Representations of the Nappi–Witten vertex operator algebra

Abstract

The Nappi–Witten model is a Wess–Zumino–Witten model in which the target space is the nonreductive Heisenberg group $H_4$. We consider the representation theory underlying this conformal field theory. Specifically, we study the category of weight modules, with finite-dimensional weight spaces, over the associated affine vertex operator algebra $\mathsf {H}_4$. In particular, we classify the irreducible $\mathsf {H}_4$-modules in this category and compute their characters. We moreover observe that this category is nonsemisimple, suggesting that the Nappi–Witten model is a logarithmic conformal field theory.

Appendices

APPENDIX A. A proof of Theorem 5

In this appendix, we prove Theorem 5 and illustrate it with a simple example. For convenience, we break the rather lengthy proof into a series of steps. Let $\left| i,j\right\rangle$ denote the (generating) highest-weight vector of $\mathcal {V}_{i,j}^+$, and let $\chi$ be a singular vector of $\mathcal {V}_{i,j}^+$, so $\chi = U \left| i,j\right\rangle$ for some $U \in \mathsf {U}\left( \mathfrak {h}_4\right)$. From Lemma 3, there exists $m \in \mathbb {Z}$ such that $J_0 \chi = (j+m) \chi$ and $L_0 \chi = (\Delta ^+_{i,j} + im) \chi$. We may (and will) assume that $i \geqslant \frac{1}{2}$, by (4.2), and hence that $m \in \mathbb {Z}_{>0}$ ($m=0$ just returns $\chi = \left| i,j\right\rangle$).

1.1 Step 1.

$\mathcal {V}_{i,j}^+$ is irreducible if $i \notin \mathbb {Z}$.

U is thus a linear combination of Poincaré–Birkhoff–Witt-ordered monomials in the $E_{-n}$, $I_{-n}$, $J_{-n}$ and $F_{-n+1}$ with $n\geqslant 1$. Consider first the condition $0 = I_n \chi = [I_n , U] \left| i,j\right\rangle$, which holds for all $n\geqslant 1$. As $I_n$ commutes with every mode except $J_{-n}$, the effect of acting with $I_n$ on $\chi$ is to replace $J_{-n}^k$ by $nk J_{-n}^{k-1}$ in each monomial of U. For $k=0$, the monomial is replaced by 0. However, making these replacements on monomials with $k>0$ results in a linear combination of linearly independent monomials. Setting $I_n \chi = 0$ therefore requires that the coefficient of every $k>0$ monomial is zero. In other words, we may assume that no J-modes appear in U.

Next, assume that some monomial of U contains a mode $F_{-n}$ and take $n\geqslant 0$ maximal such that $F_{-n}$ does appear. (The corresponding monomial comes with nonzero coefficient.) Consider $E_n \chi = 0$, $n\geqslant 0$. Then, $E_n$ commutes with every mode of U except the $F_{-n+m}$, $0 \leqslant m \leqslant n$, because there are no J-modes. However, $[E_n , F_{-n+m}] = I_m + n \delta _{m,0}$ commutes with every mode of U, again because there are no J-modes, and will annihilate $\left| i,j\right\rangle$ if $m>0$. $E_n$ will thus annihilate any monomial of $U \left| i,j\right\rangle$ that does not have an $F_{-n}$. Acting with $E_n$ therefore amounts to replacing $F_{-n}^k$ by $(n+i)k F_{-n}^{k-1}$. As before, this only gives 0 if $k=0$ (because $n+i \geqslant \frac{1}{2}$) and for $k>0$, linear independence of the resulting monomials implies that the coefficient of every monomial with an $F_{-n}$ is 0. This contradicts the maximality of n, and hence, no F-modes may appear in U.

