Suppose we have a normal projective surface $X$ over an algebraically closed field with 'nice' singularities (say canonical, or perhaps rational Gorenstein, or some other condition), with minimal resolution $Y \rightarrow X$, can we determine the arithmetic genus of a curve $C \subset X$ from numerical information about its strict transform on $Y$, for example the arithmetic genus and the intersection with the exceptional subset?


I know that by adjunction the dualizing sheaf of $C$ is given by $\omega\_C =\mathcal{Ext}^1(\mathcal{O}\_C, \omega\_X$) and that for rational singularities we have that $\omega\_X$ is the pushforward of $\omega\_Y$, so it seems that perhaps it is possible to find $H^0$ of this sheaf from information on $Y$, but I'm having trouble untwisting the definitions...


Thanks for any help!


Jordan



Arithmetic genus of curve on singular surface
First let us *not* assume that $X$ has rational singularities, just that it is a normal projective surface and $C\subset X$ a curve on $X$. Let $f:Y\to X$ be *a* resolution and
$\widetilde C=f^{-1}\_\*C\subset Y$ the strict transform of $C$ on $Y$.


Consider the following commutative diagram:
$$
0 \to \mathscr I\_C \to \mathscr O\_X \to \mathscr O\_C \to 0\quad\qquad\qquad\qquad\qquad\qquad\qquad
$$


$$\alpha\downarrow\qquad \beta\downarrow \qquad \gamma\downarrow \qquad\qquad\qquad\qquad\qquad\qquad\qquad$$


$$\qquad
0 \to f\_\*\mathscr I\_{\widetilde C} \to f\_\*\mathscr O\_Y \to f\_\*\mathscr O\_C \to R^1f\_\*\mathscr I\_{\widetilde C}\to R^1f\_\*\mathscr O\_{Y}\to 0
$$


Since $X$ is normal, $\beta$ is an isomorphism and it is clear that $\alpha$ and $\gamma$ are injective. Then by the Snake Lemma $\alpha$ is also an isomorphism and we obtain that we have an exact sequence
$$
0\to \mathscr O\_C \to f\_\*\mathscr O\_C \to R^1f\_\*\mathscr I\_{\widetilde C}\to R^1f\_\*\mathscr O\_{Y}\to 0
$$


Obviously the two $R^1$ sheaves are supported on a zero-dimensional scheme $P$, so we have
$$
\chi(\mathscr O\_C)=\chi(\mathscr O\_{\widetilde C})-\mathrm{length}(R^1f\_\*\mathscr I\_{\widetilde C}) + \mathrm{length}(R^1f\_\*\mathscr O\_{Y})
$$


Now


**1** If $(X,C)$ is log canonical, then it is a DB pair by Thm 1.4 of [this paper](https://www.ams.org/journals/jams/2010-23-03/S0894-0347-10-00663-6/home.html) and Prop 5.1 of [this paper](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-51/issue-1/Du-Bois-pairs-and-vanishing-theorems/10.1215/0023608X-2010-020.full). Then $R^1f\_\*\mathscr I\_{\widetilde C}=0$ by Cor 6.2 of the [same paper](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-51/issue-1/Du-Bois-pairs-and-vanishing-theorems/10.1215/0023608X-2010-020.full). Therefore in this case $\chi(\mathscr O\_C)=\chi(\mathscr O\_{\widetilde C})$. But of course, I just realize now that this simply means that if $(X,C)$ is lc, then at every point either $X$ is smooth or $C$ is smooth, so this is not a big surprise.


**2** If $X$ has rational singularities, then $R^1f\_\*\mathscr O\_{Y}=0$, so the question reduces to determining $R^1f\_\*\mathscr I\_{\widetilde C}$. In this case this is the same as Sasha's sheaf $T$. Clearly, it feels that one should be able to compute this by knowing the intersection of the exceptional set with $\widetilde C$.


In these situations one may try to use the Theorem on Formal Functions. That says that if $E$ denotes the pre-image of the reduced scheme supported on the singular set of $X$ (i.e., for each singular point the pre-image of the closed point defined by the maximal ideal, sometimes this is called the Artin cycle), then
$$
(R^1f\_\*\mathscr I\_{\widetilde C})\_P^{\wedge}\simeq \underset{\leftarrow}{\lim}\ H^1(mE,\mathscr I\_{\widetilde C}\otimes \mathscr O\_{mE}) 
$$


Fortunately the rest of the computation is happening on $Y$ so every divisor is Cartier.
I would try to compute the right hand side using the short exact sequences


$$
0\to \mathscr L^{m} \to \mathscr O\_{(m+1)E} \to \mathscr O\_{mE} \to 0 \tag{$\star$}
$$


where $\mathscr L=\mathscr I\_{E}/\mathscr I\_{E}^2=\mathscr N\_{E/Y}^{-1}$ the dual of the normal bundle of $E$ which is indeed a line bundle since $E$ is a local complete intersection and for the same reason $\mathscr I\_{E}^m/\mathscr I\_{E}^{m+1}\simeq \mathscr L^m$.


Again, $\mathscr I\_{\widetilde C}$ is a line bundle on $Y$ and remains that when restricted to $E$, so $(\star)$ induces the short exact sequence


$$
0\to \mathscr I\_{\widetilde C}\otimes\mathscr L^{m} \to \mathscr I\_{\widetilde C}\otimes\mathscr O\_{(m+1)E} \to \mathscr I\_{\widetilde C}\otimes\mathscr O\_{mE} \to 0 $$


Now this is where you will need some specific information. The increment in the above inverse limit is given by the cokernel of the edge map


$$
H^0(E, \mathscr I\_{\widetilde C}\otimes\mathscr O\_{mE})\to 
H^1(E, \mathscr I\_{\widetilde C}\otimes\mathscr L^{m})
$$


$\mathscr L$ is ample on $E$, so by Serre vanishing $H^1(E, \mathscr I\_{\widetilde C}\otimes\mathscr L^{m})=0$ for $m\gg 0$, so this is indeed a finite game.


Another thing that could help is that
by Thms 3 and 4 of [this paper](https://www.jstor.org/stable/2373050) $h^0(E,\omega\_E)=0$ and hence
$h^1(E, \mathscr I\_{\widetilde C}\otimes\mathscr L^{m}) =h^0(E,\omega\_E\otimes \mathscr I\_{\widetilde C}^{-1}\otimes\mathscr L^{-m})=0$ as soon as $\mathscr L^m(-\widetilde C)$ has a global section. You should be able to figure out the first $m$ for which this happens as that depends on $E^2$ (see Thm 4 of [ibid](https://www.jstor.org/stable/2373050)) and $\widetilde C\cdot E$. For the actual dimension of the intermediate steps you might be able to use some of Artin's methods from [here](https://www.jstor.org/stable/2373050).


This has certainly grown to a much longer answer than I anticipated when I started and although it does not give you a complete answer it might give you some ideas to go on.


