Quartic Hamiltonians, and higher Hamiltonians at next-to-leading order, for the affine $\bm{\mathfrak{sl}_{2}}$ Gaudin model

We define the loop algebra $\mathfrak{sl}_{2}[t,t^{-1}]=\mathfrak{sl}_{2}\otimes\mathbb{C}[t,t^{-1}]$ as the algebra of Laurent polynomials in a formal variable $t$ with coefficient in the finite-dimensional Lie algebra $\mathfrak{sl}_{2}$. The Lie brackets on this algebra are given by

$$[a\otimes f(t),b\otimes g(t)]=[a,b]_{\mathfrak{sl}_{2}}\otimes f(t)g(t),$$

(5)

where $f(t)$ and $g(t)$ are arbitrary Laurent polynomials in $\mathbb{C}[t,t^{-1}]$.

Let $(\cdot|\cdot):\mathfrak{sl}_{2}\times\mathfrak{sl}_{2}\rightarrow\mathbb{C}$ be the canonically normalized symmetric invariant bilinear form on $\mathfrak{sl}_{2}$. It is given by taking the trace in the defining representation:

$$(a|b)\vcentcolon=\Tr(ab),$$

(6)

It is possible to extend the loop algebra by a one-dimensional central element $\mathbb{C}\mathsf{k}$,

$$0\longrightarrow\mathbb{C}\mathsf{k}\longrightarrow\widehat{\mathfrak{sl}}_{2}\longrightarrow\mathfrak{sl}_{2}[t,t^{-1}]\longrightarrow 0.$$

(7)

This extension is called the affine Lie algebra $\widehat{\mathfrak{sl}}_{2}$, whose commutation relations are

	$$\displaystyle[a\otimes f(t),b\otimes g(t)]=[a,b]_{\mathfrak{sl}_{2}}\otimes f(t)g(t)-(\text{res}_{t}f\text{d}g)(a|b)\mathsf{k},$$		(8)
	$$\displaystyle[\mathsf{k},\cdot]=0.$$		(9)

We shall use the notation

$$a_{n}\vcentcolon=a\otimes t^{n},\quad\text{for $a\in\mathfrak{sl}_{2}$ and $n\in\mathbb{Z}$}.$$

(10)

The commutation relations can be equivalently written as

$$[a_{m},b_{n}]=[a,b]_{n+m}-n\delta_{n+m,0}(a|b)\mathsf{k}.$$

(11)

We can add to this algebra a one-dimensional derivation $\mathsf{d}$, such that $[\mathsf{d},\mathsf{k}]=0$ and $[\mathsf{d},a\otimes f(t)]=a\otimes t\partial_{t}f(t)$, for all $a\in\mathfrak{sl}_{2}$ and $f(t)\in\mathbb{C}[t,t^{-1}]$. It is possible to show that this algebra is isomorphic to the Kac-Moody algebra over $\mathbb{C}$ of type $\mathsf{A}_{1}^{(1)}$, see e.g. [Kac90, ch. 7],

$$\mathfrak{g}=\widehat{\mathfrak{sl}}_{2}\oplus\mathbb{C}\mathsf{d}.$$

(12)

The Cartan matrix for the Kac-Moody algebra of type ${}^{1}\mathsf{A}_{1}$ is defined as $A=(a_{i,j})_{i,j=0}^{1}=(2\delta_{i,j}-\delta_{i+1,j}-\delta_{i-1,j})_{i,j=0}^{1}$. The Cartan decomposition is given by $\mathfrak{g}=\mathfrak{n}_{-}\oplus\mathfrak{h}\oplus\mathfrak{n}_{+}$. The Chevalley-Serre generators are $\{e_{i}\}_{i=0}^{1}\subset\mathfrak{n}_{+}$, $\{f_{i}\}_{i=0}^{1}\subset\mathfrak{n}_{-}$ while $\{\check{\alpha}_{i}\}_{i=0}^{1}\subset\mathfrak{h}$ and $\{\alpha_{i}\}_{i=0}^{1}\subset\mathfrak{h^{\ast}}$ are respectively a basis for the Cartan subalgebra of simple coroots of $\mathfrak{g}$ and a basis for the dual Cartan subalgebra of simple roots of $\mathfrak{g}$. The latter are related via the canonical pairing between the Cartan algebra and its dual, $\langle\cdot,\cdot\rangle:\mathfrak{h}^{\ast}\times\mathfrak{h}\rightarrow\mathbb{C}$

