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.


1 Answer 1


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).

  • $\begingroup$ Just to be clear, the maps between various Verma modules in the "wish to be BGG resolution" exists, but the problem is that one cannot cook up any differential that would give exact complex. Right? $\endgroup$ Jun 23, 2015 at 12:14
  • $\begingroup$ @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. $\endgroup$
    – Ben Webster
    Jun 24, 2015 at 10:48
  • $\begingroup$ Ah yes, because they are actually simple. In higher rank, do we at least get a "truncated" BGG resolution? $\endgroup$ Jun 25, 2015 at 9:39
  • $\begingroup$ @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). $\endgroup$
    – Ben Webster
    Jan 14, 2021 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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