Proving this is a $2d$ submanifold

Question

I am losing my mind over this question and am nearly convinced the question itself just is wrong. I want to show $ [(a,b,c,d) : a+b-c^3+d^2=0, a^2+b^2-8d=10]$ is a 2-submanifold of $\mathbb R^4$. I have the map $$F(a,b,c,d) = (f(a,b,c,d), g(a,b,c,d))$$ where $f = a+b-c^2-d^2$ and $g=a^2+b^2-8d-10$. I am trying to show this is a submersion to $\mathbb R^2$ (and then use the preimage theorem), but after calculating the Jacobian and getting a system of $6$ equations, I find $c=0$ and from there on don't know where to go. I need to show by contradiction that $\nabla f$ and $\nabla g$ are not linearly dependent, but I seem to end up actually getting a result for the constant I'll call $k$ so that $k\nabla f = \nabla g$, which shouldn't even be possible. What am I meant to do?

Answer 1

You need only to show that $F^{-1}(0,0)$ is a submanifold. Using the pre-image theorem, it suffices to show that $(0,0)$ is a regular value of $F$. This is weaker then showing that $F$ is a submersion.

To this end, note that the Jacobian is given by (I assume $f = a+b-c^3 +d^2$) $$ \begin{pmatrix} 1 & 1& -3c^2 & 2d \\2a & 2b & 0 & -8 \end{pmatrix}$$ The first observation is that when $c\neq 0$, $$ \begin{pmatrix} -3c^2 & 2d \\ 0 & -8 \end{pmatrix}$$ is invertible and thus $J$ is of rank 2. Now assume $c=0$. If $a\neq b$, then $$ \begin{pmatrix} 1 & 1 \\2a & 2b \end{pmatrix}$$ is also invertible. So we consider the case $c=0$ and $a=b$. Taking the second and the fourth row, we have $$ \det \begin{pmatrix} 1& 2d \\2b & -8 \end{pmatrix} = -8-4bd, $$ which is zero if and only if $bd=-2$. Putting $c=0, a=b = -2/d$ into

\begin{align} a+b-c^3+d^2&=0, \\ a^2+b^2-8d&=10 \end{align} we obtain \begin{align} 2b+(-2/b)^2&=0, \\ b^2-4(-2/b)&=5 \end{align} the first equation gives $b = -2^{1/3}$, but this $b$ does not satisfy the second equation. That is, for any $(a, b, c, d)$ which is in $F^{-1}(0,0)$, $$c=0, a=b=-2/d$$ cannot be both satisfied. Hence $J$ has full rank for all $(a, b, c, d) \in F^{-1}(0,0)$. This implies that $F^{-1}(0,0)$ is a $2d$ submanifold of $\mathbb R^4$.