An $\mathfrak {sl}_2$-type tensor category for the Virasoro algebra at central charge 25 and applications

Abstract

Let ${\mathcal {O}}_{25}$ be the vertex algebraic braided tensor category of finite-length modules for the Virasoro Lie algebra at central charge 25 whose composition factors are the irreducible quotients of reducible Verma modules. We show that ${\mathcal {O}}_{25}$ is rigid and that its simple objects generate a semisimple tensor subcategory that is braided tensor equivalent to an abelian 3-cocycle twist of the category of finite-dimensional $\mathfrak {sl}_2$-modules. We also show that this $\mathfrak {sl}_2$-type subcategory is braid-reversed tensor equivalent to a similar category for the Virasoro algebra at central charge 1. As an application, we construct a simple conformal vertex algebra which contains the Virasoro vertex operator algebra of central charge 25 as a $PSL_2({\mathbb {C}})$-orbifold. We also use our results to study Arakawa’s chiral universal centralizer algebra of $SL_2$ at level $-1$, showing that it has a symmetric tensor category of representations equivalent to $\textrm{Rep}\,PSL_2({\mathbb {C}})$. This algebra is an extension of the tensor product of Virasoro vertex operator algebras of central charges 1 and 25, analogous to the modified regular representations of the Virasoro algebra constructed earlier for generic central charges by I. Frenkel–Styrkas and I. Frenkel–M. Zhu.

Appendices

Appendix A: Algebras in equivalent tensor categories

The tensor-categorical results in this appendix are straightforward and certainly known, but we include some details from their proofs to make this paper more self-contained. We also use this appendix to recall notation from, for example, [5, 38] for algebras in tensor categories and their modules.

Let ${\mathcal {C}}$ be a braided tensor category with tensor product bifunctor $\boxtimes$ and unit object ${\textbf{1}}$. We recall (from [5, 38], for example) that a (commutative) ${\mathcal {C}}$-algebra $(A,\mu _A,\iota _A)$ is an object A of ${\mathcal {C}}$ equipped with multiplication and unit morphisms

$$\begin{aligned} \mu _A: A\boxtimes A\longrightarrow A,\qquad \iota _A: {\textbf{1}}\longrightarrow A \end{aligned}$$

which satisfy natural unit and associativity (and commutativity) axioms. An isomorphism between two ${\mathcal {C}}$-algebras $(A,\mu _A,\iota _A)$ and $(B,\mu _B,\iota _B)$ is a ${\mathcal {C}}$-morphism $f: A\rightarrow B$ such that $f\circ \mu _A =\mu _B\circ (f\boxtimes f)$ and $f\circ \iota _A=\iota _B$.

Given a commutative algebra A, the tensor category ${{\,\textrm{Rep}\,}}A$ of (left) A-modules has objects $(X,\mu _X)$ where X is an object of ${\mathcal {C}}$ and $\mu _X: A\boxtimes X\rightarrow X$ is a morphism satisfying natural left unit and associativity axioms. The category ${{\,\textrm{Rep}\,}}A$ has a braided tensor subcategory ${{\,\textrm{Rep}\,}}^0A$ consisting of all objects $(X,\mu _X)$ such that

$$\begin{aligned} \mu _X\circ {\mathcal {M}}_{A,X}=\mu _X, \end{aligned}$$

where ${\mathcal {M}}_{A,X}:={\mathcal {R}}_{X,A}\circ {\mathcal {R}}_{A,X}$ is the natural double braiding, or monodromy, isomorphism in ${\mathcal {C}}$. Given an object $(X,\mu _X)$ of ${{\,\textrm{Rep}\,}}A$, an A-submodule of X is an object $(W,\mu _W)$ of ${{\,\textrm{Rep}\,}}A$ equipped with a ${{\,\textrm{Rep}\,}}A$-injection $i: W\rightarrow X$. We say that $(X,\mu _X)$ is simple if every submodule $i: W\rightarrow X$ is either 0 or an isomorphism. If A is commutative, we can identify ideals of A with A-submodules of A, so that A is simple as a ${\mathcal {C}}$-algebra if and only if it simple as an A-module.

Let ${\mathcal {F}}:{\mathcal {C}}\rightarrow {\mathcal {D}}$ be a tensor functor, so that there is an isomorphism $\varphi :{\textbf{1}}_{\mathcal {D}}\rightarrow {\mathcal {F}}({\textbf{1}}_{\mathcal {C}})$ and a natural isomorphism

$$\begin{aligned} F: \boxtimes _{\mathcal {D}}\circ ({\mathcal {F}}\times {\mathcal {F}})\longrightarrow {\mathcal {F}}\circ \boxtimes _{\mathcal {C}}\end{aligned}$$

(A.1)

which are suitably compatible with the unit and associativity isomorphisms of ${\mathcal {C}}$ and ${\mathcal {D}}$. If $(A,\mu _A,\iota _A)$ is a ${\mathcal {C}}$-algebra, then $({\mathcal {F}}(A),\mu _{{\mathcal {F}}(A)},\iota _{{\mathcal {F}}(A)})$ is a ${\mathcal {D}}$-algebra, where

$$\begin{aligned} \mu _{{\mathcal {F}}(A)}: {\mathcal {F}}(A)\boxtimes _{\mathcal {D}}{\mathcal {F}}(A)\xrightarrow {F_{A,A}} {\mathcal {F}}(A\boxtimes _{\mathcal {C}}A) \xrightarrow {{\mathcal {F}}(\mu _A)} {\mathcal {F}}(A) \end{aligned}$$

and

$$\begin{aligned} \iota _{{\mathcal {F}}(A)}: {\textbf{1}}_{\mathcal {D}}\xrightarrow {\varphi }{\mathcal {F}}({\textbf{1}}_{\mathcal {C}})\xrightarrow {{\mathcal {F}}(\iota _A)} {\mathcal {F}}(A). \end{aligned}$$

If A is commutative and ${\mathcal {F}}$ is braided or braid-reversing, then ${\mathcal {F}}(A)$ is also commutative.

If $(X,\mu _X)$ is an object of ${{\,\textrm{Rep}\,}}A$, then $({\mathcal {F}}(X),\mu _{{\mathcal {F}}(X)})$ is an object of ${{\,\textrm{Rep}\,}}{\mathcal {F}}(A)$, where

$$\begin{aligned} \mu _{{\mathcal {F}}(X)}: {\mathcal {F}}(A)\boxtimes _{\mathcal {D}}{\mathcal {F}}(X)\xrightarrow {F_{A,X}} {\mathcal {F}}(A\boxtimes _{\mathcal {C}}X)\xrightarrow {{\mathcal {F}}(\mu _X)} {\mathcal {F}}(X). \end{aligned}$$

Moreover, if $f: X_1\rightarrow X_2$ is a morphism in ${{\,\textrm{Rep}\,}}A$, then ${\mathcal {F}}(f): {\mathcal {F}}(X_1)\rightarrow {\mathcal {F}}(X_2)$ is a morphism in ${{\,\textrm{Rep}\,}}{\mathcal {F}}(A)$. If A is commutative, ${\mathcal {F}}$ is braided or braid-reversing, and $(X,\mu _X)$ is an object of ${{\,\textrm{Rep}\,}}^0A$, then $({\mathcal {F}}(X),\mu _{{\mathcal {F}}(X)})$ is an object of ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. Thus if ${\mathcal {F}}$ is braided or braid-reversing, it restricts to a functor from ${{\,\textrm{Rep}\,}}^0 A$ to ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. In the case that ${\mathcal {F}}$ is an equivalence of categories, we get:

Proposition A.1

Let ${\mathcal {F}}:{\mathcal {C}}\rightarrow {\mathcal {D}}$ be a braided or braid-reversing tensor equivalence. If A is a commutative ${\mathcal {C}}$-algebra, then ${\mathcal {F}}:{{\,\textrm{Rep}\,}}^0 A\rightarrow {{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$ is an equivalence of categories.

