- #71

PeterDonis

Mentor

- 45,278

- 22,479

Yes. What is thecianfa72 said:Yes, the identity element of the vector field vector space is the null vector field ##0##. If we "insert" it into post#68 definition we get $$\nabla_Y0 =0, \nabla_Z0=0,\ \forall Y,Z\in\Gamma(TM)$$ hence $$g(0,Z)=0, g(Y,0), \ \forall Y,Z \in\Gamma(TM)$$

*physical*meaning of the null vector field, considered as a Killing vector field?