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The differential of the exponential map on a symmetric space can be expanded (abusing some notation) as

$d{\rm Exp}_X=\sum_{n=0}^{\infty}\frac{({\rm ad}X)^{2n}}{(2n+1)!}.$

This is an old (1958) result of Helgason.

Question (EDITED): Is there any generalization to reductive homogeneous spaces?

[EDIT 2017] I managed to solve the problem myself, so now it is a published
result. (A. V. Gavrilov, The Differential of the Exponential Map for Reductive Homogeneous Spaces, J. Lie Theory 25 (2015), No. 2, 363--376). The formula is more complicated then one might expect, so I omit it here.

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  • $\begingroup$ Congratulations on solving the problem. I would like to very much read your paper but my institution doesn't have subscription to JOLT and I was unable to find preprint. $\endgroup$
    – Vít Tuček
    Commented Jul 29, 2017 at 8:22
  • $\begingroup$ Actually, I did not post a preprint. After some search, I found an old version of the paper on my old computer (in pdf) . So, you can just write me. $\endgroup$
    – Alex Gavrilov
    Commented Jul 29, 2017 at 9:33

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I think the formula will be essentially the same.

The formula you wrote is valid in general for the exponential map of analytic manifolds equipped with an analytic affine connection. It is stated and proved in this paper by Helgason (see pages 6-7 of the linked .pdf): Some remarks on the exponential mapping of an affine connection. Math. Scand. 9 (l961), l29-l46.

The existence of such connections on reductive homogeneous spaces, with an additional invariance property (that you might probably need at some point) is stated and proved in K.Nomizu, Invariant affine connections on homogeneous spaces, Amer. J. Math. 76 (1954), 33-65.

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  • $\begingroup$ Certainly I should have mentioned that I am familiar with the both those papers. I am sorry that I have not. The (very interesting) 1961 work of Helgason is not a generalization of his 1958 work, though the latter can be deduced from the former. (At least, it is my opinion). The problem is, to make sense from this formula, you have to introduce some very special vector fields and to compute their Lie brackets. I have no idea how this can be done practically in the reductive space case. Unlike this, the right hand side of the above equality is calculated instantly. $\endgroup$
    – Alex Gavrilov
    Commented Jul 28, 2012 at 9:22

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