Is this covariant derivative identity true?

Question

Trying to work through a textbook derivation of the geodesic deviation equation, I've calculated this identity:$$u_{;\beta}^{\alpha}u_{\alpha}=u_{\alpha;\beta}u^{\alpha}.$$

If this is true, I'm making progress. If it isn't, it's back to the drawing board. Does anyone know if it is correct? I've tried writing it out as$$\left(\frac{\partial u^{\alpha}}{\partial x^{\beta}}+u{}^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\right)u_{\alpha}=\left(\frac{\partial u_{\alpha}}{\partial x^{\beta}}-u_{\gamma}\Gamma_{\alpha\beta}^{\gamma}\right)u^{\alpha},$$ but I'm none the wiser.

Answer 1

The identity in question is given as,

$$(\nabla_\beta u^\alpha) u_\alpha=(\nabla_\beta u_\alpha)u^\alpha$$

Expanding the left-hand side, we find,

$$(\nabla_{\beta}u^\alpha)u_\alpha = (\nabla_\beta g^{\alpha \delta}u_\delta)u_\alpha = (u_\delta \nabla_\beta g^{\alpha\delta} + g^{\alpha\delta}\nabla_\beta u_\delta)u_\alpha$$

The Levi-Civita connection is precisely the connection which is compatible with the metric structure; as such parallel transporting two tangent vectors preserves the inner product, so we have,

$$\nabla_\beta g^{\alpha\delta} = 0$$

Hence, for the left-hand side, we are left with a term $u^\alpha \nabla_\beta u_\alpha$. Just a general tip regarding these computations: try to avoid expanding the covariant derivative explicitly in terms of the Christoffel symbols as you have done - there are usually better paths to take, just as one is usually advised not to fully expand commutators most of the time.