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Let $(M,g)$ be a Riemannian manifold and let $\exp^M:TM\to M$ denote the exponential map. Its tangent map $T\exp^M$ is a map $TTM\to TM$. The connection on $TM$ induces a canonical metric on $TM$, the Sasaki metric $S$. Denote by $\exp^{TM}$ the exponential map on $(TM,S)$, which is a map $TTM\to TM$.

Question: is there a (simple) relation between $T\exp^M$ and $\exp^{TM}$? Are they equal? If not, are there cases in which they are equal (e.g, spheres)?

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  • $\begingroup$ The two maps are closer to being comparable after pre-composing one or the other with the involution of the double tangent bundle. Jan's comment is observing that the two maps take values in involuted tangent spaces. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2023 at 4:36
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    $\begingroup$ I think your question boils down to the issue of if one varies the starting point of a geodesic, this process infinitesimally gives rise to a vector field along the geodesic. Your question asks if these vector fields are parallel along the geodesic. Presumably this isn't always true and there will be a relation with the Riemann curvature tensor. It might be something of the form $exp^{TM}-T(exp^M) \circ \iota$ being given in terms of some ODE involving the Riemann curvature tensor along the appropriate geodesic. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2023 at 8:18
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    $\begingroup$ @RyanBudney This question has also been studied. The ODE is called the Jacobi equation and these fields being parallel is equivalent to the manifold in question being flat. $\endgroup$
    – Jan Nienhaus
    Commented Oct 30, 2023 at 19:53

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These maps can never be equal unless your manifold has dimension zero.

This has a rather trivial reason. If we look at both of these maps on a neighborhood of $0$ in $T_v T_p M$ for some fixed $v\in T_p M$, we see that $d{\exp^M}$ takes values in a neighborhood of $0$ in $ T_{\exp(v)}M$, while $\exp^{TM}$ takes values near $v$ in $TM$. Since $\exp(v)$ tends to be different in $M$ from the basepoint of $v$, these have no chance of generic equality.

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  • $\begingroup$ Notation depends a bit on how you define these things. For me $TM$ is as a set the union of all tangent spaces, so an element of $TM$ is called $v$, while for you $TM$ seems to consist of pairs of a tangent vector and its basepoint. Either way, $T_v T_p M$ is not the same as $T_v TM$, one is a strict subset of the other. Of course, since we could have just evaluated only a zero to begin with, the two maps can generally agree on neither. $\endgroup$
    – Jan Nienhaus
    Commented Oct 26, 2023 at 23:17
  • $\begingroup$ Oh, you are perfectly right! I edited the answer, that was a slip of my brain. $\endgroup$
    – Jan Nienhaus
    Commented Oct 27, 2023 at 18:59
  • $\begingroup$ thanks for fixing it. I've deleted my comments which are now obsolete. $\endgroup$
    – Willie Wong
    Commented Oct 30, 2023 at 4:34

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