Isometry Invariance of Riemannian Distance Function
Jump to navigation
Jump to search
Theorem
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be connected Riemannian manifolds with or without boundaries.
Let $\phi : M \to \tilde M$ be an isometry.
Let $x, y \in M$ be points.
Let $\map {d_g} {x, y}$ be the Riemannian distance between $x$ and $y$.
Then:
- $\map {d_{\tilde g} } {\map \phi x, \map \phi y} = \map {d_g} {x, y}$
Proof
![]() | This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2018: John M. Lee: Introduction to Riemannian Manifolds (2nd ed.) ... (previous) ... (next): $\S 2$: Riemannian Metrics. Lengths and Distances