Alphabeta MathHow statement and proof provenance work
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A critical value can have a smooth level set
Statement refuted
A critical value need not have a singular level set. For , the value is critical although is the vertical line.
Facts & Assumptions
Given: The polynomial map , .
The power rule gives the continuous Jacobian row , and the continuous-partials theorem identifies it with (For a natural the function is differentiable everywhere with derivative ; for it is the constant , with derivative ; for a natural the function is differentiable at every with derivative ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).
A value is critical when some point of its fibre has nonsurjective derivative (Regular and critical points, regular and critical values, and level sets), while the graph of a map is a regular level set for a suitable defining map (The graph of a Euclidean map is a regular level set).
Counterexample
The equation is equivalent to , so , the graph of the zero function over the -axis.
By [L1], for every point of this fibre, so it is not surjective and [L2] makes a critical value of .
The same underlying set is a smooth line and, after swapping coordinates, is the graph covered by [L2]. Thus criticality of this defining function does not force singularity of the set.
Depends on
- Regular and critical points, regular and critical values, and level sets
- The graph of a $C^k$ Euclidean map is a regular level set
- For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term
- The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case
- If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- J. M. Lee, Introduction to Smooth Manifolds, critical-value discussion (standard reference, not scraped)