Rigorous version of heuristic argument for genus-degree formula?

Question

A recent MO question about non-rigorous reasoning reminded me of something I've wondered about for some time. The genus–degree formula says that genus $g$ of a nonsingular projective plane curve of degree $d$ is given by the formula $g=(d−1)(d−2)/2$. Here is a heuristic argument for the formula that someone once told me. Take $d$ lines in general position in the plane; collectively these form a (singular) degree-$d$ curve. There are $d\choose 2$ points of intersection. Now think in terms of complex numbers and visualize each line as a Riemannian sphere. If you start with $d$ disjoint spheres and then bring them together so that every one touches every other one (deforming when necessary) then you expect the genus of the resulting surface to be ${d\choose 2}−(d−1)=(d−1)(d−2)/2$, because after you connect them together in a line with $d−1$ connections, each subsequent connection increases the genus by one.

Is there any rigorous proof of the genus–degree formula that closely follows the above line of argumentation?

A standard proof of the genus–degree formula proceeds by way of the adjunction formula. This doesn't seem to me to answer my question, but perhaps I just don't understand the adjunction formula properly?

Answer 1

Yes, this argument can be made rigorous. One needs three steps.

Step 1. Show that there is at least one smooth plane curve of degree $d$ with the expected genus. Essentially, the proof is given by your heuristic topological argument (deform the union of $d$ lines in general position).

Step 2. Show that if one slightly perturbs the coefficients of a homogeneous polynomial defining a smooth curve, the genus remain unchanged. This is basically a continuity argument.

Step 3. Show that the space $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ of homogeneous polynomials of degree $d$ in three variables defining smooth curves is path-connected. This is because the complement of $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ in $\mathbb{C}[x,\,y,\,z]_d$ (the so-called "discriminant locus") has real codimension $2$.

Putting these three steps together one easily obtains the desired result. For further details you can look at Chapter 4 of Kirvan's book Plane algebraic curves.

Answer 2

Of course it seems you meant topological genus, but here is a similar argument for the arithmetic genus, that I learned from a paper of Fulton. Maybe this is related to the adjunction argument you refer to. And if you combine this with Francesco's answer, you get a proof that the two are equal for smooth plane curves, (or if you assume that result, the Hirzebruch-Riemann-Roch theorem for curves, then you get an answer to your question).

Lemma A: If $X,Y$ are two curves on a smooth surface $S$, and if $X,Y$ are linearly equivalent as divisors on $S$, then $\chi(\mathcal{O}_X) = \chi(\mathcal{O}_Y)$.

Remark: This says in some sense $\chi(\mathcal{O})$ is a deformation invariant, at least for linear deformations.

Proof: Since the line bundles $\mathcal{O}_S(-X)$ and $\mathcal{O}_S(-Y)$ are isomorphic on $S$, the invariants $\chi(\mathcal{O}_S(-X))$ and $\chi(\mathcal{O}_S(-Y))$ are equal. By the usual exact sheaf sequence $$0\to \mathcal{O}_S(-X)\to \mathcal{O}_S\to \mathcal{O}_X\to 0$$ and the analogous one for $Y$, plus the additivity of $\chi$, we get that \begin{align} \chi(\mathcal{O}_X) &= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-X)) \\ &= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-Y)) \\ &= \chi(\mathcal{O}_Y).\hspace{1cm}&\text{qed.}\end{align}

Lemma B: Now suppose that $Y, Y'$ are curves on a smooth surface $S$, and that $Y$ and $Y'$ meet transversely at precisely $n$ points. Then we claim $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$

Proof: Consider the sequence $$0\to \mathcal{O}_{Y+Y'}\to \mathcal{O}_Y + \mathcal{O}_{Y'}\to O_{Y\cdot Y'}\to 0,$$ induced by the map from the disjoint union of $Y,Y'$, to their union $Y+Y'$ on $S$, and where the map to $\mathcal{O}_{Y\cdot Y'}$ is the difference of the two restrictions, from $Y$ and from $Y'$, to the intersection of $Y$ and $Y'$. The additivity of $\chi$ then implies the desired relation, i.e. \begin{align} \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_Y') &= \chi(\mathcal{O}_Y + \mathcal{O}_{Y'}) \\ &= \chi(\mathcal{O}_{Y+Y'}) + \chi(\mathcal{O}_{Y\cdot Y'}) \\ &= \chi(\mathcal{O}_{Y+Y'}) + n. \end{align} Thus $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$ qed.

Now that we know how the function chi(O) behaves under linear degeneration, all we need is to find a formula that behaves this way, and it must be the formula for $chi(\mathcal{O})$.

Corollary: If $X$ is a smooth plane curve of degree $d$, then $\chi(\mathcal{O}_X) = 1 - \frac12 (d-1)(d-2).$

Proof: Induction on $d$. If $d = 2$, then the smooth conic $X$ moves in a linear series also containing a union $Y$ of two lines $Y_1 + Y_2$, where each line is isomorphic to $X$. Then by lemmas A,B above, we have $$\chi(X) = \chi(Y_1)+\chi(Y_2) - 1 = \chi(X)+\chi(X)-1,$$ hence $\chi(X) = 1$. This proves the case $d = 2$, and since a smooth curve of degree $d = 1$ is isomorphic to one of degree 2, we also obtain the formula for degree $d=1$.

Now assume $d \geq 3$ and that we have proved the formula for smooth curves of degree $<d$. A smooth degree-$d$ curve $X$ moves in a linear series that also contains a curve of form $Y = Y_1+Y_2,$ where $Y_1$ is smooth of degree $d-1$, and $Y_2$ is a line meeting $Y_1$ transversely in $d-1$ distinct points. Then lemmas A, B and induction give us that \begin{align} \chi(\mathcal{O}_X) &= \chi(\mathcal{O}_Y) \\ &= \chi(\mathcal{O}_{Y_1})+\chi(O_{Y_2})-(d-1) \\ &= 1-\frac12 (d-2)(d-3) + 1 - (d-1) \\ &= 1-\frac12 (d-1)(d-2),\end{align} as desired. qed.