Prove that $f$ is Morse function if an only if $det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$(2)

Question

I was trying to solve this question:

Prove that $f$ is Morse function if an only if $\mathrm{det}(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0.$

But while searching on this site I found the answer here:

Prove that $f$ is Morse function if an only if $det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$

My question is:

What is the relation between the Hessian matrix and the quantity beside it in the given formula

$$\mathrm{det}(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0?$$

What makes me confused is the answer given in the link above in which the person who answers said that "This formula is equivalent to Morse function definition", but the definition of Morse function is only stated in terms of the Hessian matrix and not the quantity beside it in the above formula, could anyone explain this for me please?

Morse Function Definition as in Guillemin and Pollack: they are functions whose critical points are all nondegenerate.

A nondegenarate critical point: is a critical point that has nonsingular Hessian matrix

Answer 1

The author of the linked question only considers that case that $f:U\rightarrow \mathbb{R}$ is a smooth function defined on a subset $U\subseteq\mathbb{R}^k$. In this case there are global coordinates, so it makes sense to talk about the Hessian $H(f)$ of $f$ as a $(k\times k)$-matrix-valued function on $U$. Then $f$ is a Morse function if the Hessian $H(f)$ is non-degenerate at each critical point of $f$. Therefore there are two things to check at an arbitrary point $x\in U$. namely

i) Is $x$ a critical point of $f$? ii) If so, then is $H(f)(x)$ non-degnerate? That is, is $\det (H(f)(x))=0$?

Then as explained in the link an arbitrary point $x\in U$ is a critical point of $f$ if and only if $\partial f/\partial x^i|_x=0$ for each $i=1,\dots,k$. This is item i) on the checklist. If this occurs, then certainly $\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2=0$, so $\det(H(f)(x))^2+\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2=\det(H(f)(x))^2$ and this is zero if and only if $\det(H(f)(x))=0$, in which case the critical point $H(f)(x)$ is degenerate, and the point $x$ is a degenerate critical point, so $f$ is not Morse.

On the other hand, if at our arbitrary point $x\in U$, it holds that $\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2=0$, then there is some index $r$, say, for which $\partial f/\partial x^r|_x\neq 0$. Thus $x$ is not a critical point of $f$. Since it always holds that $\det(H(f)(x))^2\geq 0$ we must have $\det(H(f)(x))^2+\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2\geq\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2>0$.