# Genus of a smooth curve on blowup of $\mathbb{P}^2$ at some points

Everything here takes place over $$\mathbb{C}$$.

Let $$p_1, \ldots, p_n$$ denote distinct points on $$\mathbb{P}^2$$ and let $$\pi: S\to \mathbb{P}^2$$ denote the blowup of $$\mathbb{P}^2$$ at these points. Let $$E_i = \pi^{-1}(p_i)$$ for each $$i$$. Suppose $$C\subset S$$ is a smooth (and integral) curve of degree $$d$$. How can I make sense of the genus of $$C$$.

Here are my thoughts:

If $$C$$ misses all the $$E_i$$, then it's just a regular (projective) plane curve and we have $$g = \frac{(d-1)(d-2)}{2}$$.

Now suppose $$C$$ hits at least one of the $$E_i$$. By adjunction, we have $$C\cdot (K_S + C) = 2g - 2.$$

By basic blow-up theory (say II.3 in Beauville's Complex Algebraic Surfaces), we know $$K_S = \pi^\ast K_{\mathbb{P}^2} + \sum E_i$$. Since $$K_{\mathbb{P}^2} \cong \mathcal{O}_{\mathbb{P}^2}(-3)$$, which we denote by $$-3H$$ for $$H$$ the divisor class of a hyperplane, then since $$\pi$$ is an isomorphism away from those $$n$$-points I assume (maybe (probably?) falsely) that $$\pi$$ pulls back hyperplanes to hyperplanes. In other words, $$\pi^*(-3H) = -3H$$ (by an abuse of notation where the $$H$$ on the left denotes the divisor class of a hyperplane in $$\mathbb{P}^n$$ and on the right as a hyerplane in $$S$$). Therefore, we get

$$C\cdot\left(-3H + \sum E_i + C\right) = 2g - 2.$$ Since $$H$$ is the divisor class of a hyperplane, and a curve of degree $$d$$ meets a general hyperplane at $$d$$ points then we have $$C\cdot -3H = -3d$$. Before analyzing the intersection numbers $$C\cdot E_i$$, let's first try and figure out $$C\cdot C$$. Let $$C'$$ denote the image of $$C$$ under $$\pi$$. By II.2 of Beauville, we know that $$C = \pi^\ast C' - \sum m_iE_i$$ where $$m_i$$ denotes the multiplicity with which $$C'$$ meets $$p_i$$. Therefore, $$C\cdot C = \left(\pi^\ast C' - \sum m_iE_i\right)\left(\pi^\ast C' - \sum m_iE_i\right) = \pi^\ast C'\cdot \pi^* C' + \sum_{ij} m_im_j E_iE_j.$$ By Beauville II.3, we know $$\pi^\ast C'\cdot\pi^\ast C' = C'\cdot C'$$ and $$E_i\cdot E_i = -1$$. Moreover since $$E_i$$ and $$E_j$$ don't meet for $$i\neq j$$, we know $$E_i \cdot E_j = 0$$ whenever $$i\neq j$$. Hence, we have $$C\cdot C = C'\cdot C' - \sum m_i^2.$$ Now let $$d' = \deg C'$$ (I want to say that since $$\pi$$ is an isomorphism away from finitely many points $$d' = d$$ but this feels wrong, if anybody can comment on this). So $$C'\cdot C' = \deg C' = (d')^2$$ and $$C\cdot C = (d')^2 - \sum m_i^2$$.

Lastly, we wish to compute $$C\cdot E_i$$. It's again convenient to use $$C = \pi^\ast C' - \sum m_iE_i$$. Now $$C\cdot E_i = \left(\pi^\ast C' - \sum E_i\right)\cdot E_i = \pi^\ast C'\cdot E_i - \sum_j m_jE_jE_i$$ By Beauville II.3 again, $$\pi^\ast C'\cdot E_i = 0$$ and $$\sum_j m_jE_jE_i = -m_i$$ which gives us $$C\cdot E_i = m_i$$.

Combining this all together gives $$2g - 2 = C\cdot\left(-3H + \sum E_i + C\right) = -3d + (m_1 + \cdots +m_n) + (d')^2 - \sum m_i^2.$$

Therefore,

$$g = 1 + \frac{-3d + m_1 + \cdots +m_n + (d')^2 - \sum m_i^2}{2}$$ where $$d = \deg C,$$ $$d' = \deg \pi(C),$$ $$g$$ is the genus of $$C$$, and $$m_i$$ is the multiplicity of $$\pi(C)$$ at $$p_i$$. If my thought that $$d' = d$$ is right, then we can rewrite this as $$g = 1 + \frac{d^2 -3d}{2} - \frac{\sum_i m_i^2 - m_i}{2}$$ Since $$1 + \frac{d^2 - 3d}{2} = \frac{(d-1)(d-2)}{2}$$, which is just the genus $$g'$$ of $$\pi(C)$$, we get

$$g = g' - \frac{\sum_i m_i^2 - m_i}{2}.$$ In other words, $$p_a(C) = p_a(\pi(C)) - \frac{\sum_i m_i^2 - m_i}{2}.$$

Is this analysis right, and are the assumptions that $$\deg C = \deg \pi(C)$$ and that the pullback along $$\pi$$ of a hyperplane class in $$\mathbb{P}^2$$ is a hyperplane class in $$S$$ correct?

• For $n \leq 8$ I guess you are interested in the genera of curves on Del Pezzo surfaces, but for $n > 9$ I am not sure. Commented Jul 10 at 9:08
• You have a sign error in your formula for $C$: it should be $C=\pi^\ast C^\prime - \sum_i m_i E_i$. As a check, your conclusion $C \cdot E_i=-m_i$ can't be right, since in general $C$ and $E_i$ are distinct irreducible curves. You also ask "are the assumptions...correct" but it is not clear in this context what the degree of a curve in $S$, or a hyperplane class in $S$ mean; the only reasonable choice is to define them so that those "assumptions" are true by definition. Commented Jul 10 at 9:48
• But the latter point doesn't seem to matter; once you fix the sign error the rest of the argument appears to be fine. Commented Jul 10 at 9:50
• As a final comment, in the general case when $C$ meets each of the $E_i$ everywhere transversely, then $C^\prime$ will be a plane curve with ordinary singularities and the map $C \rightarrow C^\prime$ will be the resolution of singularities. You can then compute the genus of $C$ via Noether's formula (see e.g. Section 7.3 of Kirwan, Complex Algebraic Curves) and check that the answer agrees with the (corrected) one you obtained above. Commented Jul 10 at 9:58
• @DerekAllums Here really I'm imagining $n = 6$ and that $S$ is really just a smooth cubic in $\mathbb{P}^3$, and I'm using the fact that all smooth cubics in $\mathbb{P}^3$ are the blowup of $\mathbb{P}^2$ at six points. Commented Jul 10 at 23:57