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Let $M$ be a smooth simply-connected manifold. Let $\nabla$ be a flat, symmetric connection on $M$. Let $p\in M$ and let $v,w\in T_pM$ belong to a normal neighborhood, such that the $\nabla$-geodesic triangle with vertices $p$, $\exp_p v$ and $\exp_p(v+w)$ is in $M$. I am looking for a proof that $$ \exp_p(v+w) = \exp_{\exp_p v}(\Pi_p^{\exp_p v} w), $$ where $\Pi_p^q$ denotes the (path-independent) parallel transport $T_pM\to T_qM$.

One can prove this by endowing $M$ with a metric, which is clearly Euclidean, and exploit the fact that this property holds trivially for Euclidean spaces (i.e., associativity of vector addition).

I am looking, however, for a direct proof that only uses the given properties of the connection.

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    $\begingroup$ Around $p$ your flat symmetric connection is the same as having a coordinate system i.e. the $\nabla$-parallel vector fields are the linear combinations of the coordinates vector fields hence parallel transport in such coordinate system reduce to the obvious one. Also the exponential map, w.r.t. to these coordinates, is the usual one i.e. vectors identified with points. So what you asked that $exp_p(v + w) = etc $ is obvious (If you want a detailed symbolic explanation perhaps it is better to post the question in Math StackExchange.) $\endgroup$
    – Holonomia
    Dec 18, 2016 at 14:38
  • $\begingroup$ Parallel transport defines a local coordinate system when the connection is flat. However, this property relies also on the symmetry of the connection. Where does symmetry enter in your argument? $\endgroup$
    – Raz Kupferman
    Dec 18, 2016 at 17:03
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    $\begingroup$ Symmetry enter in the fact that the bracket between $\nabla$-parallel vector fields is zero i.e. the flows of $\nabla$-parallel vector fields commute. I insist that your question is more adequated for Math StackExchange. $\endgroup$
    – Holonomia
    Dec 18, 2016 at 17:10

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