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?

  • $\begingroup$ Isn't the question a local one, so that it is sufficient to consider M=R^n equipped with some metric? $\endgroup$ – David Roberts May 23 '17 at 23:00
  • 1
    $\begingroup$ Yes thats exactly what one should do. Further, I think the problem is purely to understand what the map does on vectors tangent to $\mathcal{H}$. By looking at smooth curves which evolve in $T_x M \cap \mathcal{V}$, the curve $\alpha_h$ is the constant path with value $x$. Now as parallel transport along this path is given by the identity, one immediately sees that $T_{0_x} \theta$ is injective on all vectors in $T_{0_x} \mathcal{V}$. Now dimensional reasons tell one that it is enough to prove injectivity on the orthogonal complement. $\endgroup$ – Alexander Schmeding May 24 '17 at 6:41

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

Habib Amiri, Alexander Schmeding, A differentiable monoid of smooth maps on Lie groupoids, arXiv:1706.04816

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).


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