Prove that $V(p,q)$ is a smooth manifold

Question

Here is a homework problem for our differential geometry class.

Reference: John M. Lee, Introduction to Smooth Manifolds.

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Let $p,q$ be positive integers. Show that $$V(p,q)=\{(z,w)\in\mathbb{C}^2:|z|^2+|w|^2=1,z^p+w^q=0\}.$$

Here we are identifying $\mathbb{C}^2$ with $\mathbb{R}^4$. Then prove that $V(p,q)$ is a smooth manifold of real dimension 1.

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My attempt: I considered a map $F:\mathbb{R}^4\to \mathbb{R}^3$ defined as $$F(z,w)=(|z|^2+|w|^2,\text{Re}(z^p+w^q),\text{Im}(z^p+w^q))$$ and I have tried to prove that the differential of $F$ has rank 3, calculating the Jacobian of $F$.

The Jacobian of $F$ is given as follows:$$J=\begin{pmatrix}2x&&2y &&2u &&2v\\\text{Re}(pz^{p-1}) &&-\text{Im}(pz^{p-1}) &&\text{Re}(qw^{q-1}) && -\text{Im}(qw^{q-1})\\\text{Im}(pz^{p-1})&&\text{Re}(pz^{p-1}) && -\text{Im}(qw^{q-1})&&\text{Re}(qw^{q-1})\end{pmatrix}$$

where $z=x+iy$ and $w=u+iv$.

However, I'm not sure that the Jacobian of $F$ does not vanish on $V(p,q)$.

How should I progress here?

Thanks in advance!

Answer 1

You can simplify this a lot by using polar coordinates. Let $z=re^{i\varphi}$ and $w=se^{i\theta}$. Then the first equation simplifies to $r^2+s^2=1$ and the second one can be rearranged to $$ \frac{r^p}{s^q}=e^{i\pi+iq\theta-ip\varphi}$$ where a transformed a $-1$ into $e^{i\pi}$. Note that in this equation the left side is a positive real and the right side lies on the unit circle. This can only happen if both sides are equal to $1$. For $r$ and $s$ we have $$r^p=s^q \hspace{1cm} r^2+s^2=1$$ which gives exactly one solution for $r$ and $s$. The other equation is $$ \pi+q\theta+p\varphi = 0 \hspace{2mm} mod \hspace{2mm} 2\pi$$ which gives a $1$-dimensional solution space. There are no self intersection, hence it is a $1$-dimensional manifold.

Answer 2

I suspect this is not the point of the exercise, but if you just want the result you can do it like Milnor (Singular points of complex hypersurfaces, p. 4): first show that $V(p,q)$ is a subset of the torus $|z|^2=\eta$, $|w|^2=\xi$ for appropriate $\eta, \xi \in \mathbb{R}_+ $, then parametrize $V(p,q)$ explcitly.

Essentially the same thing can be said in different language: just look at the action of $\mathbb{C}$ on the variety $W(p,q)=\{z,w| z^p+w^q=0\}$ by $t\cdot(z,w)=(t^q z, t^p w)$ and notice that all of $W(p,q)$ is a single orbit; now the action by purely imaginary $t$ preserves levels of $|z|^2+|w|^2$ and the action of real $t$ moves between levels. So the intersection with any sphere is an orbit of the circle action by purely imaginary $t$s. (For more along these lines see Ghys's "A Singular Mathematical Promenade" http://perso.ens-lyon.fr/ghys/promenade/, particularly "The cusp and the trefoil" section.)

Answer 3

As @quarague pointed out, calculating with polar forms reduces a lot of calculation.

Calculating the determinant of the first two 3$\times$3 minors(with the first and second columns deleted, respectively) of the given Jacobian with $z=re^{i\varphi}$ and $w=se^{i\theta}$, one observes that $r,s,\varphi,\theta$ must satisfy $$\begin{cases}\frac{q}{s^2}\cos\varphi-\frac{p}{r^2}\cos(\rho-\theta)=0\\\frac{q}{s^2}\sin\varphi+\frac{p}{r^2}\sin(\rho-\theta)=0\end{cases}.$$ to make all the minors vanish. Here $\rho=(p-1)\varphi-(q-1)\theta.$ Hence, all the determinants vanish only if $\frac{q}{s^2}=\frac{p}{r^2}$. Plugging in, from the first and second equalities one gets $\varphi+\rho-\theta=2n\pi$, $n\in\mathbb{Z}$. Also note that $V(p,q)$ is contained in the open submanifold $M=\{(re^{i\varphi},se^{i\theta}):p\varphi-q\theta\notin 2\pi\mathbb{Z},r>0,s>0\}\subset\mathbb{R}^4$. Therefore, the map $F:M\to\mathbb{R}^3$ is of constant rank 3, so the result follows.