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Given a smooth closed Riemannian manifold $(M,g)$ with injectivity radius $i_M$ I can choose for every $x \in M$ an orthonormal basis of $T_x M$ and a normal coordinate system $ \phi$ in $x$, such that the Christoffel symbols satisfy $$ (\Gamma^k_{ij})_x(0)=0. $$ By differentiability we have for $p \in B_{i_M}(0)$ $$ |(\Gamma^k_{ij})_x(p)| \leq |p| \sup_{p \in B_{i_M}(0)} |D(\Gamma^k_{ij})(p)| $$ How can I bound $\sup_{p \in B_{i_M}(0)}|D(\Gamma^k_{ij})(p)|$ in terms of the curvature tensor $R$ of $(M,g)$? I found that the Christoffel symbols satisfy $$ \Gamma^k_{ij,l} = -\frac{1}{3} ( R^k_{jkl} + R^k_{kjl}). $$ Does this imply that $$ \sup_{p \in B_{i_M}(0)} |D(\Gamma^k_{ij})(p)| \leq \sup_{p \in B_{i_M}(0)} \sup_{ijkl} R^k_{jkl}(\phi^{-1}(p)). $$

djamba
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    You can express the curvature tensor in terms of Christoffel symbols, see https://en.wikipedia.org/wiki/Riemann_curvature_tensor . – quarague Aug 26 '19 at 09:20
  • But how can I use the equality $R^\ell{}{ijk}= \frac{\partial}{\partial x^j} \Gamma^\ell{}{ik}-\frac{\partial}{\partial x^k}\Gamma^\ell{}{ij} +\Gamma^\ell{}{js}\Gamma_{ik}^s-\Gamma^\ell{}{ks}\Gamma^s{}{ij}$ to find an estimate? – djamba Aug 26 '19 at 12:09
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    What makes you think this is possible? You can have very nonzero Christoffel symbols with $R=0$ — isn't polar coordinates on the plane going to give you an example? – Ted Shifrin Aug 27 '19 at 04:01

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