$$\langle\alpha_{i},\check{\alpha}_{j}\rangle=a_{i,j}.$$

(13)

The fundamental commutation relations in $\mathfrak{g}$ are

$$\begin{gathered}=\langle\alpha_{i},x\rangle e_{i},\qquad[x,f_{i}]=-\langle\alpha_{i},x\rangle f_{i},\\ [x,x^{\prime}]=0,\qquad[e_{i},f_{j}]=\check{\alpha}_{i}\delta_{ij},\end{gathered}$$

(14)

where $x,x^{\prime}\in\mathfrak{h}$ and $i,j=0,1$, together with the Serre relations

$$(\text{ad}e_{i})^{1-a_{ij}}e_{j}=0,\qquad(\text{ad}f_{i})^{1-a_{ij}}f_{j}=0.$$

(15)

The Kac-Moody algebra $\mathfrak{g}$ has a central element $\mathsf{k}=\sum_{i=0}^{1}\check{\alpha}_{i}$, which spans a one-dimensional centre. A basis for the Cartan subalgebra $\mathfrak{h}$ is given by the coroots $\{\check{\alpha}_{i}\}_{i=0}^{1}$ together with the derivation element $\mathsf{d}$, which by definition satisfies

$$\langle\alpha_{i},\mathsf{d}\rangle=\delta_{i,0}.$$

(16)

If we remove the 0th row and column from $A$, we obtain the Cartan matrix corresponding to the finite dimensional Lie algebra $\mathfrak{sl}_{2}$. This subalgebra of $\mathfrak{g}$ is generated by $e_{1}\in\mathfrak{n}_{+}$, $f_{1}\in\mathfrak{n}_{-}$ and $\check{\alpha}_{1}\in\mathfrak{h}$.

For any $k\in\mathbb{C}$, let us define $U_{k}(\widehat{\mathfrak{sl}}_{2})$ as the quotient of the universal enveloping algebra $U(\widehat{\mathfrak{sl}}_{2})$ of $\widehat{\mathfrak{sl}}_{2}$ by the two-sided ideal generated by $\mathsf{k}-k$. For each $n\in\mathbb{Z}_{\geq 0}$, let us introduce the left ideal $J_{n}=U_{k}(\widehat{\mathfrak{sl}}_{2})\cdot(\mathfrak{sl}_{2}\otimes t^{n}\mathbb{C}[t])$. The inverse limit

$$\widetilde{U_{k}}(\widehat{\mathfrak{sl}}_{2})=\varprojlim\faktor{U_{k}(\widehat{\mathfrak{sl}}_{2})}{J_{n}}$$

(17)

is a complete topological algebra, called the local completion of $U_{k}(\widehat{\mathfrak{sl}}_{2})$ at level $k$. With this definition, the elements of $\widetilde{U_{k}}(\widehat{\mathfrak{sl}}_{2})$ are possibly infinite sums of the type $\sum_{m\geq 0}a_{m}$ of elements $a_{m}\in U_{k}(\widehat{\mathfrak{sl}}_{2})$ which do truncate to finite sums when working modulo any $J_{n}$.

A module $\mathscr{M}$ over $\widehat{\mathfrak{sl}}_{2}$ is said to be smooth if, for all $a\in\mathfrak{sl}_{2}$ and all $v\in\mathscr{M}$, $a_{n}v=0$ for sufficiently large $n$. A module $\mathscr{M}$ has level $k$ if $\mathsf{k}-k$ acts as zero on $\mathscr{M}$. Any smooth module of level $k$ over $\widehat{\mathfrak{sl}}_{2}$ is also a module over the completion $\widetilde{U_{k}}(\widehat{\mathfrak{sl}}_{2})$.

