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The cusp $y^{2} = x^{3}$ has a rank drop at the origin

Statement refuted

A polynomial level curve need not be regular everywhere. The zero level of $F (x, y) = y^{2} - x^{3}$ has derivative rank $1$ away from the origin and rank $0$ at the origin.

Facts & Assumptions

Given: The polynomial $F (x, y) = y^{2} - x^{3}$ and the curve $γ (t) = (t^{2}, t^{3})$ .

[L1]

The power rule and derivative algebra give the continuous Jacobian row $(- 3 x^{2}, 2 y)$ and $γ^{'} (t) = (2 t, 3 t^{2})$ ; the continuous-partials theorem identifies that row with $D F (x, y)$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , 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, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L2]

Regularity means surjectivity of the derivative, and a regular level is locally a $C^{1}$ graph (Regular and critical points, regular and critical values, and level sets, A regular level set is locally a $C^{k}$ graph of dimension $m - n$ ).

Counterexample

technique · direct

1.1

One has $F (γ (t)) = t^{6} - t^{6} = 0$ , so the parametrized cusp lies in the zero level, and [L1] gives $D F (0, 0) = 0$ and $γ^{'} (0) = 0$ .

givenL1algebra

1.2

If $(x, y) \neq (0, 0)$ lies on $F^{- 1} (0)$ , then $(- 3 x^{2}, 2 y) \neq (0, 0)$ , so $D F (x, y)$ has rank $1$ and is surjective onto $R$ .

givenL1algebra

2.1

Hence the derivative rank drops precisely at the cusp point. The two values $y = \pm x^{3 / 2}$ for $x > 0$ prevent a graph $y = g (x)$ there, while the relation $x = ∣ y ∣^{2 / 3}$ is not differentiable at $0$ , in accord with the missing hypothesis in [L2].

step 1.1step 1.2L2algebra

3.1

The polynomial level curve therefore supplies the claimed rank-drop counterexample.

step 2.1∎

Depends on

Used by

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Sources

J. M. Lee, Introduction to Smooth Manifolds, regular-level examples (standard reference, not scraped)

The cusp y2=x3y2=x3 has a rank drop at the origin