I want to show that the standard Riemannian metric $g=\delta_{ij}d\theta^i d\theta^j$ on $\mathbb{T}^n$ is flat.
It is my understanding that I need to show that $(\mathbb{T}^n,g)$ and $(\mathbb{R}^n,\overline{g})$ are locally isometric. Where $\overline{g}$ is the Euclidean metric.
I know that $\mathbb{T}^n\cong S^1\times\cdots\times S^1$. So take $x\in \mathbb{R}^n$. We want to show that $x$ has an open neighborhood that is isometric to an open subset of $\mathbb{T}^n$. I think I can use the map $\pi:\mathbb{R}^n\to S^1\times\cdots\times S^1$, $(x_1,\dots,x_n)\mapsto (e^{2\pi i x_1},\dots, e^{2\pi i x_n})$. But I have no idea how to get an open neighborhood of $x$ or how to write this proof formally. Please help!
Let $e_1,...,e_n$ be the basis of $\mathbb{R}^n$. You can also see $\mathbb{T}^n$ as the quotient of $\mathbb{R}^n$ by the group $G$ of translation generated by $t_{e_i}, i=1,...,e_n$. Remark that the elements of $G$ are isometries, this implies that the Euclidean metric of $\mathbb{R}^n$ is induces a metric on $\mathbb{T}^n$ for which the quotient map $p:\mathbb{R}^n\rightarrow \mathbb{T}^n$ is an isometry. Since $p$ is a covering, for every $x\in \mathbb{T}^n$, there exists an open subset $x\in U$ such that $p^{-1}(U)=\bigcup_{i\in I} U_i$ and the restriction of $p$ to $U_i$ is an isometry.