Assume $X$ be a normal projective variety with $\mathbb Q$-Cartier divisor $D$, then can we extend adjunction formula on pair $(X,D)$?


If the pair $(X,D)$ is log canonical and $S$ be a component of $D$ with coefficient 1, then we have adjunction type formula as $$K_S+D_S=(K_X+D)|_S$$

[edited]: If $f:Z\to X$ be a finite Galois map, then there exists a branch divisor $B$ on $X$ s.t, $K_X+B$ is $\mathbb Q$-Cartier and $K_Z=f^*(K_X+B)$.

As an example on surfaces: Let $(X,D)$ be a surface with Kawamata log terminal singularities and $\pi:Y\to X$ be a minimal resolution of $X$. Then we can write $\pi^*(K_X+D)=K_Y+D_Y$ where $D_Y\geq 0$ and each exceptional curve of $\pi$ is a smooth rational curve.

Note: If $K_X+B$ is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $Ext$'s which is Grothendieck Duality.

See Kawamata-Kodaira canonical bundle formula also

Example: Let $f:V\to X$ is a contraction but not birational such that $K_V\sim_\mathbb Q 0$ (like Calabi-Yau model)then it often happen that $K_V\sim_\mathbb Q f^*(K_X+B)$ for some divisor $B$

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  • $\begingroup$ The pair being log canonical doesn't have anything to do with the first formula. In the second formula, the divisor B is usually the ramification divisor. $\endgroup$ – Karl Schwede Jan 15 '17 at 17:49
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    $\begingroup$ Whoops, in the second formula, that's not the ramification divisor and a divisor doesn't exist in that way generally, for that you'd want $K_Z - R = f^* K_X$ (this exists canonically for $f$ a separable finite map between normal varieties). Here $R$ is the ramification divisor. If $f$ is Galois, then such a $B$ exists (a branch divisor), but in most other cases it doesn't components of $B$ will pull back to components of $R$ with different multiplicity. $\endgroup$ – Karl Schwede Jan 15 '17 at 18:20
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    $\begingroup$ Take for example $X=\mathbb P^1\to \mathbb P^1=Y$ with ramified at two point $p_1$ and $p_2$ then $K_X=f^*(K_Y+\frac{1}{2}p_1+\frac{1}{2}p_2)$ $\endgroup$ – user21574 Jan 15 '17 at 18:24
  • $\begingroup$ yes, and that is the Galois case. The ramification divisor is $0 + \infty$ on $X$. For a more complicated example, imagine the ramification divisor on a curve $X$ is $2P + Q$ for $P$ and $Q$ both mapping to the same point $A$ on $Y$ (say in char 0, tame ramification). Then there is no multiple of $A$ that pulls back to $2P + Q$ since $f^* A = 3P + 2Q$. If the map is Galois however, then if $f * A = \sum e_i P_i$ , it follows that all $e_i$ are the same (since the Galois group moves all the $P_i$ to each other). Hopefully that makes sense. $\endgroup$ – Karl Schwede Jan 16 '17 at 0:51
  • $\begingroup$ You mean f is Galois , i.e, $f$ as etale morphism is finite and the function field extension is Galois? $\endgroup$ – user21574 Jan 16 '17 at 3:11

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