I am having a hard time solving the equations to find the $a \in \Bbb R$ for which

$$M_a = \left \lbrace {(x,y)\in \mathbb {R}^2} \mid {y^2= x^3 + a}\right \rbrace$$

is a submanifold of $\Bbb R^2$.

I defined $F: \Bbb R^2 \to \Bbb R$ with $F(x,y)=y^2-x^3-a$ such that $M_a = F^{-1}(0)$ and $F$ is smooth. By the preimage theorem $M_a$ is a $1$ dimensional submanifold if $M_a$ doesn't contain any critical values of $F$. To compute the critical values of $F$, set

$$D_{(x,y)}F = \begin{bmatrix} -3x^2 & 2y \end{bmatrix} = 0$$

and for $(x,y)$ to be a critical point of $F$ it must lie on the parabola $y=-\frac{3}{2} x^2$. So we search for the intersection points of $M_a$ and the parabola, which are all the critical points of $F$ contained in $M_a$, since we want to find $a$ such that there are no intersection points.

Substituting the parabola eq. in the defining eq. of $M_a$ we get

$$a = \frac{9}{4} x^4-x^3 = x^3(\frac{9}{4}x-1)$$

How do I proceed from here? A simple $a \lt 0$, $a \gt 0$, $a=0$ case analysis doesn't lead to anything, since for $a \lt 0$, there are $x$, namely $0 \lt x \lt \frac{4}{9}$, such that this is true and we can also find $y$ for those $x$. By playing with the plot in Mathematica I found that $a \approx -0.01$ is the turning point. For all $a$ greater than that there are two intersection points and for all $a$ smaller than that there are none. Can somebody show me how this could be extracted from the calculations above?

Here is a plot for $a=0.2$

enter image description here

And a plot for $a=-0.005$

enter image description here

  • $\begingroup$ There are formulas to solve polynomial equations of degree 4: en.wikipedia.org/wiki/… $\endgroup$ – freakish May 22 '18 at 9:53
  • $\begingroup$ For any $a$, if $-x>0$ is large, then the equation is not consistent. $\endgroup$ – HK Lee May 22 '18 at 9:54
  • 2
    $\begingroup$ Because of DF you calculated, zero point (0, 0) is the only point that can prevent subset to be correct manifold. If a != 0, then this point is not satisfies the equation, therefore all a != 0 produces correct differential manifold. If a=0 we have the equation $y^2 = x^3$ which is also manifold, but not differentiable $\endgroup$ – Serge P. May 22 '18 at 10:04
  • $\begingroup$ related: math.stackexchange.com/questions/2400631/… $\endgroup$ – philmcole May 23 '18 at 20:32

To compute the critical values of $F$, set $$D_{(x,y)}F = \begin{bmatrix} -3x^2 & 2y \end{bmatrix} = 0$$ and for $(x,y)$ to be a critical point of $F$ it must lie on the parabola $y=-\frac{3}{2} x^2$.

This is not true. $(x, y)$ is a critical point if $D_{(x, y)}F = 0$. Thus in your case the only critical point is $(0,0)$ and the only critical value is $F(0,0) = -a$. Thus $F^{-1}(0)$ is a submanifold when $a\neq 0$. When $a=0$,

$$F^{-1}(0) = \{ x^3 = y^2\}$$

is a cusp and is not a submanifold.

  • $\begingroup$ Oh, I see it now. Thanks! $\endgroup$ – philmcole May 22 '18 at 12:11

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