If \(X\) is a surface of degree \(d\) in \(\PP^3\), then \[ K^2=d(d-4)^2 .\]
If \(X\) is a product of two nonsingular curves \(C, C^{\prime}\), of genus \(g, g^{\prime}\) respectively, then \[ K^2=8(g-1)\left(g^{\prime}-1\right) .\] Cf. (II, Ex. 8.3).
Solution
In part (a), let \(H\) denote the hyperplane class on \(X\). In part (b), let \(p_1:C\times C'\to C\) and \(p_2:C\times C'\to C'\) be the projections.
1. If \(X\subseteq\PP^3\) is a nonsingular surface of degree \(d\), then\[ K_X=(d-4)H \qquad\text{and}\qquad H^2=d. \]
Hypersurface adjunction gives \[ K_X=(K_{\PP^3}+X)|_X=(-4H+dH)|_X=(d-4)H. \] The number \(H^2\) is the intersection of \(X\) with two general hyperplanes in \(\PP^3\), hence equals the degree of \(X\), namely \(d\).
2. For the surface in part (a),\[ \boxed{K_X^2=d(d-4)^2}. \]
By bilinearity of the intersection pairing and step 1, \[ K_X^2 =(d-4)^2H^2 =(d-4)^2d. \]
3. On \(C\times C'\), one has\[ K_{C\times C'}=p_1^*K_C+p_2^*K_{C'}. \]
Hartshorne II.8.3, proved on P-AGH283PRODDIFF, gives \[ \omega_{C\times C'} \cong p_1^*\omega_C\tensor p_2^*\omega_{C'}. \] Passing to divisor classes gives the displayed equality of canonical classes.
4. For divisors \(D,E\) on \(C\) and \(D',E'\) on \(C'\),\[ (p_1^*D)\cdot(p_1^*E)=0, \qquad (p_2^*D')\cdot(p_2^*E')=0, \]and\[ (p_1^*D)\cdot(p_2^*D')=(\deg D)(\deg D'). \]
Write divisors as integral linear combinations of closed points. For a point \(P\in C\), the divisor \(p_1^*P=\{P\}\times C'\) is a fibre of \(p_1\). Distinct fibres of \(p_1\) are disjoint, and the normal bundle of each fibre is the constant one-dimensional vector space \(T_PC\) tensored with its structure sheaf, hence has degree zero. Therefore every pair of pullbacks from \(C\) has intersection number zero. The same argument applies to pullbacks from \(C'\).
For points \(P\in C\) and \(Q\in C'\), the two fibres \[ \{P\}\times C' \qquad\text{and}\qquad C\times\{Q\} \] meet transversely in the single point \((P,Q)\), so their intersection number is one. Bilinearity of the intersection pairing then gives \[ (p_1^*D)\cdot(p_2^*D')=(\deg D)(\deg D'). \]
5. For \(X=C\times C'\),\[ \boxed{K_X^2=8(g-1)(g'-1)}. \]
Set \[ A=p_1^*K_C, \qquad B=p_2^*K_{C'}. \] By step 3, \(K_X=A+B\). Step 4 gives \(A^2=B^2=0\) and \[ A\cdot B =(\deg K_C)(\deg K_{C'}) =(2g-2)(2g'-2). \] Therefore \[ \begin{aligned} K_X^2 &=(A+B)^2\\ &=2A\cdot B\\ &=2(2g-2)(2g'-2)\\ &=8(g-1)(g'-1). \end{aligned} \]
6. Q.E.D.
Step 2 proves part (a), and steps <1>3–<1>5 prove part (b).