We can identify the subalgebra of positive modes $\mathfrak{sl}_{2}[t]\oplus\mathbb{C}\mathsf{k}\subset\widehat{\mathfrak{sl}}_{2}$ and introduce the one-dimensional representation $\mathbb{C}\ket{0}^{k}$ defined by

$$(\mathsf{k}-k)\ket{0}^{k}=0,\qquad\qquad a_{n}\ket{0}^{k}=0\quad\text{for all $n\geq 0$, $a\in\mathfrak{sl}_{2}$}.$$

(18)

We define $\mathbb{V}_{0}^{k}$, the vacuum Verma module at level $k$, as the induced smooth $\widehat{\mathfrak{sl}}_{2}$-module

$$\mathbb{V}_{0}^{k}=U(\widehat{\mathfrak{sl}}_{2})\otimes_{U(\mathfrak{sl}_{2}[t]\oplus\mathbb{C}\mathsf{k})}\mathbb{C}\ket{0}^{k}$$

(19)

This vector space is spanned by monomials of the form $a_{p}\cdots b_{q}\ket{0}^{k}$, with $a,\dots,b\in\mathfrak{sl}_{2}$ and strictly negative mode numbers $p,\dots,q\in\mathbb{Z}_{<0}$. We call these vectors states.

Let us denote by $[T,\cdot]$ the derivation on $U_{k}(\widehat{\mathfrak{sl}}_{2})$ defined by $[T,a_{n}]=-na_{n-1}$ and $[T,1]=0$. By setting $T(X\ket{0}^{k})=[T,X]\ket{0}^{k}$ for any $X\in U_{k}(\widehat{\mathfrak{sl}}_{2})$, one can interpret $T$ as a translation operator $T:\mathbb{V}_{0}^{k}\rightarrow\mathbb{V}_{0}^{k}$: the reason for this identification will be clear in the next section.

As we now recall, the vacuum Verma module defined in the previous section has the structure of vertex algebra. Namely, we have the state-field map $Y(\cdot,x)$, which for every state $A\in\mathbb{V}_{0}^{k}$ associates a formal power series in the variable $x$,

$$\begin{split}Y(\cdot,x):\mathbb{V}_{0}^{k}&\longrightarrow\operatorname{Hom}(\mathbb{V}_{0}^{k},\mathbb{V}_{0}^{k}((x)))\\ A&\longmapsto Y(A,x)=\sum_{n\in\mathbb{Z}}A_{(n)}x^{-n-1}\end{split}$$

(20)

where $A_{(n)}\in\operatorname{End}(\mathbb{V})$ is the $n$^th mode of $A$. By definition, if $A=a_{-1}\ket{0}^{k}$ for some $a\in\mathfrak{sl}_{2}$, then $A_{(n)}=a_{n}$ for all $n\in\mathbb{Z}$, i.e.

$$Y(a_{-1}\ket{0}^{k},x)=\sum_{n\in\mathbb{Z}}a_{n}x^{-n-1}.$$

(21)

The fields for all other states can be obtained with the following properties

$$Y(TA,x)=\partial_{z}Y(A,x),\qquad\qquad Y(A_{(-1)}B,x)=:Y(A,x)Y(B,x):$$

(22)

where we have introduced the normal ordered product between fields

$$:Y(A,x)Y(B,x):\,=\left(\sum_{m<0}A_{(m)}z^{-m-1}\right)Y(B,x)+Y(B,x)\left(\sum_{m\geq 0}A_{(m)}x^{-m-1}\right).$$

(23)

In fact, any state $C\in\mathbb{V}_{0}^{k}$ can be written as $C=a_{-n}B$, and using eqs. 22 and 23 we can always explicitly compute $Y(C,x)$. Summarising what we have introduced so far, we have

• •

  a space of states $\mathbb{V}_{0}^{k}$,

• •

  a vacuum vector $\ket{0}^{k}\in\mathbb{V}_{0}^{k}$,

• •

  a translation operator $T\in\operatorname{End}(\mathbb{V}_{0}^{k})$,

• •

  the state field map $Y(\cdot,x)$ as in eq. 20.

