# Symmetry of the Polyakov action?

Let us look at the Polyakov action for a string moving in a spacetime with metric $g_{\mu \nu}(X)$:$$S_P = -{1\over{4\pi \alpha'}} \int d^2 \sigma \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X^\nu g_{\mu\nu}(X) \tag{1}$$ and suppose there exists a Killing vector $k_\mu$ in spacetime satisfying Killing's equation $$\nabla_\mu k_\nu + \nabla_\nu k_\mu = 0.\tag{2}$$ Does this lead to a symmetry of the Polyakov action?

1. Hints: Perform an infinitesimal variation $$\delta X^{\mu}~=~\varepsilon K^{\mu}(X) \tag{A}$$ in the target space along the Killing vector field. Here $\varepsilon$ is an infinitesimal parameter.
2. Show that the induced metric $$(X^{\ast}G)_{ab}~:=~\partial_a X^{\mu} ~\partial_b X^{\nu}~ G_{\mu\nu}(X)\tag{B}$$ is invariant under the infinitesimal variation (A) $$\delta(X^{\ast}G)_{ab}~=~0.\tag{C}$$ Conclude that the Polyakov action is invariant as well.
3. Further hints to eq. (C): Use that $$\partial_a \delta X^{\lambda} ~=~ \varepsilon\partial_a X^{\mu}~\partial_{\mu}K^{\lambda}, \qquad \delta G_{\mu\nu}~=~\varepsilon K^{\lambda}~\partial_{\lambda}G_{\mu\nu}.\tag{D}$$
4. It is easier to use the Lie derivative definition of a Killing vector field $$0~=~({\cal L}_K G)_{\mu\nu}~=~ K^{\lambda}~\partial_{\lambda}G_{\mu\nu} + \partial_{\mu}K^{\lambda}~G_{\lambda\nu}+G_{\mu\lambda}~\partial_{\nu}K^{\lambda} \tag{E}$$ rather than the equivalent eq. (2).