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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.

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  • $\begingroup$ As far as visualization, this has been addresssed on MO here if that helps. For 3, I'm not sure it's really that important - it does imply some extra symmetry of the ramification points of the projection, but on the other hand $h$ and $k$ being the same isn't preserved under coordinate changes. $\endgroup$
    – KReiser
    Commented Dec 6, 2021 at 22:09
  • $\begingroup$ Thank you! The accepted answer there was exactly what I was wondering about, when thinking of the second question. :) $\endgroup$ Commented Dec 7, 2021 at 4:52
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    $\begingroup$ Glad to help. If you'd like, I could add that as a community-wiki answer here so you might have something to accept. $\endgroup$
    – KReiser
    Commented Dec 7, 2021 at 4:53
  • $\begingroup$ Yes, that would be very helpful and also (hopefully) useful for future users. $\endgroup$ Commented Dec 7, 2021 at 5:14

2 Answers 2

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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 $K_C$.

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    $\begingroup$ @hm2020 Posts on Math SE should be self-contained questions or answers, and should not contain meta information or commentary. The paragraph at the beginning of your answer is not necessary for understanding your answer, and therefore does not belong in the answer post. $\endgroup$
    – Xander Henderson
    Commented Dec 5, 2021 at 15:52
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    $\begingroup$ @XanderHenderson -Please do not alter the structure or the organization of the post - when a user asks for a clearification I usually include this request to make the thread easier to read for this particular user - when you edit this you interfere in this process. If you see spelling mistakes and wish to correct these you are of course free to do this. $\endgroup$
    – hm2020
    Commented Dec 5, 2021 at 16:18
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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.

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