We now try to repeat the previous argument for $F_n \chi = 0$, $n\geqslant 1$. So assume that some monomial of U contains an $E_{-n}$ and let $n\geqslant 1$ be maximal such that $E_{-n}$ does appear. The result of the above argument this time is that $F_n$ replaces $E_{-n}^k$ by $(n-i)k E_{-n}^{k-1}$ throughout. This is nonzero for all $k>0$, and hence, no E-modes appear in U, unless $n=i$. If $i \in \mathbb {Z}_{>0}$, then this fixes the maximal index n such that $E_{-n}$ appears. However, if $i \notin \mathbb {Z}_{>0}$, then $n=i$ is impossible and so only I-modes may appear. But, considering $J_n \chi = 0$, $n\geqslant 1$, rules these out as well. (Alternatively, a singular vector with only I-modes cannot have the eigenvalues required by Lemma 3.) It follows that there is no such singular vector $\chi$, and hence, $\mathcal {V}_{i,j}^+$ is irreducible, when $i \notin \mathbb {Z}_{>0}$. As (4.2) extends this conclusion to all $i \notin \mathbb {Z}$, this proves the first assertion of Theorem 5.

1.2 Step 2.

If $\mathcal {V}_{i,j}^+$ has a singular vector corresponding to a given $m \in \mathbb {Z}_{>0}$, then it is unique up to scalar multiples.

For this, we may assume that $i \in \mathbb {Z}_{>0}$. The arguments above prove that no monomial of U contains a $J_{-n}$ with $n\geqslant 1$, an $F_{-n}$ with $n\geqslant 0$, or an $E_{-n}$ with $n>i$. We are therefore left with I-modes and the $E_{-n}$ for $1 \leqslant n \leqslant i$. As the $J_0$-eigenvalue of $\chi$ is $j+m$, there must be precisely m E-modes in each monomial of U. Moreover, $[L_0 , U] = im$ fixes the sum of the indices of the modes in each monomial.

Let $\mathcal {P}_{n}$ denote the set of partitions of n. We write $\lambda = [\lambda _1, \lambda _2, \ldots ]$ with $\lambda _1 \geqslant \lambda _2 \geqslant \cdots$ and shall refer to the $\lambda _k$ as the parts of $\lambda$. We shall also employ the alternative notation $\lambda = [\lambda _1^{k_1}, \lambda _2^{k_2}, \ldots ]$ to indicate that the part $\lambda _1$ has multiplicity $k_1$, and so on. A subpartition of $\lambda \in \mathcal {P}_{n}$ is then a partition $\mu$ obtained from $\lambda$ by removing some of its parts. We denote by $\lambda \setminus \mu$ the partition obtained by removing the parts $\mu _k$ of $\mu$ from those of $\lambda$ (respecting multiplicities). We shall moreover employ the simplified notation $\lambda \setminus \lambda _k \equiv \lambda \setminus [\lambda _k]$ when removing a single part.

Let $\mathcal {P}(i,m;\lambda )$ be the set of subpartitions $\mu$ of $\lambda$ having precisely m parts, none of which exceeds i. Then, we may write $\chi$ in the form

$$\begin{aligned} \chi = \sum _{\lambda \in \mathcal {P}_{im}} \sum _{\mu \in \mathcal {P}(i,m;\lambda )} c(\lambda ,\mu ) I_{-(\lambda \setminus \mu )} E_{-\mu } \left| i,j\right\rangle , \end{aligned}$$

(A.1)

where the $c(\lambda ,\mu )$ are unknown coefficients and $A_{-\lambda }$ is shorthand for $A_{-\lambda _1} \cdots A_{-\lambda _{\ell }}$, with $A=E,I,J,F$ and $\lambda$ a partition of precisely $\ell$ parts. It will be convenient for what follows to set $c(\lambda ,\mu ) = 0$ if $\mu \notin \mathcal {P}(i,m;\lambda )$.

Consider now $J_n \chi = 0$, for $n\geqslant 1$. The result of applying $J_n$ to (A.1) is a sum over $\lambda$ and $\mu$ in terms of the form

$$\begin{aligned}&\displaystyle n {{\,\mathrm{mult}\,}}_{n}(\lambda \setminus \mu ) c(\lambda ,\mu ) I_{-(\lambda \setminus \mu \setminus n)} E_{-\mu } \left| i,j\right\rangle \nonumber \\&\displaystyle \text {and} \quad c(\lambda ,\mu ) I_{-(\lambda \setminus \mu )} E_{-(\mu \mathrel {-_k} n)} \left| i,j\right\rangle \quad (k=1,\dots ,m, \mu _k > n), \end{aligned}$$

