# Genus of non-complete intersections

Suppose $X\subset \mathbb{P}_k^N$ a nonsingular curve is a complete intersection of hypersurfaces $F_1, \cdots, F_{N-1}$ (of degrees $d_1, \cdots, d_{N-1}$ resp). Then, we know that the canonical divisor on $X$ is $\mathcal{O}_X(d_1 + \cdots + d_{N-1} - n - 1)$. Hence, intersection theory on projective space gives a formula for the genus of $X$ entirely in terms of the various $d_i$. Specifically, $$2g - 2 = d_1\cdots d_{N-1} (d_1 + \cdots + d_{N-1} - N - 1)$$

$\textbf{Question:}$ Suppose $X$ is a nonsingular curve in $\mathbb{P}^N$, and $X = V(F_1, \cdots, F_m)$ is not a complete intersection. Can one get a similarly simple formula for the genus of $X$, perhaps entirely in terms of the degrees of the $F_i$? Is this too much to ask? I can't even simply describe the canonical divisor.

• No. Think of it this way: being a complete intersection actually specifies a resolution of the ideal of the curve (via the Koszul complex). From this information the Euler characteristic of the structure sheaf can be determined, which gives the genus. More generally, if only generators of the ideal are known, the full structure of the resolution can be mysterious. If you know the full resolution you can determine the arithmetic genus (and hence the geometric genus since the curve is smooth), but the genus depends intimately on the resolution, i.e. on the "relations" between the $F_i$. – Jack Huizenga Nov 26 '12 at 4:06
• Jack, Thanks! Good to know that this is "too much to ask". Is it at least possible to describe the canonical divisor? – LMN Nov 26 '12 at 4:22
• how would one describe the canonical divisor without computing the genus? What properties of a divisor (that there is any hope of knowing) can you know without knowing its degree? – Will Sawin Nov 26 '12 at 4:24
• Will, you could know perhaps that it is the restriction of $\mathcal{O}(n)$. This doesn't tell you the degree. (I'm not suggesting that the canonical divisor is the restriction of a line bundle on projective space). – LMN Nov 26 '12 at 4:27
• The best you can hope for, if you do not impose some further conditions, is to bound the genus in terms of the degrees. First of all, there is a bound on the the degree $d$ of the curve $X$, namely $d\leq \text{max}(\text{deg}(F_i))^{N-1}$. For a generic linear projection of $X$ to $\mathbb{P}^2$, the image is an at worst nodal curve whose geometric genus is less than the arithmetic genus, i.e., $g(X) \leq (d-1)(d-2)/2$. Certainly if you impose further restrictions on $X$ or the equations $F_i$, then you can improve this. However, this is sharp for plane curves. – Jason Starr Nov 26 '12 at 14:44

Perhaps Liaison Theory could be useful to you. Roughly speaking two subschemes $V,W\subset\mathbb{P}^n$ are linked if there is an arithmetically Cohen-Macaulay scheme $X$ such that $X = V\cup W$. In particular they may be linked by a complete intersection $X$. If $V$, and hence $W$, is locally Cohen-Macaulay then $$deg(V)+deg(W) = deg(X).$$ Furthermore one can express the arithmetic genus of $V$ in terms of the arithmetic genera of $W$ and $X$. This theory works for schemes of arbitrary codimension but it is particularly well behaved in codimension two.
Assume to have a curve $C\subset\mathbb{P}^n$ which is not a complete intersection but such that there exists an arithmetically Cohen-Macaulay curve $X\subset\mathbb{P}^n$ with minimal free resolution $$0\mapsto S(-d)\rightarrow E_r\rightarrow E_{r-1}\rightarrow ...\rightarrow E_1\rightarrow I_X\mapsto 0$$ such that $C\cup\Gamma = X$. That is $C$ and $\Gamma$ are linked via $X$. Let $g_1$ and $g_2$ be the arithmetic genera of $C$ and $\Gamma$. Then $$p_a(X)-1-deg(\Gamma)(d-n-1) = g_1-g_2.$$ On the other hand $$p_a(X) = \frac{1}{2}deg(X)(d-n-1)+1.$$ Combining these with $$deg(X)-2deg(\Gamma) = deg(C)-deg(\Gamma)$$ we get $$g_1-g_2 = \frac{1}{2}(d-n-1)(deg(C)-deg(\Gamma)).$$ In particular, if $\deg(C)=deg(\Gamma)$ then $g_1 = g_2$. Assume $X\subset\mathbb{P}^3$ is complete intersection of two surfaces of degree $d_1,d_2$. Then we get $$g_1-g_2 = \frac{1}{2}(d_1+d_2-4)(deg(C)-deg(\Gamma)).$$ An easy example: take a twisted cibic $C\subset\mathbb{P}^4$ and two quadrics $Q_1,Q_2$ containing $C$. Then $C\cup L = X =Q_1\cap Q_2$. The residual curve $L$ is a line. Furthermore we know that $p_a(X) = 1$. From the formula we get $g_1-0 = \frac{1}{2}(4-3-1)(3-1) = 0$.
Another example: Let $C\subset\mathbb{P}^3$ be a smooth curve of degree $5$. Then $C$ is contained in a unique quadric surface $Q$. Take a cubic surface $S$ containing $C$. Then $C\cup L = Q\cap S = X$. We have $deg(X) = 6$ and $p_a(X) = 4$. Therefore $g_1 = \frac{1}{2}(2+3-4)(5-1) = 2$ and $C$ is a curve of genus two. In particular it is hyperelliptic.