I want to compute the arithmetic genus of the complete intersection of two quadrics in $\mathbb{P}^3$. I know the adjunction formula is one way of doing this, but I am keen to understand where my approach is failing (I know it is failing because I expect to get $1$ as the genus, but I am getting $0$).

My definition of arithmetic genus is $p_a(X) := (-1)^{\text{dim }X}(\chi(\mathscr{O}_X) - 1)$. For $X$ an irreducible projective curve, this turns out to be $h^1(X,\mathscr{O}_X)$.

Here is my approach: let $C := Q_1\cap Q_2$ (with inclusion $i : C \hookrightarrow \mathbb{P}^3$) be the complete intersection of two quadrics in $\mathbb{P}^3$. Assume everything is over an algebraically closed field $k$. Since $C$ is a complete intersection, the ideal sheaf $\mathscr{I}_C$ is isomorphic to $\mathscr{I}_{Q_1} + \mathscr{I}_{Q_2}$ (by Hartshorne II. Ex. 8.4). Further, we know that since $Q_1$ and $Q_2$ are hypersurfaces, $\mathscr{I}_{Q_i} \cong \mathscr{O}_{\mathbb{P}^3}(-2)$. So $\mathscr{I}_C \cong \mathscr{O}_{\mathbb{P}^3}(-2)^{\oplus 2}$. We then look at the short exact sequence of sheaves on $\mathbb{P}^3$ corresponding to the closed subscheme $C$: $$ 0 \rightarrow \mathscr{I}_C = \mathscr{O}_{\mathbb{P}^3}(-2)^{\oplus 2} \rightarrow \mathscr{O}_{\mathbb{P}^3} \rightarrow i_*\mathscr{O}_C \rightarrow 0 $$

Taking cohomology, we know that $H^i(\mathscr{O}_{\mathbb{P}^3}) = 0$ for $i = 1,2$ which leaves us with $H^1(i_*\mathscr{O}_C) \cong H^2(\mathscr{O}_{\mathbb{P}^3}(-2))^{\oplus 2}$. But I know that the latter is $0$, which means that $h^1(\mathscr{O}_C) = 0$. But this doesn't make sense since I know $C$ should be a quartic elliptic curve which implies $h^1(\mathscr{O}_C) = 1$.

What is the problem here?

  • 2
    $\begingroup$ The problem is that $\mathscr{I}_{Q_1} \cap \mathscr{I}_{Q_2}$ is not zero (i.e., their sum is not direct): there are polynomials which vanish on both quadrics. In fact, that intersection is $\mathscr{O}_{\mathbb{P}^3}(-4)$ so you get an exact sequence $0 \to \mathscr{O}(-4) \to \mathscr{O}(-3)^2 \to \mathscr{O} \to \mathscr{O}_C \to 0$. $\endgroup$ – Gro-Tsen Jan 18 '16 at 16:45
  • $\begingroup$ Oh I see. So it's not a direct sum. Thank you. Now, since the exact sequence you have written is not a short one (i.e. it has 4 terms instead of 3) how can I use it to make conclusions about cohomology and fix my analysis? $\endgroup$ – user306090 Jan 18 '16 at 16:49
  • 1
    $\begingroup$ You can still conclude that the alternating sums of the Euler characteristics of the sheaves in your sequence is $0$, and compute (most of) these Euler characteristics directly. Or (and presumably equivalently), break the four term exact sequence into two short exact sequences, the structure sequence of the subscheme $C$ which you wrote above, and the locally free resolution of your ideal $0\to \mathcal O (-4) \to \mathc $\endgroup$ – Tabes Bridges Jan 18 '16 at 16:58
  • $\begingroup$ Oh actually I see how to do it - I can split the sequence of 4 terms into two SESs. $\endgroup$ – user306090 Jan 18 '16 at 17:01
  • $\begingroup$ Yep, thanks for your help! $\endgroup$ – user306090 Jan 18 '16 at 17:02

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