(A.2)

corresponding to commuting $J_n$ with an $I_{-n}$ and an $E_{-\mu _i}$, respectively. Here, ${{\,\mathrm{mult}\,}}_{n}(\nu )$ denotes the number of parts of $\nu$ equal to n and $\mu \mathrel {-_k} n$ denotes the partition obtained from $\mu$ by subtracting n from $\mu _k$ and reordering parts if necessary. Linear independence of monomials then gives a constraint for each $\lambda \in \mathcal {P}_{im}$ and $\mu \in \mathcal {P}(i,m;\lambda )$:

$$\begin{aligned} n {{\,\mathrm{mult}\,}}_{n}(\lambda \setminus \mu ) c(\lambda ,\mu ) + \sum _{k=1}^m c\left( \lambda \cup (\mu _k + n) \setminus \mu _k \setminus n, \mu \mathrel {+_k} n\right) = 0. \end{aligned}$$

(A.3)

Here, we denote by $\lambda \cup n$ the partition obtained from $\lambda$ by including n as an additional part and $\mu \mathrel {+_k} n$ is the partition obtained from $\lambda$ to adding n to $\lambda _k$ and reordering. We also understand that a constant of the of the form $c(\lambda ' \setminus n,\mu ')$ is understood to be 0 if n is not a part of $\lambda '$.

If $\lambda \ne \mu$, then there exists $n\geqslant 1$ such that ${{\,\mathrm{mult}\,}}_{n}(\lambda \setminus \mu ) \ne 0$. We may therefore solve (A.3) for $c(\lambda ,\mu )$ in terms of constants $c(\lambda ',\mu ')$, where the number of parts of $\lambda '$ is one less than that of $\lambda$. It follows that the $c(\lambda ,\mu )$ are completely determined by the $c(\lambda ',\mu ')$ in which $\lambda '$ has the minimal possible number of parts. This obviously occurs when $\lambda ' = \mu '$. However, as $\mu '$ must have m parts and none of its parts may exceed i, but their total must be im, this forces every part to be i. In other words, every $c(\lambda ,\mu )$ is determined by the value of $c([i^m],[i^m])$. This proves that for each $m \in \mathbb {Z}_{>0}$, there is at most one singular vector $\chi$, up to scalar multiples.

1.3 Step 3.

$\mathcal {V}_{i,j}^+$ has a singular vector for each $m \in \mathbb {Z}_{>0}$.

To show that these singular vectors actually exist, it suffices to show existence for $m=1$. Indeed, if existence holds for $m=1$, then the corresponding singular vector would generate a submodule of $\mathcal {V}_{i,j}^+$ isomorphic to $\mathcal {V}_{i,j+1}^+$, since $\mathsf {U}\left( \mathfrak {h}_4\right)$ has no zero-divisors. As the structure of Verma modules is independent of j, $\mathcal {V}_{i,j+1}^+$ also has an $m=1$ singular vector that generates a submodule isomorphic to $\mathcal {V}_{i,j+2}^+$. In $\mathcal {V}_{i,j}^+$, this singular vector corresponds to $m=2$. This obviously generalises to all $m \in \mathbb {Z}_{>0}$.

For $m=1$, the constraint equations (A.3) simplify a little. Writing $C(\lambda \setminus \lambda _k, \lambda _k) = c(\lambda ,\lambda _k)$, $\mu =[\lambda _k]$ and $n=\lambda _{\ell }$, they become

$$\begin{aligned} \lambda _{\ell } {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k) C(\lambda \setminus \lambda _k, \lambda _k) + C(\lambda \setminus \lambda _k \setminus \lambda _{\ell }, \lambda _k + \lambda _{\ell }) = 0. \end{aligned}$$

(A.4)

As we have seen, these relations show, by recursively removing parts, that $C(\lambda \setminus \lambda _k, \lambda _k)$ is proportional to $C(\varnothing ,i)$, where $\varnothing$ denotes the unique partition of 0, with nonzero proportionality constant.

