The 8th lecture of algebraic curves by Karl Christ
Take \(C = S_1 \cap S_2\subseteq \mathbb P^3\) to be a complete intersection and let \(d_i = \deg S_i\). Adjunction gives us
\[\begin{aligned}
K_C = K_{\mathbb P^3} \otimes \bigwedge_{i=1}^2 N_{C / \mathbb P^3},
\end{aligned}\]
and the normal bundle of \(C\) over \(\mathbb P^3\) can be written
\[\begin{aligned}
N_{C/\mathbb P^3} = T_{\mathbb P^3}/T_C, N_{S_i} = T_{\mathbb P^3}/T_{S_i}.
\end{aligned}\]
We then have a map \(\varphi:N_C\to N_{S_1} \oplus N_{S_2}\). Since \(C\) is smooth, \(T_C = T_{S_1}\cap T_{S_2}\) which implies that \(\varphi\) is injective. Since \(\textrm{rank}(N_C) = \textrm{rk}(N_{S_1} \oplus N_{S_2}\), it follows that \(\varphi\) is an isomorphism. Thus
\[\begin{aligned}
N_C \cong \mathcal O_{\mathbb P^3}(d_1) \oplus \mathcal O_{\mathbb P^3}(d_2)
\end{aligned}\]
and
\[\begin{aligned}
\bigwedge^2 N_C = \bigwedge^2 \mathcal O_{\mathbb P^3}(d_1) \oplus \mathcal O_{\mathbb P^3}(d_2) = \mathcal O_{\mathbb P^3}(d_1 + d_2).
\end{aligned}\]
The last equality above follows from the standard trick we introduced last lecture: if you have a short exact sequence
\[\begin{aligned}
0 \to \mathcal F\to \mathcal F \to \mathcal F'' \to 0
\end{aligned}\]
where each term has rank \(r\), \(r'\) and \(r''\) respectively, then
\[\begin{aligned}
\bigwedge^{r'}\mathcal F' = \bigwedge^{r}\mathcal F \otimes \bigwedge^{r''}\mathcal F''.
\end{aligned}\]
In our case we have the somewhat trivial exact sequence
\[\begin{aligned}
0 \to \mathcal O_{\mathbb P^3}(d_1) \to \mathcal O_{\mathbb P^3}(d_1)\oplus \mathcal O_{\mathbb P^3}(d_2) \to \mathcal O_{\mathbb P^3}(d_2)\to 0,
\end{aligned}\]
which gives us the final equality above after messing around with tensors and direct sums.
Examining the formula \(2g - 2 = \deg (K_C)\) and noting that \(\deg(C) = d_1\cdot d_2\), we have
\[\begin{aligned}
2g - 2 = \deg(K_C) &= \deg\big((\mathcal_{\mathbb P^3}(-4) \otimes \mathcal )_{\mathbb P^3}(d_1 + d_2)_{|_C}\big) \\
&= d_1\cdot d_2 \cdot (d_1+d_2 - 4).
\end{aligned}\]
Thus \(g = \frac{1}{2}(d_1\cdot d_2 \cdot (d_1 + d_2 - 4)) + 1\).
Remark:
If \(d_2 = 1\) then we get the correct formula:
\[\begin{aligned}
g = \frac{1}{2}(d_1\cdot (d_1 - 3)) + 1 = \frac{(d_1 - 1)(d_1 - 2)}{2}
\end{aligned}\]
For the intersection of \(n-1\) hypersurfaces in \(\mathbb P^n\) we get analogously
\[\begin{aligned}
\bigwedge^{n-1}N_C = \mathcal O_{\mathbb P^3}\left(\sum d_i\right) \text{~ and ~} g = \frac{1}{2}\left(\prod_i d_i~\cdot~ (\sum d_i - n - 1)\right) + 1.
\end{aligned}\]
Up to \(n = 10\), the possible values of \(g\) for complete intersection curves are
\[\begin{aligned}
0,1,3,4,5,6,9,10.
\end{aligned}\]
For a plane curve, they are \(0,1,3,6\).
Example: Take \(C = Q_1\cap Q_2 \subseteq \mathbb P^3\) to be a smoooth curve where each \(Q_i\) is a smooth quadric. Our formula gives us
\[\begin{aligned}
g(C) = \frac12 (4 \cdot ( 0 ) ) + 1 = 1,
\end{aligned}\]
so we get a genus 1 curve.
Conversely, any genus 1 curve can be realized as the intersection of two quadrics in \(\mathbb P^3\). If \(X\) is a genus 1 curve, then Riemann-Roch gives us
\[\begin{aligned}
h^1(\mathcal O_X(4p)) = d - g + 1 = 4
\end{aligned}\]
which means \(\mathcal O_X(4p)\) gives an embedding into \(X\to \mathcal P^3\). We therefore have a map
\[\begin{aligned}
H^0(\mathbb P^3,\mathcal O_{\mathbb P^3}(2)) \to H^0(X,\mathcal O_X(2)).
\end{aligned}\]
The left term above has dimension \(\binom{3+2}{2} = 10\) while the right term has dimension \(8 - 1 + 1\) (by Riemann-Roch), so the kernel of this map is at least 2. So \(X\) is the intersection of at least \(2\) quadrics in \(\mathbb P^3\), and by dimension considerations we can choose two of those and they will still cut out \(X\).
Example 2: Consider \(\mathbb P^1 \to \mathbb P^3\) given by \(V = \{t&4, t^3s, ts^3, s^4\} \subseteq H^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(4))\). Observe that there are smooth curves of genus 0 and genus 1 in \(\mathbb P^3\). Note that the image of this map must lie on a quadric as a bidegree \((1,4)\) curve; we can see this by first passing through an embedding \(\mathbb P^1\to \mathbb P^1\times \mathbb P^1\).
Observation: No smooth non-degenerate rational curve \(C\subseteq \mathbb P^3\) can be a complete intersection. Remember that \(2g - 2 = d_1\cdot d_1 \cdot (d_1 + d_2 - 4)\), if \(g = 0\), then this means \(d_1 + d_2 \leq 3\). The only options are \(d_1 = d_2 = 1\) which implies we have a line or \(d_1 = 1\) and \(d_2 = 2\) which implies we have a cubic in \(\mathbb P^2\), both of which are degenerate.
A cover (or covering) is a surjective map \(f:Y\to X\) between two smooth curves \(X\) and \(Y\).
The degree of \(f\) is the number of preimages in a general fibre; or more algebraically, its the degree of the field extension \(K(Y)/K(X)\).
The ramification index of \(f\) at \(p\in Y\) is \(v_p(t) - 1\) where \(t\) is a local parameter at \(f(p)\). A "local parameter" at \(f(p) = q\) is a function which vanishes to order \(1\) at \(q\). It lives in \(\mathcal O_{X,q}\) and we take the valuation in \(K(Y)\) under the embeddings
\[\begin{aligned}
\mathcal O_{X,q} \hookrightarrow K(X) \hookrightarrow K(Y).
\end{aligned}\]