# Understanding the Weyl Character Formula

Let $G$ be a compact (connected) Lie group with a maximal torus $T$. For each (analytically) integral weight $\lambda$ the Weyl character formula $$\Theta_{\lambda}(H)=\frac{\sum_{w\in W(G)}\epsilon(w)e^{w(\lambda+\rho)(H)}}{\sum_{w\in W(G)}\epsilon(w)e^{w(\rho)(H)}}$$ defines a function that descends to the set of regular elements in $T$.

We also know, by the highest weight theory, that there is a correspondence between dominant integral weights and the irreducible representations of $G$.

Weyl proved that when $\pi$ is an irreducible representation with the highest weight $\lambda$, its character is given by $\Theta_\lambda$.

My question is whether the right-hand side of Weyl's formula has an interesting meaning or interpretation when $\lambda$ is not a dominant integral weight.

Of course, if $\lambda$ is not integral Weyl's formula doesn't descend to the group $G$, but I wonder if Weyl's formula reflects a property of the (representations of the) group or its Lie algebra in this case.

One way to interpret the Weyl character formula is as the Euler characteristic of the BGG resolution. If $\lambda$ is not integral, then the terms in the BGG resolution still make sense (they are Verma modules), but the maps used to define the complex structure no longer exist (if $\lambda$ is generic, there are no $\mathfrak{g}$-equivariant maps from one term of the complex to another).
• @VítTuček No, there are really no maps. Think about $\mathfrak{sl}_2$. You have the Verma $M(n)$ and the Verma $M(-n-2)$. The space $Hom(M(-n-2),M(n)$ is 1-d if $n$ is an integer with $n\geq -1$, and 0 dimensional otherwise. – Ben Webster Jun 24 '15 at 10:48
• @VítTuček For generic $\lambda$ in any semi-simple Lie algebra, the terms of the BGG resolution are all simple, and there are no non-trivial maps from one to any other. For $\lambda$ having integral inner product with some roots but not others, then yes, there are some maps, but not enough to make all the non-zero cohomology all concentrate in one degree (so "what goes wrong" is not exactness: it's just components of the maps stop existing). – Ben Webster Jan 14 at 14:45