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.

  • $\begingroup$ 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 $[0,1)$. $\endgroup$
    – Alex M.
    Jan 13, 2023 at 17:51


You must log in to answer this question.