1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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.... $\endgroup$ – reuns Jun 4 '16 at 7:23
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.