The canonical Euclidean metric on $\mathbb{R}^n$ is as $\sum_{i=1}^ndx_i^2$. We also know that every Riemannian space of zero sectional curvature is Euclidean. Therefore, for example, $\sum_{i=1}^na_idx_i^2$, where $a_i>0$, is a non-canonical Euclidean metric on $\mathbb{R}^n$. Now my doubt is as following: How are the isometries of this non-canonical metric? I considered a rotation $D$ on $\mathbb{R}^3$ and tried to show that it is an isometry for $(\mathbb{R}^3,\sum_{i=1}^na_idx_i^2)$. Indeed I wanted to show that $||X||=||DX||$. However I failed!! It means $D$ is not an isomtry for this non-canonical Euclidean space?
For example,
$ D=\left(\begin{matrix}
1 & 0 & 0\\
0 & \cos \alpha(p) & -\sin \alpha(p)\\
0 & \sin \alpha(p) & \cos \alpha(p)
\end{matrix}\right)$ with the metric $dx^2+2dy^2+3dz^2$.
Will appreciate any comment or corrections!
Best Answer
Yes, a standard rotation on $\mathbb{R}^3$ is in general not an isometry of $\mathbb{R}^3$ with the metric $\sum_{i=1}^3 a_i dx_i^2$.
More generally, let's say you have a real inner product on $\mathbb{R}^n$, denoted by $\left< \cdot, \cdot \right>$ which is not necessary the standard inner product and also the standard inner product $$\left< x, y \right>_{\textrm{euc}} = x^T \cdot y .$$
It is true that $(\mathbb{R}^n, \left< \cdot, \cdot \right>)$ and $(\mathbb{R}^n, \left< \cdot, \cdot \right>_{\textrm{euc}})$ are isometric (choose an orthonormal basis and map it to an orthonormal basis) but it doesn't mean that the isometries of $(\mathbb{R}^n, \left< \cdot, \cdot \right>)$ and $(\mathbb{R}^n, \left< \cdot, \cdot \right>_{\textrm{euc}})$ are the same . In fact, if $\varphi \colon (\mathbb{R}^n, \left< \cdot, \cdot \right>) \rightarrow (\mathbb{R}^n, \left< \cdot, \cdot \right>_{\textrm{euc}})$ is an isometry you can easily verify that $g \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is an isometry of $(\mathbb{R}^n, \left< \cdot, \cdot \right>)$ iff $\varphi \circ g \circ \varphi^{-1}$ is an isometry of $(\mathbb{R}^n, \left< \cdot, \cdot \right>_{\textrm{euc}})$. This means that the group of isometries of $(\mathbb{R}^n, \left< \cdot, \cdot \right>)$ and $(\mathbb{R}^n, \left< \cdot, \cdot \right>_{\textrm{euc}})$ are conjugate, not identical.