My question is inspired from this older question and the discussion in the comments there.
I know that isometries of Riemmanian manifold form a Lie group. I was wondering if something more concrete can be said in the following case:
Let $(M, g), (N, h)$ two smooth Riemannian 2-dimensional manifolds (with or without boundary).
Let $\phi :N \to \mathbb{R}^3$ be smooth isometric embedding of $N$ in Euclidean space $\mathbb{R}^3$.
- Consider an embedding of $M$ in Euclidean space $\mathbb{R}^3$: $$ \mu: M \to \mathbb{R}^3, $$ which also maps $M$ isometrically to $\phi(N)$, i.e., to the embedding of $N$ in $\mathbb{R}^3$.
(So really, we have $\mu: M \to \phi(N) \subset \mathbb{R}^3$.)
My question: Can one think of such $\mu(\cdot)$ as, necessarily, built of composition of 3 types of maps:
- (i) A rigid transformation of $\mathbb{R}^3$,
- (ii) a flow of a Killing vector field of $M$ (if exists),
- and perhaps (iii) a map that realizes other discrete symmetries of $M$ (i.e., other symmetries that do not arise from a Killing vector field).
My knowledge of Riemmanian geometry is rather limited and I'm not familiar with Lie groups. So even if a full answer is involved, I would appreciate any intuition or anyone pointing to problems in the formulation of the question itself, why it is ill posed or why it does not make sense.