We first demonstrate that the proportionality constants do not depend on the order in which one removes parts. If they did, then this would force $C(\varnothing ,i) = 0$ and the singular vector $\chi$ would not exist. Removing $\lambda _{\ell }$ and then $\lambda _{\ell '}$ from $\lambda$, we have

$$\begin{aligned} C(\lambda \setminus \lambda _k, \lambda _k) = \frac{C(\lambda \setminus \lambda _k \setminus \lambda _{\ell } \setminus \lambda _{\ell '}, \lambda _k + \lambda _{\ell } + \lambda _{\ell '})}{\lambda _{\ell } \lambda _{\ell '} {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k) {{\,\mathrm{mult}\,}}_{\lambda _{\ell '}}(\lambda \setminus \lambda _k \setminus \lambda _{\ell })}, \end{aligned}$$

(A.5)

which is symmetric under $\ell \leftrightarrow \ell '$ if

$$\begin{aligned} {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k) {{\,\mathrm{mult}\,}}_{\lambda _{\ell '}}(\lambda \setminus \lambda _k \setminus \lambda _{\ell }) = {{\,\mathrm{mult}\,}}_{\lambda _{\ell '}}(\lambda \setminus \lambda _k) {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k \setminus \lambda _{\ell '}). \end{aligned}$$

(A.6)

If $\lambda _{\ell } = \lambda _{\ell '}$, then (A.6) obviously holds. But, $\lambda _{\ell } \ne \lambda _{\ell '}$ implies that ${{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k) = {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus \lambda _k \setminus \lambda _{\ell '})$ and ${{\,\mathrm{mult}\,}}_{\lambda _{\ell '}}(\lambda \setminus \lambda _k \setminus \lambda _{\ell }) = {{\,\mathrm{mult}\,}}_{\lambda _{\ell '}}(\lambda \setminus \lambda _k)$, and hence, (A.6) also holds in this case. We conclude that $C(\varnothing ,i)$ is a free parameter in the solutions of (A.4).

To finish the proof of existence of an $m=1$ singular vector, we show that a (nonzero) solution of (A.4) corresponds to a $\chi$ that is annihilated by $\mathfrak {h}_4^+$. As (A.4) was deduced from $J_n \chi = 0$, $n\geqslant 1$, and $\mathfrak {h}_4^+$ is generated by $E_0$, $F_1$ and these $J_n$, we only have to check that the solution of (A.4) we have obtained gives a $\chi = U \left| i,j\right\rangle$ satisfying $E_0 \chi = F_1 \chi = 0$. Since U contains only I- and E-modes, the former is trivially satisfied.

We therefore consider $-F_1 \chi = 0$ (adding the minus sign for convenience). Writing the $m=1$ version of (A.1) in the form

$$\begin{aligned} \chi = \sum _{\lambda \in \mathcal {P}_{i}} \sum _{\lambda _k = 1}^i C(\lambda \setminus \lambda _k,\lambda _k) I_{-(\lambda \setminus \lambda _k)} E_{-\lambda _k} \left| i,j\right\rangle , \end{aligned}$$

(A.7)

we see that acting with $-F_1$ amounts to replacing each $E_{-\lambda _k}$ by $I_{-(\lambda _k-1)}$, if $\lambda _k>1$, and by $(i-1)$, if $\lambda _k=1$. The constraint equations derived from linear independence are therefore

$$\begin{aligned} (i-1) C(\lambda \setminus 1,1) + \sum _{\lambda _{\ell }=1}^{i-1} C(\lambda \setminus \lambda _{\ell } \setminus 1, \lambda _{\ell }+1) = 0. \end{aligned}$$

(A.8)

Now, substitute $\lambda _k=1$ into (A.4) to obtain

$$\begin{aligned} \lambda _{\ell } {{\,\mathrm{mult}\,}}_{\lambda _{\ell }}(\lambda \setminus 1) C(\lambda \setminus 1, 1) + C(\lambda \setminus \lambda _{\ell } \setminus 1, \lambda _{\ell } + 1) = 0. \end{aligned}$$

(A.9)

Summing over $\lambda _{\ell }$ from 1 to $i-1$ then gives (A.8), demonstrating that the constraints derived from $-F_1 \chi = 0$ already follow from those derived from $J_n \chi = 0$. This proves that there is indeed a unique singular vector $\chi$ corresponding to $m=1$. Again, (4.2) extends this conclusion from $i \in \mathbb {Z}_{>0}$ to all $i \in \mathbb {Z}$ and so we have established the second assertion of Theorem 5.

