think about this is the boson-fermion correspondence, see Frenkel and Ben-Zvi's book. $\endgroup$
    – Pulcinella
    Commented Jul 15, 2019 at 14:54

1 Answer 1


Kac's book Infinite dimensional Lie algebras contains the general version of the Weyl character formula. It is

$$\mathrm{ch}(L(\lambda))=\frac{1}{\prod_{\alpha \in R^+} (1-e^{-\alpha})^{\mathrm{mult}(\alpha)}} \sum_{w \in W} (-1)^{\ell(w)} e^{w \cdot \lambda},$$

where we write $w \cdot \lambda=w(\lambda+\rho)-\rho$ for the dot action of the Weyl group, and $\mathrm{mult}(\alpha)$ is the dimension of the $\alpha$-root space in the Kac-Moody algebra $\mathfrak{g}$. This is Theorem 10.4 (page 173 in my third edition). Evidently, if we obtain another formula for the character of $L(\lambda)$, this equality produces a (possibly) interesting identity.

The "trivial" case is already interesting. When $\lambda=0$, so $L(\lambda) \cong \mathbf{C}$ is one-dimensional, this implies a generalization of the triple product identity. The triple product identity occurs for the Kac-Moody algebra $\widehat{\mathfrak{sl}}_2$, which is (essentially) a certain central extension of the loop algebra of $\mathfrak{sl}_2$ (see example 12.1 on page 218).

The Jacobi triple product identity is, of course, not the only $q$-series identity that may be obtained using representation theory. For others, one needs another way to calculate the character of $L(\lambda)$. One route, via vertex operators, leads to the Rogers-Ramanujan identities, where the relevant representations of $\widehat{\mathfrak{sl}}_2$ are of level 3. You might see e.g. A new family of algebras underlying the Rogers-Ramanujan identities and generalizations, by Lepowsky and Wilson, for a sketch of the ideas involved and references to more details.

  • $\begingroup$ Thanks a lot for this! I'll see if I can work out the details. $\endgroup$
    – Pulcinella
    Commented Apr 21, 2018 at 0:06
  • $\begingroup$ Could any of this be used to find a product formula for $\sum_{n\in\mathbb Z}x^ny^{n^2}z^{n^3}$ (or higher-order generalizations)? Or is there something special about $2$ here? $\endgroup$
    – mr_e_man
    Commented Jan 27, 2020 at 2:13
  • 1
    $\begingroup$ @mr_e_man There's something special about $2$: ultimately, it comes down to the way in which the affine Weyl group acts, involving a certain quadratic form. $\endgroup$
    – Stephen
    Commented Jan 27, 2020 at 14:18

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