Proof

To show that ${\mathcal {F}}$ is essentially surjective onto ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, suppose $(W,\mu _W)$ is an object of ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. Since ${\mathcal {F}}$ is essentially surjective onto ${\mathcal {D}}$, there is an isomorphism $f: W\rightarrow {\mathcal {F}}(X)$ in ${\mathcal {D}}$ for some object X in ${\mathcal {C}}$, and then $f: (W,\mu _W)\rightarrow ({\mathcal {F}}(X),\mu _{{\mathcal {F}}(X)})$ is an isomorphism in ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$ where

$$\begin{aligned} \mu _{{\mathcal {F}}(X)}=f\circ \mu _W\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}f^{-1}). \end{aligned}$$

Using the full faithfulness of ${\mathcal {F}}$, we define $\mu _X: A\boxtimes _{\mathcal {C}}X\rightarrow X$ to be the unique ${\mathcal {C}}$-morphism such that

$$\begin{aligned} {\mathcal {F}}(\mu _X)=\mu _{{\mathcal {F}}(X)}\circ F_{A,X}^{-1}. \end{aligned}$$

It is then straightforward to check that $(X,\mu _X)$ is an object of ${{\,\textrm{Rep}\,}}A$ such that ${\mathcal {F}}(X,\mu _X)=({\mathcal {F}}(X),\mu _{{\mathcal {F}}(X)})\cong (W,\mu _W)$. For example, to prove the associativity of $\mu _X$, the faithfulness of ${\mathcal {F}}$ implies it is enough to show

$$\begin{aligned} {\mathcal {F}}(\mu _X\circ (\textrm{Id}_A\boxtimes _{\mathcal {C}}\mu _X)) = {\mathcal {F}}(\mu _X\circ (\mu _A\boxtimes _{\mathcal {C}}\textrm{Id}_X)\circ {\mathcal {A}}_{A,A,X}), \end{aligned}$$

which follows from the definitions and the associativity of $({\mathcal {F}}(X),\mu _{{\mathcal {F}}(X)})$. Similarly, since W, equivalently ${\mathcal {F}}(X)$, is an object of ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, then

$$\begin{aligned} {\mathcal {F}}(\mu _X)={\mathcal {F}}(\mu _X\circ {\mathcal {M}}_{A,X}^{\pm 1}) \end{aligned}$$

since ${\mathcal {F}}$ is braided or braid-reversing, and it follows that X is an object of ${{\,\textrm{Rep}\,}}^0A$.

Now since ${\mathcal {F}}: {\mathcal {C}}\rightarrow {\mathcal {D}}$ is faithful, ${\mathcal {F}}$ remains faithful when restricted to ${{\,\textrm{Rep}\,}}^0A$. Then to show that the restriction to ${{\,\textrm{Rep}\,}}^0 A$ is full, let $g: {\mathcal {F}}(X_1)\rightarrow {\mathcal {F}}(X_2)$ be a morphism in ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, where $X_1$ and $X_2$ are objects of ${{\,\textrm{Rep}\,}}^0 A$. Then $g={\mathcal {F}}(f)$ for some morphism $f: X_1\rightarrow X_2$ in ${\mathcal {C}}$, and

$$\begin{aligned} {\mathcal {F}}(f\circ \mu _{X_1})&=g\circ \mu _{{\mathcal {F}}(X_1)}\circ F_{A,X_1}^{-1} =\mu _{{\mathcal {F}}(X_2)}\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}g)\circ F_{A,X_1}^{-1} \\&= \mu _{{\mathcal {F}}(X_2)}\circ F_{A,X_2}^{-1}\circ {\mathcal {F}}(\textrm{Id}_A\boxtimes _{\mathcal {C}}f) ={\mathcal {F}}(\mu _{X_2}\circ (\textrm{Id}_A\boxtimes _{\mathcal {C}}f)). \end{aligned}$$

Since ${\mathcal {F}}$ is faithful, it follows that f is a morphism in ${{\,\textrm{Rep}\,}}^0 A$. $\square$

Since equivalences of abelian categories preserve zero objects and thus also zero morphisms, if ${\mathcal {F}}:{\mathcal {C}}\rightarrow {\mathcal {D}}$ is an equivalence, then a morphism i in ${\mathcal {C}}$ is injective if and only if ${\mathcal {F}}(i)$ is injective in ${\mathcal {D}}$. Consequently, ${\mathcal {F}}$ also preserves simple objects. Thus:

Corollary A.2

Let ${\mathcal {F}}:{\mathcal {C}}\rightarrow {\mathcal {D}}$ be a braided or braid-reversing tensor functor. If A is a commutative ${\mathcal {C}}$-algebra, then an object X of ${{\,\textrm{Rep}\,}}^0 A$ is simple if and only if ${\mathcal {F}}(X)$ is a simple object of ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. In particular, A is a simple ${\mathcal {C}}$-algebra if and only if ${\mathcal {F}}(A)$ is a simple ${\mathcal {D}}$-algebra.

The next result is that if ${\mathcal {F}}: {\mathcal {C}}\rightarrow {\mathcal {D}}$ is a braided or braid-reversing tensor functor and A is a commutative ${\mathcal {C}}$-algebra, then ${\mathcal {F}}$ is also braided or braid-reversing on ${{\,\textrm{Rep}\,}}^0A$. The proof is straightforward but somewhat long, so we only include the more non-trivial details:

Theorem A.3

Let ${\mathcal {F}}: {\mathcal {C}}\rightarrow {\mathcal {D}}$ be a right exact braided, respectively braid-reversing, tensor functor. If A is a commutative ${\mathcal {C}}$-algebra, then ${\mathcal {F}}:{{\,\textrm{Rep}\,}}^0 A\rightarrow {{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$ has the structure of a braided, respectively braid-reversing, tensor functor.

Proof

Letting $\boxtimes _A$ and $\boxtimes _{{\mathcal {F}}(A)}$ denote the tensor products in ${{\,\textrm{Rep}\,}}^0 A$ and ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, respectively, we need to obtain a natural isomorphism

$$\begin{aligned} F^A: \boxtimes _{{\mathcal {F}}(A)}\circ ({\mathcal {F}}\times {\mathcal {F}})\longrightarrow {\mathcal {F}}\circ \boxtimes _A. \end{aligned}$$

from the original natural isomorphism F of (A.1). To do so, we recall (from [5, Section 2.3] for example) that for objects $X_1$ and $X_2$ in ${{\,\textrm{Rep}\,}}^0 A$, $X_1\boxtimes _A X_2$ is the cokernel of $\mu ^{(1)}-\mu ^{(2)}$, where $\mu ^{(1)}=\mu _{X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2}$ and $\mu ^{(2)}$ is the composition

$$\begin{aligned} (A\boxtimes _{\mathcal {C}}X_1)\boxtimes _{{\mathcal {C}}} X_2&\xrightarrow {{\mathcal {R}}_{A,X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2}} (X_1\boxtimes _{\mathcal {C}}A)\boxtimes _{{\mathcal {C}}} X_2 \\&\xrightarrow {{\mathcal {A}}_{X_1,A,X_2}^{-1}} X_1\boxtimes _{\mathcal {C}}(A\boxtimes _{\mathcal {C}}X_2)\xrightarrow {\textrm{Id}_{X_1}\boxtimes _{\mathcal {C}}\mu _{X_2}} X_1\boxtimes _{\mathcal {C}}X_2. \end{aligned}$$

We use $\eta _{X_1,X_2}: X_1\boxtimes _{\mathcal {C}}X_2\rightarrow X_1\boxtimes _A X_2$ to denote the cokernel morphism. Similar definitions and notation apply to objects in ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. We would like to define morphisms $F^A_{X_1,X_2}$ and $G^A_{X_1,X_2}$ in ${\mathcal {D}}$ such that the diagrams

and

commute.