Before tackling the third and last assertion, we detail the existence of singular vectors in the case $i=4$ to illustrate the arguments used above. In this case, the singular vector generating the maximal submodule of $\mathcal {V}_{4,j}^+$ has the form $\chi = U \left| i,j\right\rangle$ with

$$\begin{aligned} U&= C(\varnothing ,4) E_{-4} + C([1],3) I_{-1} E_{-3} + C([2],2) I_{-2} E_{-2} + C([1,1],2) I_{-1}^2 E_{-2} \nonumber \\&\quad + C([3],1) I_{-3} E_{-1} + C([2,1],1) I_{-2} I_{-1} E_{-1} + C([1,1,1],1) I_{-1}^3 E_{-1}. \end{aligned}$$

(A.10)

It is clear that $E_0 \chi = 0$. From $J_3 \chi = 0$, we obtain

$$\begin{aligned} C(\varnothing ,4) + 3 C([3],1) = 0, \end{aligned}$$

(A.11a)

while $J_2 \chi = 0$ gives

$$\begin{aligned} C(\varnothing ,4) + 2 C([2],2) = 0 \quad \text {and} \quad C([1],3) + 2 C([2,1],1) = 0 \end{aligned}$$

(A.11b)

and $J_1 \chi = 0$ yields instead

$$\begin{aligned} \begin{aligned} C(\varnothing ,4) + C([1],3)&= 0,&C([1],3) + 2 C([1,1],2)&= 0, \\ C([2],2) + C([2,1],1)&= 0&\text {and} \quad C([1,1],2) + 3 C([1,1,1],1)&= 0. \end{aligned}\nonumber \\ \end{aligned}$$

(A.11c)

On the other hand, $F_1 \chi = 0$ instead results in

$$\begin{aligned} C(\varnothing ,4) + 3 C([3],1)= & {} 0, \quad C([1],3) + C([2],2) + 3 C([2,1],1) = 0 \quad \nonumber \\&\text {and} \quad C([1,1],2) + 3 C([1,1,1],1) = 0. \end{aligned}$$

(A.12)

Note that there is only one part to remove from [3], hence the first equation of (A.12) matches (A.11a). Similarly, there is only one way to remove a part from [1, 1, 1], and hence, the third equation of (A.12) matches the fourth equation of (A.11c). Finally, there are two ways to remove a part from [2, 1], and hence, the second equation of (A.12) is the sum of the second equation of (A.11b) and the third equation of (A.11c). For completeness, the singular vector $\chi = U \left| i,j\right\rangle$ is explicitly determined (when $C(\varnothing ,4) = 1$) by taking

$$\begin{aligned} U = E_{-4} - I_{-1} E_{-3} - \frac{1}{2} I_{-2} E_{-2} + \frac{1}{2} I_{-1}^2 E_{-2} - \frac{1}{3} I_{-3} E_{-1} + \frac{1}{2} I_{-2} I_{-1} E_{-1} - \frac{1}{6} I_{-1}^3 E_{-1}.\nonumber \\ \end{aligned}$$

(A.13)

Every singular vector of $\mathcal {V}_{4,j}^+$ therefore has the form $U^m \left| i,j\right\rangle$, for $m \in \mathbb {Z}_{\geqslant 0}$.

As an aside, it is actually quite easy to solve the constraints (A.4) in general. Writing $\lambda \setminus \lambda _k = [\mu _1^{k_1}, \mu _2^{k_2}, \ldots ]$, we find that

$$\begin{aligned} C(\lambda \setminus \lambda _k, \lambda _k) = \frac{(-1)^{k_1+k_2+\cdots }}{k_1! k_2! \cdots \, \mu _1^{k_1} \mu _2^{k_2} \cdots } C(\varnothing ,i). \end{aligned}$$

(A.14)

Substituting into (A.7) then gives a closed-form formula for $\chi$ when $i \in \mathbb {Z}_{>0}$. (4.2) may then be used to obtain a similar formula for $i \in \mathbb {Z}_{\leqslant 0}$. It is easy to verify that (A.14) reproduces (A.13) when $i=4$ (and $C(\varnothing ,4) = 1$).