This structure, together with some additional properties (see e.g. [FB01, §2.4.4]), defines a vertex algebra on $\mathbb{V}_{0}^{k}$.

Let us introduce a set of complex numbers $\bm{k}=\{k_{i}\}_{i=1}^{N}$, where $N\in\mathbb{Z}_{>0}$ and $k_{i}\neq-2$ for all $i=1,\dots,N$. Consider the following tensor product of vacuum Verma modules

$$\mathbb{V}_{0}^{\bm{k}}=\mathbb{V}_{0}^{k_{1}}\otimes\dots\otimes\mathbb{V}_{0}^{k_{N}}.$$

(24)

This space can be interpreted as a module over the direct sum of $N$ copies of $\widehat{\mathfrak{sl}}_{2}$. Let us denote by $A^{(i)}\in\widehat{\mathfrak{sl}}_{2}^{\oplus N}$ the copy of $A\in\widehat{\mathfrak{sl}}_{2}$ in the $i$^th direct summand.

Let us denote by $\mathbb{C}\ket{0}^{\bm{k}}$ the one-dimensional vacuum representation of the “positive modes” Lie subalgebra $(\mathfrak{sl}_{2}[t]\oplus\mathbb{C}\mathsf{k})^{\oplus N}\subset\widehat{\mathfrak{sl}}_{2}^{\oplus N}$, defined by $(\mathsf{k}^{(i)}-k_{i})\ket{0}^{\bm{k}}=0$ and $a_{n}^{(i)}\ket{0}^{\bm{k}}=0$ for all $n\geq 0$, $a\in\mathfrak{sl}_{2}$ and $i=1,\dots,N$. Therefore $\mathbb{V}_{0}^{\bm{k}}$ is the induced $\widehat{\mathfrak{sl}}_{2}^{\oplus N}$-module, namely

$$\mathbb{V}_{0}^{\bm{k}}=U(\widehat{\mathfrak{sl}}_{2}^{\oplus N})\otimes_{U(\mathfrak{sl}_{2}[t]\oplus\mathbb{C}\mathsf{k})^{\oplus N}}\mathbb{C}\ket{0}^{\bm{k}}.$$

(25)

Repeating similar arguments to those of the previous sections, we can define $U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$ as the quotient of $U(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$ by the two-sided ideal generated by $\mathsf{k}^{(i)}-k_{i}$ for all $i=1,\dots,N$. We have the isomorphism

$$U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})\cong U_{k_{1}}(\widehat{\mathfrak{sl}}_{2})\otimes U_{k_{2}}(\widehat{\mathfrak{sl}}_{2})\otimes\dots\otimes U_{k_{N}}(\widehat{\mathfrak{sl}}_{2}).$$

(26)

Thanks to this fact, $A^{(i)}\in\widehat{\mathfrak{sl}}_{2}^{\oplus N}\subset U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$ can be presented as

$$A^{(i)}=\mathds{1}\otimes\dots\otimes\mathds{1}\otimes A\otimes\mathds{1}\otimes\dots\otimes\mathds{1},$$

(27)

where $A$ is acting as the $i$th tensor factor. Again, we can introduce the left ideals $J_{n}^{N}\in U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$ generated by $a_{r}^{(i)}$ for all $r\geq n$, $a\in\mathfrak{sl}_{2}$ and $i=1,\dots,N$. Let $\widetilde{U}_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})=\varprojlim U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})/J_{n}^{N}$ be the inverse limit. This space is a complete topological algebra and

$$\widetilde{U}_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})\cong\widetilde{U}_{k_{1}}(\widehat{\mathfrak{sl}}_{2})\widehat{\otimes}\cdots\widehat{\otimes}\widetilde{U}_{k_{N}}(\widehat{\mathfrak{sl}}_{2}),$$

(28)

where $\hat{\otimes}$ denotes the completed tensor product. This space $\widetilde{U}_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$ is called the algebra of observables of the Gaudin model.