From the cokernel definitions of ${\mathcal {F}}(X_1)\boxtimes _A{\mathcal {F}}(X_2)$ and $X_1\boxtimes _A X_2$ and by the right exactness of ${\mathcal {F}}$ (so that ${\mathcal {F}}(X_1\boxtimes _A X_2)$ is still a cokernel), the existence and uniqueness of $F^A_{X_1,X_2}$ and $G^A_{X_1,X_2}$ is equivalent to the identities

$$\begin{aligned} {\mathcal {F}}( \eta _{X_1,X_2})\circ F_{X_1,X_2}\circ \mu ^{(1)}&={\mathcal {F}}(\eta _{X_1,X_2})\circ F_{X_1,X_2}\circ \mu ^{(2)},\\ \eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\circ F^{-1}_{X_1,X_2}\circ {\mathcal {F}}(\mu ^{(1)})&=\eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\circ F^{-1}_{X_1,X_2}\circ {\mathcal {F}}(\mu ^{(2)}). \end{aligned}$$

When ${\mathcal {F}}$ is braid-reversing (the more interesting case), the first identity is proved as follows:

$$\begin{aligned}&{\mathcal {F}}( \eta _{X_1,X_2}) \circ F_{X_1,X_2}\circ (\textrm{Id}_{{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}\mu _{{\mathcal {F}}(X_2)})\circ {\mathcal {A}}_{{\mathcal {F}}(X_1),{\mathcal {F}}(A),{\mathcal {F}}(X_2)}^{-1}\circ ({\mathcal {R}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)}) \\&\quad = {\mathcal {F}}(\eta _{X_1,X_2}) \circ F_{X_1,X_2}\circ (\textrm{Id}_{{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}{\mathcal {F}}(\mu _{X_2}))\circ (\textrm{Id}_{{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}F_{A,X_2})\circ \\&\quad \quad \circ {\mathcal {A}}_{{\mathcal {F}}(X_1),{\mathcal {F}}(A),{\mathcal {F}}(X_2)}^{-1}\circ ({\mathcal {R}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)}) \\&\quad ={\mathcal {F}}(\eta _{X_1,X_2}\circ (\textrm{Id}_{X_1}\boxtimes _{\mathcal {C}}\mu _{X_2}))\circ F_{X_1,A\boxtimes _{\mathcal {C}}X_2}\circ (\textrm{Id}_{{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}F_{A,X_2})\circ \\&\quad \quad \circ {\mathcal {A}}_{{\mathcal {F}}(X_1),{\mathcal {F}}(A),{\mathcal {F}}(X_2)}^{-1}\circ ({\mathcal {R}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)}) \\&\quad ={\mathcal {F}}(\eta _{X_1,X_2}\circ (\textrm{Id}_{X_1}\boxtimes _{\mathcal {C}}\mu _{X_2})\circ {\mathcal {A}}_{X_1,A,X_2}^{-1}\circ ({\mathcal {R}}^{-1}_{X_1,A}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2}))\circ \\&\quad \quad \circ F_{A\boxtimes _{\mathcal {C}}X_1,X_2}\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2}\circ (\textrm{Id}_{X_1}\boxtimes _{\mathcal {C}}\mu _{X_2})\circ {\mathcal {A}}_{X_1,A,X_2}^{-1}\circ ({\mathcal {R}}_{A,X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2}))\circ \\&\quad \quad \circ {\mathcal {F}}({\mathcal {M}}_{A,X_1}^{-1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2})\circ F_{A\boxtimes _{\mathcal {C}}X_1,X_2}\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2}\circ (\mu _{X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2})\circ ({\mathcal {M}}_{A,X_1}^{-1}\boxtimes \textrm{Id}_{X_2}))\circ F_{A\boxtimes _{\mathcal {C}}X_1,X_2}\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2})\circ F_{X_1,X_2}\circ ({\mathcal {F}}(\mu _{X_1})\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2})\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2})\circ F_{X_1,X_2}\circ \mu ^{(1)}, \end{aligned}$$

where the next to last equality uses the assumption that $X_1$ is an object of ${{\,\textrm{Rep}\,}}^0A$. The second identity is proved similarly, so we get the desired morphism $F^A_{X_1,X_2}$ in ${\mathcal {D}}$ with inverse $G^A_{X_1,X_2}$.

To prove that $F^A_{X_1,X_2}$ is an isomorphism in ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, we use the definitions of $\mu _{X_1\boxtimes _A X_2}$ and $\mu _{{\mathcal {F}}(X_1)\boxtimes _{{\mathcal {F}}(A)}{\mathcal {F}}(X_2)}$ from [5, Section 2.3], along with compatibility of F with the associativity isomorphisms, to calculate

$$\begin{aligned}&\mu _{{\mathcal {F}}(X_1\boxtimes _A X_2)} \circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}F^A_{X_1,X_2})\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}\eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)})\\&\quad ={\mathcal {F}}(\mu _{X_1\boxtimes _A X_2})\circ F_{A,X_1\boxtimes _A X_2}\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}{\mathcal {F}}(\eta _{X_1,X_2}))\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}F_{X_1,X_2})\\&\quad ={\mathcal {F}}(\mu _{X_1\boxtimes _A X_2}\circ (\textrm{Id}_A\boxtimes _{\mathcal {C}}\eta _{X_1,X_2}))\circ F_{A,X_1\boxtimes _{\mathcal {C}}X_2}\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}F_{X_1,X_2})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2})\circ {\mathcal {F}}(\mu _{X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2})\circ {\mathcal {F}}({\mathcal {A}}_{A,X_1,X_2})\circ F_{A,X_1\boxtimes _{\mathcal {C}}X_2}\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}F_{X_1,X_2})\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2})\circ {\mathcal {F}}(\mu _{X_1}\boxtimes _{\mathcal {C}}\textrm{Id}_{X_2})\circ F_{A\boxtimes _{\mathcal {C}}X_1,X_2}\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\circ {\mathcal {A}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\\&\quad ={\mathcal {F}}(\eta _{X_1,X_2})\circ F_{X_1,X_2}\circ ({\mathcal {F}}(\mu _{X_1})\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\circ (F_{A,X_1}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\circ {\mathcal {A}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\\&\quad = F^A_{X_1,X_2}\circ \eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\circ (\mu _{{\mathcal {F}}(X_1)}\boxtimes _{\mathcal {D}}\textrm{Id}_{{\mathcal {F}}(X_2)})\circ {\mathcal {A}}_{{\mathcal {F}}(A),{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}\\&\quad = F^A_{X_1,X_2}\circ \mu _{{\mathcal {F}}(X_1)\boxtimes _{{\mathcal {F}}(A)}{\mathcal {F}}(X_2)}\circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}\eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}). \end{aligned}$$

Because $\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}\eta _{{\mathcal {F}}(X_1),{\mathcal {F}}(X_2)}$ is surjective, it follows that

$$\begin{aligned} \mu _{{\mathcal {F}}(X_1\boxtimes _A X_2)} \circ (\textrm{Id}_{{\mathcal {F}}(A)}\boxtimes _{\mathcal {D}}F^A_{X_1,X_2}) =F^A_{X_1,X_2}\circ \mu _{{\mathcal {F}}(X_1)\boxtimes _{{\mathcal {F}}(A)}{\mathcal {F}}(X_2)} \end{aligned}$$

as required. The proof that $F^A$ defines a natural transformation is also straightforward from the definitions in [5, Section 2.3] and the fact that F is a natural transformation.