It remains to prove the final assertion of Theorem 5, namely that the maximal submodule of $\mathcal {V}_{i,j}^+$ is generated by the singular vector $\chi$ corresponding to $m=1$. Let M denote the submodule of $\mathcal {V}_{i,j}^+$ generated by $\chi$.

1.4 Step 4.

The highest-weight module $\mathcal {V}_{i,j}^+ / M$ has no singular vectors, except multiples of the image $\overline{\left| i,j\right\rangle }$ of the highest-weight vector $\left| i,j\right\rangle$ of $\mathcal {V}_{i,j}^+$, and hence, it is irreducible.

So, suppose that $\overline{\psi }$ is a singular vector of $\mathcal {V}_{i,j}^+ / M$. Without loss of generality, we may choose a representative $\psi \in \mathcal {V}_{i,j}^+$ of $\overline{\psi }$ that is a weight vector. $\psi$ is then a subsingular vector of $\mathcal {V}_{i,j}^+$ satisfying $\psi \notin M$ and we have $\psi = U \left| i,j\right\rangle$, for some $U \in \mathsf {U}\left( \mathfrak {h}_4\right)$. Since $\overline{\psi }$ has $J_0$-eigenvalue $j+m$ and $L_0$-eigenvalue $\Delta ^+_{i,j} + im$ for some $m \in \mathbb {Z}_{\geqslant 0}$, by Lemma 3, the same is true for $\psi$.

The argument now generalises that used above to analyse the existence of $\chi$. We start by assuming that U has a J-mode and let $n\geqslant 1$ be maximal such that $J_{-n}$ appears. Choosing an appropriate Poincaré–Birkhoff–Witt-ordering, there exists $k>0$ such that we may write $U = J_{-n}^k V + W$, where V is a linear combination of monomials with no $J_{-n}$-modes and W is a linear combination of monomials with fewer than k $J_{-n}$-modes. Applying $I_n^k$, we have $I_n^k \overline{\psi } = 0$ and $I_n^k \psi = I_n^k U \left| i,j\right\rangle = n^k k! V \left| i,j\right\rangle$, from which we conclude that $V \left| i,j\right\rangle \in M$ and so $J_{-n}^k V \left| i,j\right\rangle \in M$. In other words, $W \left| i,j\right\rangle$ is another representative of $\overline{\psi }$ in which all monomials have fewer than k $J_{-n}$-modes. Iterating this argument shows that one can choose the representative $\psi$ so that it contains no J-modes.

Repeating this argument with $J_{-n}$ replaced by $F_{-n}$, $n\geqslant 0$, and $I_n$ replaced by $E_n$, we see as before (because $n+i\ne 0$) that $\psi$ may be chosen so that it also contains no F-modes. With $J_{-n}$ replaced by $E_{-n}$, $n>i$, and $I_n$ replaced by $F_i$, we similarly learn (from $n-i\ne 0$) that $\psi$ may be refined to eliminate any $E_{-n}$ modes with $n>i$. If we can likewise eliminate any $E_{-i}$-modes, then continuing this argument will rule out all E-modes. But, the $J_0$-eigenvalue of $\psi$ is then only consistent with $m=0$. Thus, U is constant and $\psi$ is proportional to $\left| i,j\right\rangle$, as desired.

