In a flat Euclidean 2d plane, the metric tensor is given by:

$g_{\mu\nu} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

If we imagine this plane as curving into an external 3rd dimension according to $f(x,y)=z$, then the metric tensor would be:

$g_{\mu\nu} = \begin{pmatrix} 1+(\partial_xf)^2 & \partial_xf\partial_yf \\ \partial_xf\partial_yf & 1+(\partial_yf)^2 \end{pmatrix}$

Now if we imagined the metric tensor in General Relativity as arising from such curvature into external dimensions:

$g_{ij}=\eta_{ij}+\partial_i f_a\partial_j f_a$

What is the least number of $f_a$ that ensures that we can always find $f_a$ that solve Einsteins equations?

I would think 10, the number of independent metric tensor components. I'm not sure if the Bianchi-Identities or something else can decrease this number...


1 Answer 1


I think you are after some form of the Nash embedding theorem.


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