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Say the first and second fundamental forms of a surface (a and b) in 2D are incompatible (i.e. they do not satisfy the Codazzi-Mainardi equations), then the "surface" cannot be embedded in 3D. Is this surface embeddable in some (albeit unknown) higher dimension? I feel this may be related to Nash's theorem in differential geometry, but I am not confident.

FWIW - I am not a mathematician, I am somewhere between an engineer and a physicist.

dpholmes
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    You're perfectly correct that every smooth surface with a smooth Riemannian metric embeds isometrically in some Euclidean space by the Nash embedding theorem. The second fundamental form, however, is usually defined with respect to an embedding; are you working with some abstract structure formally equivalent to the second fundamental form? – Andrew D. Hwang Jun 29 '16 at 23:09
  • @AndrewD.Hwang: in codimension one, the second fundamental form can be viewed as a bilinear form on the tangent bundle of the surface, and thus specified purely in terms of data on the surface. In higher codimension the second fundamental form is valued in the normal bundle, which is not a line bundle any more; so I don't think this kind of embedding problem even makes sense any more, and if it does then you certainly can't use the same type of data you tried for the 3D case. – Anthony Carapetis Jun 29 '16 at 23:48

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Let $g$ be the first form and $h$ another qdratic form. Consider, for $\alpha$ small enough, the product $S\times ]-\alpha, \alpha[$ endowed with the metric $g = g_0+th+ (dt)^2$. In this three manifold, the normal vector to the horizontal $S\times \{0\}$ is $\partial _t$. So its flow is just the addition by $t$, and in this three manifold, the second form of $S$ is $h$. recall that the second form (the shape operator) is excatly the derivative at the time $0$ of the surface you obtained by deforming $S$ using the unit normal vector. This is an embedding in dimension 3 in something non flat. In higher dimension, the second form is not a quadratic form.

Thomas
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  • I'm sorry, I'm unable to connect what you're saying here to the question. Maybe I posed it poorly, and I'm surely ignorant to much of the terminology. Would you mind translating these ideas for me? – dpholmes Jul 01 '16 at 02:29
  • I just said that you can embed your 2D surface , with the prescribed second fundamental form, in a 3D Riemanian manifold, but this 3D manifold is not flat . – Thomas Jul 01 '16 at 03:48
  • Ok, thank you for that clarification. I thought to satisfy the fundamental theorem of surfaces the first and second fundamental forms must satisfy the Gauss equation and the Codazzi-Mainardi equations. Does your comment imply that by not satisfying the Codazzi-Mainardi equations that just means the 3D Riemanian manifold will not be flat? – dpholmes Jul 01 '16 at 12:59