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The connection between is written in wikipedia:

"We represent the surface by the implicit function theorem as the graph of a function, $f$, of two variables, in such a way that the point $p$ is a critical point, i.e., the gradient of $f$ vanishes (this can always be attained by a suitable rigid motion). Then the Gaussian curvature of the surface at $p$ is the determinant of the Hessian matrix of $f$ (being the product of the eigenvalues of the Hessian)."

I am searching for an example to better understand it.

C. Falcon
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1 Answers1

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An example.

Let us work with the $2$-sphere of radius $r$, its north hemisphere is the graph of the following map: $$f\colon (x,y)\mapsto\sqrt{r^2-x^2-y^2}.$$ The north pole is a critical point of $f$, since $\nabla_{(0,0)}f=(0,0)$. Furthermore, one has: $$\textrm{Hess}_{(0,0)}f=\begin{pmatrix}-\frac{1}{r} & 0\\0&-\frac{1}{r}\end{pmatrix}.$$ Hence, the hessian determinant of $f$ at $(0,0)$ is $1/r^2$ which is indeed the Gaussian curvature of the $2$-sphere at the north pole (and at any point).

The general setup.

Let us first compute the Gaussian curvature of a graph.

Let $S$ be the surface given by $z=f(x,y)$, where $f\colon\mathbb{R}^2\rightarrow\mathbb{R}$ is smooth. The first fundamental form of $S$ is: $$g(X,Y)=\left(1+(\partial_xf)^2\right)X^2+2(\partial_xf)(\partial_yf)XY+\left(1+(\partial_yf)^2\right)Y^2.$$ Besides, the second fundamental form of $S$ is: $$h(X,Y)=\frac{1}{\sqrt{1+(\partial_xf)^2+(\partial_yf)^2}}\left((\partial_{xx}f)X^2+2(\partial_{xy}f)XY+(\partial_{yy}f)Y^2\right).$$ Hence, the Gaussian curvature of $S$ is given by: $$K=\frac{\det(h)}{\det(g)}=\frac{(\partial_{xx}f)(\partial_{yy}f)-(\partial_{xy}f)^2}{(1+(\partial_xf)^2+(\partial_yf)^2)^2}.$$

[You can compute the Gaussian curvature using the definition, but I did not want to differentiate a quotient.]

Now, at a critical point, one has $\partial_xf=0=\partial_yf$, so that the Gaussian curvature boils down to: $$K=(\partial_{xx}f)(\partial_{yy}f)-(\partial_{xy}f)^2=\det(\textrm{Hess}_{\cdot}f).$$

C. Falcon
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