# Does Nash's theorem allow an embedded representation of the Riemann tensor without loss of generality?

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.

Your $\Delta^{\alpha\beta}$ is the coordinate values of the pushforward of the inverse metric on your embedded manifold. This can be interpreted as the ambient metric minus projections unto the normal bundle. Which means that...

Congratulations! You have re-discovered an equivalent form of the Gauss equation for embeddings into Euclidean spaces.

• Thanks for the answer. Is this representation of the Riemann tensor useful in teaching others how the Riemann tensor measures the existence of curavture, or is there a simplier method of teaching the Riemann tensor? Nov 20, 2014 at 23:56
• Yes and no. Yes: it is crutch that can be leaned on if all else fails. No: the great discovery of Gauss was that curvature can be described intrinsically, so philosophically extrinsic descriptions of Riemann curvature should be in subservient roles. // The Riemann tensor measures non-commutativity of covariant derivatives. So yes, it manifests itself as unavailability of "flat" embeddings, and this picture can be an auxiliary guide. But emphasis is better placed on angle defects and things like that. Nov 21, 2014 at 13:20