"Vertex operator algebra (VOA)"의 두 판 사이의 차이

2012년 11월 1일 (목) 14:14 판

vertex operator algbera is a quadruple $(V,Y,\mathbf{1},\omega)$ with the following axioms
$V=\bigoplus_{n\in\mathbb{Z}}V_{(n)}$ vector space
$\dim V_{(n)} <\infty$ for $n\in \mathbb{Z}$
$\dim V_{(n)}=0$ for $n<<0$
vertex operator
$V\to (\operatorname{End})[[x,x^{-1}]]$
$v\mapsto Y(v,x)=\sum_{n\in \mathbb{Z}}v_{n}x^{-n-1}$
two distinguished vectors $\mathbf{1}\in V_{(0)}$ and $\omega\in V_{(2)}$

$u_{n}v=0$ for $n>>0$
$Y(\mathbf{1},z)=\operatorname{id}_{V}$
(creation property)
$Y(v,z).\mathbf{1}=v+\cdots$
conformal vector
$Y(\omega,z)=L(z)=\sum L(n)z^{-n-2}$ satisfies
$[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m)\delta_{m+n}$
$L(0)v=nv$ for $n\in\mathbb{Z}$ and $v\in V_{(n)}$
translation covariance
$[D, Y(v,z)]=\sum_{n}[D,V_n]z^{-n-1}=\partial Y(v,z)$
Jacobi identity
$ $z_0^{-1}\delta(\frac {z_1-z_2}{z_0})Y(u,z_1)Y(v,z_2)-z_0^{-1}\delta(\frac {z_2-z_1}{-z_0})Y(v,z_2)Y(u,z_1)=z_2^{-1}\delta\left(\frac {z_1-z_0}{z_2}\right)Y(Y(u,z_0)v,z_2)$$

Jacobi identity for Lie algebra says
$(\operatorname{ad} u)(\operatorname{ad} v)-(\operatorname{ad} v)-(\operatorname{ad} u)=(\operatorname{ad}(\operatorname{ad} u) v)$

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 * http://eom.springer.de
 * http://www.proofwiki.org/wiki/
-* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])