Introduction to Vertex Operator Algebras

--- Base version+++ Proposed version@@ -112,6 +112,18 @@ If $\mathbf1\in I$, creation gives $a=\left.Y(a,x)\mathbf1\right|_{x=0}\in I$ for every $a\in V$. The assertions about sums and intersections follow from linearity. ::: +::: remark [The ordering of ideals] {@rem:ideal-lattice}+Ideals are partially ordered by inclusion. By @prop:ideal-properties, they form a complete lattice: the greatest lower bound of a family is its intersection, and its least upper bound is its sum,++$$+\bigwedge_\alpha I_\alpha=\bigcap_\alpha I_\alpha,+\qquad+\bigvee_\alpha I_\alpha=\sum_\alpha I_\alpha.+$$++The least and greatest ideals are $0$ and $V$. Among proper ideals, **maximal** means that no strictly larger proper ideal contains it, whereas **largest** means that it contains every proper ideal. A largest proper ideal need not exist merely because sums of ideals exist: their sum might be $V$.+:::+ ## Quotients and simple quotients  ::: proposition [Quotient VOA] {@prop:quotient-voa}@@ -145,12 +157,26 @@  **Exercise.** Apply $f$ to $Y(a,x)b$ for $b\in\ker f$, and use preservation of the vacuum. For the general quotient and kernel constructions, see [Joyce, lectures 1–2](https://people.maths.ox.ac.uk/joyce/VertexAlgebras2021/VA1%2B2.pdf). -::: proposition [Unique simple quotient for CFT type] {@prop:unique-simple-quotient}-If $V$ is of CFT type, the sum $J$ of all proper ideals is its unique maximal proper ideal, and $V/J$ is simple.+::: proposition [Largest proper ideal and simple quotient for CFT type] {@prop:unique-simple-quotient}+Let $V$ be a nonzero VOA of CFT type, so that $V=\bigoplus_{h\geq0}V_h$ and $V_0=\mathbb C\mathbf1$. Then++$$+J(V)=\sum_{\substack{I\subsetneq V\\I\text{ an ideal}}}I+$$++is the largest proper ideal of $V$: every proper ideal $I$ satisfies $I\subseteq J(V)$. In particular, $J(V)$ is the unique maximal proper ideal, and $V/J(V)$ is the unique simple quotient, up to the isomorphism induced by a quotient map. Moreover, $J(V)=0$ if and only if $V$ is simple. :::  ::: proof-Every proper ideal is graded and excludes $\mathbf1$, so its weight-zero part is zero because $V_0=\mathbb C\mathbf1$. Their sum $J$ still has zero weight-zero part and is therefore proper. It contains every proper ideal. The inverse image of a nonzero proper ideal of $V/J$ would be a proper ideal strictly containing $J$, a contradiction.+By @prop:ideal-properties, every proper ideal is graded and excludes $\mathbf1$. Since $V_0=\mathbb C\mathbf1$, every proper ideal is contained in $\bigoplus_{h>0}V_h$. Their sum $J(V)$ is therefore an ideal with zero weight-zero part, so it is proper. By construction it contains every proper ideal.++The inverse image of a nonzero proper ideal of $V/J(V)$ would be a proper ideal strictly containing $J(V)$, a contradiction. Thus $V/J(V)$ is simple. Conversely, the kernel of any surjective VOA homomorphism from $V$ onto a nonzero simple VOA is a maximal proper ideal, hence equals $J(V)$. The uniqueness follows from @prop:voa-isomorphism. Finally, $J(V)=0$ precisely when $V$ has no nonzero proper ideals.+:::++::: remark [What CFT type guarantees] {@rem:cft-type-largest-ideal}+This argument applies to every nonzero VOA of CFT type, including universal Virasoro VOAs; it is not specific to minimal-model central charges. A unique largest proper ideal may itself contain smaller nonzero ideals, so the proposition does not assert that there is only one nonzero proper ideal.++For comparison, let $W$ be a nonzero simple VOA of CFT type and equip $V=W\oplus W$ with componentwise fields, vacuum $(\mathbf1_W,\mathbf1_W)$, and conformal vector $(\omega_W,\omega_W)$. The proper ideals $W\oplus0$ and $0\oplus W$ sum to $V$, so no largest proper ideal exists. Here $\dim V_0=2$, and $V$ is not of CFT type. :::