I recently happened upon a very interesting construction for adapted local additions (used in manifolds of mappings constructions). However, the construction requires a piece of information on parallel transport which I do not get. Here are the details:


Consider a Riemannian manifold $M$ with Riemannian exponential map $\exp\_M \colon \Omega\_M \rightarrow M$. 
Assume that $TM = \mathcal{V} \oplus \mathcal{H}$ as smooth vector bundles, where over each $x$, $\mathcal{V}\_x$ and $\mathcal{H}\_x$ are orthogonal.
Now for a smooth curve $\alpha \colon [0,1] \rightarrow M$ and $v \in \mathcal{V}\_{\alpha (0)}$, we let $P (\alpha , v) \in \mathcal{V}\_{\alpha (1)}$ be the parallel transport in $\mathcal{V}$ induced by the metric over $\alpha$.


For $v \in \Omega\_M$ define the smooth curve $\alpha\_v \colon [0,1] \rightarrow M,t\mapsto \exp\_M (tv)$ and consider the smooth map
$$\theta \colon (\mathcal{V} \oplus \mathcal{H}) \cap \Omega\_X \rightarrow TM, (v,h) \mapsto P(\alpha\_h,v).$$ 
Then the source claims that for every $x \in M$ the tangent map $T\_{0\_x} \theta \colon T\_{0\_x} \Omega\_M \rightarrow T\_{0\_x} T\_x M$ is (perhaps up to canonical identifications) the identity.
However, I do not see how one would prove this. Actually it would be sufficient for me to see that $T\_{0\_x} \theta$ is an isomorphism. Does somebody have an idea? 



Tangent of Parallel transport with respect to the curve
It turns out that the tangent map is indeed the identity. 


To see this, note that per the comments above the question is a local matter, whence without loss of generality we ma assume that our manifold is $U \subseteq \mathbb{R}^n$ open and there is some Riemannian metric on it together with the above described smooth splitting of the tangent bundle.


Not lets consider an element $(x,0,Z,w) \in T\_{(x,0)} TU$. Using the splitting of the tangent bundle, we may further split $w = w\_v + w\_h$ into a part tangent to $\mathcal{V}$ and tangent to $\mathcal{H}$.
Using linearity of $T\_{(x,0)} \theta$ it is enough to consider two cases:


1. Case $w\_h=0$. Then we can choose a smooth curve $c$ which takes its image in $\mathcal{V}$ and represents the tangent vector (i.e. passes through $(x,0)$ at $t=0$ whose derivative coincides with $(Z,w\_v)$). Plugging this into $\theta$ we see that we are asking to parallel transport a vertical vector along constant curves (= identity by definition), so passing to the derivative we recover $(Z,w\_v)$ exactly.
2. Case $Z=0, w\_v=0$. We can choose a curve which takes its image in $T\_x U \cap \mathcal{H}$ and represents the tangent vector. By definition of $\theta$ we are asking for the solution of the parallel transport equation along $\exp (t c(s))$ of the zero vector. Thus the curve we obtain in $TU$ is $(\exp (c(s)), 0)$. Taking the $s$-derivative in $0$ we use now that the derivative of $\exp$ in $(x,0)$ is the identity, whence the result equals $(x,0,0,w\_h)$.


Combining both cases we obtain that the derivative of $\theta$ in $(x,0)$ is the identity as desired.
For more details see the preprint



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> Habib Amiri, Alexander Schmeding, *A differentiable monoid of smooth maps on Lie groupoids*, arXiv:[1706.04816](https://arxiv.org/abs/1706.04816)
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The computation is Step 4 in the proof of Lemma C.4 on page 32.The preprint also contains an application of this result to manifolds of mappings (this is a crucial step in the proof of what we call the Stacey-Roberts Lemma).


