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Many if not most authors use the term “$n$-dimensional Euclidean space” as synonymous with “$n$-dimensional real space”, $\mathbb{R}^n$. Some, however chose to be more rigorous and use the term Euclidean space and the symbol $\mathbb{E}^n$ to specifically refer the metric real space equipped with the usual Euclidean distance function, whereas the general real space $\mathbb{R}^n$ is the set of real $n$-tuples, which is not necessarily equipped with additional structure such as a metric.

Considering this distinction, are $n$-dimensional manifolds locally homeomorphic to $\mathbb{E}^n$ or just to $\mathbb{R}^n$? In other words, do the codomains of the coordinate maps of charts in a manifold’s atlas have to be open subsets of a metric space with a distance function?

user175324
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    If your local identification is by homeomorphisms, then Euclidean space is equivalent to $\mathbb R^n$. The topological category knows nothing about the metric. If you really want to have Euclidean charts (with Euclidean transition functions), then you're getting into geometric structures. – Ethan Dlugie Nov 25 '23 at 16:54
  • Got it, of course. If the open set of the chart is homeomorphic to $\mathbb{R}^n$ it is also homeomorphic to $\mathbb{E}^n$. The metric is irrelevant so saying that the manifold is locally Euclidean doesn’t mean the manifold has a corresponding metric. Thanks! – user175324 Nov 25 '23 at 17:12
  • More precisely, they mean $\mathbb R^n$ equipped with the standard topology. – Moishe Kohan Nov 25 '23 at 17:54
  • Thanks. Isn’t “$\mathbb{R}^n$ equipped with the standard topology” precisely the $n$-dimensional Euclidean metric space since the standard topology’s basis consists of open balls (or equivalently open cubes) defined using the Euclidean distance function? – user175324 Nov 26 '23 at 07:08
  • Topology is not the same as metric (you cannot say "topology is a metric"). You can say that the standard topology is induced by the Euclidean metric. – Moishe Kohan Nov 27 '23 at 04:09
  • Understood. Thanks. – user175324 Nov 27 '23 at 12:46

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Thanks to the comments, and chewing on this a bit more, I now see the gap in my thinking on what may be a somewhat nitpicky point.

A homeomorphism is by definition a function between topological spaces. So when we say that every point in a manifold is an element of an open set which is homeomorphic to $\mathbb{R}^n$ we mean to an open set of the topological space “$\mathbb{R}^n$ with the standard topology”. The standard topology is induced by the Euclidean metric so this topological space is in fact the Euclidean metric space $\mathbb{E}^n$. So the shorthand description of a manifold as “locally Euclidean” is not as loose as I had thought. One does in fact need the Euclidean metric to induce a the standard topology on plain old $\mathbb{R}^n$, but this does not imply that the manifold itself is equipped with a metric.

user175324
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