# Generalization of Gaussian Curvature?

Consider a 2 dimensional manifold M parametrized by coordinates (x,y) embedded in $\mathbb{R}^{3}$.

Thee is a smooth curve in the manifold given by $(\gamma_{1}(t),\gamma_{2}(t))$ with $t\in\mathbb{R}$. There is a smooth function $\Phi$ : $M \to \mathbb{R}$ with the property that $\Phi(\gamma_{1}(t),\gamma_{2}(t))=constant$ $\forall t\in\mathbb{R}$. Furthermore we can decompose $\partial_t = g_{1}(x,y)\ \partial_{x} + g_{2}(x,y)\ \partial_{y}$ for some smooth functions $g_{1}$ and $g_{2}$. If we apply this smooth vectorfield to $\Phi(\gamma_{1}(t),\gamma_{2}(t))$ we obtain that $0=\partial_{t}\Phi(\gamma_{1}(t),\gamma_{2}(t)) = g_{1}(\gamma_{1}(t),\gamma_{2}(t))\ \partial_{x}\Phi(\gamma_{1}(t),\gamma_{2}(t)) + g_{2}(\gamma_{1}(t),\gamma_{2}(t))\ \partial_{y}\Phi(\gamma_{1}(t),\gamma_{2}(t))$. This means that for any point (x,y) on the curve the vectors $(g_{1}(x,y),g_{1}(x,y))$ and $\nabla\Phi(x,y)$ are orthogonal.

My goal is now to give an interpretation of the formula $A_{ij} \cdot \nabla\Phi(x,y)$. With the vectorfield $A_{ij} := (\partial_{x_{i}}\partial_{x_j}g_{1}(x,y),\partial_{x_{i}}\partial_{x_j}g_{2}(x,y))$. Here " $\cdot$ " denotes the ordinary dot product and $x_{i},x_{j}\in \{x,y\}$

My initial guess was to relate it to some kind of Gaussian curvature. Consider for example the sketch on page 3 of the following reference http://www.cs.iastate.edu/~cs577/handouts/surface-curvature.pdf In the present case $V_{ij}$ would corresponds to some "acceleration" of the curve.

Any thoughts and comments are highly appreciated:)

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All the simplest combinations of such quantities for surfaces in $\mathbb R^3$ have been accounted for. I recommend the book Elementary Topics in Differential Geometry by John A. Thorpe. He is very careful, plus he consistently takes a manifold as the level set $f^{-1}(c)$ of a single smooth function $f$ on $\mathbb R^{n+1}.$ –  Will Jagy Nov 18 '12 at 20:21
I'm not saying that this is going to be some kind of curvature. I'm just speculating at the moment because I cannot give a meaningful interpretation of the quantity described above. –  MrLee Nov 18 '12 at 20:47