To show that the natural isomorphism $F^A$ gives ${\mathcal {F}}:{{\,\textrm{Rep}\,}}^0 A\rightarrow {{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$ the structure of a braided (or braid-reversing) tensor functor, we also need $F^A$ to be compatible with the unit, associativity and braiding isomorphisms. Since the compatibility proofs are straightforward, we only discuss compatibility with the right unit and braiding isomorphisms in the case that ${\mathcal {F}}$ is braid-reversing. For the right unit isomorphisms, we need to show

$$\begin{aligned} {\mathcal {F}}(r^A_X)\circ F^A_{X,A}=r_{{\mathcal {F}}(X)}^{{\mathcal {F}}(A)} \end{aligned}$$

for any object X of ${{\,\textrm{Rep}\,}}^0 A$, where $r^A$ and $r^{{\mathcal {F}}(A)}$ are the right unit isomorphisms in ${{\,\textrm{Rep}\,}}^0 A$ and ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$, respectively. Indeed, from the definitions in [5, Section 2.3] and the assumption that X is an object of ${{\,\textrm{Rep}\,}}^0A$, we get

$$\begin{aligned} {\mathcal {F}}(r^A_X)\circ F^A_{X,A}\circ \eta _{{\mathcal {F}}(X),{\mathcal {F}}(A)}&= {\mathcal {F}}(r^A_X)\circ {\mathcal {F}}(\eta _{X,A})\circ F_{X,A} \\&={\mathcal {F}}(\mu _X)\circ {\mathcal {F}}({\mathcal {R}}_{A,X}^{-1})\circ F_{X,A} \\&={\mathcal {F}}(\mu _X)\circ {\mathcal {F}}({\mathcal {R}}_{X,A})\circ F_{X,A} \\&={\mathcal {F}}(\mu _X)\circ F_{A,X}\circ {\mathcal {R}}_{{\mathcal {F}}(A),{\mathcal {F}}(X)}^{-1} \\&=\mu _{{\mathcal {F}}(X)}\circ {\mathcal {R}}_{{\mathcal {F}}(A),{\mathcal {F}}(X)}^{-1} = r^{{\mathcal {F}}(A)}_{{\mathcal {F}}(X)}\circ \eta _{{\mathcal {F}}(X),{\mathcal {F}}(A)}. \end{aligned}$$

So the desired equality follows from surjectivity of $\eta _{{\mathcal {F}}(X),{\mathcal {F}}(A)}$. Similarly, the braid-reversing properties of F and the definitions in [6, Section 2.6] show that

$$\begin{aligned} {\mathcal {F}}({\mathcal {R}}^A_{X_1,X_2})\circ F^A_{X_1,X_2}= F^A_{X_2,X_1}\circ {\mathcal {R}}^{-1}_{{\mathcal {F}}(X_2),{\mathcal {F}}(X_1)} \end{aligned}$$

for objects $X_1$ and $X_2$ of ${{\,\textrm{Rep}\,}}^0 A$, so that ${\mathcal {F}}$ defines a braid-reversed tensor functor from ${{\,\textrm{Rep}\,}}^0 A$ to ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$. $\square$

Since equivalences between abelian categories are automatically exact, we get the following corollary of Proposition A.1 and Theorem A.3:

Corollary A.4

Let ${\mathcal {F}}:{\mathcal {C}}\rightarrow {\mathcal {D}}$ be a braided, respectively braid-reversing, tensor equivalence. If A is a commutative ${\mathcal {C}}$-algebra, then ${\mathcal {F}}$ induces a braided, respectively braid-reversing, tensor equivalence between ${{\,\textrm{Rep}\,}}^0 A$ and ${{\,\textrm{Rep}\,}}^0{\mathcal {F}}(A)$.

Appendix B: Uniqueness of $L_1(\mathfrak {sl}_2)$

In this appendix, we present some uniqueness results for the simple affine vertex operator algebra $L_1(\mathfrak {sl}_2)$ associated to $\mathfrak {sl}_2$ at level 1, and also for its non-trivial irreducible module. These results are perhaps known to experts, but we provide proofs here for completeness. The automorphism group of $L_1(\mathfrak {sl}_2)$ is $PSL_2({\mathbb {C}})$, with fixed-point subalgebra the Virasoro vertex operator algebra $V_1$ [10, 46]. As a $PSL_2({\mathbb {C}})\times V_1$-module,

$$\begin{aligned} L_1(\mathfrak {sl}_2)\cong \bigoplus _{n=0}^\infty V(2n)\otimes {\mathcal {L}}_{2n+1,1}. \end{aligned}$$

(B.1)

We first establish the uniqueness of a simple vertex operator algebra with this decomposition as a $V_1$-module, while ignoring $PSL_2({\mathbb {C}})$-actions:

Theorem B.1

If V is a simple vertex operator algebra extension of $V_1$ such that $V\cong \bigoplus _{n=0}^\infty V(2n)\otimes {\mathcal {L}}_{2n+1,1}$ as $V_1$-modules (with V(2n) considered just as a $(2n+1)$-dimensional vector space), then $V\cong L_1(\mathfrak {sl}_2)$ as vertex operator algebras.

Proof

For $n\in {\mathbb {N}}$, we use $v_{2n+1,1}$ to denote a generating vector of ${\mathcal {L}}_{2n+1,1}$ of minimal conformal weight, and we use $\pi _{2n+1}: V\rightarrow V(2n)\otimes {\mathcal {L}}_{2n+1,1}$ to denote the ${\mathcal {V}}ir$-module projection.

By assumption, the conformal-weight-1 space $V_{(1)}=V(2)\otimes v_{3,1}$ is three-dimensional, and as in [17, Remark 8.9.1] it is a Lie algebra with bracket $[u,v]=u_0 v$ and invariant symmetric bilinear form $\langle \cdot ,\cdot \rangle$ defined by $u_1 v=\langle u,v\rangle {\textbf{1}}$ for $u,v\in V_{(1)}$. We claim that $\langle \cdot ,\cdot \rangle$ is non-degenerate. Indeed, consider $a\otimes v_{3,1}, b\otimes v_{3,1}\in V_{(1)}$ for non-zero $a,b\in V(2)$. Then we have an identification of $V_1$-module intertwining operators

$$\begin{aligned} \pi _1\circ Y_V\vert _{(a\otimes {\mathcal {L}}_{3,1})\otimes (b\otimes {\mathcal {L}}_{3,1})} =\langle a\otimes v_{3,1},b\otimes v_{3,1}\rangle {\mathcal {Y}}_{33}^1 \end{aligned}$$

where ${\mathcal {Y}}_{33}^1$ is the unique $V_1$-module intertwining operator of type $\left( {\begin{array}{c}{\mathcal {L}}_{1,1}\\ {\mathcal {L}}_{3,1}\,{\mathcal {L}}_{3,1}\end{array}}\right)$ such that

$$\begin{aligned} {\mathcal {Y}}_{33}^1(v_{3,1},x)v_{3,1}\in x^{-2}({\textbf{1}}+x{\mathcal {L}}_{1,1}[[x]]). \end{aligned}$$

Since there is a non-zero intertwining operator of type $\left( {\begin{array}{c}{\mathcal {L}}_{1,1}\\ {\mathcal {L}}_{r,1}\,{\mathcal {L}}_{3,1}\end{array}}\right)$ only for $r=3$, it follows that if $\langle a\otimes v_{3,1}, b\otimes v_{3,1}\rangle =0$ for all $a\in V(2)$, then the V-submodule generated by $b\otimes v_{3,1}$ is contained in $\bigoplus _{n= 1}^\infty V(2n)\otimes {\mathcal {L}}_{2n+1,1}$. Since V is simple, this is impossible for $b\ne 0$, and thus for such b there exists $a\in V(2)$ such that $\langle a\otimes v_{3,1}, b\otimes v_{3,1}\rangle \ne 0$. This proves that $\langle \cdot ,\cdot \rangle$ is non-degenerate.

Since $\langle \cdot ,\cdot \rangle$ is non-degenerate, we can take $h\in V_{(1)}$ such that $\langle h,h\rangle =1$. We would like to show that $\omega =L_{-2}{\textbf{1}}$ is a multiple of $h_{-1}^2{\textbf{1}}$. Because $h_{2n+1,1}=n^2>2$ if $n>1$, we have

$$\begin{aligned} h_{-1}^2{\textbf{1}}=c\cdot \omega +L_{-1}u \end{aligned}$$

for some $c\in {\mathbb {C}}$ and $u\in V_{(1)}$. We first claim that $u=0$. To show this, observe that vertex operator modes acting on V satisfy

$$\begin{aligned} 2\sum _{m=0}^\infty h_{-m}h_{m+1}&=\textrm{Res}_x\,x^2 Y_V(h_{-1}^2{\textbf{1}},x) \\&=c\cdot \textrm{Res}_x\,x^2 Y_V(\omega ,x)+\textrm{Res}_x\,x^2\frac{d}{dx}Y_V(u,x) =c\cdot L_1-2 u_1. \end{aligned}$$

The left side of this equation annihilates $V_{(1)}$ because

$$\begin{aligned} h_0h_1 v =\langle h,v\rangle h_0{\textbf{1}}=0 \end{aligned}$$

for all $v\in V_{(1)}$, and $L_1$ annihilates $V_{(1)}$ as well since $V_{(1)}$ consists of Virasoro primary vectors. Thus $u_1$ also annihilates $V_{(1)}$, that is, $\langle u,v\rangle =0$ for all $v\in V_{(1)}$. Since $\langle \cdot ,\cdot \rangle$ is non-degenerate, this proves the claim.

