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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).

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up vote 2 down vote accepted

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.

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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). – Pavel Safronov Aug 5 '13 at 15:10
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. – Antun Milas Aug 5 '13 at 23:34

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