The tensor product $\mathbb{V}_{0}^{\bm{k}}$ is again a vertex algebra. The state-field map $Y(\cdot,x):\mathbb{V}_{0}^{\bm{k}}\rightarrow\operatorname{Hom}(\mathbb{V}_{0}^{\bm{k}},\mathbb{V}_{0}^{\bm{k}}((x)))$ is just as in section 2.4 but decorated with the extra index ⁽ⁱ⁾.

Let us introduce the map $\Delta:\widehat{\mathfrak{sl}}_{2}\hookrightarrow\widehat{\mathfrak{sl}}_{2}^{\oplus N}$, which is the diagonal embedding of $\widehat{\mathfrak{sl}}_{2}$ into $\widehat{\mathfrak{sl}}_{2}^{\oplus N}$, defined as

$$\Delta x=\sum_{i=1}^{N}x^{(i)},\qquad\text{for all }x\in\widehat{\mathfrak{sl}}_{2}.$$

(29)

It extends to an embedding $\Delta:U(\widehat{\mathfrak{sl}}_{2})\hookrightarrow U(\widehat{\mathfrak{sl}}_{2}^{\oplus N})\cong U(\widehat{\mathfrak{sl}}_{2})^{\otimes N}$. It is easy to check that

$$[\Delta X_{m},\Delta Y_{n}]=\Delta[X,Y]_{n+m}-n\delta_{n+m,0}(X|Y)\sum_{i=1}^{N}\mathsf{k}^{(i)},$$

(30)

where $(\cdot|\cdot)$ is the usual Killing form as in eq. 6.

Therefore $\Delta$ descends to an embedding of the quotients $\Delta:U_{|\bm{k}|}(\widehat{\mathfrak{sl}}_{2})\hookrightarrow U_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N})$, where $|\bm{k}|=\sum_{i=1}^{N}k_{i}$, and hence of their completions

$$\Delta:\widetilde{U}_{|\bm{k}|}(\widehat{\mathfrak{sl}}_{2})\hookrightarrow\widetilde{U}_{\bm{k}}(\widehat{\mathfrak{sl}}_{2}^{\oplus N}).$$

(31)

Let $\{I^{a}\}_{a=1}^{3}$ be a basis of $\mathfrak{sl}_{2}$, and let $\{I_{a}\}_{a=1}^{3}$ be its dual basis with respect to the non-degenerate Killing form of eq. 6. Let $f^{ab}{}_{c}$ denote the structure constants, so that

$$[I^{a},I^{b}]=f^{ab}{}_{c}I^{c}.$$

(32)

Here and in what follows we employ the summation convention on Lie algebra indices. Thanks to the non-degeneracy of the bilinear form, we may suppose our basis is chosen in such a way that

$$(I^{a}|I^{b})=\delta^{ab}.$$

(33)

By doing this, we no longer have to distinguish between upper and lower indices. The structure constants are then

$$f^{ab}{}_{c}=f^{abc}=i\sqrt{2}\epsilon^{abc},$$

(34)

where $\epsilon^{abc}$ is the usual Levi-Civita symbol.

(Concretely, in the defining representation we have

$$I^{1}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\qquad I^{2}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\qquad I^{3}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}.$$

(35)

Using (6), it is easy to check that eq. 33 holds.)

