Questions on the Hessian matrix on the unit sphere $\mathcal S^2$

differential-geometryhessian-matrixmultivariable-calculus

Let $\mathcal S^2:=\{(x,y,z)\;\vert \; x^2+y^2+z^2=1\}$ be the unit sphere and $f\in C^2(\mathcal S^2)$ with $f(x,y,z)=f(x)$. Moreover, assume that $N=(1,0,0)$ and $S=(-1,0,0)$ denote respectively the North and the South Pole of the sphere. Our professor wrote the other day this:

$\begin{equation*} \nabla^2 f \big \vert_{N,S}=\begin{pmatrix}
\frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y
\partial z}\\ \frac{\partial^2 f}{\partial z \partial y} &
\frac{\partial^2 f}{\partial z^2} \end{pmatrix} \end{equation*}\quad$
since $\nabla f\big \vert_{N,S}=0 \quad (*)$

However, I have some trouble understanding $(*)$. My questions are:

  1. The fact that $\nabla f\big \vert_{N,S}=0 $ indicates that the North and the South Pole are critical points for $f$. How do we know that?
  2. Why does the Hessian matrix has this form? It seems to me that some sort of parametrization has been used here but I don't see how we conclude this matrix.
  3. Is the particular form of the Hessian related to the fact that $\nabla f\big \vert_{N,S}=0 $ or is it independent on that?

EDIT (after @Arctic Char comments): Using the parametrization

$x=\pm \sqrt{1-y^2-z^2}, \quad y=y,\quad z=z$

we write $f(x,y,z)=\tilde f(y,z).\;$ Then, the gradient of $\tilde f$ is given by: $\nabla \tilde f=(\frac{\partial \tilde f}{\partial y},\frac{\partial \tilde f}{\partial z})=(\frac{-y}{\sqrt{1-y^2-z^2}},\frac{-z}{\sqrt{1-y^2-z^2}})$ which vanishes at $(\pm 1,0,0)$. The Hessian matrix is actually the Hessian of $\tilde f$.

Is this correct or am I missing something?

I apologize if these questions are too silly but it's been ages since I did some calculations of that kind and I have forgotten some stuff. Any help is much appreciated.

Many thanks in advance!

Best Answer

It is given that $f(x, y, z) = f(x)$ depends only on $x$. Around $N = (1,0,0)$, one can parameterize the sphere as $$ (y, z) \mapsto (\sqrt{1-y^2-z^2} , y, z),$$ so locally $f$ is given by $$ \tilde f (y, z) = f(x(y, z)) = f( \sqrt{1-y^2-z^2}).$$ Hence

$$ \left( \frac{\partial \tilde f}{\partial y} , \frac{\partial \tilde f}{\partial z}\right) = -f' \left( \frac{y}{\sqrt{1-y^2-z^2}}, \frac{z}{\sqrt{1-y^2-z^2}}\right)$$ is zero when $(y, z) = (0,0)$.

For the Hessian: away from a critical point, the Hessian of a function is actually not well defined, unless you specify some Riemannian metric. Please see this post, or this article for more details.

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