Vertex algebra confusion

Question

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.

Answer 1

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.