If $C$ is an irreducible plane curve we have the well known formula relating the airthmetic (obtained via the degree-genus formula) and the geometric genus

$$\frac{(d-1)(d-2)}{2} - \sum \frac{r(r-1)}{2},$$

where $d$ is the degree of the curve and the sum is taken over the ordinary singularities with multiplicity $r$. But what happens if the curve is reducible? Does one have to take into account also the number of components?


1 Answer 1


The arithmetic genus of a curve in a surface is given by $$-2+2g_a=C(C+K),$$ where $K$ is the canonical divisor.

In the case where the surface is $\mathbb{P}^2$ and the curve has degree $d$ we have $C=dL$ and $K=-3L$ where $L$ is the class of a line. Hence $-2+2g_a=d(d-3)$, which yields $$g_a=\frac{(d-1)(d-2)}{2}$$ as you know.

The geometric genus of an irreducible curve is however $g_a-\sum m_i(m_i-1)/2$ where $m_i$ is the multiplicity of the points, including all infinitely near points. This can be checked by blowing-up the points and checking that the arithmetric genus decreases by $m_i(m_i-1)/2$ when you blow-up a point of multiplicity $m_i$.

Now, if you have a reducible curve $C=C_1+C_2$, you can compute the arithmetic genus of $C$ by the formula $-2+2g_a(C)=C(C+K)$ (which is basically a definition):

$$\begin{array}{rcl} -2+2g_a(C)&=&C(C+K)\\ &=&(C_1+C_2)(C_1+C_2+K)\\ &=&C_1(C_1+K)+C_2(C_2+K)+2C_1C_2\\ &=&(-2+2g_a(C_1))+(-2+2g_a(C_2))+2C_1\cdot C_2\end{array}$$

Hence $$g_a(C)=g_a(C_1)+g_a(C_2)+C_1\cdot C_2-1.$$

Proceeding by induction, you can do the case with more components.


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