# Vertex algebra confusion

In Blumenhagen's book on CFT, the authors have defined $$\bar{v}(\bar{z})$$ to be the antiholomorphic part of the vertex operator for a free bosonic CFT, $$V(z,\bar{z})=:\exp{(\alpha X(z,\bar{z})}):$$ where $$X$$ is the field.

Then on page 52, right after the 3rd equation they claim that $$[L_0, \bar{v}(\bar{z})]=0$$. $$L_n$$ are the Laurent modes of the EM tensor. I don't understand why this is true.

$$\bar{v}$$ is composed of the operators, $$\bar{j_n}$$, the Laurent modes of the operator $$\bar{j}=i\bar{\partial}{X}$$. Using the OPE of primary anti chiral fields with the EM tensor (equation (2.40) in the same book) I tried to prove that the EM tensor modes commute with the anti chiral modes but it's not working.

They have put all the anti-holomorphic dependence into $$\bar{v}(\bar{z})$$. So for holomorphic modes of stress energy tensor $$[L_n, \bar{v}(\bar{z})]=0$$. For anti-holomorphic generators $$[\bar{L}_n, \bar{v}(\bar{z})]\neq0$$.
Field $$X(z, \bar{z})$$ have holomorphic and anti-holomorphic parts in Laurent expansion (2.89). Then from equation (2.40) with $$h=\bar{h}=0$$ you can find commutation relations for $$L_n$$ and anti-holomorphic modes of $$X(z,\bar{z})$$. It's straightforward to show, that result is zero.
• Even the holomorphic component of the stress tensor, $T(z)$ has an OPE with a antichiral field $\bar{\phi}(\bar{z})$ (see equation (2.40) and assume $\phi$ depends only on $\bar{z}$). So how can the holomorphic generators of the EM tensor commute with the Laurent modes of an anti chiral field? If you follow the calculation on page 27 with one of the EM tensor modes replaced by a field mode, shouldn't you get a non zero answer? Commented Jan 1, 2020 at 10:18