Recall that for any finite-dimensional Lie algebra $\mathring{\mathfrak{g}}$, a tensor $t:\mathring{\mathfrak{g}}\times\dots\times\mathring{\mathfrak{g}}\rightarrow\mathbb{C}$ is invariant if

$$t([a,x],y,\dots,z)+t(x,[a,y],\dots,z)+\dots+t(x,y,\dots,[a,z])=0,\qquad\text{for all }x\in\mathring{\mathfrak{g}},$$

(36)

or equivalently, if its components $t^{a_{1}\dots a_{n}}:=t(I^{a_{1}},\dots,I^{a_{n}})$ satisfy

$$f^{ca_{1}}{}_{b}t^{ba_{2}\dots a_{n}}+f^{ca_{2}}{}_{b}t^{a_{1}b\dots a_{n}}+\dots+f^{ca_{n}}{}_{b}t^{a_{1}a_{2}\dots b}=0,$$

(37)

where the indices take values from $1$ to $\dim\mathring{\mathfrak{g}}$. In our case of $\mathfrak{sl}_{2}$, the ring of invariant tensors is generated by $\delta^{ab}$ and $f^{abc}$. We shall need the following syzygy relations between them:

$$\begin{gathered}f^{abc}f^{cde}=2(\delta^{ae}\delta^{bd}-\delta^{ad}\delta^{be}),\qquad\qquad f^{abc}f^{abd}=-4\delta^{cd},\\ f^{abc}\delta^{de}-f^{bcd}\delta^{ae}+f^{cda}\delta^{be}-f^{dab}\delta^{ce}=0.\end{gathered}$$

(38)

Note in particular the last of these, which will play a crucial role in the explicit calculations of the next sections. It can also be generalized to higher rank tensors (see e.g. [Itō93, §369 F]).

Let us introduce a set $\{z_{1},\dots,z_{N}\}$ of $N\in\mathbb{Z}_{>0}$ points $z_{i}\in\mathbb{C}$ in the complex plane, chosen to be pairwise distinct, $z_{i}\neq z_{j}$ whenever $i\neq j$. For any element $A\in\widehat{\mathfrak{sl}}_{2}$ we introduce the $\widehat{\mathfrak{sl}}_{2}^{\oplus N}$-valued meromorphic functions

$$A(z):=\sum_{i=1}^{N}\frac{A^{(i)}}{z-z_{i}}.$$

(39)

We are allowed to take derivatives of such functions, which will be denoted by $A^{\prime}(z)$ or, in general, for each $p\geq 0$,

$$A^{[p]}(z):=\left(\frac{d}{dz}\right)^{p}A(z)=\sum_{i=1}^{N}(-1)^{p}p!\frac{A^{(i)}}{(z-z_{i})^{p+1}}.$$

(40)

Considering two of these functions with different spectral parameters, we get the following commutation relations

$$\begin{split}[A^{[p]}(z),B^{[q]}(w)]=&(-1)^{p+1}(p+q)!\frac{[A,B](z)-[A,B](w)}{(z-w)^{p+q+1}}\\ &+\sum_{k=1}^{p}(-1)^{p+1-k}\binom{p}{k}(p+q-k)!\frac{[A,B]^{[k]}(z)}{(z-w)^{p+q+1-k}}\\ &-\sum_{k=1}^{q}(-1)^{p+1}\binom{q}{k}(p+q-k)!\frac{[A,B]^{[k]}(w)}{(z-w)^{p+q+1-k}}.\end{split}$$

(41)

By taking the limit $w\to z$, we get the commutation relations for the same spectral parameter, namely

$$[A^{[p]}(z),B^{[q]}(z)]=-[A,B]^{[p+q+1]}(z).$$

(42)

We see that these $A^{[p]}(z)$, for $A\in\widehat{\mathfrak{sl}}_{2}$ and $p\geq 0$, span a Lie algebra of $\widehat{\mathfrak{sl}}_{2}^{\oplus N}$-valued meromorphic functions of $z$ with poles at the marked points.

