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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 10 at 9:08
  • $\begingroup$ 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. $\endgroup$ Commented Jul 10 at 9:48
  • $\begingroup$ But the latter point doesn't seem to matter; once you fix the sign error the rest of the argument appears to be fine. $\endgroup$ Commented Jul 10 at 9:50
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    $\begingroup$ 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. $\endgroup$ Commented Jul 10 at 9:58
  • $\begingroup$ @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. $\endgroup$ Commented Jul 10 at 23:57

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