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Given a germ of manifolds and compatible Riemannian metrics, can we construct a new Hausdorff manifold using the exponential map?

A germ of manifolds at a point $m$ is a series of manifolds $U_i$ containing the point $m$ such that each $U_i$ agrees with $U_j$ in a smaller open set $(m \in ) U_{ji} \subset U_i$ by $x \sim f_{ji}(x)$, with $f_{ji} :U_i \rightarrow U_j$ satisfying the cocycle condition $f_{kj} \circ f_{ji} = f_{ki}$.

A compatible Riemannian metric of a germ of manifolds consists of a Riemannian metric $g_i$ on each $U_i$ such that two such Riemannian metrics $g_i$ and $g_j$ on $U_i$ and $U_j$ agree with each other in the sense that $g_i(x) = g_j(f_{ji}(x))$ in a smaller open set (possibly a subset of $U_{ji}$).

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    $\begingroup$ What's a "germ of manifolds", and what are "compatible Riemann metrics"? Also, the manifold you construct, do you want it to have any properties or could we just say it is the empty set? $\endgroup$
    – Ryan Budney
    Commented Mar 22, 2021 at 7:03
  • $\begingroup$ What does it mean that $U_i$ agrees with $U_j$? If $f_{ji}$ is defined on all of $U_i$, and a diffeomorphism, it seems you are just describing diffeomorphic manifolds $U_i$. $\endgroup$
    – Ben McKay
    Commented Mar 23, 2021 at 14:48
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    $\begingroup$ Why not just take any one of these $U_i$ as the Hausdorff manifold you want to construct? What other conditions should it satisfy? $\endgroup$
    – Ben McKay
    Commented Mar 23, 2021 at 14:50

2 Answers 2

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Let me suggest a possible formulation of the problem. Let $M$ be a Riemannian manifold, and let $p$ be a point of $M$. Does there exist an open set $U\subseteq M$ containing $p$, and a Riemannian manifold $N$, and an isometric embedding $f\colon U\to N$, such that the exponential map can be defined on all of $T_{f(p)}N$?

This would certainly hold if $M$ could be isometrically embedded in a complete Riemannian manifold. This paper shows that such an embedding is not always possible. It would also be enough if we could choose a neighbourhood $U$ as above and an isometric embedding of $U$ in a complete Riemannian manifold. I don't know if that is always possible.

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    $\begingroup$ Are you sure you want to phrase it in this way? If $U$ is a small enough geodesic ball, you can easily construct a complete $N$ - closed even - into which it embeds isometrically. $\endgroup$
    – Leo Moos
    Commented Mar 23, 2021 at 21:08
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    $\begingroup$ Your formulation has a positive answer. Take a bump function, equal to one near $p$ and to zero outside a compact set around $p$, with the compact set contained inside the domain of a coordinate chart. Pick $U$ inside the set where the bump function is one. Use the bump function to glue the given metric to the Euclidean metric in the coordinates, bump times g plus (one minus bump) times Euclid. $\endgroup$
    – Ben McKay
    Commented Mar 23, 2021 at 21:08
  • $\begingroup$ @BenMcKay Fair enough. Perhaps the OP can explain whether that answers their question. $\endgroup$
    – Neil Strickland
    Commented Mar 23, 2021 at 21:13
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This is too long for a comment, but not an answer.

Consider the set $Z$ of connected Riemannian manifolds, each with one marked point. Consider two equivalent if there is a diffeomorphism matching the metrics and points. Let $X$ be the set of equivalence classes.

Now consider the same set $Z$ of connected Riemannian manifolds, each with one marked point. Consider two equivalent in a different sense if there is a diffeomorphism of some open neighborhoods of the points, matching the metrics and points. Let $Y$ be the set of equivalence classes for this equivalence relation.

The identity map $Z \to Z$ descends to the obvious map $X\to Y$.

I wonder if the question you are asking can be stated as asking if there is a continuous section $Y\to X$?

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