Lectures on String Theory

– 18 –
3.2.3 Conformal gauge
Consider the conformal Killing equation
ξ γ ∂ γ h αβ + h αγ ∂ β ξ γ + h βγ ∂ α ξ γ = Λh αβ (3.15)
and solve it assuming the conformal gauge
( ) −1 0
h αβ = e 2φ 0 1
(3.16)
This equation can be split as follows. First we take α = τ and β = σ and using
h τσ = 0 we find
h ττ ∂ σ ξ τ + h σσ ∂ τ ξ σ = 0 → ∂ σ ξ τ − ∂ τ ξ σ = 0
Second, we take α, β = τ and then α, β = σ. We get
i.e.
Subtracting one from the other we get
ξ γ ∂ γ h ττ + 2h ττ ∂ τ ξ τ = Λh ττ
ξ γ ∂ γ h σσ + 2h σσ ∂ σ ξ σ = Λh σσ
ξ γ h ττ ∂ γ h ττ + 2∂ τ ξ τ = Λ
ξ γ h σσ ∂ γ h σσ + 2∂ σ ξ σ = Λ .
∂ τ ξ τ − ∂ σ ξ σ = 0 .
Thus, the conformal Killing equations reduce in the conformal gauge to
or equivalently
∂ σ ξ τ − ∂ τ ξ σ = 0
∂ τ ξ τ − ∂ σ ξ σ = 0 (3.17)
(∂ τ + ∂ σ )(ξ τ − ξ σ ) = 0
(∂ τ − ∂ σ )(ξ τ + ξ σ ) = 0 (3.18)
By using the world-sheet light-cone coordinates σ ± = τ ± σ this can be reformulated
as 4 ∂ + ξ − = 0 = ∂ − ξ + .
4 The basic relations are ∂ τ = ∂ + + ∂ − , ∂ σ = ∂ + − ∂ − , ∂ ± = 1 2
(∂ τ ± ∂ σ
)
.