To eliminate $E_{-i}$-modes, suppose that there is one and let $k\geqslant 1$ be the maximal power with which $E_{-i}$ appears. Then, we may write $U = V E_{-i}^k + W$, where V is a linear combination of monomials with no $E_{-i}$ and W is a linear combination of monomials with fewer than k $E_{-i}$-modes. We next recall that the singular vector corresponding to $m=k$ has the form $\chi ^k = U^k \left| i,j\right\rangle$ with $U^k = E_{-i}^k + W'$, where $W'$ is a linear combination of monomials in the $E_{-n}$-modes, with $1 \leqslant n \leqslant i$ and with fewer than k $E_{-i}$-modes, and the I-modes. Since $\chi ^k \in M$, $V \chi ^k \in M$ and so

$$\begin{aligned} \overline{\psi } = \overline{\psi - V \chi ^k} = \overline{(W - VW') \left| i,j\right\rangle }, \end{aligned}$$

(A.15)

that is we can replace the representative $\psi = U \left| i,j\right\rangle$ by $U' \left| i,j\right\rangle$, noting that $U' = W-VW'$ is a linear combination of monomials each of which has fewer than k $E_{-i}$-modes. Iterating this construction therefore allows us to find a representative with no $E_{-i}$-modes at all. As noted above, this proves that $\mathcal {V}_{i,j}^+ / M$ is irreducible for $i\in \mathbb {Z}_{>0}$. Because (4.2) easily extends this conclusion to all $i\in \mathbb {Z}$, the proof of Theorem 5 is complete.

APPENDIX B. An irreducibility proof

This appendix is devoted to proving that the relaxed highest-weight $\mathsf {H}_4$-modules $\mathcal {R}_{0,[j]; h}$ are irreducible, for all $[j] \in \mathbb {C}/ \mathbb {Z}$ and $h \ne 0$. The proof is similar in spirit to that of the irreducibility of the $\mathcal {L}_{0,j}$, itself a corollary of Lemma 3, but is slightly more involved. We note that $\Delta _{0;h} = h$ is the conformal weight of the generating relaxed highest-weight vectors $\left| 0,j';h\right\rangle \in \mathcal {R}_{0,[j]; h}$, $j' \in [j]$.

Suppose that $\mathcal {R}_{0,[j]; h}$ is reducible. Then, there is a relaxed highest-weight vector $v \in \mathcal {R}_{0,[j]; h}$ of conformal weight strictly greater than h. Since $L_0$ acts on relaxed highest-weight vectors as $Q_0 = F_0 E_0 + I_0 J_0$ and $I_0$ acts as 0, it follows that v is an eigenvector of $F_0 E_0$ with eigenvalue greater than h. However, we shall show that the only eigenvalue of $F_0 E_0$ on $\mathcal {R}_{0,[j]; h}$ is h, a contradiction.

Write v as a linear combination of Poincaré–Birkhoff–Witt-ordered monomials of the form

$$\begin{aligned} F_{-\lambda } E_{-\mu } J_{-\nu } I_{-\rho } \left| 0,j';h\right\rangle , \end{aligned}$$

(B.1)

where $j' \in [j]$ and $\lambda$, $\mu$, $\nu$ and $\rho$ are partitions. (We use the same notational conventions here as in Appendix A, see (A.1) and the surrounding text.) Acting with $F_0 E_0$ on such a monomial returns h times the monomial plus a number of commutator terms. These fall into two classes for which we observe the following simple facts:

• Commuting either $F_0$ with an E-mode or $E_0$ with an F-mode increases the number of I-modes by 1. These I-modes commute with every negative mode, so the number of I-modes strictly increases when the result is written as a linear combination of monomials (B.1).

• Commuting either $F_0$ or $E_0$ with a J-mode decreases the number of J-modes by 1. The result may require further commutation to represent it as a linear combination of monomials (B.1). However, this will never increase the number of J-modes because J is not in $[\overline{\mathfrak {h}}_4 , \overline{\mathfrak {h}}_4]$. The number of J-modes thus strictly decreases in the each summand of the result, when written as a linear combination of monomials (B.1).

Noting that any leftover $F_0$ or $E_0$ modes may be commuted to the right and thus change $j'$, this completely accounts for the action of $F_0 E_0$ on the monomials (B.1).

Order these monomials so that the number of J-modes weakly increases and, when the number of J-modes is the same, so that the number of I-modes weakly decreases. Then, the matrix representing $F_0 E_0$ in the weight space of $\mathcal {R}_{0,[j]; h}$ containing v is upper-triangular, with h as every diagonal entry. This is the desired contradiction, and hence, the proof is complete.