3
$\begingroup$

Please reference this paper for notation in this question.

I'm trying to understand two claims made in the above paper (they may be related). First, in the construction of $\mathcal{H}_\lambda$ on page 4, we start with the usual irreducible representation $V_\lambda$ of the semisimple Lie algebra $\mathfrak{g}$, declare the positive powers of $z$ to act trivially and the central element $c$ to act by multiplication by $l$, and then induce an action of the rest of the affine Lie algebra $\hat{\mathfrak{g}}$ on $V_\lambda$. The result is called $\mathcal{V}_\lambda$ and it is isomorphic to $U(\hat{\mathfrak{g}}_-) \otimes_{\mathbb{C}} V_\lambda$ as a $\hat{\mathfrak{g}}_-$-module. As a consequence, it is a highest weight module, and by theory analogous to that of semisimple Lie algebras, it has a maximal submodule.

I understand all of that. What I do not understand is the claim that this maximal submodule $\mathcal{Z}_\lambda$ is generated by $(X_\theta \otimes z^{-1})^{l - \lambda(H_\theta) + 1} v_\lambda$. Following Remark 3.6 on page 8, I managed to prove that this element is annihilated by $\hat{\mathfrak{g}}_+$, and indeed that no lower nonzero power of $X_\theta \otimes z^{-1}$ acting on $v_\lambda$ is annihilated by all of $\hat{\mathfrak{g}}_+$. Thus I'm lead to believe that this should somehow imply that $\mathcal{Z}_\lambda$ is a highest weight module with $(X_\theta \otimes z^{-1})^{l - \lambda(H_\theta) + 1} v_\lambda$ its highest weight vector, and hence the claim follows, but I can't see why this is the case.

The second claim is labeled (*) on page 7: The endomorphism $X_{-\theta} \otimes f$ of $\mathcal{H}$ is locally nilpotent for all $f \in \mathcal{O}(U)$. From what I can tell, this is equivalent to the claim that $X_{-\theta} \otimes f$ is a locally nilpotent endomorphism of $\mathcal{H}_\lambda$ for any $f \in \mathcal{C}((z))$. Following the given hint, I can see why this is true for $f \in \mathcal{C}[[z]]$, but I don't know about negative powers of $z$. I think this might be related to the first claim: Perhaps once we have pushed any element of $\mathcal{V}_\lambda$ into a suitably low weight space by acting on it by $X_{-\theta} \otimes z^{-1}$ repeatedly, it actually falls into some proper submodule of $\mathcal{V}_\lambda$, and is therefore quotiented out when we pass to $\mathcal{H}_\lambda$. Again, I can't see why this is actually true though.

Any ideas on either claim are greatly appreciated.

$\endgroup$
2
$\begingroup$

I answer the first question only.

The space $\mathcal{V}_\lambda$ is highest weight for the affine Lie algebra, hence a quotient of a Verma $\Delta(\mu)$ of the affine Lie algebra. The maximal submodule $\mathcal{Z}_\lambda$ is thus the image of the maximal submodule of the Verma.

The maximal submodule of a Verma is known to be generated by the vectors $f_i^{\langle \mu,\alpha_i^\vee\rangle+1}v_\mu$ as $\alpha_i$ runs over the simple roots for the affine root system. In your case, the vectors in this set corresponding to finite simple roots all map to zero in the quotient $\mathcal{V}_\lambda$ by its construcion from a finite dimensional irreducible. Therefore the maximal submodule of $\mathcal{V}_\lambda$ is generated by the image of the vector for the affine root, which is the generator you have written down.

$\endgroup$

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.