# Algebraic content of Gauss's Theorema Egregium

Let me first recall what Gauss's Theorema Egregium says. Consider a surface isometrically embedded in $$\mathbb{R}^3$$. In some local coordinates, let the first and second fundamental forms be $$E \text{dx}^2 + 2F \text{dx} \text{dy} + G \text{dy}^2$$ and $$L \text{dx}^2 + 2M \text{dx} \text{dy} + N \text{dy}^2.$$ The Gaussian curvature is $$K=\frac{LN-M^2}{EG-F^2}.$$ Gauss's theorem says that despite this formula, $$K$$ only depends on the first fundamental form.

The proof of this basically algebraic, and comes down to some remarkable formulas (the Gauss Equations) arising from the equality of iterated mixed partial derivatives.

Question: What is the right algebraic setting for this? The only relationships between the first and second fundamental forms are the Codazzi–Mainardi equations, so it must be that somehow these equations force $$K$$ to be independent of the second fundamental form. These equations involve some partial derivatives, but given that no real analysis goes into Gauss's theorem I assume there must be a more general algebraic setting for this sort of result (but I have no idea what buzzwords to search for).

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• Since Gauss+Codazzi-Mainardi can be seen as integrability conditions (Frobenius type -thus in terms of Lie brackets) of a distribution given rise to the surface having prescibed 1st and 2nd fundamental form I guess that one could show in a purely algebraic manner that when you have a pos.def. symmetric matrix G and a symmetric matrix L satisfy the compatibility conditions then $det L$ depends only on elements of $G$. I haven't ventured into making this explicit though. – Nicola Ciccoli 2 days ago
• If you like the notion of connection, you can prove that the Levi-Civita connection depends only on the metric, and then that the Gauss curvature is the curvature of the Levi-Civita connection. Clearly all steps are algebraic, and you can count derivatives to see that the algebra is on the 2-jet bundle of metrics. – Ben McKay 2 days ago
• Good question! This is a good example of why I dislike using $E, F, G, L, M, N$ to denote the coefficients of the metric tensor and second fundamental form. As you suspect, the proof is indeed algebraic. It follows from tensor algebraic identities. This is more easily seen when doing the proof in $n$ dimensions and using coefficients with indices to represent a tensor with respect to local coordinates. I have two short notes that might address your question using minimal machinery (specifically, I avoid mention of the Levi-Civita connection and Christoffel symbols). – Deane Yang yesterday
• The core fact that drives all of this is that partial derivatives commute. The two notes are: math.nyu.edu/~yangd/papers/riemann.pdf and math.nyu.edu/~yangd/papers/gauss.pdf. The notes are pretty terse, so feel free to post questions or email me. I would add that in these notes, the definition of Gauss curvature arises naturally from the calculation and does not need to be stated in advance. So it provides a plausible way it was originally discovered. – Deane Yang yesterday
• @DeaneYang: That's a very attractive way of presenting the argument! I'll think about it, and might follow up with some questions. – Tara yesterday