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For $X \subseteq \mathbb{P}^n$ a smooth hypersurface, the canonical divisor $K_X$ can be computed as $$ K_X = (K_{\mathbb{P}^n} + X)|_X. $$ Is there a similar formula where $X$ is of higher codimension?

If necessary we can assume $X$ to be a surface over the complex numbers, in particular it is a complex analytic manifold.

Thanks!

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    $\begingroup$ Yes, this is essentially the content of Proposition II.8.20 of Hartshorne, although that result is more general. You'll notice that the "adjunction formula" for curves on surfaces of Proposition V.1.5 is presented as a direct application of the general formula. $\endgroup$
    – Andrew
    Mar 12, 2013 at 20:25

2 Answers 2

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The adjunction formula is a very general, pleasant, result valid for any complex submanifold $X\subset Z$ of codimension $r$ of any complex manifold $Z$ (projectivity is irrelevant).
It computes the canonical bundle $K_X=\wedge ^{dim X}(TX)^*$ of the submanifold $X$ and reads $$ K_X=K_Z|X\otimes \wedge^r N $$ where $N=N_Z(X)$ is the normal bundle of $X$ in $Z$, defined by $N=\frac{TZ|X}{TX}$.

Ah, you will say, such a beautiful, extremely general theorem, must be quite difficult to prove, must it not?
Not at all! As is surprisingly often the case, the theorem boils down to (multi)linear algebra: any exact sequence of vector spaces $0\to F\to E\to Q\to$ gives rise to a canonical isomorphism between determinants $$det E=det F\otimes det Q$$ ( where $det E=\wedge ^{dim E}E$, etc.)
We can globalize this formula to vector bundles, apply it to the exact sequence $$0\to N^*\to T^*Z|X\to T^*X\to 0$$ (which is the dual of the sequence defining the normal bundle) and obtain $$ K_Z|X=\wedge^rN^*\otimes K_X $$ Tensoring both sides with $\wedge^r N$, the dual line bundle to $\wedge^rN^*$, then immediately gives the promised adjunction formula $K_X=K_Z|X\otimes \wedge^r N$.
Et voilà!

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The principle of adjunction is much more general than just a formula for the canonical class, it holds at the level of total Chern classes for any submanifold $X$ of a complex manifold $Z$ (the canonical class is just the negative of the first Chern class of $X$). More precisely, the more general form of adjunction at the level of Chern classes is $$c(X)=c(Z)/c(N),$$ where $N$ is the normal bundle to $X$ in $Z$ (this follows from the exact sequence of vector bundles in Georges' answer along with the fact that Chern classes are multiplicative with respect to exact sequences). In the special case $X\subset \mathbb{P}^n$ is a complete intersection (i.e., $X$ is the zero locus of the system $F_1=\cdots =F_k=0$ and $X$ has codimension $k$), adjunction easily yields a nice formula for the total Chern class of $X$, from which can you can extract (among other invariants) not only the canonical class of $X$ but it's topological Euler characteristic as well. In particular (if you know a little bit about Chern classes, or if not see chapter 3 of Fulton's "Intersection theory"), if we denote the class of a hyperplane in $\mathbb{P}^n$ by $H$ and the degree of the hypersurface $F_i=0$ by $d_i$, then $c(\mathbb{P}^n)=(1+H)^{n+1}$ and $c(N)=(1+d_1H)\cdots(1+d_kH)$, thus $$c(X)=\frac{(1+H)^{n+1}}{(1+d_1H)\cdots(1+d_kH)}\cdot d_1\cdots d_kH^k$$ (we multiply by $d_1\cdots d_kH^k$ as this is the class of $X$ in $\mathbb{P}^n$). The $i$th Chern class of $X$ is then just the term of degree $k+i$ in the power series expansion of $\frac{(1+H)^{n+1}}{(1+d_1H)\cdots(1+d_kH)}\cdot d_1\cdots d_kH^k$. In particular, $K_X=-(n+1-d_1-\cdots - d_k)H^{k+1}$ and the topological Euler characteristic is the degree of the $(n-k)$th Chern class of $X$, namely the coefficient of $H^n$ in the power series expansion of $c(X)$.

As I stated earlier, all of this can be found in much greater detail in chapter 3 of Fulton's book.

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  • $\begingroup$ Very nice answer, +1 $\endgroup$
    – Andrew
    Mar 13, 2013 at 18:54
  • $\begingroup$ I wish i could accept both, thanks! $\endgroup$
    – Joachim
    Mar 19, 2013 at 11:59

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