# Covariant derivative of a tensor field

Let $F$ be a tensor field of type $(0,2)$ on a Riemannian manifold (like a Riemannian metric). Let $\gamma$ be a geodesic on $M$ and let $e(t)$ be a parallel transport along $\gamma$. I want to find a formula for $\frac{d}{dt} F_{\gamma(t)}(e(t),\gamma'(t))$. How can I derivate this quantity. I understand calculcations when $F$ is the Riemanian metric (this is zero using compatibility between the metric and Levi Civita connection). But I don't see how to do for general $F$. Thanks in advance for your help.

• you can go back to the very low-level definition : en.wikipedia.org/wiki/… $\nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - V_{\gamma(0)}}{h} = \left.\frac{d}{dt}\Gamma(\gamma)_t^0V_{\gamma(t)}\right|_{t=0}$ so from the covariant derivative you can deduce the order $1$ Taylor expansion of.... – reuns Jun 4 '16 at 7:23

I'm not sure I understand precisely what you want, but the covariant derivative of $F$ along $\gamma$ is given by

$$\frac{DF}{dt}(e(t),\dot{\gamma}(t)) = \frac{d}{dt} F(e(t),\dot{\gamma}(t)) - F \left( \frac{De}{dt}(t), \dot{\gamma}(t) \right) - F \left( e(t), \frac{D \dot{\gamma}}{dt}(t) \right) = \frac{d}{dt} F(e(t),\dot{\gamma}(t))$$

since $e(t), \dot{\gamma}(t)$ are parallel along $\gamma$. The expression $F(e(t),\dot{\gamma}(t))$ is a scalar depending on $t$ so you can differentiate it just like you differentiate a regular function. If $e_i(t)$ is a frame alone $\gamma$ and you expand

$$e(t) = a^i(t) e_i(t), \,\, \dot{\gamma}(t) = \gamma^i(t)e_i(t), \,\, F(e_i(t),e_j(t)) = f_{ij}(t)$$

then

$$F(e(t), \dot{\gamma}(t)) = F(a^i(t)e_i(t), \gamma^j(t)e_j(t)) = a^i(t)\gamma^j(t) f_{ij}(t)$$

and so

$$\frac{d}{dt} F(e(t), \dot{\gamma}(t)) = \dot{a}^{i}(t)\gamma^j(t)f_{ij}(t) + a^i(t)\dot{\gamma}^j(t)f_{ij}(t) + a^i(t)\gamma^j(t)\dot{f_{ij}}(t).$$