I've got a question about local affine diffeomorphisms between affine manifolds.
There ist a good characterisation about affine diffeomorphisms of connected affine Mannifolds:
Let $f,g\colon (M,\nabla^M) \rightarrow (N,\nabla^N)$ be affine diffeomorphisms between connected affine manifolds, $p\in M$ a Point with $f(p) = g(p)$ and $df_p = dg_p$. Then $f = g$.
Does this characterisation also hold for local affine diffeomorphisms?
The main Point in the proof is that given an affine diffeomorphism $f\colon M\rightarrow N$, $f\circ exp_p = exp_{f(p)}\circ df_p$ holds for every $p\in M$. I see no reasons why this shouldn't hold for local affine diffeomorphisms.
Any help would be appreciated as I don't see the point.
I don't believe one needs any "diffeomorphism" assumption to prove a result like this. In particular the following seems to be true:
Theorem. Let $M, N$ be affine manifolds with $M$ connected, and let $f$ and $g$ be (smooth) affine maps $M \to N$. Suppose there exists $p \in M$ with $f(p) = g(p)$ and $df_p = dg_p$. Then $f = g$.
For this we'll prove something like your formula:
Lemma. Let $f : M \to N$ be affine. Then given $p \in M$, there exists a neighborhood $U$ of $0 \in T_pM$ such that for $v \in U$ $$ f(\exp_p(v)) = \exp_{f(p)}(df_p(v)). $$
Proof of Lemma. Let $V$ be a neighborhood of $0 \in T_{f(p)}N$ such that $\exp_{f(p)}$ is a diffeomorphism on $V$ and so that $V$ is star-convex with respect to the origin. Let $U$ be a star-convex neighborhood of $0 \in T_pM$ contained in $df_p^{-1}(V)$.
Given $v \in U$, $f$ must map the the geodesic $t \mapsto \exp_p(tv)$ to a geodesic (since $f$ is affine) and thus this geodesic must be $\exp_{f(p)}(df_p(tv))$, q.e.d.
Proof of Theorem. Let $S \subseteq M$ be the set of $q$ with $f(q) = g(q)$ and $df_q = dg_q$. Obviously $S$ is closed since $f$ and $g$ are smooth. We claim $S$ is also open; since it is nonempty and $M$ is connected, this would complete the proof.
Fix $q \in S$ and a neighborhood $U$ of $0 \in T_qM$ as in the lemma that works for both $f$ and $g$. Then for $v \in U$, $$ f(\exp_q(v)) = \exp_{f(q)}(df_q(v)) = \exp_{g(q)}(dg_q(v)) = g(\exp_q(v)), $$ the middle equality following since $q \in S$; so $f = g$ on $\exp_q(U)$, and hence $\exp_q(U) \subseteq S$, q.e.d.