# Manifold being locally euclidean vesus Manifold being locally homeomorphic to an open set in $R^n$.

I was reading the definition of smooth manifold and i am little bit of confused. Informally it says

A smooth manifold is a topological manifold (i.e. a topological space locally homeomorphic to a Euclidean space) equipped with an equivalence class of atlases whose transition maps are all smooth. Here transition maps are from Euclidean space to Euclidean space.

Now what if I replace the word Euclidean space by topological vector space $\mathbb{R}^n$. Still we will get some object as we can talk about smoothness in topological vector space $\mathbb{R}^n$ without using any reference of coordinate system. So how much difference is there if I do the replacement? And my other question is

What are the properties of Euclidean space, we use to study Smooth manifolds, which are not present in $\mathbb{R}^n$ just as a topological vector space.

Till now what I have found that to describe local coordinates in manifold or to describe basis in tangent space one need a coordinate system in $\mathbb{R}^n$ so that you can pull it back to the manifold to define local coordinate there. Apart from these, where do we use the properties of Euclidean space to study smooth manifold?

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Any n-dimensional Euclidean space is isomorphic to $\mathbb{R}^n$, so the terms are used interchangebly. –  Nick Alger Jun 20 '13 at 21:54
@ Nick: Of which isomorphism between $\mathbb{R^n}$ and Euclidean space you are talking about.Is it vector space isomorphism you are talking about?Is it true $\mathbb{R^n}$ with $l^1$ norm and Euclidean space are Isomorphic? –  timon Jun 21 '13 at 5:42

## 1 Answer

Euclidean spaces are, by definition, $\mathbb{R}^{n}$. See if this clears things up for you. If not, come back and I'll try to help you some more.

Also, as the previous comment noted, any open set homeomorphic to an open ball will also be homeomorphic to all of $\mathbb{R}^{n}$ since open balls are homeomorphic to $\mathbb{R}^{n}$. Hence, in the definition of locally Euclidean, it does not matter if we a priori decided that our spaces should be locally homeomorphic to all of $\mathbb{R}^{n}$, an open ball, or an open set.

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This is roughly Exercise 1.1 in Lee's, "Introduction to Smooth Manifolds". –  Dylan Yott Jun 20 '13 at 22:58
@ Alex: I think there is a distinction between R^n as just topological vector space and Euclidean Space i.e. R^n with inner product structure. So how locally Euclidean and locally R^n could be same? Although topologically they are same but then why in the definition of manifold it asks to be locally Euclidean? –  timon Jun 21 '13 at 5:10
@ Alex: If you don't have a standard co-ordinate in R^n ,how you will talk about local coordinate in manifold? –  timon Jun 21 '13 at 5:12
@ Yott: You are mistaken.May be I was not able to pose my question properly.But I am not asking the Exercise 1.1 mentioned in Lee's book. I want to know why call it locally Euclidean instead of locally R^n.What property of Euclidean space (where you can talk about angle and distance) we are going to use in future to study smooth manifold. –  timon Jun 21 '13 at 5:27
Eventually, when you go deep enough to get to Riemmanian Geometry, you will starting putting inner product structures on your manifolds, but that's not for a while. For now, the words "locally Euclidean" and "locally $\mathbb{R}^{n}$" are synonymous. Also, locally Euclidean sounds better. The book you're reading doesn't make the distinction between $\mathbb{R}^{n}$ as a topological vector space and as an inner product space. The reason we use $\mathbb{R}^{n}$ as opposed to any topological vector space is that we already know how to do calculus on $\mathbb{R}^{n}$. –  Alex Lapanowski Jun 21 '13 at 15:12