Does a differential-geometry-isometry map a (sub)vector space to a (sub)vector space? Given an isometric embedding of the tangent bundle $TM$ of a Riemannian manifold with Sasaki metric into $ℝ^N$ with standard metric.
My intuition tells me that the vector spaces $T_pM$ ($p ∈ M$) are mapped to affine sub spaces of $ℝ^N$ but cannot find a rigourous proof. Can you help?

Additional thoughts:
The existence of the isometric embedding is guaranteed by the Nash embedding theorem.
It is unclear to me though how this embedding looks like (see my other question).
Here isometry does not mean that the distance function is preserved (that would be isometry in the metric space sense) but the Riemannian metric is preserved: $⟨dι_v A, dι_v B⟩_{ℝ^N} = ⟨A, B⟩_{G_s}$ for any $v ∈ TM$, $A, B ∈ T_vTM$, $ι$ is the isometric embedding.
The Sasaki metric restricted to $T_pM$ is what one would expect: the vertical part of $T_vTM$ ($v ∈ T_pM$) can be identified with $T_pM$ but answer my question one probably has to be very careful how to exploit this identification.
 A: $\newcommand{\Reals}{\mathbf{R}}$Every Riemannian manifold can be isometrically embedded in a sphere; for a Sasaki metric the fibres map to bounded sets, which are therefore not affine.
To flesh out the comment, IIRC, a flat cylinder may be isometrically embedded into a bounded subset of Euclidean four-space by choosing a bounded, complete plane spiral $C$, such as the polar graph $r = 2 + \tanh\theta$, and mapping $S^{1} \times \Reals$ to $S^{1} \times C$ by the identity in the first factor and an arclength parametrization in the second.
Added: I overlooked that the linked question asks about closed (i.e., compact) Riemannian manifolds. The linked answer does not assume closedness, however. For self-containedness: If $(M, g)$ is a Riemannian manifold, there exists an isometric embedding into some Euclidean space $\Reals^{N}$ by the Nash embedding theorem. The real line embeds isometrically in a flat torus $S^{1} \times S^{1}$ (using for example the curve below from the linked post), so $\Reals^{N}$ embeds isometrically in $(S^{1} \times S^{1})^{N} \subset \Reals^{4N}$. Composing gives an isometric embedding of $(M, g)$ in a torus (which incidentally embeds isometrically in a Euclidean sphere).
In particular, a Sasaki metric embeds isometrically into a bounded subset of some Euclidean space, so its fibres do not generally map to affine subsets.
In case it's useful for posterity, these curves can be chosen real-analytically, so the resulted bounded embedding preserves whatever real smoothness is possessed by the Nash embedding.

