Introduction to Vertex Operator Algebras

--- Base version+++ Proposed version@@ -73,6 +73,51 @@ **Exercise.** Prove @prop:homomorphism-compatibility. *Hint:* Apply the homomorphism identity to $Y(u,x)\mathbf1$ and $Y(\omega,x)\omega$.  The inclusion of a VOA subalgebra is always a vertex-algebra homomorphism; it is a VOA homomorphism in the above sense precisely when the subalgebra is full.++### Inner automorphisms++::: proposition [Exponentiating weight-one zero modes] {@prop:voa-inner-automorphisms}+Suppose $V$ is of CFT type. For $a\in V_1$, the operator $a_{(0)}$ preserves every $V_h$ and satisfies++$$+[a_{(0)},Y(b,x)]=Y(a_{(0)}b,x)\quad(b\in V),+\qquad a_{(0)}\mathbf1=a_{(0)}\omega=0.+$$++Its exponential, defined on each finite-dimensional $V_h$ by the ordinary matrix exponential, is a VOA automorphism:++$$+e^{a_{(0)}}Y(b,x)e^{-a_{(0)}}=Y(e^{a_{(0)}}b,x).+$$+:::++::: proof+The grading assertion follows from the weight of $a$, and the commutator identity follows from the Jacobi identity. Creation gives $a_{(0)}\mathbf1=0$. To check the conformal vector, skew-symmetry gives++$$+a_{(0)}\omega=-Ta+T L_0a-\frac12T^2L_1a=0:+$$++here $L_0a=a$, $L_1a\in\mathbb C\mathbf1$, and $L_ja=0$ for $j\ge2$. Exponentiating the commutator identity proves the field identity. All exponentials are taken on finite-dimensional graded pieces, and the vacuum and conformal vector are fixed.+:::++::: definition [Inner automorphism group] {@def:voa-inner-automorphisms}+For a VOA of CFT type, its **inner automorphism group** is++$$+G_{\mathrm{inn}}(V)+:=\big\langle e^{a_{(0)}}\mid a\in V_1\big\rangle+\subseteq\operatorname{Aut}(V).+$$+:::++More precisely, the Lie algebra action $a\mapsto a_{(0)}$ integrates on each $V_h$ to the simply connected complex Lie group $\widetilde G$ with Lie algebra $V_1$. These actions give a homomorphism $\rho:\widetilde G\to\operatorname{Aut}(V)$ with image $G_{\mathrm{inn}}(V)$. Thus++$$+G_{\mathrm{inn}}(V)\cong\widetilde G/\ker\rho.+$$++The action need not be faithful: even a nonzero $a\in V_1$ may have $a_{(0)}=0$ on all of $V$. Consequently, one should distinguish $V_1$ from its image in the derivations of $V$, and $\widetilde G$ from its image in $\operatorname{Aut}(V)$. See [van Ekeren–Möller–Scheithauer, §1.6](https://www.mathematik.tu-darmstadt.de/media/algebra/homepages/scheithauer/publications/Dimension_formulae_in_genus_zero.pdf#page=8).  ## Ideals and simplicity