Recall that the arithmetic genus of a projective scheme \(D\) of dimension 1 is defined as \[ p_a=1-\chi\left(\mathcal{O}_D\right) .\] See III, Ex. 5.3.
If \(D\) is an effective divisor on the surface \(X\), use (1.6) to show that \[ 2 p_a-2= D.(D+K) .\]
\(p_a(D)\) depends only on the linear equivalence class of \(D\) on \(X\).
More generally, for any divisor \(D\) on \(X\), we define the virtual arithmetic genus (which is equal to the ordinary arithmetic genus if \(D\) is effective) by the same formula: \(2 p_a-2=D .(D+K)\). Show that for any two divisors \(C, D\) we have \[ p_a(-D)=D^2-p_a(D)+2 \] and \[ p_a(C+D)=p_a(C)+p_a(D)+C . D-1 . \]
Solution
Let \(K\) be a canonical divisor on \(X\).
1. If \(D\) is effective, then\[ \chi(\OO_D) = -\frac12D\cdot(D+K). \]
Since \(D\) is an effective Cartier divisor on the nonsingular surface \(X\), there is an exact sequence \[ 0 \longrightarrow \OO_X(-D) \longrightarrow \OO_X \longrightarrow \OO_D \longrightarrow0. \] Additivity of Euler characteristic gives \[ \chi(\OO_D) = \chi(\OO_X)-\chi(\OO_X(-D)). \]
Apply surface Riemann–Roch T-COHRRS to the divisor \(-D\): \[ \begin{aligned} \chi(\OO_X(-D)) &= \chi(\OO_X) +\frac12(-D)\cdot(-D-K)\\ &= \chi(\OO_X) +\frac12D\cdot(D+K). \end{aligned} \] Substitution yields the stated formula.
2. For an effective divisor \(D\),\[ \boxed{2p_a(D)-2=D\cdot(D+K)}. \]
Because \(D\) has dimension one, its arithmetic genus is \[ p_a(D)=1-\chi(\OO_D) \] by D-COHEULER. Step 1 therefore gives \[ \begin{aligned} 2p_a(D)-2 &= -2\chi(\OO_D)\\ &= D\cdot(D+K). \end{aligned} \] This proves part (a).
3. For effective divisors, \(p_a(D)\) depends only on the linear equivalenceclass of \(D\).
Suppose \[ D\sim D'. \] The surface intersection pairing D-SRFINT depends only on linear equivalence classes, so \[ D^2=(D')^2, \qquad D\cdot K=D'\cdot K. \] Hence \[ D\cdot(D+K)=D'\cdot(D'+K). \] Applying step 2 to \(D\) and \(D'\) gives \[ 2p_a(D)-2=2p_a(D')-2, \] and therefore \[ p_a(D)=p_a(D'). \] This proves part (b).
4. For an arbitrary divisor \(D\), the virtual arithmetic genus is\[ p_a(D) = 1+\frac12D\cdot(D+K). \]
This is exactly the definition in part (c), rewritten from \[ 2p_a(D)-2=D\cdot(D+K). \] By step 2 it agrees with the ordinary arithmetic genus whenever \(D\) is effective.
5. For every divisor \(D\),\[ \boxed{p_a(-D)=D^2-p_a(D)+2}. \]
Using step 4, \[ \begin{aligned} p_a(-D) &= 1+\frac12(-D)\cdot(-D+K)\\ &= 1+\frac12D^2-\frac12D\cdot K. \end{aligned} \] On the other hand, \[ \begin{aligned} D^2-p_a(D)+2 &= D^2- \left(1+\frac12D^2+\frac12D\cdot K\right)+2\\ &= 1+\frac12D^2-\frac12D\cdot K. \end{aligned} \] The two expressions agree.
6. For arbitrary divisors \(C,D\),\[ \boxed{ p_a(C+D) = p_a(C)+p_a(D)+C\cdot D-1 }. \]
By bilinearity and symmetry of the intersection pairing, \[ \begin{aligned} p_a(C+D) &= 1+\frac12(C+D)\cdot(C+D+K)\\ &= 1 +\frac12C\cdot(C+K) +\frac12D\cdot(D+K) +C\cdot D. \end{aligned} \] Step 4 gives \[ \frac12C\cdot(C+K)=p_a(C)-1 \] and \[ \frac12D\cdot(D+K)=p_a(D)-1. \] Substituting these identities yields \[ p_a(C+D) = p_a(C)+p_a(D)+C\cdot D-1. \]
7. Q.E.D.
Step 2 proves part (a), step 3 proves part (b), and steps <1>4–<1>6 prove both identities in part (c).