Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra and $\tilde{\mathfrak{g}}=\mathfrak{g}((t))\oplus \mathbf{C}K\oplus \mathbf{C}d$ its Kac-Moody extension ($K$ is the level and $d$ is the energy operator). $\tilde{\mathfrak{h}}=\mathfrak{h}\oplus \mathbf{C}K\oplus \mathbf{C}d$ is the Cartan subalgebra.

Pick an element $\sigma$ of the Weyl group $W$ of order $k$. Then the Heisenberg subalgebra $\hat{\mathfrak{a}}\subset \tilde{\mathfrak{g}}$ is the subalgebra of fixed points of $\mathfrak{h}((t^{1/k}))\oplus \mathbf{C}K$ under the action of $\sigma$, where $\sigma$ acts on $t$ by $t^{1/k}\mapsto \exp(2\pi i/k) t^{1/k}$. For example, for $\sigma=e$ one gets the homogeneous Heisenberg subalgebra $\widehat{\mathfrak{a}}\cong\mathfrak{h}((t))\oplus \mathbf{C} K$, while for $\sigma$ the Coxeter element one gets the principal Heisenberg subalgebra. Let $\mathfrak{a}_+\subset\hat{\mathfrak{a}}$ be the subalgebra containing nonnegative powers of $t$.

Consider $L(\lambda)$, an irreducible integrable highest-weight representation of $\tilde{\mathfrak{g}}$ of highest weight $\lambda\in \tilde{\mathfrak{h}}^*$. What is known about the restriction of $L(\lambda)$ to the Heisenberg $\hat{\mathfrak{a}}$? For example, is it true that the space of invariants $L(\lambda)^{\mathfrak{a}_+}$ is finite-dimensional?

If $\mathfrak{g}$ is simply-laced, $\hat{\mathfrak{a}}$ the homogeneous Heisenberg and $L(\lambda)$ is the level 1 basic representation, then a theorem of Frenkel-Kac and Segal identifies the restriction with a direct sum of Fock modules over the root lattice; in particular, $L(\lambda)^{\mathfrak{a}_+}$ is one-dimensional. It is also one-dimensional for the principal Heisenberg (using the principal realization of Kac-Kazhdan-Lepowsky-Wilson).


1 Answer 1


The answer to your second question is no. Take for instance $\Lambda=k \Lambda_0$, where the level $k$ is two or more. I'll consider only the homogeneous Heisenberg subalgebra. In this case, the space of invariants you're interested is very important in 2-dimensional conformal field theory and is known as the $parafermionic$ subspace $K({ \frak g},k)$ (actually a vertex algebra). It is known that it contains the conformal vector


so it carries a representation of the Virasoro algebra of central charge $c_{paraf}=\frac{dim({\frak g})k}{k+h^\vee}-rank({\frak g})$. As long as $\omega_{paraf} \neq 0$ the parafermionic space is infinite-dimensional. I believe this happens precisely if the level is at least two for simply-laced algebras. For non-simply laced, it can be infinite-dimensional even if the level is one (e.g. $G_2$). As $\omega_{paraf} \neq 0$ can be difficult to check directly, you can simply look at $c_{paraf}$. For example, let $\frak{g}$$=A_n$. Then $c_{paraf}=\frac{k(n^2+2n)}{n+1+k}-n$. This is nonzero and positive for all $n$ and $k \geq 2$. But there are no finite dimensional Virasoro modules with nonzero central charge so you're done.

For a general dominant weight $\Lambda$, it is more interesting to consider a slightly bigger space $L(\Lambda)^{t \mathbb{C}[t] \otimes \frak{h} }$ (the coset subspace), a $K({\frak g},k)$-module. Then you study decomposition of $L(\Lambda)$ as $K({\frak g},k) \otimes M(1)$-module (where $M(1)$ is the vacuum space for the Heisenberg, also a VOA). Thus $L(\Lambda)=\oplus_{\mu} M_{\mu} \otimes M(1,\mu)$, where $\mu \in \frak{h}^*$. If $\mu=0$ appears in the decomposition you get $L(\lambda)^{\mathbb{C}[t] \otimes \frak{h}}=M_0$ and infinite-dimensional if $c_{paraf} \neq 0$.

Same thing happens if $\sigma$ is the Coxeter element. Already for $k=2$ and ${\frak g}={\frak sl}_2$ the invariant space is infinite-dimensional.

I should also mention that $K({\frak g},k)$ is quite difficult to describe in terms of generators.

  • $\begingroup$ Thanks! Is there a conformal vector on the space of invariants for non-homogeneous Heisenbergs? In other words, can one modify the Sugawara construction for $\hat{\mathfrak{g}}$ so that it acts in the right way on any Heisenberg? By the way, $dim(\mathfrak{g}) = rank(\mathfrak{g}) (h+1)$, which proves that $c_{paraf}\neq 0$ for all $k\geq 2$ (and even $k=1$ for non simply-laced algebras). $\endgroup$ Aug 5, 2013 at 15:10
  • $\begingroup$ Once you fix the level, I think these "twisted" invariant spaces can be viewed as $twisted$ $modules$ (hope you're familiar with this terminology) for a certain subalgebra of the vertex algebra $L(k \Lambda_0)$. So naturally they do come equipped with an action of Vir with $c=c_{Paraf}$. You would have to choose a pair of $\sigma$-homogeneous dual bases in the Sugawara construction for this to work. $\endgroup$
    – user3775
    Aug 5, 2013 at 23:34

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.