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Let $f:\mathbb{R}^n\to\mathbb{R}^m$ be given such that it is continuously differentiable. According to Wikipedia, a "critical point" of $f$ is a point $p\in\mathbb{R}^n$ such that:

1) According to one paragraph it is the condition that $\partial_j f_i = 0$ for all $j\in\{1,\dots,n\}$ and $i\in\{1,\dots,m\}$.

2) According to another paragraph, it is the condition that the matrix with the $(i,j)$ component $\partial_j f_i$ (i.e. the Jacobian matrix) has rank less than $m$.

It seems like these two definitions are not equivalent. Why are there two different definitions? I can understand why the first one is called "critical point": it gives exactly the points which are local maximum, minimum or saddle points. Why is the second one called "critical point"? What does it give, except for saying that the function at this point is not surjective?

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  • $\begingroup$ In wikipedia, it says that a critical point is where the derivative is not surjective. I think you misunderstood your definition 1. $\endgroup$
    – user40276
    Commented Apr 8, 2015 at 19:50
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    $\begingroup$ Oh! Now I see the problem. In wikipedia, it's assumed that "a function of several real variables" is a function with codomain $\mathbb{R}$ $\endgroup$
    – user40276
    Commented Apr 8, 2015 at 19:52

1 Answer 1

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$\newcommand{\Reals}{\mathbf{R}}$Condition (1) refers to real-valued functions, and therefore specializes condition (2): If all the partials vanish at a point, the Jacobian has rank $0 < 1 = m$.

As for "why is (2) the general definition?", one reason originates with the implicit function theorem: If $f:\Reals^{n} \to \Reals^{m}$ is continuously-differentiable, and if $p \in \Reals^{n}$ is a regular point of $f$ (i.e., if $Df(p)$ is surjective; $p$ is not a critical point), then (a sufficiently small piece of) the level set of $f$ through $p$, i.e., the set $\{x\text{ in } \Reals^{n} : f(x) = f(p)\}$ is a manifold of dimension $(n - m)$.

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  • $\begingroup$ Thanks for your answer. So is it correct to say, when the domain has dimension more than 1, to require that the Jacobian matrix is zero (and not just of rank-less-than-maximal) is too strong for a critical point? That is, Jacobian matrix = 0 is meaningful only when the domain has dimension 1? $\endgroup$
    – PPR
    Commented Apr 8, 2015 at 22:45
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    $\begingroup$ @PPR: Yes. :) For instance, the polar coordinates mapping $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ has critical points along the line $r = 0$; or, if $k > 1$ is an integer, the mapping $(x, y) \mapsto (x^{k}, y)$ has critical points along the line $x = 0$. (Generally, the condition $Df(p) = 0$ is sufficient for $p$ to be a critical point, but not necessary if $m > 1$.) $\endgroup$
    – Andrew D. Hwang
    Commented Apr 8, 2015 at 23:06

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