Is there anything special about the hyperelliptic curve that is obtained by gluing the same algebraic curve?

Question

According to Rick Miranda's Algebraic curves and Riemann surfaces, a hyperelliptic curve is defined as the Riemann surface obtained by gluing two algebraic curves, $y^2=h(x)$ and $w^2 = k(z)$ (where $h$ has distinct roots and $k(z) := z^{2g+2} h(1/z)$) through the map $(x,y) \mapsto (z,w) := (1/x, y/x^{g+1})$. He also mentions that this is topologically a genus $g$ surface; and has a notion of involution $\sigma : (x,y)\mapsto(x,-y)$

I am trying to visualise how all these actually looks like, and I have a couple of questions:

1. Is there any geometric way to visualise the construction? Formally, is there any embedding/immersion of this construction into $\mathbb{R}^3$, that helps see, atleast topologically, atleast for specific examples, what parts of genus $g$ surface are being glued together?

2. What does the $\sigma$ 'looks like' as a map on genus $g$ surface, $\Sigma_g$, imagined as embedded in $\mathbb{R}^3$ in the standard way? Does it look like a reflection? Or maybe a $180^\circ$ rotation?

3. I think this brings about some sort of duality between the polynomials $h$ and $k$. Is this somehow important? Specifically, is there anything special about the Riemann surfaces that are obtained by gluing the same algebraic curve? As an example for such a curve, if $x\neq 0$, for $g=3$, $$y^2 = 3x^8 +10x^4 +3 \Leftrightarrow \bigg(\frac{y}{x^{3+1}}\bigg)^2 = 3\bigg(\frac{1}{x}\bigg)^8 +10\bigg(\frac{1}{x}\bigg)^4 +3 $$

Sorry if the questions are a bit vague. I am just trying to get more intuition about the subject.

Answer 1

This question has been addressed on MathOverflow. Here's the accepted answer there, from user stankewicz:

One example: lay your $g$-holed torus $T$ out flat and draw a line the long way through each hole. It hits the torus in $2g + 2$ points. Consider the 180 degree rotation $w$ through that line. Now consider the space $T/w$ formed by identifying two points $P$ and $Q$ if $P = wQ$ (since $w^2 = 1$ we also have $Q = wP$). I claim that it's not too hard to see that $T/w$ is isomorphic to the projective line, and the $2g+2$ points which hit the line are the ramification points.

edit: This is of course more general than you were asking, but the picture is completely general when you're talking about topology.

edit 2: A picture of this (albeit approached from the perspective of starting on the projective line and cutting slits) can be found in section 20e of Fulton's Algebraic Topology book.

Answer 2

Question1: "Is there any geometric way to visualise the construction? Formally, is there any embedding/immersion of this construction into R3, that helps see, atleast topologically, atleast for specific examples, what parts of genus g surface are being glued together?"

Answer: About question 1: When $E$ is an elliptic curve over the complex number field, there is a result in Hartshorne (Thm.IV.4.16) that realizes E topologically as a quotient $\mathbb{C}/\Gamma$. Maybe

"Farkas; Riemann Surfaces. Graduate Texts in Mathematics. Springer-Verlag 1980."

for the general case.

Note: Naively when an algebraic group $G$ acts on a commutative ring $A$ and an affine scheme $X:=Spec(A)$, we may construct the "quotient" $X/G:=Spec(A^G)$ using the invariant ring. The above "topological quotient" $\mathbb{C}/\Gamma$ is projective, hence we cannot naively construct $\mathbb{C}/\Gamma$ using $Spec(A^G)$ for some $G$ acting on $A$.

Question2: "What does the map $\sigma$ 'looks like' as a map on genus g surface, Σg, imagined as embedded in R3 in the standard way? Does it look like a reflection? Or maybe a 180∘ rotation?"

Answer question 2: Since $C$ is hyper-elliptic, there is a degree 2 separable map

$$\phi_K:C \rightarrow \mathbb{P}^1_{\mathbb{C}}$$

coming from the canonical divisor. You involution $\sigma$ permutes the points in a fiber: If $p,q \in \phi_K^{-1}(x)$ it follows $\sigma(p)=q, \sigma(q)=p$. You find this discussed in the same book HH Chapter IV.

Note: There is a classical result saying the following: Let $C \subseteq \mathbb{P}^n_k$ be a smooth projective curve over an algebraically closed field $k$. If $g(C)=g \geq 2$, let $K_C$ be the canonical divisor with associated morphism

$$\phi_C: C \rightarrow \mathbb{P}^{g-1}.$$

Then either $\phi$ is a closed immersion or $Im(\phi_C) \cong \mathbb{P}^1$ and $deg(\phi_C)=2$. Moreover: $C$ is hyperelliptic iff there is an element $\sigma \in Aut_k(C)$ with $\sigma^2=Id$ and $C/<\sigma> \cong \mathbb{P}^1_k$.

Being a hyper elliptic curve is an "intrinsic property" (independent of any embedding into projective space), and when the base field is algebraically closed this property is detected by the canonical divisor $K_C$. Hence if someone "gives you such a curve" that is not embedded into a projective space and asks: "Is it hyperelliptic?" you can check this using $K_C$.