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?


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.

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    $\begingroup$ More details than Kirwan gives, are in chap. 6 of C.T.C.Wall's book, Singular Points of Plane curves. The essential point in making step 1 rigorous is to describe the topology of the Milnor fiber, a neighborhood of the singular point, on a nearby smooth curve. It is a smooth connected manifold with as many boundary components as the singularity has local branches, and first homology of rank equal to the Milnor number, which is 1 for a transverse union of lines. Hence indeed smoothing the node replaces it by plumbing in a cylinder, as expected. Step 2 is implied by Ehresmann's theorem. $\endgroup$ – roy smith Jun 9 '14 at 19:10

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.


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