It is helpful to be able to treat this as an abstract Lie algebra. Thus, let $\mathfrak{L}$ denote the Lie algebra over $\mathbb{C}$ with basis consisting of $I_{n}^{a[p]}(z)$ and $\mathsf{k}^{[p]}(z)$, for $n\in\mathbb{Z}$, $p\in\mathbb{Z}_{\geq 0}$ and $a\in\{1,2,3\}$ with the non-vanishing Lie brackets given by

$$[I^{a[p]}_{m}(z),I^{b[q]}_{n}(z)]=-\frac{p!q!}{(p+q+1)!}(f^{ab}_{c}I^{c[p+q+1]}_{m+n}(z)-n\delta^{ab}\delta_{m+n,0}\mathsf{k}^{[p+q+1]}(z)).$$

(43)

Let $\mathfrak{L}_{+}$ denote subalgebra generated by $I_{n}^{a[p]}(z)$ for $n\geq 0$, $p\in\mathbb{Z}_{\geq 0}$ and $a\in\{1,2,3\}$, and let

$$\mathcal{V}:=U(\mathfrak{L})\otimes_{U(\mathfrak{L}_{+})}\mathbb{C}\ket{0}$$

(44)

denote the module over $\mathfrak{L}$ induced from the trivial one-dimensional module $\mathbb{C}\ket{0}$ over $\mathfrak{L}_{+}$.

We call the $\mathcal{V}$ the space of meromorphic states. It is again a vertex algebra, with the state-field map as in section 2.4 but decorated with extra indices. For each $z\in\mathbb{C}\setminus\{z_{1},\dots,z_{N}\}$, one has the homomorphism of Lie algebras $\mathfrak{L}\to\widehat{\mathfrak{sl}}_{2}^{\oplus N}$ given by evaluating at $z$. It gives rise to a map $\mathcal{V}\to\mathbb{V}_{0}^{\bm{k}}$ of vertex algebras.

There is a bi-gradation of $\mathfrak{L}$ in which $X_{-n}^{[p]}(z)$ (for any $X\in\mathfrak{sl}_{2}$) has weight $(n,p+1)$ and $\mathsf{k}^{[p]}(z)$ has weight $(0,p+1)$. This yields a bi-gradation of $\mathcal{V}$

$$\mathcal{V}=\bigoplus_{n\geq 0,p\geq 0}\mathcal{V}_{n,p}.$$

(45)

For each $n$, let $\mathcal{V}_{n}:=\mathcal{V}_{n,n}$ denote the subspace of grade $(n,n)$. We call elements of $\mathcal{V}_{n}$ homogeneous meromorphic states of degree $n$.

There is an evident diagonal action of the Lie algebra $\widehat{\mathfrak{sl}}_{2}$ on the $\mathfrak{L}$-module $\mathcal{V}$, defined in the same way as the action on $\mathbb{V}_{0}^{\bm{k}}$ in eq. 29. In particular, for any $X\in\mathfrak{sl}_{2}$, the zero modes stabilize each subspace $\mathcal{V}_{n,p}$, namely

$$\Delta X_{0}:\mathcal{V}_{n,p}\rightarrow\mathcal{V}_{n,p}.$$

(46)

An important fact is that every state in $\mathcal{V}_{n}$ properly contracted with an $\mathfrak{sl}_{2}$-invariant tensor vanishes under the diagonal action of the zero modes. This follows directly from the defining property of invariant tensors in eq. 37. Let denote with $\mathcal{V}_{n}^{\mathfrak{sl}_{2}}$ the invariant subspace. We can characterize this space for small $n$:

• •

  for $n=0$, $\mathcal{V}_{0}^{\mathfrak{sl}_{2}}=\mathcal{V}_{0}=\mathbb{C}\ket{0}$.

• •

  for $n=1$, $\mathcal{V}_{1}^{\mathfrak{sl}_{2}}=\{0\}$. Indeed, elements of $\mathcal{V}_{1}$ are of the form $t_{a}I^{a}_{-1}(z)\ket{0}$. Such an element is in $\mathcal{V}^{\mathfrak{sl}_{2}}_{1}$ if and only if $t_{a}$ are the components of an $\mathfrak{sl}_{2}$-invariant tensor of rank 1. But there are no nonzero such tensors.

