2

I have been studying String Theory and CFT for a while, and I am sad to say I do not know why we treat $z$ and $\bar z$ as independent variables, and why we go on to consider the algebra $Vir\oplus\bar {Vir}$ instead of simply $Vir$. The literature I followed never really explained this, just simply did it. Is there a clear explanation of this fact?

Qmechanic
  • 220,844
soap
  • 817

1 Answers1

4

For simplicity let us consider global conformal symmetry rather than the full Virasoro symmetry. Global symmetry corresponds to the group $SL(2,\mathbb{C})$ acting on $z$ as $z\to \frac{az+b}{cz+d}$. So far $\bar z$ is the complex conjugate of $z$, not independent.

In a quantum theory we want the symmetry algebra to act on a complex space of states, not just on the geometry. So we have to complexify the symmetry algebra. In two dimensions, it turns out that the complexified symmetry algebra factorizes, $$ sl(2,\mathbb{C})^\mathbb{C} = sl(2,\mathbb{C}) \oplus sl(2,\mathbb{C}) $$ The two factors lead to Ward identities that involve $z$ and $\bar z$ independently: this can be interpreted as $z$ and $\bar z$ becoming independent. But this independence only holds at the level of Ward identities. Physical quantities (such as correlation functions) are functions of $z\in \mathbb{C}$, not $(z,\bar z)\in\mathbb{C}^2$.

For more details along these lines, see my review article: https://arxiv.org/abs/1406.4290 .

Sylvain Ribault
  • 2,906