One of the first big formula's you learn in algebraic geometry is the genus-degree formula which states that an irreducible homogeneous polynomial in $f \in \mathbb{C}[x,y,z]$ of degree $d$ gives a genus $(d-1)(d-2)/2$ curve. Unfortunately, this does not help with constructing genus 2 curves since we have the following table of degrees and genera for $f$ $$ \begin{matrix} \text{degree} & 1 & 2 & 3 & 4 & 5 & \cdots \\ \text{genus} & 0 & 0 & 1 & 3 & 6 & \cdots \end{matrix} $$ How can I find a generalization of the genus-degree formula so that I can construct curves in some projective space with the desired genus?

  • 1
    $\begingroup$ You can similarly get, using adjunction, formulas for the genera of curves on the quadric surface $\mathbb P^1 \times \mathbb P^1$ (something of type $(3,2)$ for example) and complete intersections in $\mathbb P^3$ in terms of degrees. $\endgroup$ – Hoot Aug 12 '16 at 20:16
  • $\begingroup$ Maybe more direct: all such curves are hyperelliptic, so you really just need to write down $6$ points in $\mathbb P^1$. $\endgroup$ – Hoot Aug 13 '16 at 17:45

Let $f \in \mathbb{C}[x_0, x_1, y_0, y_1]$ be a polynomial such that $f(\lambda x_0, \lambda x_1, y_0, y_1) = \lambda^af(x_0, x_1, y_0, y_1)$ and $f(x_0, x_1, \mu y_0, \mu y_1) = \mu^bf(x_0, x_1, y_0, y_1)$, then $$X = \{([x_0, x_1], [y_0, y_1]) \in \mathbb{CP}^1\times\mathbb{CP}^1 \mid f(x_0, x_1, y_0, y_1) = 0\}$$ is a curve. If $X$ is smooth, it has genus $(a-1)(b-1)$, so every genus can be realised. As $\mathbb{CP}^1\times\mathbb{CP}^1$ embeds in $\mathbb{CP}^3$ via the Segre embedding, $X$ is a curve in $\mathbb{CP}^3$.

Another way of constructing curves in a projective space is via complete intersections. Let $f_1, \dots, f_{n-1} \in \mathbb{C}[x_0, \dots, x_n]$ be homogeneous polynomials of degrees $d_1, \dots, d_{n-1}$ respectively, then

$$Y = \{[x_0, \dots, x_n] \in \mathbb{CP}^n \mid f_1(x_0, \dots, x_n) = \dots = f_{n-1}(x_0, \dots, x_n) = 0\}$$

is a curve. If $Y$ is smooth, it has genus $1 - \frac{1}{2}(n + 1 - d_1 - \dots - d_{n-1})d_1\dots d_{n-1}$. This construction gives rise to many genera that don't appear in the degree-genus formula, but not all of them: see this sequence. For example, there is no choice of dimension $n$ and degrees $d_1, \dots, d_{n-1}$ which give rise to a genus two curve, i.e. a genus two curve is not a complete intersection.

  • $\begingroup$ Where can I find the proof that the genus is $(a-1)(b-1)$? Also, can this be done using GRR? $\endgroup$ – 54321user Aug 12 '16 at 20:25
  • $\begingroup$ I don't know of a reference for this. It follows from the adjunction formula, just as in the case of the degree-genus formula for $\mathbb{CP}^2$. I don't know if you can use GRR to prove it. $\endgroup$ – Michael Albanese Aug 12 '16 at 20:29
  • $\begingroup$ Isn't the adjunction formula, Riemann-Hurwitz and the degree genus formula often derived from the usual RR theorem (It's been a LONG time, but I think Hartshorne does this.)? This is all only about curves so why you would need some generalization like GRR? $\endgroup$ – PVAL-inactive Aug 12 '16 at 21:05
  • 3
    $\begingroup$ The proof of $g=(a-1)(b-1)$ is Hartshorne, III, Ex. 5.6 (c). $\endgroup$ – Jürgen Böhm Aug 12 '16 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.