# Euclidean space as a manifold

Let $\mathbbmss{E}^{n}$ be $n$-dimensional Euclidean space  , and let $(\mathbbmss{V},\langle\cdot,\cdot\rangle)$ be the corresponding $n$-dimensional inner product space of translation  isometries  . Alternatively, we can consider Euclidean space as an inner product space that has forgotten which point is its origin. Forgetting even more information, we have the structure of $\mathbbmss{E}^{n}$ as a differential manifold. We can obtain an atlas with just one coordinate chart, a Cartesian coordinate system $(x^{1},\ldots,x^{n})$ which gives us a bijection between $\mathbbmss{E}^{n}$ and $\mathbbmss{R}^{n}$. The tangent bundle  is trivial, with $\operatorname{T}\mathbbmss{E}^{n}\cong\mathbbmss{E}^{n}\times\mathbbmss{V}.$ Equivalently, every tangent space  $\operatorname{T}_{p}\mathbbmss{E}^{n},\;p\in\mathbbmss{E}^{n}$. is isomorphic to $\mathbbmss{V}$.

We can retain a bit more structure, and consider $\mathbbmss{E}^{n}$ as a Riemannian manifold  by equipping it with the metric tensor

 $\displaystyle g$ $\displaystyle=$ $\displaystyle dx^{1}\otimes dx^{1}+\cdots+dx^{n}\otimes dx^{n}$ $\displaystyle=$ $\displaystyle\delta_{ij}dx^{i}\otimes dx^{j}.$

We can also describe $g$ in a coordinate-free fashion as

 $g(u,v)=\langle u,v\rangle,\quad u,v\in\mathbbmss{V}.$

## Properties

Conversely, we can characterize Eucldiean space as a connected, complete Riemannian manifold with vanishing curvature and trivial fundamental group  .

Title Euclidean space as a manifold EuclideanSpaceAsAManifold 2013-03-22 15:29:48 2013-03-22 15:29:48 matte (1858) matte (1858) 9 matte (1858) Definition msc 53B21 msc 53B20