Consider the set $ \Sigma = \{ (x, y, z) \in \mathbb R ^3 : xyz=1 \} > $. Show that $ \Sigma$ is a surface and evaluate the Gaussian curvature at a general point $(x,y,z) \in \Sigma $

I've managed to do the first part using the regular value theorem, that is defining the function $\Phi : \mathbb R ^3 \to \mathbb R, \; \Phi(x,y,z) = xyz $ and then demonstrating that the preimage $ \Phi^{-1}(1) $ contains no singular points.

Since we have defined a surface as the level set of a function, the unit normal is given by the (normalised) gradient vector $$ N : \Sigma \to S^2, \\ N(x,y,z) = \frac{\nabla \Phi}{\| \nabla \Phi \| } = \dfrac{( yz, xz, xy )}{\sqrt{x^2y^2 + y^2z^2 + z^2x^2}}. $$

To find the Gaussian curvature, we could of course just calculate the required partial derivatives of this map to construct the shape operator and then evaluate the resulting determinant but this approach seems unnecessarily messy considering the differentiation involved, let alone expanding the determinant and I was wondering if there was an easier approach to this problem instead?


In this we can actually recognize the graph of a smooth function $f\colon U \subseteq \Bbb R^2 \to \Bbb R$, and the Gaussian curvature of its graph $\Gamma_f$ is given by $$K = \frac{\det(\nabla^2f)}{(1+\|\nabla f\|^2)^2},$$where $\nabla^2f$ denotes the Hessian of $f$. I mean... the more direct way I can see to check this is just sucking it up and doing the computation using the Monge chart $(x,y)\mapsto (x,y,f(x,y))$, but if you do it with general $f$ you won't risk missing anything essential because of particularities of your function $z = 1/(xy)$ there. Do note that $xyz=1$ says that $x$, $y$ and $z$ are all non-zero, so this last step is allowed.

Now, if you cannot recognize any convenient $f$ and need to express $K$ in terms of $\Phi$ instead, you can do it, but I would not say that the formulas are exactly simple. I think that the paper Curvature formulas for implicit curves and surfaces by Ron Goldman is exactly what you're looking for.


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