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  1. For genus > 1 hyperelliptic Riemann surface the definition guarantees that there is a degree 2 map from that to $\mathbb{P}^1$. Under this map the inverse image of the "point at infinity" has to be two distinct points - why?

    Why couldn't it be that the map has a double pole? (..as wanted that would also not have the injectivity and hence biholomorphism crisis..)

  2. Is the canonical map on a compact Riemann surface holomorphic?

    (..for the hyperelliptic compact Riemann surfaces one knows that it factors as a composition of the definition guaranteed degree 2 map (to $\mathbb{P}^1$) and the rational canonical curve and hence its not holomorphic..)

  3. The above question is motivated from my difficulty in piecing together the argument that all compact Riemann surfaces at genus 2 are hyperelliptic. To prove that one wants to show that there the canonical curve is not injective. For a genus 2 compact Riemann surface the canonical map is a map to $\mathbb{P}^1$ (which is genus 0) which if were injective then something absurd would have to happen. But what?

One possibile way I can think of the argument is if the canonical map were holomorphic then one could say that this map is surjective - since continuous image of a compact surface is compact and hence its closed in $\mathbb{P}^1$ and but its also open by being the image of a holomorphic map and hence by the connectedness of $\mathbb{P}^1$ it is surjective. So the canonical map for genus 2 Riemann surface would be a bijective holomorphism and which would in turn imply that topologically we have a homeomorphism from genus 2 to genus 0 and which is impossible. Hence proved by reductio absurdum. But I am not at all sure of the above argument!

4 One knows how the canonical map for hyperelliptic Riemann surfaces factors (as stated in question 2) and one also knows as to how they occur as normalizations of the algebraic curve $\{y^2 - \prod _{1}^{2g+2} (x-a_i) = 0\} U [0,0,1]$ (as a subset of $\mathbb{P}^2$) (..where the $a_i$ are distinct complex numbers..)

From the above two facts how does it follow that if all these $a_i$ are permuted or affected by some $SL(2,\mathbb{Z})$ then the compact Riemann surface up there doesn't change?

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(4): you have to add two points at the infinity, they correspond to $y/x^{g+1}=\pm 1$. –  user18119 Nov 23 '11 at 20:21
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1 Answer 1

  1. Is false. There may be one or two preimages of the point at infinity. If your compact Riemann surface is written in affine coordinates as $y^2 = f(x)$, then the point at $\infty$ is a branch point if and only if $\deg f$ is odd.

  2. The canonical map is ALWAYS holomorphic. This is clear from its definition.

  3. Your argument is good.

  4. First a nitpick: the point at infinity has coordinates $[0:1:0]$. If I understand you correctly, you wonder why you get an isomorphic curve when you permute the $a_i$ or act on them by an element $f$ of $\rm{SL}(2,\bf{Z})$. This is clear if you write down the two subsets of $\mathbf P^2$ that you get: you can then also write down an isomorphism in coordinates. If you permute the $a_i$, you get the same equation back and no isomorphism is needed.

If you move the $a_i$ around by an automorphism $f$ of $\mathbf P^1$ you have to be slightly careful, since in particular your $a_i$ might be moved off to infinity. But rationally you can write $$ x \mapsto \frac{ax+by}{cx+dy} \qquad y \mapsto \frac{y}{(cx+d)^{g+1}}$$ where $g$ is the genus of your curve, for the isomorphism. Verify this!

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So you mean that for the hyperelliptic Riemann surfaces the inverse image of "infinity" under the definition guaranteed degree 2 map is always a ramified point? If I say look at Page 137-138 of Griffith's "Introduction to Algebraic Curves" then it seems that they say that the inverse images of $\infty$ don't contribute to the ramification divisor. These seem to say that there are two distinct unramified points which map to "infinity". Hence I am getting confused. –  Anirbit Nov 23 '11 at 19:50
    
@DanThis is unlike the genus 1 case where the inverse image of infinity is ramified and does contribute a $1$ to the ramification divisor. Can you kindly tell me as to what I am missing? –  Anirbit Nov 23 '11 at 19:50
    
Why do you say that the canonical map is always holomorphic if for the hyperelliptics it is factoring through that degree 2 map to P^1? That degree 2 map does have poles. Please tell me as to what I am confusing! –  Anirbit Nov 23 '11 at 19:52
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