The question is: given a Riemannian manifold with a non-degenerate metric g and an inner product $u_{\beta}u^{\beta}=1$, is $\nabla_{\mu} (u_{\alpha}u_{\beta})=0$ without demanding the trivial solution $\nabla_{\mu}u_{\alpha}=0$? The equivalent question can be raised on a Lorentzian manifold with $u_{\beta}u^{\beta}=-1$. I recently asked this seemingly simple question on Math Exchange but am getting little feedback and would appreciate the expertise on this site.
Given the norm as stated, it is clear that $\nabla_{\mu}(u_{\beta}u^{\beta})=0$ which means $ g^{\alpha\beta}\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $. Since g is non-degenerate, $ \mid \nabla_{\mu}(u_{\alpha}u_{\beta})\mid=0 $ and $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ is a solution (the singular solution of the corresponding matrix which would be represented by a sum of linear dependent terms) that does not require the trivial solution $\nabla_{\mu}u_{\alpha}=0 $. The trivial solution is sufficient but not necessary. In fact, $ u_{\beta}\nabla_{\mu}u_{\alpha}+u_{\alpha}\nabla_{\mu}u_{\beta}=0 $ means $ u^{\beta}\nabla_{\mu}u_{\beta}=0 $ so in general, the vector is orthogonal to the covariant derivative of its covector, and not constant.
The only comment I have had against taking $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ to be consistent with $u_{\beta}u^{\beta}=1$ and with the vector u being orthogonal to the covariant derivative of its covector, is to require the vector u to be constant. That makes no sense to me. Please give me your valued opinions on this issue.
