Hartshorne Exercise V.1.5
(a) If $X$ is a surface of degree $d$ in $\PP{k}{3}$, then $K^2=d(d-4)^2$.
(b) If $X$ is a product of two nonsingular curves $C,C’$ of genus $g,g’$ respectively, then $K^2=8(g-1)(g’-1)$
(a) Let $\iota: X\ra \PP{k}{3}$ be a smooth surface of degree $d$. Then by the adjunction formula,
$$\omega_X\cong\iota^*\OO_{\PP{k}{3}}(d-4)$$
Let $H.X=\iota^*H$ be a very ample divisor on $X$ corresponding to its embedding. Then we see that,
$$K_X=(d-4)H|_X.$$
Then
$$ K_X^2=(d-4)^2H|_X^2.$$
Like in 1.2, it is no loss of generality to assume $H|_X$ is smooth and irreducible. By 1.2, $H|_X^2=\deg H|_X$ as a curve in $\PP{k}{n}$. By Bezout’s theorem (the generalized version from II.6), $\deg H|_X=(\deg H)\cdot(\deg X)=d$, hence, $K_X^2=d^2(d-4)^2$.
(b) Suppose $X=C\times C’$ and let $p,p’$ be the first and second projections. Then
$$ K_X= p^*K_C+ p^{‘*}K_{C’}.$$
Before we proceed, note that the intersection product between 2 distinct irreducible curves is just the sum over their intersection points, counting multiplicities. Our strategy will be to reduce to this and use the fibered product structure of $X$ to our advantage.
First, assume neither of the curves are rational nor elliptic. Then both $K_C$ and $K_{C’}$ are linearly equivalent to effective divisors (as we will soon see) that are sums of $2g-2$ and $2g’-2$ points $p_i$ and $q_i$ on $C$ and $C’$ respectively. Their pullbacks are thus sums of curves isomorphic to $C$ and $C’$:
$$p^*K_C= \displaystyle\sum_{i=1}^{(2g-2)}C’_i\:\text{ and }\:p^{‘*}K_{C’}=\displaystyle\sum_{i=1}^{(2g’-2)}C_i,$$
where we identify $C_i\cong C\times{q_i}$ and $C’\cong {p_i}\times C’$ as curves on $X$. Moreover, both $K_C$ and $K_{C’}$ are basepoint free, and $h^0(\omega_C),h^0(\omega_{C’})\ge2$. Therefore, we can find effective divisors $\tilde K_C\sim K_C$ and $\tilde K_{C’}\sim K_{C’}$ with $\tilde K_C\cap K_C=\varnothing$ and $\tilde K_{C’}\cap K_{C’}=\varnothing$.
It follows that,
$$ K_X^2= (p^*K_C)^2+2(p^*K_C).(p^{‘*}K_{C’})+(p^{‘*}K_{C’})^2=2(\displaystyle\sum_{i=1}^{(2g-2)}C’_i)(\displaystyle\sum_{i=1}^{(2g’-2)}C_i),$$
where we have moved $p^*K_C$ and $p^{‘*}K_{C’}$ to linearly equivalent divisors disjoint from the originals. We can do this because pullbacks of linearly equivalent divisors are linearly equivalent (via the pullback of the corresponding rational function), and pulling back does not change whether a divisor is effective. Lastly, I claim that $C_i.C_j’=1\:\: \forall i\ne j$. Once we show this, it follows that,
$$K_X^2=2(\displaystyle\sum_{i=1}^{(2g-2)}C’_i)(\displaystyle\sum_{i=1}^{(2g’-2)}C_i)=8(g-1)(g’-1),$$
as desired. But this claim is immediate because,
$$C_i\cap C_j’=(C\times\{q_i\})\cap(\{p_i\}\times C’)=\{p_i\}\times\{q_i\}.$$
It remains to show the cases where at least one of $C$ and $C’$ is rational or elliptic. First we do the elliptic case. We must show $K_X^2=0$. WLOG $C’$ is an elliptic curve. Then $K_{C’}\sim 0$, so
$$ K_X=p^*K_C.$$
If $C$ is also elliptic then $K_X\sim 0$ and if $g\ge 2$ then by our work above, $K_X^2=(p^*K_C)^2=0$. This solves the elliptic case. The final case is the rational case, and here we must not exclude if one of the curves is rational and the other elliptic. In this case,
$$K_X=p^*(-2q),$$
for any point q on the rational curve, because $K_{\PP{k}{2}}=-2q$. Then $-K_X$ is the pullback of an effective divisor, $-K_{\PP{k}{n}}=2q$. Every point on $\PP{k}{1}$ is linearly equivalent to any other point (in fact this is an equivalent condition for a curve to be rational), so we can pick any other point $\tilde q\in\PP{k}{1}$, and then $p^*(2q)\sim p^*(2\tilde q’)$. But, $p^*(2q)\cap p^*(2\tilde q’)=\varnothing $ because $p^*(2q)=2({q}\times C’)$ and $({q}\times C’)\cap({\tilde q}\times C’)=\varnothing$. Hence,
$$K_X^2=(-K_X)^2=(p^*(2q))^2=0.$$
In the case that both curves are rational, it is easy to see that we can apply the exact same reasoning to find that,
$$K_X^2=(-K_X)^2=2(p^*(2q)).(p^*(2q’))=8(p^*q).(p^*q’)=8=8(-1)(-1)=8(g-1)(g’-1),$$
as desired. The final case is when one curve $C’\cong\PP{k}{1}$ is rational and the other has genus $g\ge 2$. Then, again using notation and reasoning from before,
$$K_X^2=2(\displaystyle\sum_{i=1}^{(2g-2)}C’_i)(p^{‘*}(-2q))+(p^{‘*}(-2q))^2$$
$$=-4(\displaystyle\sum_{i=1}^{(2g-2)}C’_i)(p^{‘*}(q))+(-2p^{‘*}(q))^2=-4(2g-2)+4\cdot 0=8(g-1)(0-1),$$
as desired.
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