We now have $h_{-1}^2{\textbf{1}}=c\cdot \omega$, and it remains to calculate c. Let $(\cdot ,\cdot )$ be the unique non-degenerate invariant bilinear form on ${\mathcal {L}}_{1,1}\subseteq V$ such that $({\textbf{1}},{\textbf{1}})=1$. Thus

$$\begin{aligned} (L_{-2}{\textbf{1}},L_{-2}{\textbf{1}})=({\textbf{1}},L_2 L_{-2}{\textbf{1}})=\frac{1}{2}, \end{aligned}$$

so that using (2.3) and the $L_{-1}$-derivative property,

$$\begin{aligned} \frac{c}{2}&= (L_{-2}{{\textbf {1}}},c\cdot {L}_{-2}{{\textbf {1}}}) =(L_{-2}{{\textbf {1}}},h_{-1}^2{{\textbf {1}}})\\ {}&=\text {Res}_x\,x^{-1}(L_{-2}{{\textbf {1}}},(\pi _1\circ Y_V)(h,x)h) =\text {Res}_x\,x^{-1}({{\textbf {1}}},L_2(\pi _1\circ Y_V)(h,x)h\rangle \\ {}&=\text {Res}\,x\left( {{\textbf {1}}}, x\frac{d}{dx} (\pi _1\circ Y_V)(h,x)h+3(\pi _1\circ Y_V)(L_0 h,x)h\right) \\ {}&=({{\textbf {1}}}, -2h_1 h+3h_1 h) =\langle h,h\rangle ({{\textbf {1}}},{{\textbf {1}}})=1. \end{aligned}$$

Thus $c=2$ and we get $\omega =\frac{1}{2}h_{-1}^2{\textbf{1}}$.

We will now use [41, Corollary 3.15] to show that V is isomorphic to the lattice vertex operator algebra $V_L$ where L consists of all $\alpha \in {\mathbb {C}}$ such that for some non-zero $v\in V$, $h_0 v=\alpha v$. To apply this result, we need to verify Conditions (1) – (3) of [41]: Condition (1) is satisfied because

$$\begin{aligned} L_n h=\delta _{n,0} h,\qquad h_n h=\delta _{n,1}{\textbf{1}}\end{aligned}$$

for $n\ge 0$. Condition (2) is that the algebra generated by the operators $h_n$, $n\ge 0$, acts locally finitely on V. This is clear because V is a vertex operator algebra. Condition (3) is that $G_0 ={\mathbb {C}}{\textbf{1}}$, where $G_0$ is the intersection of the generalized $h_0$-eigenspace with generalized eigenvalue 0 with the Heisenberg vacuum space

$$\begin{aligned} G=\lbrace v\in V\,\vert \,h_nv =0\,\,\text {for}\,\,n>0\rbrace . \end{aligned}$$

Indeed, if $v\in G$, then $L_0 v=\frac{1}{2} h_0^2 v$, so if v is additionally a generalized eigenvector for h with generalized eigenvalue 0, then $L_0^N v=0$ for some $N\in {\mathbb {Z}}_+$. Thus v has conformal weight 0 and it follows that $G_0={\mathbb {C}}{\textbf{1}}$.

It now follows from [41, Corollary 3.15] that V is isomorphic to a rank-one lattice vertex operator algebra $V_L$. Since V has the same character as the $\mathfrak {sl}_2$-root lattice vertex operator algebra, which in turn is isomorphic to $L_1(\mathfrak {sl}_2)$, it follows that V is isomorphic to $L_1(\mathfrak {sl}_2)$. $\square$

We now consider what happens when a simple vertex operator algebra with decomposition (B.1) does have a $PSL_2({\mathbb {C}})$-action:

Theorem B.2

Suppose V is a simple vertex operator algebra extension of $V_1$ with an action of $PSL_2({\mathbb {C}})$ by vertex operator algebra automorphisms such that $V\cong \bigoplus _{n=0}^\infty V(2n)\otimes {\mathcal {L}}_{2n+1,1}$ as a $PSL_2({\mathbb {C}})\times V_1$-module. Then the vertex operator algebra isomorphism $L_1(\mathfrak {sl}_2)\rightarrow V$ guaranteed by Theorem B.1 can be chosen to be an isomorphism of $PSL_2({\mathbb {C}})$-modules.

Proof

By assumption, the conformal-weight-1 space $V_{(1)}=V(2)\otimes v_{3,1}$ is a $PSL_2({\mathbb {C}})$-module isomorphic to $\mathfrak {sl}_2$, so we may fix a standard basis $\lbrace e, f, h\rbrace$ for $V_{(1)}$ as a $PSL_2({\mathbb {C}})$-module. Moreover, because $PSL_2({\mathbb {C}})$ acts on V by vertex operator algebra automorphisms, the map $V_{(1)}\otimes V_{(1)} \rightarrow V_{(1)}$ given by $u\otimes v\mapsto u_0 v$ is a $PSL_2({\mathbb {C}})$-module homomorphism, and thus it must be a multiple of the Lie bracket on $\mathfrak {sl}_2$, that is,

$$\begin{aligned} e_0 f= c\cdot h,\qquad f_0 h=2c\cdot f,\qquad h_0 e=2c\cdot e \end{aligned}$$

(B.2)

for some $c\in {\mathbb {C}}$. Since $V_{(1)}$ is also a Lie algebra with bracket $[u,v]=u_0 v$, and since this Lie algebra is isomorphic to $\mathfrak {sl}_2$ by Theorem B.1, we have $c\ne 0$ (but note that we cannot immediately identify the $\lbrace e, f, h\rbrace$ basis for $V_{(1)}$ considered as a $PSL_2({\mathbb {C}})$-module with the corresponding basis for $V_{(1)}$ as a Lie algebra).

We can now replace the vertex operator $Y_{V}$ with $\varphi ^{-1}\circ Y_{V}\circ (\varphi \otimes \varphi )$, where $\varphi$ is the linear isomorphism that acts as the identity on $V(2n)\otimes {\mathcal {L}}_{2n+1,1}$ for $n\ne 1$ and by the scalar c on $V(2)\otimes {\mathcal {L}}_{3,1}$. Note that

$$\begin{aligned} \varphi : (V,\varphi ^{-1}\circ Y_{V}\circ (\varphi \otimes \varphi ),{\textbf{1}},\omega ) \longrightarrow (V, Y_{V}, {\textbf{1}},\omega ) \end{aligned}$$

is a vertex operator algebra isomorphism that is also a $PSL_2({\mathbb {C}})$-module isomorphism. Thus we may assume that the simple vertex operator algebra structure on V satisfies (B.2) with $c=1$. As in the proof of Theorem B.1, $(u,v)\rightarrow u_1 v$ defines a non-degenerate invariant bilinear form on $V_{(1)}\cong \mathfrak {sl}_2$:

$$\begin{aligned} u_1 v=k\langle u,v\rangle {\textbf{1}}\end{aligned}$$

for some non-zero $k\in {\mathbb {C}}$, where $\langle \cdot ,\cdot \rangle$ is the invariant bilinear form on $\mathfrak {sl}_2$ such that $\langle h,h\rangle =2$. Thus $h_1 h=2k{\textbf{1}}$, so by the proof of Theorem B.1, $\omega =\frac{1}{4k} h_{-1}^2{\textbf{1}}$, and then

$$\begin{aligned} L_0 =\frac{1}{4k}h_0^2 +\frac{1}{2k}\sum _{n=1}^\infty h_{-n} h_n. \end{aligned}$$

Using $h_0 e=2e$, $h_1 e=k\langle h,e\rangle {\textbf{1}}=0$, and $h_n e=0$ for $n\ge 2$, we get $e=L_0 e=\frac{1}{k}e$, and thus $k=1$.

