First Cohomology of a Subscheme of Projective Plane

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Let $k$ be a field, and let $X$ be a subscheme of $\mathbb{P}_k^2$ defined by a single homogeneous equation $f(x_0, x_1, x_2) = 0$ of degree $d$. Show that $$\dim_k H^1(X, \mathcal{O}_X) = \frac{(d-1)(d-2)}{2}$$

 

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Spoiler

Let $i : X \to \mathbb{P}^2_k$ be the inclusion. If $S = k[x_0,x_1,x_2]$, there is a short exact sequence of graded $S$-modules
$$0 \to S(-d) \xrightarrow{f} S \to S/(f)\to 0$$ which induces a short exact sequence of sheaves $$0 \to \mathcal{O}_{\mathbb{P}^2_k}(-d) \to \mathcal{O}_{\mathbb{P}^2_k}\to i_*\mathcal{O}_X \to 0$$ Since $H^1(\mathbb{P}^2_k, \mathcal{O}_{\mathbb{P}^2_k}) = 0 = H^2(\mathbb{P}^2_k, \mathcal{O}_{\mathbb{P}^2_k})$, the boundary map from the long exact sequence in cohomology gives an isomorphism of $k$-vector spaces $H^1(X, \mathcal{O}_X) \xrightarrow[\approx]{\delta} H^2(\mathbb{P}^2_k, \mathcal{O}_{\mathbb{P}^2_k}(-d))$. Since $H^2(\mathbb{P}^2_k, \mathcal{O}_{\mathbb{P}^2_k}(-d)) \approx k^{\binom{d-1}{2}}$ it follows that $$\dim_k H^1(X, \mathcal{O}_X) = \binom{d-1}{2} = \frac{1}{2}(d-1)(d-2)$$ as claimed.

 

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Here is an alternative approach that computes at least $\dim H^0(O) - \dim H^1(O) = \chi(O)$, and without knowing the value of the cohomology groups of the projective plane. This reduces the problem to showing $\dim H^0(O) = 1$, i.e. the only global sections of $O$ are constants. I learned this idea from a lovely paper of W. Fulton: A note on the arithmetic genus, where he cites Severi (1909) and Zariski (1952), a nice instance of cohomological insights preceding the introduction of cohomology!

To show that $\chi(O) = 1 - (d-1)(d-2)/2 = 1-g$, we will compute how $\chi(O)$ varies as a curve moves in a linear series.

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 $\chi(O_X) = \chi(O_Y)$.

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

Proof: Since the line bundles $O_S(-X)$ and $O_S(-Y)$ are isomorphic on $S$, the invariants $\chi(O_S(-X))$ and $\chi(O_S(-Y))$ are equal. By the usual exact sheaf sequence
$0 \to O_S(-X) \to O_S \to O_X \to0$, and the analogous one for $Y$, plus the additivity of $\chi$ we get that $\chi(O_X) = \chi(O_S) - \chi(O_S(-X)) = \chi(O_S) - \chi(O_S(-Y)) = \chi(O_Y)$. qed.

Thus if the formula holds for one curve of degree $d$, it holds for all. Thus we may compute it for a smooth curve, or for a reducible curve.

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 $\chi(O_{Y+Y’}) = \chi(O_Y) + \chi(O_Y’) - n$.

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

[Remark: If we combine lemma B with (the proof of) lemma A, we get a formula for the intersection number of two curves on a surface, in terms of invariants of the surface and the curves, i.e. Bezout's theorem.]

Corollary: If $X$ is a smooth plane curve of degree $d$, then $\chi (O_X) = 1 - (d-1)(d-2)/2$.

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 $d ≥ 3$ 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 that $$\chi(O_X) = \chi(O_Y) = \chi(O_{Y_1})+\chi(O_{Y_2})-(d-1) = 1-(d-2)(d-3)/2 + 1 - (d-1) = 1-(d-1)(d-2)/2$$ as desired.
q e d .

Note also, that over the complex numbers, a line and a conic are both homeomorphic to a sphere, hence have topological genus zero, so adding a transverse line to a smooth curve of degree $d$, and smoothing out the intersection points, adds $(d-1)$ to the topological genus. Hence by the same induction argument, the formula $(1/2)(d-1)(d-2)$ is also the topological genus g of a smooth curve of degree d.

Since sheaf sequences make it trivial to show (by induction) that $\chi(D) - \chi(O) = \deg(D)$, for any divisor $D$ on a curve, these two calculations together prove the weak Riemann Roch theorem for a plane curve: $\chi(D) = \deg(D) + \chi(O) = \deg(D) + 1-g$.