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Suppose you have a Riemannian manifold $(M,g)$ and a point $p\in M$ fixed. Let $v: s\mapsto v(s)$ be a curve in $T_pM$. Now consider the map $f(s):=\exp_p(v(s))$.

Can one get an explicit formula for $f'(s)$? Maybe that is confusing but I thought of something like $f'(s)=d(\exp_p)_{v(s)}(\nabla_sv(s))$, where $\nabla_s$ denotes the covariant derivative.

Formally, $d(\exp_p)_{v(s)}$ is a map from $T_{v(s)}T_pM$ to $T_{\exp_p(v(s))}M$ and one can identify $T_{v(s)}T_pM\cong T_pM$. I would like to know if the formula above is true and how this identification is related to the Levi-Civita-connection on $M$. Thanks for any explanation.

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1 Answer 1

Since $T_pM$ is a vector space, the derivative of $v(s)$ is defined without need to resort to "covariant derivative", so in your formula $\nabla_s v(s)$ can be replaced simply by the derivative $v'(s)$. The formula you wrote down is true by chain rule. In fact, it would be true for any map and not just the exponential map. To define the exponential map, one certainly needs the Levi-Civita connection.

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Thank you very much for your answer. So you say the usual and the covariant derivative are the same in this case? –  gfq May 27 '13 at 12:47
    
In a vector space we have a natural covariant derivative which coincides with the "usual" derivative. In a manifold of course we don't have any covariant derivative without additional data such as a metric. As I said, they are the same in the tangent space at $p$. –  user72694 May 27 '13 at 12:49

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