We can now conclude that there is a vertex algebra homomorphism from the universal affine vertex algebra $V^1(\mathfrak {sl}_2)$ at level 1 to V which sends the generators $e(-1){\textbf{1}}$, $f(-1){\textbf{1}}$, and $h(-1){\textbf{1}}$ of $V^1(\mathfrak {sl}_2)$ to e, f, and h, respectively. Since V is simple and generated by e, f, and h (because V is abstractly isomorphic to $L_1(\mathfrak {sl}_2)$), this homomorphism descends to a vertex operator algebra isomorphism $L_1(\mathfrak {sl}_2)\rightarrow V$. By construction, this isomorphism commutes with the $PSL_2({\mathbb {C}})$-actions on the weight-one subspaces of $L_1(\mathfrak {sl}_2)$ and V. Then because both algebras are generated by their weight-one subspaces, and because $PSL_2({\mathbb {C}})$ acts by automorphisms on both algebras, the isomorphism $L_1(\mathfrak {sl}_2)\rightarrow V$ is an isomorphism of $PSL_2({\mathbb {C}})$-modules, as desired. $\square$

The vertex operator algebra $L_1(\mathfrak {sl}_2)$, has a unique (up to isomorphism) irreducible module which is not isomorphic to $L_1(\mathfrak {sl}_2)$. We denote this module by X. The action of $\mathfrak {sl}_2$ on X by zero-modes of the vertex operators for elements of $L_1(\mathfrak {sl}_2)_{(1)}\cong \mathfrak {sl}_2$ exponentiates to an action of $SL_2({\mathbb {C}})$ on X such that

$$\begin{aligned} g\cdot Y_{X}(v,x)w = Y_{X}(g\cdot v,x)g\cdot w \end{aligned}$$

(B.3)

for $g\in SL_2({\mathbb {C}})$, $v\in L_1(\mathfrak {sl}_2)$, and $w\in X$. In particular, the $SL_2({\mathbb {C}})$-action and $V_1$-action on X commute; we have

$$\begin{aligned} X=\bigoplus _{n=1}^{\infty } V(2n-1)\otimes {\mathcal {L}}_{2n,1} \end{aligned}$$

as an $SL_2({\mathbb {C}})\times V_1$-module. We now have a uniqueness result for the $SL_2({\mathbb {C}})$-action on X:

Theorem B.3

Suppose W is a simple $L_1(\mathfrak {sl}_2)$-module with an $SL_2({\mathbb {C}})$-action such that

$$\begin{aligned} g\cdot Y_W(v,x)w= Y_W(g\cdot v,x)g\cdot w \end{aligned}$$

(B.4)

for all $g\in SL_2({\mathbb {C}})$, $v\in L_1(\mathfrak {sl}_2)$, and $w\in W$, and such that $W\cong \bigoplus _{n=1}^\infty V(2n-1)\otimes {\mathcal {L}}_{2n,1}$ as an $SL_2({\mathbb {C}})\times V_1$-module. Then there is an $L_1(\mathfrak {sl}_2)$-module isomorphism $X\rightarrow W$ which is also an $SL_2({\mathbb {C}})$-module isomorphism.

Proof

Let T(W) denote the top level of W, that is, the lowest conformal weight space $V(1)\otimes v_{2,1}$. The assumed compatibility of the $SL_2({\mathbb {C}})$- and $L_1(\mathfrak {sl}_2)$-actions on W implies that the linear map $L_1(\mathfrak {sl}_2)_{(1)}\otimes T(W)\rightarrow T(W)$ give by $v\otimes w\mapsto v_0 w$ is an $SL_2({\mathbb {C}})$-module homomorphism, and thus must be a multiple of the $\mathfrak {sl}_2$-action on V(1). That is, if $\lbrace e, f, h\rbrace$ is the standard basis of $L_1(\mathfrak {sl}_2)_{(1)}\cong \mathfrak {sl}_2$, then there is a basis $\lbrace v_1, v_{-1}\rbrace$ of T(W) such that

$$\begin{aligned} \begin{array}{lll} e_0 v_1 =0, &{} h_0 v_1 =c\cdot v_1, &{} f_0 v_1=c\cdot v_{-1},\\ e_0 v_{-1} = c\cdot v_1, &{} h_0 v_{-1} =-c\cdot v_{-1}, &{} f_0 v_{-1} =0\\ \end{array} \end{aligned}$$

(B.5)

for some $c\in {\mathbb {C}}$.

On the other hand, the axioms for vertex operator algebras and modules show that the action of $L_1(\mathfrak {sl}_2)_{(1)}$ on T(W) by zero-modes gives T(W) the structure of an $\mathfrak {sl}_2$-module. The formulas (B.5) describe an $\mathfrak {sl}_2$-module structure only when $c=0$ or $c=1$, since if $c\ne 0$, then $v_1$ is the unique (up to scale) highest-weight vector in T(W) and thus must have h-eigenvalue 1 as T(W) is two-dimensional. In fact, we may conclude $c\ne 0$ because the classification of simple $L_1(\mathfrak {sl}_2)$-modules shows that W must be isomorphic to X as an $L_1(\mathfrak {sl}_2)$-module, and we know that $L_1(\mathfrak {sl}_2)_{(1)}$ acts non-trivially on the top level of X. Thus (B.5) holds with $c=1$, showing that the assumed $SL_2({\mathbb {C}})$-action on T(W) is the same as that obtained by exponentiating the zero-mode action of $L_1(\mathfrak {sl}_2)_{(1)}\cong \mathfrak {sl}_2$ on T(W).

We have now shown that any $L_1(\mathfrak {sl}_2)$-module isomorphism $X\rightarrow W$ restricts to an $SL_2({\mathbb {C}})$-module isomorphism on top levels, since the $SL_2({\mathbb {C}})$-actions on both top levels are obtained by exponentiating the zero-mode actions of $L_1(\mathfrak {sl}_2)_{(1)}$. Then because both X and W are generated by their top levels, and because (B.3) and (B.4) both hold, we conclude that any $L_1(\mathfrak {sl}_2)$-module isomorphism $X\rightarrow W$ is also an $SL_2({\mathbb {C}})$-module isomorphism. $\square$

Appendix C: Uniqueness of the chiral universal centralizer

In this appendix, we prove the uniqueness assertion in Theorem 8.1 using Theorems B.2 and B.3. Thus let V be any simple conformal vertex algebra containing $V_1\otimes V_{25}$ as a vertex operator subalgebra and such that

$$\begin{aligned} V \cong \bigoplus _{r\in {\mathbb {Z}}_+} {\mathcal {L}}_{r,1}^{(1)}\otimes {\mathcal {L}}_{r,1}^{(25)} \end{aligned}$$

as a $V_1\otimes V_{25}$-module. The $\mathfrak {sl}_2$-type fusion rules for $V_1$- and $V_{25}$-modules imply that V has a conformal vertex algebra involution which acts as $(-1)^{r-1}$ on ${\mathcal {L}}_{r,1}^{(1)}\otimes {\mathcal {L}}_{r,1}^{(25)}$. Let $V^0$ be the fixed-point subalgebra of this involution and $V^1$ the eigenspace with eigenvalue $-1$, so that

$$\begin{aligned} V^0 \cong \bigoplus _{n=0}^\infty {\mathcal {L}}_{2n+1,1}^{(1)}\otimes {\mathcal {L}}_{2n+1,1}^{(25)},\qquad V^1\cong \bigoplus _{n = 1}^\infty {\mathcal {L}}_{2n,1}^{(1)}\otimes {\mathcal {L}}_{2n,1}^{(25)} \end{aligned}$$

