Chapter 5 of 11

§G-orbifolds

We construct VOAs by taking fixed points of automorphism groups, using the subalgebra and ideal properties from the preceding chapter.

§5.1 The fixed-point VOA

Definition 5.1 (Fixed-point subalgebra and orbifold)

Let $G\leq\operatorname{Aut}(V)$ $Aut (V)$ . The fixed-point subalgebra is

V^G=\{a\in V:ga=a\text{ for every }g\in G\}, \qquad Y_{V^G}(a,x)=Y_V(a,x)|_{V^G}.

When $G$ is finite, we call $V^G$ the $G$ $G$ -orbifold of $V$ $V$ .

Proposition 5.2 (Inherited conformal structure)

For any $G\leq\operatorname{Aut}(V)$ $Aut (V)$ , $V^G$ is a full VOA subalgebra of $V$ $V$ , with

\mathbf1_{V^G}=\mathbf1_V,\qquad \omega_{V^G}=\omega_V,\qquad (V^G)_h=(V_h)^G,\qquad c_{V^G}=c_V.

CFT type is inherited. In particular, $V$ is an extension of $V^G$ $V^{G}$ , proper whenever $G\ne\{1\}$ ${1}$ .

Exercise. Prove Proposition 5.2 using Proposition 4.2 and the identity

g\bigl(Y(a,x)b\bigr)=Y(ga,x)gb=Y(a,x)b \quad(a,b\in V^G).

Thus taking fixed points restricts the states while retaining the conformal field; it does not divide the central charge by $|G|$ $∣ G ∣$ .

§5.2 Averaging

For the rest of this chapter, assume $G$ is finite.

Proposition 5.3 (Averaging projection)

The averaging operator

R_G=\frac1{|G|}\sum_{g\in G}g

preserves weights and satisfies

R_G^2=R_G,\qquad \operatorname{im}R_G=V^G, \qquad V=V^G\oplus\ker R_G.

For $u\in V^G$ and $v\in V$ $V$ ,

R_G\bigl(Y(u,x)v\bigr)=Y(u,x)R_Gv, \qquad R_G\bigl(Y(v,x)u\bigr)=Y(R_Gv,x)u.

Consequently, both summands are preserved by the fields of $V^G$ $V^{G}$ .

Exercise. Prove Proposition 5.3 by averaging the automorphism identity. The projection $R_G$ is generally not a VOA homomorphism: the identities above require one state to be invariant.

§5.3 Simplicity

Theorem 5.4 (Simplicity of finite orbifolds)

If $V$ is simple and $G\leq\operatorname{Aut}(V)$ is finite, then $V^G$ is simple.

Proof.

Complicated.

§5.4 Fixed points in stages

Proposition 5.5 (Subgroups and successive orbifolds)

If $H\leq G$ $G$ , then

V^G\subseteq V^H\subseteq V

are full VOA inclusions. If $H\trianglelefteq G$ $G$ , the quotient $G/H$ acts on $V^H$ by VOA automorphisms through $(gH)a=ga$ $g a$ , and

(V^H)^{G/H}=V^G.

Exercise. Check that normality makes the quotient action well-defined and that being fixed by $G/H$ inside $V^H$ is equivalent to being fixed by $G$ $G$ . The quotient action need not be faithful for a general $V$ $V$ ; the displayed equality uses its fixed vectors.

§5.5 References

Chongying Dong, Haisheng Li, Geoffrey Mason, Compact automorphism groups of vertex operator algebras, Proposition 2.1 and Theorem 2.4: cyclicity and simplicity of fixed-point VOAs.

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