Hyperellipticity (or not!) of a Riemann surface and the singularities of the curve

Largely I want to know as to how does one say anything about the hyperellipticity or the genus of the Riemann surface by looking at the algebraic curve and its singularities.

• To give a specific example, what is the meaning of the statement that, "a curve of genus 2 can be expressed as a fourth degree plane curve possessing one double point" ?

Does this mean that any Riemann surface of genus 2 is a normalization of a fourth degree algebraic curve in $\mathbb{P}^2$ with one double point?

In general the proof says that any compact hyperelliptic Riemann surface of genus $g$ is a normalization of a an algebraic curve of degree $2g+2$ of the form $y^2 = \prod _{i = 1}^{2g+2} (x-a_i)$

So I would have naively thought that a genus $2$ Riemann surface (which is always hyperelliptic) will need a $2\times 2 +2 = 6$ degree algebraic curve. Hence I am not clear as to what to read of the quoted statement. Is something very special happening for genus $2$? Is the general theorem not a sharp statement?

• The general statement seems to tell me that the $a_i$ being distinct guarantees the smoothness of the algebraic curve except may be at the points at infinity. Now if there is a lower degree curve that can equally well represent the genus $2$ surface then is that necessarily going to be a curve with singularities?

• If the general statement is not a sharp statement and one can in cases do with lower degree curves than $2g+2$ then how does one derive the genus of the Riemann surface by looking at the algebraic curve and may be its singularities. Is there a "generalized" genus formula that works always?

• I recommend the book Riemann Surfaces by Farkas and Kra, it has a lot of the answers to your questions. If that is too advanced, Miranda's book answers some of these questions, but not as well IMO – AnonymousCoward Nov 25 '11 at 20:27
• @GottfriedLeibniz I currently have access to only the book on Riemann surfaces by Griffiths. Thats the book that I am reading. It would be great if you can sketch the argument or give enough hints on these matters on which I can build on. – Anirbit Nov 25 '11 at 21:10

A smooth degree $4$ plane curve has genus $3$. If you now degenerate this curve so that it obtains one ordinary double point, this corresponds to pinching off one loop on the genus $3$ Riemann surface. If you visualize this, you will see that it now looks like a genus $2$ Riemann surface with two points identified. Desingularizing this, you obtain a genus $2$ curve.
Now hyperelliptic curves are also traditionally represented as $y^2 = f(x)$, where $f(x)$ has degree $2g+1$ or $2g+2$. When the degree is $> 3$, these are singular equations, but the singularity is not an ordinary double point. So this is simply a different way of representing a hyperelliptic curve (which also puts the hyperelliptic involution in evidence: it is the map $(x,y) \mapsto (x,-y)$). Note that in the model of the first paragraph, the hyperelliptic involution is not so evident.
• Thanks for the answer but I am not clear about many of the things that you are seeing. Though I can see your intuition about the pinching of the genus - I don't see how the guaranteed degree 6 curve is related to this claimed degree 4 curve apparently describing the same genus 2 surface. Can you elaborate more on this? Also I did not completely get your point about degree > 3 being singular equations - I though we had already proven that the distinctness of $a_i$ makes the curve smooth as a variety. And how does the hyperelliptic involution play into this? – Anirbit Nov 26 '11 at 17:58
• @Anirbit: Dear Anirbit, The singularity when deg $> 3$ is at infinity. As for how the degree $4$ curve relates to the degree $6$ one, there is some (non-linear) change of coordinates that relates them; I don't know it off the top off my head. If you want to find it, you should look for the hyperelliptic involution of the degree $4$ curve; I don't know how to describe it off the top of my head either. Regards, – Matt E Nov 26 '11 at 19:49