as $V_1\otimes V_{25}$-modules. It is easy to see that $V^0$ is a simple conformal vertex algebra as follows: For any non-zero $v\in V^0$,

$$\begin{aligned} V =\textrm{span}\lbrace u_n v\,\vert \,u\in V, n\in {\mathbb {Z}}\rbrace = \textrm{span}\lbrace u_n v\,\vert \,u\in V^0, n\in {\mathbb {Z}}\rbrace +\textrm{span}\lbrace u_n v\,\vert \,u\in V^1, n\in {\mathbb {Z}}\rbrace \end{aligned}$$

by [39, Proposition 4.5.6] since V is simple. But since $V=V^0\oplus V^1$ and since

$$\begin{aligned} \textrm{span}\lbrace u_n v\,\vert \,u\in V^i, n\in {\mathbb {Z}}\rbrace \subseteq V^i \end{aligned}$$

for $i=0,1$, it follows that

$$\begin{aligned} V^0=\textrm{span}\lbrace u_n v\,\vert \,u\in V^0, n\in {\mathbb {Z}}\rbrace \end{aligned}$$

for any non-zero $v\in V^0$, and thus $V^0$ is simple. A similar argument shows that $V^1$ is a simple $V^0$-module.

Proposition C.1

Any simple conformal vertex algebra extension of $V_1\otimes V_{25}$ which is isomorphic to $V^0$ as a $V_1\otimes V_{25}$-module is isomorphic to $V^0$ as a conformal vertex algebra. Moreover, any simple $V^0$-module which is isomorphic to $V^1$ as a $V_1\otimes V_{25}$-module is isomorphic to $V^1$ as a $V^0$-module.

Proof

From Proposition 6.1, ${\mathcal {O}}_{25}^0$ is braided tensor equivalent to $({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau$ with a suitable braiding. Thus using [7, Theorem 7.5], uniqueness of the simple conformal vertex algebra structure on $V^0=\bigoplus _{n=0}^\infty {\mathcal {L}}_{2n+1,1}^{(1)}\otimes {\mathcal {L}}_{2n+1,1}^{(25)}$ extending $V_1\otimes V_{25}$ is equivalent to uniqueness of the simple commutative algebra structure on the object

$$\begin{aligned} A^0=\bigoplus _{n=0}^\infty V(2n)\otimes {\mathcal {L}}_{2n+1,1}^{(1)} \end{aligned}$$

in the Ind-category $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_1^0)$. Moreover, uniqueness of the simple $V^0$-module structure on $V^1$ is equivalent to uniqueness of the simple $A^0$-module structure on the object

$$\begin{aligned} A^1 =\bigoplus _{n=1}^\infty V(2n-1)\otimes {\mathcal {L}}_{2n,1}^{(1)} \end{aligned}$$

in $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_{1}^0)$.

Since the $\mathfrak {sl}_2$-modules V(2n) for $n\in {\mathbb {N}}$ are objects of the maximal symmetric tensor subcategory ${{\,\textrm{Rep}\,}}PSL_2({\mathbb {C}})\subseteq ({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau$, we can use [7, Theorem 7.5] again to see that a simple commutative algebra structure on $A^0$ in $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_1^0)$ amounts to a vertex operator algebra structure $(A^0, Y_{A^0},{\textbf{1}},\omega )$ such that:

• (1) The $PSL_2({\mathbb {C}})$-action on the V(2n) factors in $A^0$ gives a $PSL_2({\mathbb {C}})$-action by vertex operator algebra automorphisms on $A^0$.

• (2) The only $PSL_2({\mathbb {C}})$-invariant ideals of $A^0$ are 0 and $A^0$.

Moreover, uniqueness of the simple commutative algebra structure on $A^0$ in $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_1^0)$ amounts to uniqueness of the vertex operator algebra structure on $A^0$ up to an isomorphism that is also a $PSL_2({\mathbb {C}})$-module isomorphism.

Similarly, because $({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau$ is equivalent to ${{\,\textrm{Rep}\,}}\mathfrak {sl}_2$ when considered as a module category for ${{\,\textrm{Rep}\,}}PSL_2({\mathbb {C}})$, a simple $A^0$-module structure on $A^1$ (where $A^0$ is considered as a commutative algebra in $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_1^0)$) is equivalent to an $A^0$-module structure $(A^1, Y_{A^1})$ (where $A^0$ is considered as a vertex operator algebra) such that:

• (3) The $SL_2({\mathbb {C}})$-action on the $V(2n-1)$ factors in $A^1$ and the $PSL_2({\mathbb {C}})$-action on $A^0$ are compatible in the sense that

  $$\begin{aligned} g\cdot Y_{A^1}(v,x)w =Y_{A^1}(g\cdot v,x)g\cdot w \end{aligned}$$

  for all $g\in SL_2({\mathbb {C}})$, $v\in A^0$, and $w\in A^1$.

• (4) The only $SL_2({\mathbb {C}})$-invariant $A^0$-submodules of $A^1$ are 0 and $A^1$.

Moreover, uniqueness of the $A^0$-module structure on $A^1$ (where $A^1$ is considered as a commutative algebra in $\textrm{Ind}(({{\,\textrm{Rep}\,}}\mathfrak {sl}_2)^\tau \otimes {\mathcal {O}}_1^0)$) is equivalent to uniqueness of the $A^0$-module structure on $A^1$ (where $A^0$ is considered as a vertex operator algebra) up to an $A^0$-module isomorphism that is also an $SL_2({\mathbb {C}})$-module isomorphism.

We claim that Conditions (1) and (2) imply that $A^0$ is a simple vertex operator algebra, and we claim that $A^1$ is a simple $A^0$-module. Then the desired uniqueness assertions will follow from Theorems B.2 and B.3: $A^0$ will be isomoprhic to the simple affine vertex operator algebra $L_1(\mathfrak {sl}_2)$ with its standard $PSL_2({\mathbb {C}})$-action by automorphisms, and $A^1$ will be isomorphic to the unique non-trivial simple $L_1(\mathfrak {sl}_2)$-module with its standard $SL_2({\mathbb {C}})$-action.

To prove the claims, first let $I\subseteq A^0$ be a non-zero ideal. Since $A^0$ is a semisimple ${\mathcal {V}}ir$-module, so is I, and thus I contains $v\otimes {\mathcal {L}}_{2n+1,1}^{(1)}$ for some $n\in {\mathbb {N}}$ and non-zero $v\in V(2n)$. Now consider the ideal generated by $V(2n)\otimes {\mathcal {L}}_{2n+1,1}^{(1)}$; it is $PSL_2({\mathbb {C}})$-invariant by Condition (1), and thus it equals $A^0$ by Condition (2). In particular, the ideal generated by $V(2n)\otimes {\mathcal {L}}_{2n+1,1}^{(1)}$ contains $V(0)\otimes {\mathcal {L}}_{1,1}^{(1)}$. Since there is a non-zero $PSL_2({\mathbb {C}})$-module homomorphism $V(2m)\otimes V(2n)\rightarrow V(0)$ if and only if $m=n$, it follows that the Virasoro intertwining operator

$$\begin{aligned} \pi _1\circ Y_{A^0}\vert _{(V(2n)\otimes {\mathcal {L}}_{2n+1,1}^{(1)})\otimes (V(2n)\otimes {\mathcal {L}}_{2n+1,1}^{(1)})}, \end{aligned}$$

where $\pi _1: A^0\rightarrow V(0)\otimes {\mathcal {L}}_{1,1}^{(1)}$ is the projection, is non-zero. Thus this intertwining operator has the form $b\otimes {\mathcal {Y}}$ where $b: V(2n)\otimes V(2n)\rightarrow V(0)$ is a non-zero (and thus non-degenerate) invariant bilinear form and ${\mathcal {Y}}$ is a non-zero intertwining operator of type $\left( {\begin{array}{c}{\mathcal {L}}_{1,1}^{(1)}\\ {\mathcal {L}}_{2n+1,1}^{(1)}\,{\mathcal {L}}_{2n+1,1}^{(1)}\end{array}}\right)$.