• •

  for $n=2$, $\mathcal{V}_{2}^{\mathfrak{sl}_{2}}$ has dimension 1 and it is spanned by the state

  $$\varsigma_{1}(z)=\delta_{ab}I^{a}_{-1}(z)I^{b}_{-1}(z)\ket{0}.$$

  (47)

• •

  for $n=3$, $\mathcal{V}^{\mathfrak{sl}_{2}}_{3}$ has dimension 2 and it is spanned by the states

  $$\begin{split}f^{abc}I^{a}_{-1}(z)I^{b}_{-1}(z)I^{c}_{-1}(z)\ket{0}=&f^{abc}\frac{1}{2}(I^{a}_{-1}(z)I^{b}_{-1}(z)-I^{b}_{-1}(z)I^{a}_{-1}(z))I^{c}_{-1}(z)\ket{0}\\ =&f^{abc}f^{abd}I^{d\prime}_{-2}(z)I^{c}_{-1}(z)\ket{0}\\ =&-4I^{c\prime}_{-2}(z)I^{c}_{-1}(z)\ket{0},\end{split}$$

  (48)

  and

  $$I^{c}_{-2}(z)I^{c}_{-1}(z)\mathsf{k}(z)\ket{0}.$$

  (49)

• •

  for $n=4$, $\mathcal{V}_{4}^{\mathfrak{sl}_{2}}$ has dimension 14. Below, we will make use of the following explicit choice of basis:

  $$\begin{gathered}\mathsf{v}_{1}:=\delta^{(ab}\delta^{cd)}I^{a}_{-1}(z)I^{b}_{-1}(z)I^{c}_{-1}(z)I^{d}_{-1}(z),\qquad\mathsf{v}_{2}:=f^{abc}I^{a}_{-2}(z)I^{b\prime}_{-1}(z)I^{c}_{-1}(z),\\ \mathsf{v}_{3}:=I^{a\prime\prime}_{-3}(z)I^{a}_{-1}(z),\qquad\mathsf{v}_{4}:=I^{a}_{-3}(z)I^{a\prime\prime}_{-1}(z),\qquad\mathsf{v}_{5}:=I^{a\prime\prime}_{-2}(z)I^{a}_{-2}(z),\\ \mathsf{v}_{6}:=I^{a\prime}_{-3}(z)I^{a\prime}_{-1}(z),\qquad\mathsf{v}_{7}:=I^{a\prime}_{-2}(z)I^{a\prime}_{-2}(z),\qquad\mathsf{v}_{8}:=I^{a}_{-3}(z)I^{a}_{-1}(z)\mathsf{k}^{\prime}(z),\\ \mathsf{v}_{9}:=I^{a}_{-2}(z)I^{a}_{-2}(z)\mathsf{k}^{\prime}(z),\qquad\mathsf{v}_{10}:=I^{a\prime}_{-3}(z)I^{a}_{-1}(z)\mathsf{k}(z),\\ \mathsf{v}_{11}:=I^{a}_{-3}(z)I^{a\prime}_{-1}(z)\mathsf{k}(z),\qquad\mathsf{v}_{12}:=I^{a\prime}_{-2}(z)I^{a}_{-2}(z)\mathsf{k}(z),\\ \mathsf{v}_{13}:=I^{a}_{-3}(z)I^{a}_{-1}(z)\mathsf{k}(z)^{2},\qquad\mathsf{v}_{14}:=I^{a}_{-2}(z)I^{a}_{-2}(z)\mathsf{k}(z)^{2}.\end{gathered}$$

  (50)

Note that to write these terms we have to choose an ordering prescription. In this work we sort level first in ascending order from left to right and after that, for a given level, we sort derivatives in descending order from left to right. For example $f^{abc}I^{a}_{-2}I^{b\prime\prime}_{-2}I^{c\prime}_{-3}=f^{abc}I^{c\prime}_{-3}I^{b\prime\prime}_{-2}I^{a}_{-2}+\text{terms obtained from commutations}$.