Let $X$ be a variety over an algebraically closed field $k$ and let $\Omega_X^n=\bigwedge_{i=1}^n\Omega_X$, where $\Omega_X$ is the canonical bundle.
I'm trying to understand the well-known fact that plurigenera $P_n:=h^0(X,\Omega_X^n)$ are birational invariants.
For convenience, I'll write invertible sheaves as divisors, so $\Omega_X$ will be replace by the canonical divisor $K_X$.
Let $\varphi:X\dashrightarrow Y$ be a birational map. My fist idea was to prove that for every divisor $D\in\text{Div}(Y)$ we should have $h^0(X,\varphi^*D)=h^0(Y,D)$, which is the same as finding an isomorphism of $k$-vector spaces $\Gamma(Y,D)\simeq \Gamma(X,\varphi^*D)$, which I'm not sure can be done.
Maybe this is true when $D=K_Y^n$, but I'm also not sure. For example, suppose $X,Y$ are smooth surfaces and $\varphi$ is the blowup at a point with exceptional divisor $E$ and $D=K_Y$. In this case $K_X=\varphi^*K_Y+E$, and I don't know what to do with $E$.
Any help will be appreciated. Thank you!
Best Answer
I had a quick look at Beauville's book "Complex Algebraic surfaces" where a proof for surfaces is given Proposition 3.20, I think I got a feeling for the crux of argument. Lets restrict to surfaces first, note that it is only neccesary to check it for point blow-ups.
Let $Y \rightarrow X $ be the a blow-up at $p \in X$. Note that there is a zariski open subset in $Y$ that is isomorphic to $X \setminus p$. Suppose we have a differential form on $Y$, then we get a form on $X \setminus p$. This section can be extended to a rational section on $X$ (by a general fact which holds for any coherent sheaf). but the singularities of a rational section are always a divisor, so it is in fact a section. Then one may check that the morphism defined in this was is in fact injective, i.e. if two differential forms agree on a zariski open then they agree everywhere. This is just a fancy version of the identity theorem from complex analysis.
Now for higher dimensional, there is a theorem that birational varieties may be blown up along sequences of smooth centres with codimension at least $2$ to get a common variety. Then because the codimension is atleast $2$ we can apply the above argument; i.e. producing a rational section then seeing it has to be an actual section.