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Let $f\!:M\!\rightarrow\!\mathbb{R}$ be a smooth function on a manifold and $p\!\in\!M$. Is there any way to geometrically/visually characterize the conditions

  1. $p$ is a critical point (i.e. $D(f)_p\!=\!0$) and

  2. $p$ is a nondegenerate critical point (i.e. $\det D^2(f)_p\!\neq\!0$)?

If $f(x)=\langle x,a\rangle$ is a height function on a surface, then $f$ is linear, so $D(f)_p\!=\!f\!:\, T_pM\rightarrow T_p\mathbb{R}\!=\!\mathbb{R}$. Thus $f$ is the zero map at those $p$ for which $T_pM$ is perpendicular to $a$, i.e. the critical points of $f$ are those points at which $T_pM=a^\bot$. But what about higher dimensions and different $f$s?

And what about nondegeneracy?

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1 Answer 1

up vote 5 down vote accepted

Consider the case where $M$ is an open subset of $\mathbb{R}^2$ and visualize $f$ by visualizing its graph in $\mathbb{R}^3$. In the neighborhood of a point $p$, the function $f$ has an expansion into Taylor series. Then

  • $p$ is a critical point if this expansion has no linear terms. This occurs whenever $f$ attains a local minimum or a local maximum but it can also occur e.g. if $p$ is a saddle point.
  • $p$ is a nondegenerate critical point if the quadratic terms of the Taylor series expansion give a nondegenerate quadratic form. There are three cases: $p$ is a local minimum (so think locally $f(x, y) = x^2 + y^2$) $p$ is a local maximum (so think locally $f(x, y) = - x^2 - y^2$, or $p$ is a saddle point (so think locally $f(x, y) = x^2 - y^2$). This follows from the three possible distributions of signs of eigenvalues of the Hessian.

The reason we care about nondegeneracy is that if $p$ is a degenerate critical point then the behavior of $f$ in a neighborhood of $p$, even in a very qualitative sense and in a very small neighborhood, may depend on higher-order terms in the Taylor series (e.g. consider $f(x, y) = x^2 + y^4$ vs. $f(x, y) = x^2 - y^4$). Nondegeneracy allows us to only consider the linear and quadratic terms.

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When is a critical point of the height function $f(x)=\langle x,a\rangle$ on a closed surface degenerate? –  Leon Aug 11 '12 at 21:26
@Leon: I am not sure how to interpret this question. Examples have already been given in my answer (criticality is a local question so there's no reason to restrict to closed surfaces). –  Qiaochu Yuan Aug 11 '12 at 22:27
The classical example is a torus lying on a plane. –  Chris Gerig Aug 11 '12 at 22:28
(nondegenerate critical points are isolated) –  Chris Gerig Aug 11 '12 at 23:22
@Qiaochu Yuan: You are right, criticality (actually, degeneracy) is a local question. E.g. the graphs of the maps $(x,y)\!\mapsto\!x^2+y^2$ and $(x,y)\!\mapsto\!x^4+y^4$ have roughly the same look, yet the first has a nondegenerate critical point 0, whilst the second has a degenerate one. Thank you! –  Leon Sep 8 '12 at 14:20

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