Complex Scalar Fields and Killing Vectors

In a stationary and axisymmetric spacetime, there are two Killing vectors, say $$\zeta^\mu$$ and $$\xi^\mu$$, one timelike and one space like.

I understand that for a real scalar field, $$\phi$$, that obeys the symmetry of this spacetime one can say

$$\zeta^\mu \nabla_\mu \phi = \xi^\mu \nabla_\mu \phi = 0$$

and therefore since the gradient of $$\phi$$ is orthogonal to both Killing vectors, $$\nabla_\mu \phi \nabla^\mu \phi \geq 0$$.

Now, my question is in regards to how this logic translates across if $$\phi$$ becomes a complex scalar field, where $$\Phi = \phi_1 + i\phi_2$$.

$$\nabla_\mu \Phi \nabla^\mu \Phi \geq 0$$ no longer makes sense to me, since the gradient is complex, however is it a valid comment to say that the real components of the complex field must be spacelike or null, if the complex field is to still obey the spacetime symmetries, i.e. $$\nabla_\mu \phi_i \nabla^\mu \phi_i \geq 0$$

You're right to conclude $$\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$$ for $$j\in\{1,\,2\}$$, so $$\nabla_\mu\Phi^\ast\nabla^\mu\Phi=\sum_{j=1}^2\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$$.
• Okay thanks, just as a (potentially trivial) follow up, can I check that can that be generalised to $$\sum^2_{i,j=1} \nabla_{\mu} \phi_i \nabla^{\mu} \phi_{j} \geq 0$$. – Smp03 Aug 15 at 22:37
• @Smp03 You can rewrite that as $\nabla_\mu\chi\nabla^\mu\chi\ge0$ with $\chi:=\sum_j\phi_j$. – J.G. Aug 16 at 7:05