Now recall that our non-zero ideal $I\subseteq A^0$ contains $v\otimes {\mathcal {L}}_{2n+1,1}^{(1)}$ for some non-zero $v\in V(2n)$. Thus letting $v_{2n+1,1}$ denote a non-zero lowest-conformal-weight vector in ${\mathcal {L}}_{2n+1,1}^{(1)}$, we have now shown that there is some $v'\in V(2n)$ such that

$$\begin{aligned} Y_{A^0}(v'\otimes v_{2n+1,1},x)(v\otimes v_{2n+1,1})\in I((x)) \end{aligned}$$

has non-zero projection to $V(0)\otimes {\mathcal {L}}_{1,1}^{(1)}$. Then because I is a semisimple ${\mathcal {V}}ir$-module, this shows that I contains $V(0)\otimes {\mathcal {L}}_{1,1}^{(1)}$ and thus contains the vacuum vector ${\textbf{1}}$. Since ${\textbf{1}}$ generates $A^0$ as an $A^0$-module, we conclude that $I=A^0$, proving $A^0$ is simple.

To show that $A^1$ is a simple $A^0$-module, we can now use Theorem B.1, which shows that $A^0$ is isomorphic to $L_1(\mathfrak {sl}_2)$ as a vertex operator algebra. Then $A^1$ must be simple because the only $L_1(\mathfrak {sl}_2)$-module with the same $V_1$-module decomposition as $A^1$ is the non-trivial simple $L_1(\mathfrak {sl}_2)$-module. $\square$

We can now prove uniqueness of the simple conformal vertex algebra structure on $V^0\oplus V^1$, and thus the uniqueness assertion in Theorem 8.1:

Proof

Let $(V,Y_V,{\textbf{1}},\omega )$ and $(V,{\widetilde{Y}}_V,{\textbf{1}},\omega )$ be two simple conformal vertex algebra structures on V such that

$$\begin{aligned} Y_V\vert _{({\mathcal {L}}^{(1)}_{1,1}\otimes {\mathcal {L}}_{1,1}^{(25)})\otimes V} ={\widetilde{Y}}_V\vert _{({\mathcal {L}}^{(1)}_{1,1}\otimes {\mathcal {L}}_{1,1}^{(25)})\otimes V}. \end{aligned}$$

Since $V=V^0\oplus V^1$ where $V^0$ is a simple subalgebra and $V^1$ is a simple $V^0$-module with respect to either conformal vertex algebra structure on V, Proposition C.1 shows that we may assume

$$\begin{aligned} Y_V\vert _{V^0\otimes V} = {\widetilde{Y}}_V\vert _{V^0\otimes V}. \end{aligned}$$

Then skew-symmetry dictates that

$$\begin{aligned} Y_V\vert _{V^1\otimes V^0}=e^{xL_{-1}}Y_V(\cdot ,-x)\cdot \vert _{V^0\otimes V^1} =e^{xL_{-1}}{\widetilde{Y}}_V(\cdot ,-x)\cdot \vert _{V^0\otimes V^1}={\widetilde{Y}}_V\vert _{V^1\otimes V^0}, \end{aligned}$$

so it remains to consider ${\widetilde{Y}}_V\vert _{V^1\otimes V^1}$.

The $\mathfrak {sl}_2$-type fusion rules of $V_1$- and $V_{25}$-modules show that $\textrm{Im}\,{\widetilde{Y}}_V\vert _{V^1\otimes V^1}\subseteq V^0$. We claim the space of $V^0$-module intertwining operators of type $\left( {\begin{array}{c}V^0\\ V^1\,V^1\end{array}}\right)$ is one dimensional, so that

$$\begin{aligned} {\widetilde{Y}}_V\vert _{V^1\otimes V^1}=\alpha \cdot Y_V\vert _{V^1\otimes V^1} \end{aligned}$$

for some $\alpha \in {\mathbb {C}}$ (note that $Y_V\vert _{V^1\otimes V^1}\ne 0$ since otherwise $V^1$ would be a proper ideal of $(V,Y_V,{\textbf{1}},\omega )$). Assuming the claim, we have $\alpha \ne 0$ since $(V,{\widetilde{Y}}_V,{\textbf{1}},\omega )$ is a simple conformal vertex algebra, so then

$$\begin{aligned} \sigma :=\textrm{Id}_{V^0}\oplus \sqrt{\alpha }\cdot \textrm{Id}_{V^1} \end{aligned}$$

is a linear isomorphism of V such that $\sigma \circ {\widetilde{Y}}_V=Y_V\circ (\sigma \otimes \sigma )$, showing that the two simple conformal vertex algebra structures on V are isomorphic (this is a special case of the uniqueness proof for simple current extensions in [12, Proposition 5.3]).

To prove the claim, let $q_r: {\mathcal {L}}_{r,1}^{(1)}\otimes {\mathcal {L}}_{r,1}^{(25)}\rightarrow V$ and $\pi _r: V\rightarrow {\mathcal {L}}_{r,1}^{(1)}\otimes {\mathcal {L}}_{r,1}^{(25)}$ for $r\in {\mathbb {Z}}_+$ denote the natural inclusion and projection, respectively. It is enough to show that the linear map

$$\begin{aligned} {\mathcal {Y}}\mapsto \pi _1\circ {\mathcal {Y}}\circ (q_{2}\otimes q_{2}) \end{aligned}$$

from the space of $V^0$-module intertwining operators of type $\left( {\begin{array}{c}V^0\\ V^1\,V^1\end{array}}\right)$ to the one-dimensional space of $V_1\otimes V_{25}$-module intertwining operators of type $\left( {\begin{array}{c}{\mathcal {L}}_{1,1}^{(1)}\otimes {\mathcal {L}}_{1,1}^{(25)}\\ {\mathcal {L}}_{2,1}^{(1)}\otimes {\mathcal {L}}_{2,1}^{(25)}\,{\mathcal {L}}_{2,1}^{(1)}\otimes {\mathcal {L}}_{2,1}^{(25)}\end{array}}\right)$ is injective. Indeed, let

$$\begin{aligned} {\mathcal {Y}}: V^1\otimes V^1&\rightarrow V^0((x))\\ u_1\otimes v_1&\mapsto {\mathcal {Y}}(u_1,x)v_1=\sum _{n\in {\mathbb {Z}}} (u_1)_n v_1\,x^{-n-1} \end{aligned}$$

be a non-zero intertwining operator, and fix a non-zero $v_1\in {\mathcal {L}}_{2,1}^{(1)}\otimes {\mathcal {L}}_{2,1}^{(25)}$. Then

$$\begin{aligned} \textrm{span}\lbrace (u_1)_n v_1\,\vert \,u_1\in V^1, n\in {\mathbb {Z}}\rbrace \end{aligned}$$

is a $V^0$-submodule of $V^0$ by the easy intertwining operator generalization of [39, Proposition 4.5.7]. Using the commutator formula for intertwining operators, we see that this submodule is non-zero, since $v_1$ generates $V^1$ as a $V^0$-module and ${\mathcal {Y}}$ is non-zero (see [11, Proposition 11.9]). So in fact

$$\begin{aligned} {\mathcal {L}}_{1,1}^{(1)}\otimes {\mathcal {L}}_{1,1}^{(25)} =\pi _1(V^0) =\pi _1(\textrm{span}\lbrace (u_1)_n v_1\,\vert \,u_1\in V^1, n\in {\mathbb {Z}}\rbrace ) \end{aligned}$$

since $V^0$ is simple. That is, there exists $m\in {\mathbb {Z}}_+$ such that

$$\begin{aligned} \pi _1\circ {\mathcal {Y}}\circ (q_{2m}\otimes q_2)\ne 0. \end{aligned}$$

Finally, the $\mathfrak {sl}_2$-type fusion rules of $V_1$- and $V_{25}$-modules force $m=1$, completing the proof of the claim and also of the theorem. $\square$