# How to express the covariant derivative on $TM$ by the covariant derivative on $M\times M$?

Let $$(M,g,\Gamma)$$ be a Riemannian manifold with the Levi-Civita connection. $$M\times M$$ and $$TM$$ have natural metric structrures inherited from $$(M,g,\Gamma)$$. Let $$\phi:TM \rightarrow M \times M$$ be defined as follows: $$\phi(z,u) = \big(\exp_z(u),\exp_z(-u)\big)$$ where $$\exp_z$$ is the exponential map $$\exp_z : T_zM \rightarrow M$$. At least in some neighbourhood of the section $$u=0$$ it is invertible.

Let $$F$$ be a tensor field on $$M\times M$$. Let $$G$$ be a tensor field on $$TM$$ defined as $$G = \phi^*F$$ where $$\phi^*$$ is the pullback/pushfoward of a tensor field (at least in the domain where $$\phi$$ is invertible). Let $$v \in T_{(z,u)}TM$$. My question is: how to write the covariant derivative $$\nabla_{v} G$$ in terms of the covariant derivative of $$F$$? My first thought was to just write $$\nabla_{v} G = \phi^*(\nabla_{d\phi(v)} F)$$ but since $$\phi$$ doesn't seem to be a homomorphism of the metric structures on $$TM$$ and $$M\times M$$, the covariant derivative on $$TM$$ is not a pullback of the covariant derivative on $$M\times M$$, so I don't think that's the correct formula. Still, the metric structures on $$TM$$ and $$M\times M$$ both originate from the same structure on $$M$$, so I think there must be some relation.

EDIT: I was able to conclude that

$$\nabla_{v} G - \phi^*(\nabla_{d\phi(v)} F) = -\frac{d}{dt}\Big|_{t=0} \Big( (\phi^*\tilde P^{\phi(\gamma)}_t) (P^\gamma_t)^{-1} G\Big)$$

where $$\gamma$$ is any curve on $$TM$$ such that $$\gamma'(0) = v$$, $$P^\gamma_t$$ is the parallel transport on $$TM$$ along $$\gamma$$ from point $$\gamma(t)$$ to point $$\gamma(0)$$, and $$\tilde P^{\phi(\gamma)}_t$$ is the parallel transport on $$M\times M$$ along the curve $$\phi(\gamma)$$ from point $$\phi(\gamma(t))$$ to $$\phi(\gamma(0))$$. I still don't know how to calculate this derivative $$\frac{d}{dt}$$ and express it in terms of, for example, the Riemann tensor, or at least the Synge's function.

• First, there is no natural metric on $TM$ stemming from $g$. There exist, for instance, the Sasaki metric, and the Cheeger-Gromoll metric, but they are not the only ones. Second, $\exp_z (u)$ may not even be defined if $u$ is too far away from $0$. Remember that $\exp_z (u)$ is the point $\gamma(1)$ where $\gamma$ is the unique geodesic determined by $\gamma (0) = z$ and $\dot \gamma (0) = u$. Nobody guarantees that $\gamma$, which is defined on some interval $[0, \varepsilon)$, may be extended to $[0,1)$. Jan 13, 2023 at 17:51