Why is the set $\displaystyle \{ [z_1,z_2,z_3] \in \mathbb C P^2 : z_2^2 z_3 = 4 z_1^3 - g_2 z_1 z_3^2 - g_3 z_3^3 \}$ connected?

Question

I am wondering why the set $\displaystyle \{ [z_1,z_2,z_3] \in \mathbb C P^2 : z_2^2 z_3 = 4 z_1^3 - g_2 z_1 z_3^2 - g_3 z_3^3 \}$ is connected, where $g_2$ and $g_3$ are some fixed constants (for those of you who know, the constants come from a differential equation involving $\wp$, but this fact is tangential to this question). I wrote the set as $\pi(P^{-1}(0))$ where $\pi : \mathbb C^3 - \{0\} \to \mathbb C P^2$ is the projection and $P(z) := 4 z_1^3 - g_2 z_1 z_3^2 - g_3 z_3^3 - z_2^2 z_3$ but I am not sure how I can go from here.

Answer 1

Here’s a basically elementary argument.

Let $P(z)=4z^3-g_2z-g_3$, and let $Z$ be the set of roots of $P$.

Let $x_0,y_0,x_1,y_1\in \mathbb{C}$ such that $y_i^2=P(x_i)$. I claim that, if $y_0\neq 0$, $y_1 \neq 0$, then there are continuous maps $x,y:[0,1] \rightarrow \mathbb{C}$ such that $x(i)=x_i,y(0)=y_0, y(1) = \pm y_1$, and $y^2=P(x)$.

Indeed, we can choose a continuous path $x$ from $x_0$ to $x_1$ that avoids $Z$. Then $P \circ x:[0,1] \rightarrow \mathbb{C}^{\times}$ is well-defined, and we can thus consider its unique continuous square root $y$ such that $y(0)=y_0$. Then $y(1)$ is a square root of $P(x_1)=y_1^2$.

Now, let $a$ have a large module, and $b$ be one square root of $P(x)$. I claim that there are continuous paths $x,y:[0,1] \rightarrow \mathbb{C}$ such that $x(0)=x(1)=a$ and $y(0)=-y(1)=b$ and $y^2=P(x)$.

Proof: take $x(t)=ae^{2i\pi t}$ (and assume that $|a|$ is very large wrt $P$), and $y$ as the unique continuous square root of $P\circ x$ such that $y(0)=b$. Let’s show that $y(1)=-b$.

Indeed, $P(x(t))=4a^3e^{6 i \pi t}(1-R(t))$, where $R(t)=\frac{g_2}{4a^2}e^{-4i \pi t}+\frac{g_3}{4a^3}e^{-6i\pi t}$, and $|R(t)| < 1$ when $|a|$ is large enough.

So we have $y(t)=2\beta^3 e^{6i\pi t}\sum_{n \geq 0}{\binom{1/2}{n}(-R(t))^n}$ (where $\beta$ is a certain square root of $a$), which implies that $y(1)=-y(0)$.

Let $E_a=\{(x,y) \in \mathbb{C}^2,\, y^2=P(x)\}$. What we have seen is that if $(x,y) \in E_a$ with $|x|$ large enough, then $(x,y)$ and $(x,-y)$ are in the same path-connected component.

We’ve also seen that for any two points $(x,y),(x’,y’) \in E_a$ with $yy’ \neq 0$, $(x,y)$ is in the same path-connected component as $(x’,y’)$ or $(x’,-y’)$ (or both).

By setting a fixed $(x’,y’)$ with large enough $|x’|$ and letting $(x,y)$ vary, we see that $E_a \backslash \{y=0\}$ is path-connected.

Let $z$ be a root of $P$, it’s easy to see that for any $x$ close to $z$, there is some $(x,y) \in E_a$ with $y$ being arbitrarily close to zero. In other words, $E_a \backslash \{0\}$ is dense in $E_a$, hence $E_a$ is connected.

Now, it’s easy to see that $\{[x:y:1],\,(x,y) \in E_a\}$ is dense in the set you’re interested in, hence it’s connected.