Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $\alpha : [a,b] \rightarrow M$ be a smooth curve on $M$, and let $\frac{D}{dt}$ be the induced covariant derivative on $\alpha$. I want to prove that for all $X,Y$ smooth vector fields on $\alpha$,
$$\frac{d}{dt}\big{(}g(X,Y)\big{)} = g(\frac{D}{dt}X,Y) + g(X,\frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).
I've tried using substituting coordinates (for a chart with coordinate functions $x^1,…,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $\frac{D}{dt}X = \left( \frac{dX^i}{dt} + \Gamma^i_{jk}\frac{d(x^j\circ\alpha)}{dt}X^k\right)\partial_i$. But I am stuck:
$$\frac{d}{dt}\big{(}g(X,Y)\big{)} = \frac{dg_{ij}}{dt}X^iY^j + g_{ij}\frac{dX^i}{dt}Y^j + g_{ij}X^i\frac{dY^j}{dt}$$
but
$$ g(\frac{D}{dt}X,Y) + g(X,\frac{D}{dt}Y) = g_{ij}\frac{dX^i}{dt}Y^j + g_{ij}X^i\frac{dY^j}{dt} + g_{ij}(\Gamma^i_{kl}\frac{d(x^k\circ\alpha)}{dt}X^lY^j + \Gamma^i_{kl}\frac{d(x^k\circ\alpha)}{dt}X^jY^l) $$
I do not see how these two expressions are equal.
Is there something wrong? Is there something I am missing?
Thanks!
Best Answer
From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}\Gamma^{i}_{kl}\dot{x}^k X^l Y^j + g_{il}\Gamma^{i}_{kj}\dot{x}^k X^lY^j = (g_{ij}\Gamma^i_{kl} + g_{il}\Gamma^i_{kj})\dot{x}^kX^lY^j$$
Since $g_{ij}\Gamma^{i}_{kl} = \dfrac{1}{2}(\partial_k g_{lj} + \partial_l g_{jk} - \partial_j g_{kl})$ and $g_{il}\Gamma^{i}_{kj} = \dfrac{1}{2}(\partial_k g_{jl} + \partial_j g_{lk} - \partial_l g_{kj})$, then $g_{ij}\Gamma^{i}_{kl} + g_{il}\Gamma^{i}_{kj} = \partial_k g_{lj}$. By the chain rule, $\partial_k g_{lj}\, \dot{x}^k = \dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}\Gamma^{i}_{kl} + g_{il}\Gamma^{i}_{kj})\dot{x}^k X^l Y^j = \frac{d g_{lj}}{dt}X^l Y^j = \frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.