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Given a smooth algebraic surface $X$, and a group $G$ acting on it and letting $Y := X / G$, how can we compute $K_Y^2$ from from $K_X^2$?

Current progress: In Borisov and Fatighenti - New explicit constructions of surfaces of general type, during the proof of theorem 2.2 this situation arises, and we have $K_X^2 = 42$, $K_Y^2 = 3$, and $G = C_{14}$, leading me to guess that there is some formula like $K_X^2 = \lvert G\rvert K_Y^2$. However, in some situations like Keum - Quotients of fake projective planes Theorem 1.1 (2), we have a surface with $K_X^2 = 9$ and $G = C_7$, so such a formula can't hold because the order of the group doesn't divide $K_X^2$.

Also, I'm looking for a "down-to-earth" (i.e. not full of étale jargon) reference for Galois quotients of varieties, especially the behavior of invariants like $K_Y^2$. I would even settle for a jargon-filled reference that shows how to precisely do computations.

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    $\begingroup$ If $G$ acts freely, then $K_X^2 = |G|K_Y^2$. Note that in the example from Keum's paper, the quotient of the action has singular points (because $G$ does not act freely). $\endgroup$ Jul 21, 2021 at 17:43
  • $\begingroup$ You can have a smooth quotient also if the action is not free: this happens when the fixed locus is purely divisorial. Of course, the formula for the canonical divisor must be corrected by adding the ramification term. $\endgroup$ Jul 21, 2021 at 18:25

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When the action is a Galois cover then if $\pi$ is the quotient map then $$\pi^*TY \cong TX$$ by an isomorphism covering the identity on $X$. To see this, note that $\pi$ is a biholomorphism on a small neighbourhood of every $p \in X$, so $\pi$ induces an isomorphism between these bundles.

This implies in cohomology $$ c_{1}(TX) = c_{1}(\pi^*(TY)) = \pi^*(c_{1}(TY))$$ by functoriality of Chern classes. Furthermore since pull back in cohomology respects cup product, $$\pi^*(c_{1}(TY)^2) = \pi^*(c_{1}(TY)) . \pi^*(c_{1}(TY))= c_{1}(TX)^2.$$ Next, since $\pi^{*}$ will induce multiplication by the degree of $\pi$ (i.e. |G|) on $H^4$, this gives the formula. $\int_{X} c_{1}(TX)^2 = |G|\int_{Y} c_{1}(TY)^2$.

Finally for any complex surface $S$ we have $\int_{S} K_{S}^2 = \int_{S} c_{1}(TS)^2$, becuase for any rank $2$ bundle $E$, $c_{1}(\wedge^2E) = c_1(E)$.

For non-free actions this is clearly false.

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