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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

Daniele Tampieri
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Intuition
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1 Answers1

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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 $\det(H(f)(x))^2+\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2\geq\Sigma^k_{i=1}(\partial f/\partial x^i|_x)^2>0$.

Tyrone
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  • I think in the line after the first paragraph it is $det(H(f)(x)) \neq 0$, correct? – Intuition Nov 10 '18 at 16:15
  • but your answer did not answer my question ..... I am asking about how is the last line in the answer in the link mentioned correct ...... what is the relation between grad (f) and the Hessian matrix of f? – Intuition Nov 10 '18 at 16:46
  • Well, if $\det(H(f)(x))=0$ then $H(f)(x)$ is not non-degenerate. It seemed like the most concise way to express things at the time for some reason. – Tyrone Nov 10 '18 at 17:01
  • Sorry ...... I can not see what do you mean? – Intuition Nov 10 '18 at 17:03
  • The gradient is the vector $\nabla f=(\partial f/\partial x^1,\partial f/\partial x^2,\dots,\partial f/\partial x^k)$. The only way that this is related to the Hessian $H(f)$ is through the equation you wrote down, since $|\nabla f|^2=\Sigma^k_{i=1}(\partial f/\partial x^i)^2$. The equation is the relation between the two. – Tyrone Nov 10 '18 at 17:04
  • what about my first comment? – Intuition Nov 10 '18 at 17:06
  • @hopefully, I mean that asking whether $\det(H(f)(x))=0$ or not is enough to determine whether or not $H(f)(x)$ is degerate or non-degenerate. – Tyrone Nov 10 '18 at 17:06
  • you mean is singular or nonsingular ..... correct? – Intuition Nov 10 '18 at 17:07
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    Singular and degenerate mean the same thing, see https://en.wikipedia.org/wiki/Invertible_matrix, for example. – Tyrone Nov 10 '18 at 17:09
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    As for your first comment, "How is if gad (f) is zero then det (H) is nonzero ...... could u explain this please?" (I think), if $\nabla(f)(x)=0$, then $x$ is a critical point of $f$. Then if $\det(H(f)(x))=0$ it is a degenerate critical point. If $\det(H(f)(x))\neq 0$ then it is a non-degerate critical point. The point is that $\nabla(f)(x)=0$ does not imply anything about $\det(H(f)(x))$ on its own. – Tyrone Nov 10 '18 at 17:14
  • I mean my first comment here under your answer – Intuition Nov 10 '18 at 17:16
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    I think you mean your question "I am asking about how is the last line in the answer in the link mentioned correct". The last line in the answer linked reads "This is the definition of a Morse function.", and it is. Take a point $x\in U$ and ask if $|\nabla f(x)|^2=0$. If $|\nabla f(x)|^2\neq 0$, then $x$ is a regular point of $f$ and there is nothing more to consider. If on the other hand $|\nabla f(x)|^2=0$, then $x$ is a critical point. In this case, if $\det(H(f)(x)=0$ then it is a degenerate critical point, whilst if $\det(H(f)(x)\neq0$ then it is a non-degenerate critical point. – Tyrone Nov 10 '18 at 17:24
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    If all points of $U$ are either regular points of $f$ or non-degenerate critical points, then $f$ is Morse. If there exists one point $x$ which is both critical, $|\nabla f(x)|^2=0$, and degenerate $\det(H(f)(x))=0$, then $f$ is not Morse. – Tyrone Nov 10 '18 at 17:26