Introduction to Vertex Operator Algebras

--- Base version+++ Proposed version@@ -0,0 +1,116 @@+# G-orbifolds++We construct VOAs by taking fixed points of automorphism groups, using the subalgebra and ideal properties from the preceding chapter.++## The fixed-point VOA++::: definition [Fixed-point subalgebra and orbifold] {@def:orbifold}+Let $G\leq\operatorname{Aut}(V)$. 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 **$G$-orbifold** of $V$.+:::++::: proposition [Inherited conformal structure] {@prop:orbifold-structure}+For any $G\leq\operatorname{Aut}(V)$, $V^G$ is a full VOA subalgebra of $V$, 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 $V^G$, proper whenever $G\ne\{1\}$.+:::++**Exercise.** Prove @prop:orbifold-structure using @prop:inherited-va 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 $|G|$.++## Averaging++For the rest of this chapter, assume $G$ is finite.++::: proposition [Averaging projection] {@prop:orbifold-averaging}+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 $v\in V$,++$$+R_G\bigl(Y(u,x)v\bigr)=Y(u,x)R_Gv,+\qquad+R_G\bigl(Y(v,x)u\bigr)=Y(R_Gv,x)u.+$$++Consequently, both summands are preserved by the fields of $V^G$.+:::++**Exercise.** Prove @prop:orbifold-averaging by averaging the automorphism identity. The projection $R_G$ is generally not a VOA homomorphism: the identities above require one state to be invariant.++## Simplicity++::: theorem [Simplicity of finite orbifolds] {@thm:orbifold-simple}+If $V$ is simple and $G\leq\operatorname{Aut}(V)$ is finite, then $V^G$ is simple.+:::++::: proof+We use the cyclicity result that, for any nonzero $u$ in a simple VOA $V$, the coefficients of $Y(a,x)u$, as $a$ ranges over $V$, span $V$; see [Dong–Li–Mason, Proposition 2.1](https://arxiv.org/pdf/q-alg/9608009).++Let $0\ne I\subseteq V^G$ be an ideal and choose $0\ne u\in I$. By cyclicity, there are finitely many $a_i\in V$ and Laurent polynomials $f_i(x)$ such that++$$+\mathbf1=\sum_i\operatorname{Res}_x f_i(x)Y(a_i,x)u.+$$++Apply $R_G$ and use @prop:orbifold-averaging:++$$+\mathbf1=\sum_i\operatorname{Res}_x f_i(x)Y(R_Ga_i,x)u\in I.+$$++Since $R_Ga_i\in V^G$, each residue lies in $I$. Creation now gives $a=\left.Y(a,x)\mathbf1\right|_{x=0}\in I$ for every $a\in V^G$. Thus $I=V^G$.+:::++## Fixed points in stages++::: proposition [Subgroups and successive orbifolds] {@prop:orbifold-stages}+If $H\leq G$, then++$$+V^G\subseteq V^H\subseteq V+$$++are full VOA inclusions. If $H\trianglelefteq G$, the quotient $G/H$ acts on $V^H$ by VOA automorphisms through $(gH)a=ga$, 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 $G$. The quotient action need not be faithful for a general $V$; the displayed equality uses its fixed vectors.++The construction above requires no rationality assumption. Questions about modules, rationality, and extensions involving twisted sectors will be treated after module theory is introduced.++## References++- Chongying Dong, Haisheng Li, Geoffrey Mason, [Compact automorphism groups of vertex operator algebras](https://arxiv.org/abs/q-alg/9608009), Proposition 2.1 and Theorem 2.4: cyclicity and simplicity of fixed-point VOAs.