Symmetry of the Polyakov action?

Question

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?

Answer 1

The answer is: Yes, the Polyakov action is invariant under a Killing symmetry in the target space.

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).

Answer 2

Killing vector fields correspond to infinitesimal isometry generators of the spacetime manifold and any physical action including the Polyakov action should be preserved under it. In fact, any physical action should be invariant under the (infinitely) larger group of diffeomorphisms of a manifold. Isomotry transformations are just a finite subset of these diffeomorphisms.