Does Nash's theorem allow an embedded representation of the Riemann tensor without loss of generality?
Based on what is found here Nash embedding theorem: "The Nash embedding theorems (or imbedding theorems), named after John Forbes Nash, state that every Riemannian manifold can be isometrically embedded into some Euclidean space."
In the following post : How to prove the covariant derivative cannot be written as an eigendecomposition of the partial derivative?, I discovered an embedded representation of the Riemann tensor
\begin{align}\tag{1}
R_{adbc}
=
\dfrac{\partial y_\alpha}{\partial x^a}
\dfrac{\partial \Delta^{\alpha \beta} }{\partial x^b}
\dfrac{\partial^2 y_\beta}{\partial x^c \partial x^d}
\bigg|_{[b,c]}
=
\left(
\partial_b \Gamma_{acd} - \Gamma_{kab}\Gamma^{k}{}_{cd}{}
\right)
\bigg|_{[b,c]}
,
\end{align}
where $x^b$ is embedded into $y^\beta$,
\begin{align}\tag{2}
\Gamma_{abc} = \tfrac{1}{2}\left( \partial_b g_{ac} + \partial_c g_{ba} - \partial_a g_{bc} \right)
=
\dfrac{\partial y_\alpha}{\partial x^a}
\dfrac{\partial^2 y^\alpha}{\partial x^b \partial x^c}
\end{align}
is the Christoffel symbol,
\begin{align}
g_{mn}
=
\dfrac{\partial y^\alpha}{\partial x^m}
\dfrac{\partial y_\alpha}{\partial x^n}
\end{align}
is the metric tensor,
\begin{align}\tag{5}
\partial_b
=
\dfrac{\partial }{\partial x_b}
\end{align}
is the partial derivative,
\begin{align}\tag{3}
A_{bc}\big|_{[b,c]} = A_{bc}-A_{cb}
\end{align}
is the index commutator, and
\begin{align}\tag{4}
\Delta^{\alpha \beta}
=
\dfrac{\partial y^\alpha}{\partial x^m}
g^{mn}
\dfrac{\partial y^\beta}{\partial x^n}
\end{align}
where $g^{mn}$ is the matrix inverse of $g_{mn}$.
The tensor $\Delta^{\alpha \beta}$ can be summarized as a curved identity matrix with the following properties: \begin{align} \Delta^{\alpha\beta} &=& \lim_{g\rightarrow\text{flat}} \partial_{b}\Delta^{\alpha \beta} &=& \partial_b \delta^{\alpha \beta} &=& \partial_b (\text{constant}) &=& 0 \\ \Delta^{\alpha\beta} &=& \lim_{g\rightarrow\text{curved}} \partial_{b}\Delta^{\alpha \beta} &=& \partial_{b}\Delta^{\alpha \beta} &=& \partial_b (\text{variable}) &\neq& 0 \end{align} where $\delta^{\alpha\beta}$ is the identity matrix.
In my opinion, the embedded representation of $R_{adbc}$ using $\Delta^{\alpha\beta}$ is a powerful representation which explicitly illustrates how the Riemann tensor is a measure of the existence of curvative in the metric in question. However, I do not know if Eq. (1) is a representation of the Riemann tensor that losses generality (does not cover all examples of metrics to be evaluated), and whether if Nash's theorem allows Eq. (1) to be general. Also, I have never seen the Riemann tensor in the form in Eq. (1), and I wonder if this representation has been published elsewhere.
