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6 deleted 4 characters in body

This isn't an answer its a conjecture. Nice question.

Suppose that $M, N$ are a Riemannian manifolds and $M\subset N$ is an open subset and $N$ is complete. Lets assume that $M$ is path connected, so that there is no funny business in defining the distance between $p, q \in M$ to be the infimum of the length of a path joining $p$ to $q$. Also lets assume that that path metric is bounded, so you don't have infinite ends.

There is a map from the metric space completion of $M$ into $N$ and its image will be the closure of $M$ in $N$. There is now a plethora of obstructions to the embedding, derived from this map.

For instance: Let $CI(\overline{M})$ be those continuous functions on the metric space completion of $M$ whose restriction to $M$ is smooth. Let $I$ be the ideal of all functions in $CI(\overline{M})$ that vanish at a point $p$ of the completion. It should be the case that $T=(I/I^2)^*$ is isomorphic to $\mathbb{R}^n$ where $n$ is the dimension of the manifold. Next, the metric tensor should extend to the completion, where you interpret it a point at infinity as a tensor on $T$, and its coefficients should be elements of $CI(\overline{M})$. Next you should be able to extend the Riemann curvature connection tensor appropriately as a map from the tensor square of T to itself, and the coefficients of the extension should also be in $CI(\overline{M})$ and they should satisfy all the restrictions on the tensor that the Riemann curvature tensor of a smooth manifold satisfies.

Here is my conjecture : The condition above is necessary and sufficient. The reason is you should be able to build a candidate piece of the manifold $N$ with normal coordinates, and those normal coordinate patches should glue together coherently.

5 added 4 characters in body

This isn't an answer its a conjecture. Nice question.

Suppose that $M, N$ are a Riemannian manifolds and $M\subset N$ is an open subset and $N$ is complete. Lets assume that $M$ is path connected, so that there is no funny business in defining the distance between $p, q \in M$ to be the infimum of the length of a path joining $p$ to $q$. Also lets assume that that path metric is bounded, so you don't have infinite ends.

There is a map from the metric space completion of $M$ into $N$ and its image will be the closure of $M$ in $N$. There is now a plethora of obstructions to the embedding, derived from this map.

For instance: Let $CI(\overline{M})$ be those continuous functions on the metric space completion of $M$ whose restriction to $M$ is smooth. Let $I$ be the ideal of all functions in $CI(\overline{M})$ that vanish at a point $p$ of the completion. It should be the case that $T=(I/I^2)^*$ is isomorphic to $\mathbb{R}^n$ where $n$ is the dimension of the manifold. Next, the metric tensor should extend to the completion, where you interpret it a point at infinity as a tensor on $T$, and its coefficients should be elements of $CI(\overline{M})$. Next you should be able to extend the Riemann curvature connection appropriately as a map from the tensor square of T to itself, and the coefficients of the extension should also be in $CI(\overline{M})$ and they should satisfy all the restrictions on the tensor that the Riemann curvature tensor of a smooth manifold satisfies.

Here is my conjecture : The condition above is necessary and sufficient. The reason is you should be able to build a candidate piece of the manifold $N$ with normal coordinates, and those normal coordinate patches should glue together coherently.

4 added 85 characters in body

This isn't an answer its a conjecture. Nice question.

Suppose that $M, N$ are a Riemannian manifolds and $M\subset N$ is an open subset and $N$ is complete. Lets assume that $M$ is path connected, so that there is no funny business in defining the distance between $p, q \in M$ to be the infimum of the length of a path joining $p$ to $q$. Also lets assume that that path metric is bounded, so you don't have infinite ends.

There is a map from the metric space completion of $M$ into $N$ and its image will be the closure of $M$ in $N$. There is now a plethora of obstructions to the embedding, derived from this map.

For instance: Let $CI(\overline{M})$ be those continuous functions on the metric space completion of $M$ whose restriction to $M$ is smooth. Let $I$ be the ideal of all functions in $CI(\overline{M})$ that vanish at a point $p$ of the completion. It should be the case that $T=(I/I^2)^*$ is isomorphic to $\mathbb{R}^n$ where $n$ is the dimension of the manifold. Next, the metric tensor should extend to the completion, where you interpret it a point at infinity as a tensor on $T$, and its coefficients should be elements of $CI(\overline{M})$. Next you should be able to extend the Riemann curvature connection appropriately as a map from the tensor square of T to itself, and the coefficients of the extension should also be in $CI(\overline{M})$ and they should satisfy all the restrictions on the tensor that the Riemann curvature tensor of a smooth manifold satisfies.

Here is my conjecture : The condition above is necessary and sufficient. The reason is you should be able to build a candidate piece of the manifold with normal coordinates, and those normal coordinate patches should glue together coherently.